LUMO_Math_G10 by Uolo

Mathematics (Subject Code–041)

About the Book LUMO is a comprehensive and student-friendly one-stop resource for Mathematics (CBSE subject code 041) for Grades IX–X. It offers a brief overview of the theory and a wide variety of practice questions, designed to promote conceptual clarity, core skill development and exam readiness. The book presents the entire NCERT curriculum in a clear, accessible, and exam-oriented format to support students at every stage of their learning journey.

Special Features • Integrated AI tutor for evaluation, effective feedback and doubt clarification • Includes latest solved CBSE papers for 2024–25 • 4 Practice Test For CBSE board examination

a+b=c

MATHEMATICS (Subject Code – 041)

• Complete NCERT Exemplar Solutions

NCERT Textbook Solutions

• Expert video content on exam strategy and time management

• Suitable for classroom instruction, self-study and revision

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LUMO for Grade 10 10

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SOCIAL SCIENCE

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Competency-based Questions

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SCIENCE (Subject Code – 086)

CH4 + 2O

F=ma

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� CO + 2H 2

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E= mc

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(Subject Code–184)

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ENGLISH LANGUAGE AND LITERATURE

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x= c + 25 - 8

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2H + 2

F=m

O2 = 2H

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+ b=(

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E= mc

x= b x = - + b2b + 22a 4ac b 2a -4ac

Mathematics

x 2

a=2ab + b=(a+b) 2

(Subject Code–041)

c b

Acknowledgements Academic Authors: Parveen Sharma, Nagesh Pathak Book Production: Surendra Prajapati Project Management: Vishesh Agarwal VP, Learning: Abhishek Bhatnagar

All products and brand names used in this book are trademarks, registered trademarks or trade names of their respective owners. © Lumo Learning Private Limited First impression 2025 This book is sold subject to the condition that it shall not by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of both the copyright owner and the above-mentioned publisher of this book. Book Title: LUMO Mathematics 10 ISBN: 978-93-49697-89-8 Published by Lumo Learning Private Limited Corporate Office Address: 91Springboard, 3rd Floor 145, Sector 44, Gurugram, Haryana 122003 CIN: U74999DL2017PTC322986 Illustrations and images: www.shutterstock.com, www.stock.adobe.com and www.freepik.com All suggested use of the internet should be under adult supervision.

Mathematics (Subject Code-041) COURSE STRUCTURE CLASS X (2025-26) (Annual Examination) Units

S. No.

Unit Name

Marks

Number Systems

Algebra

III

Coordinate Geometry

Geometry

Trigonometry

Mensuration

VII

Statistics & Probability

Total

Content

Competencies

Explanation

UNIT I: NUMBER SYSTEMS 1.

REAL NUMBERS 1. Fundamental Theorem of Arithmetic – statements after reviewing work done earlier and after illustrating and motivating through examples 2. Proofs of irrationality of 2 , 3 , 5

• Develops understanding of • Describes Fundamental numbers, including the set of real Theorem of Arithmetic with numbers and its properties. examples. • Extends the understanding of powers (radical powers) and exponents. • Applies Fundamental Theorem of Arithmetic to solve problems related to real life contexts.

• Prove algebraically the irrationality of numbers like 2 , 3 , 5 , 3 + 2 5 etc.

UNIT II: ALGEBRA 1.

POLYNOMIALS 1. Zeros of a polynomial 2. Relationship between zeros and coefficients of quadratic polynomials.

• develops a relationship between algebraic and graphical methods of finding the zeros of a polynomial.

• Find the zeros of polynomial graphically and algebraically and verifying the relation between zeros and coefficients of quadratic polynomials.

iii

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 1. Pair of linear equations in two variables and graphical method of their solution, consistency/ inconsistency. 2. Algebraic conditions for number of solutions. 3. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situationalproblems.

QUADRATIC EQUATIONS 1. Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). 2. Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. 3. Situational problems based on quadratic equations related to day-to-day activities to be incorporated

ARITHMETIC PROGRESSIONS 1. Motivation for studying Arithmetic Progression 2. Derivation of the nth term and sum of the first n terms of AP and their application in solving daily life problems.

• Describes plotting a pair of linear • Find the solution of pair equations and of linear equations in two graphically finding the solution. variables graphically and algebraically (substitution • Models and solvescontextualised and elimination method) problems using equations (e.g., simultaneous line are quations in twovariables).

• Demonstrates strategies of finding roots and determining the nature of roots of a quadratic equation.

• Solves quadratic equations using factorization and quadratic formula • Determines the nature of roots using discriminant • Formulates and solves problems based on real life context

• develops strategies to apply the concept of A.P. to daily life situations.

• Applies concepts of AP to find the nth term and sum of n terms. • Application of AP in real life problems

UNIT III: COORDINATE GEOMETRY 1.

Coordinate Geometry

• Derives formulae to establish • Solves problems using relations for geometrical shapes in distance formula and 1. Review: Concepts of coordinate the context of a coordinate plane, section formula geometry. Distance formula. Section such as, finding the distance formula (internal division). between two given points, to determine the coordinates of a point between any two given points. UNIT IV: GEOMETRY

TRIANGLES Definitions, examples, counter examples of similar triangles.

• works out ways to differentiate between congruent and similar figures.

• Prove Basic Proportionality theorem and applying the theorem and its converse in solving questions

1. (Prove) If a line is drawn parallel to • establishes properties for • Prove similarity of triangles one side of a triangle to intersect the similarity of two triangles logically using different similarity other two sides in distinct points, using different geometric criteria criteria the other two sides are divided in established earlier such as, Basic the same ratio. Proportionality Theorem, etc. 2. State (without proof) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3. State (without proof ) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. 4. State (without proof) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. 5. State (without proof ) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar. 2.

CIRCLES

• derives proofs of theorems related • Prove the theorems based to the tangents of circles. on the tangent to a circle. Tangent to a circle at point of contact. • Applies the concept of 1. (Prove) The tangent at any point of a tangents of circle to solve circle is perpendicular to the radius various problems. through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.

UNIT V: TRIGONOMETRY 1.

INTRODUCTION TO TRIGONOMETRY • Understands the definitions of • Evaluates trigonometric the basic trigonometric functions ratios 1. Trigonometric ratios of an acute (including the introduction of the • Describes trigonometric angle of a right-angled triangle. sine and cosine functions). ratios of standard angles Proof of their existence (well and solving related defined) expressions 2. Motivate the ratios whichever are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45° and 60°. 3. Relationships between the ratios.

TRIGONOMETRIC IDENTITIES 1. Proof and applications of the identity sin2A + cos2A = 1. 2. Only simple identities to be given.

HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. 1. Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, and 60°.

• Uses Trigonometric identities to solve problems.

• Proves trigonometric identities using sin2A + cos2 A = 1 and other identities

• Applies Trigonometric ratios in solving problems in daily life contexts like finding heights of different structures or distance from them.

• Find heights and distances in real life word problems using trigonometric ratios

UNIT VI: MENSURATION 1.

AREAS RELATED TO CIRCLES 1. Area of sectors and segments of a circle. 2. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only.

• Derives and uses formulae to calculate areas of plane figures.

• Visualises and evaluates areas of sector and segment of a circle

SURFACE AREAS AND VOLUMES 1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.

• Visualises and uses mathematical thinking to discover formulae to calculate surface areas and volumes of solid objects (cubes, cuboids, spheres, hemispheres, right circular cylinders/cones, and their combinations).

• Evaluates the surface areas and volumes of combinations of solids by visualisation.

UNIT VII: STATISTICS AND PROBABILITY 1.

STATISTICS

PROBABILITY

• calculates mean, median and mode for different sets of data 1. Mean, median and mode of grouped related with real life contexts. data (bimodal situation to be avoided).

1. Classical definition of probability. 2. Simple problems on finding the probability of an event.

• Computes the mean, of a grouped frequency distribution using direct, assumed mean and step deviation method. • Computes the median and mode of grouped frequency distribution by algebraic method

• Applies concepts from probability • Determines the to solve problems on the probabilities in simple likelihood of everyday events. real-life problems

MATHEMATICS - STANDARD (Code - 041) QUESTION PAPER DESIGN CLASS - X (2025-26) Time: 3 Hours

S. No.

Max. Marks: 80

Typology of Questions Remembering: Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. Understanding: Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas Applying: Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

Total Marks

% Weightage (approx.)

100

Analysing: Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Evaluating: 3

Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Creating: Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions Total

INTERNAL ASSESSMENT

20 MARKS

Pen Paper Test and Multiple Assessment (5+5)

10 Marks

Portfolio

05 Marks

Lab Practical (Lab activities to be done from the prescribed books)

05 Marks

vii

MATHEMATICS-BASIC (Code - 241) QUESTION PAPER DESIGN CLASS – X (2025-26) Time: 3 Hours

Max. Marks: 80

S. No. Typology of Questions

Total Marks

% Weightage (approx.)

Applying: Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

Analysing: Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations

100

Remembering: Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. 1

Understanding: Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas

Evaluating: Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Creating: Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions Total

INTERNAL ASSESSMENT

20 MARKS

Pen Paper Test and Multiple Assessment (5+5)

10 Marks

Portfolio

05 Marks

Lab Practical (Lab activities to be done from the prescribed books)

05 Marks

PRESCRIBED BOOKS: 1. Mathematics - Textbook for class IX - NCERT Publication 2. Mathematics - Textbook for class X - NCERT Publication 3. Guidelines for Mathematics Laboratory in Schools, class IX - CBSE Publication 4. Guidelines for Mathematics Laboratory in Schools, class X - CBSE Publication 5. Laboratory Manual - Mathematics, secondary stage - NCERT Publication 6. Mathematics exemplar problems for class IX, NCERT publication 7. Mathematics exemplar problems for class X, NCERT publication.

viii

Chapter Walk Through 1

QR Code: Provides access to the digital content of each chapter.

Real Numbers

Chapter at a Glance Fundamental Theorem of Arithmetic Every composite number can be uniquely expressed as a product of primes, except for the order in which these prime factors occur. Composite number = Product of prime numbers This method is also called Prime Factorisation Method. Example: 12 = 2 × 2 × 3 Which is the same as: 3 × 2 × 2 or 2 × 3 × 2 = 22 × 3 We can write it as: 12 Which is still a unique combination (2, 2, and 3).

Chapter Number and Chapter Name

Relation Between HCF and LCM of Two Numbers For two positive integers a and b, we have HCF (a, b) × LCM (a, b) = a × b a ×b ⇒ HCF (a, b) = LCM ( a , b ) a ×b Or LCM (a, b) = HCF ( a , b )

Chapter at a Glance: A short and crisp summary of a chapter with all the important ideas/concepts/relations.

16. If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is (a) 4 (b) 2 (c) 1 (d) 3

Example: a = 8 and b = 6 LCM ( 8, 6 ) = 23 × 3 = 24 HCF ( 8, 6 ) = 2 HCF ( 8, 6 ) × LCM ( 8, 6 ) = 8×6

17. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is (a) 13 (b) 65 (c) 875 (d) 1750

2 × 24 =8 × 6 48 = 48

18. The product of a non-zero rational and an irrational number is (a) always irrational (b) always rational

If ‘p’ is a prime and p divides a 2, then p divides ‘a’ also, where a, is a positive integer. Example: 5 divides 100 i.e., (102) ⇒ 5 divides 10.

Irrational Numbers Numbers that cannot be expressed in the form 3,

Example: 2,

5, π are irrational numbers.

19. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is (a) 13 (b) 65 (c) 875 (d) 1750

p , where p and q are integers and q ≠ 0. q

1. Express each number as a product of its prime factor: (a) 140 (b) 156 (c) 3825 (d) 5005

(d) 5005 = 5 × 7 × 11 × 13 (e) 7429 = 17 × 19 × 23 2. Find the LCM and HCF of following pairs of integers and verify that LCM × HCF = product of the two numbers.

(e) 7429

Sol. (a) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(a) 26 and 91

(b) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

(b) 510 and 92

1. (c) 3, 420

2. (d) 5

3. (a) 28 × 32

4. (d) 2

5. (c) ( 3, 27)

6. (b) 63

10. (a) 1650

11. (b) 19 m/min

12. (d) 10:12 a.m.

Constructed Response Questions: A set of solved questions (Very Short Answers, Short Answers, Long Answers), that are based on constructed response.

LCM × HCF = Product of two numbers.

( LCM )( 3 ) = 180 3 + 5 is irrational.

3 + 5 = a, where a is rational.

Let

3 = a − 5 on squaring both sides we

(b)

(d)

k 2

9. The remainder when the square of any prime number greater than 3 is divided by 6, is (a) 1 (b) 3 (c) 2 (d) 4

48 = 2 × 2 × 2 × 2 × 3 108 = 2 × 2 × 3 × 3 × 3 Therefore, LCM of 48, 72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

5 is irrational, which is

Hence, after every 432 seconds, the light will change simultaneously. 5

5/8/2025 2:26:33 PM

(d) Cannot be determined without knowing n

(iii) k is an even number

6 4 2. If a =6 − 5, b =6 + 6 5, c = 6 + 6 5, d = 6 + 4 5, = e 4 6 + 4 5, then which of the following is a rational number? (b) abde (c) ab (d) cd (a) abcde

(b) only (i), (ii) and (iii) (c) only (i), (iii) and (iv)

(d) all (i), (ii), (iii) and (iv)

Additional Practice Questions: A set of unsolved question, covering the entire chapter for practice.

5 + 3 + 2 is (a) a natural number

(b) an integer

(d) an irrational number

(CBSE 2023)

multiply the numerator and denominator by the

conjugate of the denominator, which is n − n + 1 ) 4. If X= 28 + (1 × 2 × 3 × 4 ×…× 16 × 28 ) and Y= 17 + (1 × 2 × 3 ×…× 17 ), then which of the following is/are true?

4. How many prime numbers are of the form 10n + 1, where n is a natural number such that 1 ≤ n < 10? (a) 5 (b) 6 (c) 4 (d) 3 5. LCM of 18 and M is 36. M is a natural number. Which of these can be true for some values of M ? (i) M is a multiple of 3 (ii) M is a multiple of 18 (a) only (i) and (ii)

(b) only (i) and (iii)

(d) only (ii) and (iii)

6. The pair of positive integers, where sum is 91 and HCF is 13, is/are (i) (13, 78) (ii) (26, 65) (iii) (39, 52) (b) only (i) and (ii)

(d) all (i), (ii) and (iii)

7. Find the values of x, y, z and w in the following factor tree. Also write the prime factorisation of x.

(ii) Y is a prime number

(d) Both (i) and (ii)

w 13

10. (a) (iv) 233 and 377

1. (b) 3

2. (c) 338

3. (d) an irrational number

4. (a) 5

5. (b) only (i) and (iii)

6. (d) all (i), (ii) and (iii)

= = y 1639, = z 273, = w 91 7. x 3276, 9. 17

Scan me for Exemplar Solutions 27

13 8

(b) (i) 2m × 5n (d) (i) 1

34 + 21 = 55 55 + 34 = 89 89 + 55 = 144 Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... Based on the given sequence, answer the following questions: (a) The next two numbers in the sequence are (i) 223 and 367 (ii) 233 and 367 (iii) 223 and 377

(iv) 233 and 377

(b) A rational number has a terminating decimal expansion, if the denominator is of the form m

2 ×5

(ii) 3m × 5n

(iii) 2m × 3n

(iv) 3m × 4n

(c) Which of the following rational numbers (formed by taking the ratio of a term to its previous term) has a terminating decimal expansion?

8. Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers?

(i)

5 3

(ii)

21 13

(iii)

13 8

1. (a) L = 21P

2. (b) abde

3. 7

4. (a) Both (i) and (iv)

Complete Exemplar Solutions: Scan the QR code at end of each chapter for complete NCERT Exemplar solutions.

Math Grade 10 Exam Book_Ch01.indd 15

SELF-ASSESSMENT 1

Scan me for Exemplar Solutions

Self-Assessments: Two Assessments at the end of each chapter to help learners prepare for tests and exams. Complete Solutions will be available on QR code.

Max. Marks: 40

5/8/2025 2:26:41 PM

[3 × 1 = 3 Marks]

1. If p and q are two distinct prime numbers, then LCM (p, q) is (a) 1 (b) p (c) q

(d) pq

2. The sum of exponents of prime factors in the prime factorisation of 196 is (a) 3 (b) 4 (c) 5

(d) 2

3. The product of a non-zero rational and an irrational number is: (a) always irrational. (b) always rational. (c) rational or irrational.

(d) one.

Assertion-Reason Based Questions

Scan me for Solutions

Time: 60 mins

16-04-2025 15:54:58

34 21

5/8/2025 2:26:40 PM

Multiple Choice Questions

Math Grade 10 Exam Book_Ch01.indd 16

(iv)

(d) HCF of two consecutive numbers in the given sequence (except the first 3 terms) is (i) 1 (ii) 0 (iii) 2 (iv) 3

Chapter 1 • Real Numbers

Challenge Yourself

21 + 13 = 34

Answers to Unsolved Problems: Solution to objective/numerical questions from the Additional Practice Questions and Challenge Yourself sections. Complete solutions can be accessed on scanning the QR code. 9. Find the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 13 respectively.

Answers Additional Practice Questions

8+5= 13

where m, n are non-negative integers.

z 3

3 5 12

13 + 8 = 21

819

(iv) X + Y is a composite number (b) Both (ii) and (iii)

3+2 = 5

(i)

(a) Both (i) and (iv)

2 +1 = 3 5+3= 8

(iii) X − Y is a prime number

3. Find the correct value of 1 1 1 + + +… upto 91 terms. 9 + 10 10 + 11 11 + 12 1 (Hint: Rationalise each term of form n + n +1

0 +1 = 1

Challenge Yourself: A unique set of higher order questions that are designed to challenge you and your understanding of the chapter. Additional Questions can be found on scanning the QR code. (a) only (i)

5/8/2025 2:26:37 PM

(i) X is a composite number

10. Given figure represents the golden ratio spiral. The sequence 0, 1, 1, 2, 3, 5, 8, 13 is called a Fibonacci sequence, in which each term (except the first two terms) is obtained by adding the previous two terms of the sequence. 1+1 = 2

(iii) M is an even number

Challenge Yourself

Math Grade 10 Exam Book_Ch01.indd 11 8 8 8

1. If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, is (a) 2 (b) 3 (c) 4 (d) 7 2. If HCF(26, 169) = 13, then LCM(26, 169) = (a) 26 (b) 52 (c) 338 (d) 13

Assertion-Reason Based Questions

6. If the LCM of two prime numbers p and q (p > q ) is tweight, rust-resistant metal used in vehicle 221,bodies. then the value of 3p − q is Corrosion-resistant, used for exhaust systems. (a) 4 (b) 28 (c) 38 (d) 48 erials: Durable, corrosion-proof non-metallic materials like carbon fiber. 7. Three bells ring at intervals of 3, 5 and 18 minutes. and Manufacturing: All the three rang at 6 a.m. izing crevices and tight joints where water and dirt accumulate. When will they ring together again? eatments: (a) 6:25 a.m. (b) 6:30 a.m. (c) 7:30 a.m. (d) 9:00 a.m. Wax or oil applied to critical areas. 8. corrosion. If n is a natural number, then 92n − 42n is always tion: Sacrificial anodes to prevent metal divisible by a metal that will be preferentially corroded instead of iron. These metals are normally (a) 5 aluminium and zinc. They (b) 13protect reactivity series than iron, such as magnesium,

el as these metals are corroded first, are widely employed corrosion (c) both 5 and 13 to mitigate(d) none ofby these s to protect the steel structures. 1. P is the LCM of 2, 4, 6, 8, 10, Q is LCM of 1, 3, 5, 7, 9 Chapter 1 • Real Numbers and L is the LCM of P and Q. Then, which of the following is true? (a) L = 21P (b) L = 4Q (c) L = 63P (d) L = 16Q

Additional Practice Questions

(iv) k is an odd number

(a) only (ii) and (iii)

(d) 15 − 10

ant Materials:

ions and Equations

8. 30

To proceed with the given problem, we need to find the LCM of 48, 72, and 108.

(b) only (i) and (ii)

4. Which of the following is the rationalised form 5 of (CBSE QB) ? 3+ 2 3+ 2 (a) 5 (b) 5 (c) 15 − 2

Sol. We will be using the concept of LCM (Least Common Multiple) to solve this.

Competency Based Questions: A set of questions (Multiple Choice, Assertion-Reason and Case-based questions), that are based on competencies defined by CBSE.

2. The HCF of k and 93 is 31, where k is a natural number. Which of these can be true for some values of k ? (i) k is a multiple of 31 (ii) k is a multiple of 93

(b) only k and 7k k (d) all , k , 7k and k 3 2

contradiction. Since, 3 and 5 are prime numbers. Hence, 3 + 5 is irrational. 3. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Math Grade 10 Exam Book_Ch01.indd 5

n is a natural number such that n > 1. Which of these can definitely be expressed as a product of primes? n (i) n (ii) n (iii) 2 (a) only (i)

HOTS: “High Order Thinking Skills” Questions. (a) only k

22. (d) four decimal place

Chapter 1 • Real Numbers

Select Response Type MCQs

2a 5 = a2 + 2 a2 + 2 5= 2a From the above, as the right-hand side is rational number while

3. Let p be a prime number and k be a positive integer. If p divides k 2, then which of these is definitely divisible by p?

20. (b) xy2

21. (d) 2520

72 = 2 × 2 × 2 × 3 × 3

( 3 ) = (a- 5 ) ⇒ 3 = a + 5 - 2a 5

Competency Based Questions

(NCERT Exemplar)

Sol. Let us suppose that 3 + 5 is rational.

Hence, there is no positive integer n for which n − 1 + n + 1 is rational.

(CBSE 2024)

18. (a) always irrational

19. (a) 13

(CBSE 2023)

LCM = 60 2. Prove that

(a)

(CBSE 2021)

Sol. We know that,

Therefore, get,

2. The smallest irrational number by which 20 should be multiplied, so as to get a rational number, is

(NCERT Exemplar)

16. (b) 2

17. (a) 13

9. (c) 4

(d) four decimal place

14. (a) co-prime

15. (b) 4

8. (c) 3

Very Short Answer Questions

5/8/2025 2:26:29 PM

= 22 × 3x , b = 22 × 3 × 5, c = 22 × 3 × 7 and 1. If a LCM (a, b, c) = 3780, then x is equal to (CBSE 2024) (a) 1 (b) 2 (c) 3 (d) 0

13. (b) 16

7. (c) 20

1. If xy = 180 and HCF ( x , y ) = 3 then find the LCM ( x , y ). Math Grade 10 Exam Book_Ch01.indd 1

Multiple Choice Questions

(NCERT Exemplar)

22. The decimal expansion of the rational number 14587 will terminate after 1250 (a) one decimal place (b) two decimal places

Constructed Response Questions

Chapter 1 • Real Numbers

However, this contradicts the fact that two perfect squares differ at least by 3.

(NCERT Exemplar)

21. The least number that is divisible by all the numbers from 1 to 10 (both inclusive). (a) 10 (b) 100 (c) 504 (d) 2520

NCERT Zone: Detailed model answers to NCERTAnswers questions (Intext and Exercises) given in simple, easy to understand language.

NCERT Zone NCERT Exercise 1.1

(d) one

20. If two positive integers a and b are written as a = x 3 y 2 and b = xy 3 , where x , y are prime numbers, then HCF (a, b) is (c) x3y3 (d) x2y2 (a) x y (b) xy2

[2 × 1 = 2 Marks]

Reason (R): Product of two irrational numbers is always irrational. 5. Assertion (A): If HCF of 510 and 92 is 2, then the LCM of 510 and 92 is 32460. Reason (R): As HCF (a, b) × LCM (a, b) = a × b.

Very Short Answer Questions 6. Prove that 6 − 4 5 is an irrational number, given that 5 is an irrational number. 7. Show that 11 × 19 × 23 + 3 × 11 is not a prime number.

[2 × 2 = 4 Marks]

Table of Contents PART A

1. Real Numbers �� 1 2. Polynomials �� 21 3. Pair of Linear Equations in Two Variables �� 43 4. Quadratic Equations �� 76 5. Arithmetic Progressions �� 108 6. Triangles �� 143 7. Coordinate Geometry �� 174 8. Introduction to Trigonometry �� 200 9. Some Applications of Trigonometry �� 232 10. Circles �� 256 11. Areas Related to Circles �� 286 12. Surface Areas and Volumes �� 309 13. Statistics �� 334 14. Probability �� 373 Self-Assessment 1 Answers �� 396 Self-Assessment 2 Answers �� 400

PART B

CBSE Question Paper (Standard) 2024–25 �� 407 Sample Question Paper 1 (Solved) Basic �� 425 Sample Question Paper 2 (Solved) Standard �� 438 Sample Question Paper 3 (Unsolved) Standard �� 450 Sample Question Paper 4 (Unsolved) Standard �� 455

Real Numbers

Relation Between HCF and LCM of Two Numbers For two positive integers a and b, we have HCF (a, b) × LCM (a, b) = a × b a ×b ⇒ HCF (a, b) = LCM ( a , b ) a ×b Or LCM (a, b) = HCF ( a , b ) Example: a = 8 and b = 6 LCM ( 8, 6 ) = 23 × 3 = 24 HCF ( 8, 6 ) = 2 HCF ( 8, 6 ) × LCM ( 8, 6 ) = 8×6 2 × 24 =8 × 6 48 = 48

If ‘p’ is a prime and p divides a 2, then p divides ‘a’ also, where a, is a positive integer. Example: 5 divides 100 i.e., (102) ⇒ 5 divides 10.

Irrational Numbers Numbers that cannot be expressed in the form Example: 2,

5, π are irrational numbers.

p , where p and q are integers and q ≠ 0. q

NCERT Zone NCERT Exercise 1.1 1. Express each number as a product of its prime factor: (a) 140 (b) 156 (c) 3825 (d) 5005

(e) 7429

Sol. (a) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 (b) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13 Chapter 1 • Real Numbers

(c) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17 (d) 5005 = 5 × 7 × 11 × 13 (e) 7429 = 17 × 19 × 23 2. Find the LCM and HCF of following pairs of integers and verify that LCM × HCF = product of the two numbers. (a) 26 and 91

(b) 510 and 92

Sol. (a) 26 and 91 26= 2 × 13

(b) 17, 23 and 29

91= 7 × 13

23= 1 × 23

HCF of 26 and 91 is = 13

29 = 1 × 29

LCM of 26 and 91 is = 2 × 7 ×13 = 182

HCF of 17, 23 and 29 = 1

Product of the two numbers = 26 ×91 = 2366

HCF × LCM =13 × 182 =2366

LCM of 17, 23 and 29 = 17 × 23 × 29

Therefore, product of two numbers is product of their HCF and LCM. (b) 510 and 92

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

HCF of 510 and 92 is = 2

LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of the two numbers = 510 × 92 = 46920

2 × 23460 = 46920 HCF × LCM =

17= 1 × 17

= 11339

8 = 2×2×2

9= 3 × 3

25= 5 × 5

HCF of 8, 9 and 25 = 1

LCM of 8, 9 and 25 = 2 × 2 × 2 × 3 × 3 × 5 × 5

= 1800

4. Given that HCF(306, 657) = 9. Find LCM(306, 657). Sol. HCF(306, 657) = 9

Therefore, product of two numbers is product of their HCF and LCM.

As we know that,

LCM × HCF = Product of two numbers

336 = 2 × 2 × 2 × 2 × 3 × 7

= 306 × 657 LCM × HCF

= 24 × 3 × 7

54 = 2 × 3 × 3 × 3

= 2 × 33

HCF of 336 and 54 = 2 × 3 = 6

LCM of 336 and 54 = 24 × 33 × 7 = 3024

Product of the two numbers = 336 × 54 = 18144

6 × 3024 = 18144 HCF × LCM =

Therefore, product of two numbers is product of their HCF and LCM. 3. Find the LCM and HCF of the following integers by applying prime factorisation method. (a) 12, 15 and 21 (b) 17, 23 and 29 (c) 8, 9 and 25

LCM =

306 × 657 9

LCM = 34 × 657 LCM = 22338 5. Check whether 6n can end with digit 0 for any natural number n. Sol. If any number end with the digit 0, it should be divisible by 10 or in other words, it will be divisible by 2 and 5 as 10 = 2 × 5 Prime factorisation of 6n=

( 2 × 3)n

Hence, for any value of n, 6n will not be divisible by 5.

12 = 22 × 3

15= 3 × 5

21= 3 × 7

HCF of 12, 15 and 21 = 3

LCM of 12, 15 and 21 = 22 × 3 × 5 × 7

306 × 657 HCF

It can be observed that 5 is not in the prime factorisation of 6n.

Sol. (a) 12, 15 and 21

LCM =

= 420

Therefore, 6n can’t end with digit 0 for any natural number n.

6. Explain why 7 × 11 ×13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Sol. We have, ⇒ 1014 = 2 × 3 × 13 × 13

Let p = 5k , where k is an integer

It is the product of prime factors.

(5k)2 = 5q 2

⇒ 2 × 3 × 13 × 13.

⇒ 5k 2 = q2

Hence this is a composite number.

This means that q2 is divisible by 5 and hence, q is divisible by 5. This implies that p and q have 5 as a common factor. And this is a contradiction to the fact that p and q are co-prime. Hence, 5 a cannot be expressed as or it can be said that b 5 is irrational.

⇒ ( 7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 ⇒ ( 42 × 20 × 6 ) + 5 ⇒ ( 42 × 120 ) + 5 ⇒ ( 5040 + 5 ) ⇒ 5045 5045 can be written as product of 5 × 1009. Therefore, given expression has factor 5 and 1009. Hence, it is a composite number. 7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at same point and at the same time, and go in same direction. After how many minutes will they meet again at the starting point? Sol. We can say that Ravi takes lesser time than Sonia for completing 1 round of circular path. They will meet again when Ravi will have completed 1 round with respect to Sonia. Which is calculated by LCM of time taken by Ravi and Sonia to complete 1 circular round. ⇒ 18 = 2 × 3 × 3 ⇒ 12 = 2 × 2 × 3 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36 minutes Therefore, Ravi and Sonia will meet at starting point after 36 minutes.

NCERT Exercise 1.2

5 be a rational number.

Therefore, we can find two integers p, q ( q ≠ 0) such p that 5 = q Let p and q have a common factor apart from 1. Then we can divide them by the common factor, and assume that p and q are co-prime. Chapter 1 • Real Numbers

2. Prove that ( 3 + 2 5 ) is irrational. Sol. Let 3 + 2 5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that a 3+2 5 = b a ⇒ 2 5 =− 3 b 1a  ⇒ 5=  − 3 2b  1 a  Since a and b are integers,  − 3  will also 2b  be rational and therefore, 5 is rational. This contradicts the fact that 5 is irrational. Hence, our assumption that 3 + 2 5 is rational is false.

(

)

Therefore, 3 + 2 5 is irrational. 3. Prove that the following are irrationals: 1 (a) (b) 7 5 (c) 6 + 2 2 1 Sol. (a) Let is rational. 2 Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 1 a = 2 b

OR b 2= a b is rational as a and b are integers. Therefore, a 2 is rational which contradicts to the fact that 2 is irrational. Hence, our assumption is false 1 and is irrational. 2

5 is irrational. (CBSE 2025)

Sol. Let

⇒ p2 = 5q 2 Therefore, p2 is divisible by 5 and it can be said that p is divisible by 5.

⇒ 7 × 11 × 13 + 13= 1001 + 13= 1014

1. Prove that

p = 5q

(b) Let 7 5 is rational.

Therefore, we can find two co-prime integers a, a b (b ≠ 0) such that 7 5 = b a ⇒ 5 = 7b a is rational as a and b are integers. Therefore, 7b 5 should be rational.

Let 6 + 2 be rational. Therefore, we can find two co-prime integers a , b (b ≠ 0) such that a 6 + 2 = b a ⇒ 2= −6, b a Since, a and b are integers, − 6 is also b rational and hence, 2 should be rational. This contradicts the fact that 2 is irrational. Therefore, our assumption is false and hence, 6 + 2 is irrational.

This contradicts the fact that 5 is irrational. Therefore, our assumption that 7 5 is rational is false. Hence, 7 5 is irrational.

Multiple Choice Questions 1. The HCF and the LCM of 12, 21, 15 respectively are (a) 3, 140

(b) 12, 420

(d) 420, 3

9. If LCM (x, 18) = 36 and HCF (x, 18) = 2, then x is (a) 2 (b) 3 (c) 4 (d) 5

(CBSE 2020)

(CBSE 2021, 2022)

2. The smallest irrational number by which 20 should be multiplied, so as to get a rational number, is

10. Given HCF ( 2520, 6600 ) = 40 and LCM (2520, 6600) = 252 × k , then the value of k is (a) 1650 (b) 1600 (c) 165 (d) 1625

(a)

(b) 2

(d)

(CBSE 2024)

3. The prime factorisation of the number 2304 is (a) 28 × 32 (b) 27 × 33 (c) 28 × 3 (d) 27 × 32 4. If n is a natural number, then 2(5n + 6n) always ends with (a) 1 (b) 4 (c) 3 (d) 2 (CBSE 2022)

5. A pair of irrational numbers whose product is a rational number is (a) ( 16, 4)

(b) ( 5, 2)

(d) ( 36, 2)

(

6. HCF of 34 × 22 × 73 and ( 32 × 5 × 7 ) is (a) 630

(b) 63

(d) 567 (CBSE 2023)

7. The LCM of smallest two-digit number and smallest composite number is (a) 12 (b) 4 (c) 20 (d) 40 (CBSE 2023)

8. If a = 2 (a, b, c) = 3780, then x is equal to (a) 1 (b) 2 (c) 3 2

× 3x, b = 22 × 3 × 5, c = 22 × 3 × 7 and LCM (d) 0 (CBSE 2024)

11. What is the greatest possible speed at which a girl can walk 95 m and 171 m in an exact number of minutes? (a) 17 m/min (b) 19 m/min (c) 23 m/min

(d) 13 m/min

(CBSE 2022)

12. Three bulb red, green and yellow flash at intervals of 80 seconds, 90 seconds and 110 seconds. All three flash together at 8:00 a.m. At what time will the three bulbs flash altogether again? (a) 9:00 a.m. (b) 9:12 a.m. (c) 10:00 a.m.

(CBSE 2024)

)

(CBSE 2024)

(d) 10:12 a.m.

13. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. The total number of sections thus formed is: (a) 22 (b) 16 (c) 36 (d) 21 14. If a and b are two co-prime numbers, then a3 and b3 are (a) co-prime (b) not co-prime (c) even

(d) odd

15. The exponent of 5 in the prime factorisation of 3750 is (a) 3 (b) 4 (c) 5 (d) 6 (CBSE 2022)

16. If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is (a) 4 (b) 2 (c) 1 (d) 3 17. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is (a) 13 (b) 65 (c) 875 (d) 1750 18. The product of a non-zero rational and an irrational number is (a) always irrational (b) always rational (c) rational

(d) one

19. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is (a) 13 (b) 65 (c) 875 (d) 1750

20. If two positive integers a and b are written as a = x 3 y 2 and b = xy 3 , where x , y are prime numbers, then HCF (a, b) is (a) x y (b) xy2 (c) x3y3 (d) x2y2

(NCERT Exemplar)

21. The least number that is divisible by all the numbers from 1 to 10 (both inclusive). (a) 10 (b) 100 (c) 504 (d) 2520

(NCERT Exemplar)

22. The decimal expansion of the rational number 14587 will terminate after 1250 (a) one decimal place (b) two decimal places (c) three decimal places

(d) four decimal place

(NCERT Exemplar)

Answers 1. (c) 3, 420

2. (d) 5

3. (a) 28 × 32

4. (d) 2

5. (c) ( 3, 27)

6. (b) 63

7. (c) 20

8. (c) 3

9. (c) 4

10. (a) 1650

11. (b) 19 m/min

12. (d) 10:12 a.m.

13. (b) 16

14. (a) co-prime

15. (b) 4

16. (b) 2

17. (a) 13

18. (a) always irrational

19. (a) 13

20. (b) xy2

21. (d) 2520

22. (d) four decimal place

Constructed Response Questions Very Short Answer Questions 1. If xy = 180 and HCF ( x , y ) = 3 then find the LCM ( x , y ). (CBSE 2021)

Sol. We know that, LCM × HCF = Product of two numbers. ( LCM )( 3) = 180 3 + 5 is irrational.

(NCERT Exemplar)

Sol. Let us suppose that 3 + 5 is rational. Let

3 + 5 = a, where a is rational.

Therefore, get,

( )

3 = a − 5 on squaring both sides we

3 = (a- 5 )2 ⇒ 3 = a2 + 5 - 2a 5 2a 5 = a2 + 2 a2 + 2 5= 2a From the above, as the right-hand side is rational number while Chapter 1 • Real Numbers

3. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again? (CBSE 2023)

LCM = 60 2. Prove that

contradiction. Since, 3 and 5 are prime numbers. Hence, 3 + 5 is irrational.

5 is irrational, which is

Sol. We will be using the concept of LCM (Least Common Multiple) to solve this. To proceed with the given problem, we need to find the LCM of 48, 72, and 108. 48 = 2 × 2 × 2 × 2 × 3 72 = 2 × 2 × 2 × 3 × 3 108 = 2 × 2 × 3 × 3 × 3 Therefore, LCM of 48, 72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 Hence, after every 432 seconds, the light will change simultaneously. 5

Thus, the required time is 432 seconds which can be converted into minutes and seconds as 7 minutes 12 seconds past 7 a.m. 4. Can two numbers have 18 as their HCF and 380 as their LCM? Give reason. Sol. No, because HCF is always a factor of LCM but here 18 is not a factor of 380. 5. Using prime factorisation, find HCF and LCM of 96 and 120. (CBSE 2023) Sol. 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3 120 = 2 × 2 × 2 × 3 × 5 = 23 × 3 × 5 Therefore, HCF = 23 × 3 = 24 and LCM = 25 × 3 × 5 = 480 6. For what value of n, 2n × 5n ends with 5?

7. Prove that 2 + 3 is an irrational number, given that 3 is an irrational number. (CBSE 2023) Sol: Let us assume that 2 + 3 is a rational. p ; q ≠ 0 and p , q are integers Let 2 + 3= q p − 2q ⇒ 3= q  p and q are integers ∴ p - 2q is an integer p − 2q ⇒ is a rational number. q

⇒ 3 is a rational number which contradicts with the question that 3 is an irrational number. Hence proved 2 + 3 is an irrational number.

Sol. Clearly, n

2n × 5n =( 2 × 5 ) =10n Also, all the values of n will make 10n end in 0. Thus, there is no value of n for which 2n × 5n ends in 5. 8. If HCF of 65 and 117 is expressible in the form 65n - 117, then find the value of n.

(CBSE 2019)

Topper’s Answer

9. Two numbers are in the ratio 21:17. If their HCF is 5, find the numbers. Sol. Let numbers are 21x and 17x. Now, common factor of 21x and 17x = x

Also, HCF = 5 ⇒x=5 Therefore, numbers are 21 × 5 and 17 × 5 i.e., 105 and 85.

10. Can the number (15)n, n being a natural number, end with the digit 0? Give reasons. (CBSE 2018)

Topper’s Answer

Short Answer Questions 1. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

(CBSE 2018)

Topper’s Answer

2. If HCF of 144 and 180 is expressed in the form 13m – 6. Find the value of m. (CBSE 2020) Sol. Given, HCF of 144 and 180 is in form of 13m − 16. Now, HCF of 144 and 180: 144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32 180 = 2 × 2 × 3 × 3 × 5 = 22 × 32 × 5 Therefore, HCF (144, 180): 22 × 32 = 36 Thus, HCF of 144 and 180 is given in the form of 13m − 16, 13m − 16 = 36

Chapter 1 • Real Numbers

13m = 36 + 16 13m = 52 m=

52 13

m=4

3. Prove that ( 2 + 3)2 is an irrational number, given that 6 is an irrational number. (CBSE 2024)

(

)

Sol. On the contrary, let 2 + 3 is rational number such that 2 a 2+ 3 = , where a and b are co-prime, b ≠ 0. b

(

)

a 2+3+2 6 = b a ⇒5+2 6 = b a − 5b ⇒ 6= 2b Now, a, b, 2 and 5 are integers.

(

)

a − 5b is a rational number while it is given that 2b 6 is an irrational number.

⇒

But rational number ≠ irrational number, which contradicts our assumption. Hence proved

( 2 + 3 ) is an irrational number. 2

4. Prove that 3 + 5 is irrational. Sol. Let us assume 3 + 5 is rational. Let

3+ 5= a , where a is rational

Therefore,

3= a − 5

Also, HCF + LCM = 495 x + 10x = 495 ⇒ 11x = 495 ⇒ x = 45 450 HCF = 45, LCM = 10 × 45 = Now, HCF × LCM = Product of two numbers 45 × 450 =90 × other number

45 × 450 = other number 90 Other number = 225

7. Find the largest number which divides 615 and 963 leaving remainder 6 in each case. Sol. According to the question, Number divides 615 − 6 and 963 − 6 i.e., 609 and 957 Required number = HCF of 609 and 957. 609 = 3 × 7 × 29 957 =3 × 11 × 29

On squaring both sides we get,

Therefore, HCF of 609 and 957 =3 × 29 =87

2 ( 3)= (a − 5)2

Required number = 87

3 = a + 5 − 2a 5 2a = 5 a2 + 2 a2 + 2 which is a contradiction. 2a As the right-hand side is a rational number while 5 is irrational. Since 3 and 5 are prime numbers, hence 3 + 5 is an irrational number. 5=

2+ 3 is an irrational number. Given 5 (CBSE 2019) 3 is an irrational number.

5. Prove that that

2+ 3 a = (b ≠ 0, a and b are integers) Sol. Let us assume 5 b 5a − 2b ⇒ 3= b ⇒ a , b are integers 5a − 2b is a rational number b i.e., 3 is a rational number which contradicts the fact that 3 is irrational number. 2+ 3 is an irrational number. 5 6. LCM of two numbers is 10 times their HCF. Sum of HCF and LCM is 495. If one number is 90, then find the other number.

Therefore,

Therefore, the largest number which divides 615 and 963 leaving remainder 6 in each case is 87. 8. A forester wants to plant 66 mango trees, 88 orange trees and 110 apple trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of trees (i.e., only one type of trees in one row). Find the number of minimum rows required. Sol. According to the question, Mango Trees, 66 = 2 × 3 × 11 Orange Trees, 88 = 2 × 2 × 2 × 11 Apple Trees, 110 = 2 × 5 × 11 Frome the above prime factorisation, we get, HCF of 66, 88 and 110 =2 × 11 =22 Therefore, number of trees in each row = 22 To get the number of distinct rows: Number of rows for Mango Trees =

66 = 3 22

Number of rows for Orange Trees =

88 = 4 22

Number of rows for Apple Trees =

110 = 5 22

Sol. Let HCF = x

Therefore, total number of distinct rows = 3 + 4 + 5 = 12

LCM = 10x

Hence, number of minimum rows required is 12.

Long Answer Questions

(

)

a 3 =6m + 1, where m =36q 3 + 18q 2 + 3q is an integer.

1. Prove that one of every three consecutive positive integers is divisible by 3. (CBSE 2020) = then putting r 2 in eq. (1) , we get Case 3: r 2,=

n 3q + r , where Sol. Let n be any positive integer. = r = 0, 1, 2

a 3= 216q 3 + 216q 2 + 72q + 8

Putting r = 0, n = 3q + 0 = 3q , which is divisible by 3 r 1,= n 3q + 1, which is divisible by 3 Putting= r 2,= n 3q + 2, which is divisible by 3 Putting=

Sol. Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that a = 6q + r , where 0 ≤ t < 6 ⇒ a 3 = (6q + r )3 = 216q 3 + 108q 2r + 18qr 2 + r 3,, ⇒= a

where r = 0, 1, 2, 3, 4, 5 and putting r in eq.(1), we get: = then putting r 0 in eq. (1) , we get Case 1: r 0,= 3

(

)

= a 216 = q 6 36 = q 6m 3

where, m = 36q is an integer.

= then putting r 1 in eq. (1) , we get Case 2: r 1,=

(

)

(

)

= a 3 6 36q 3 + 54q 2 + 27q + 4 + 3

(

)

a 3 =6m + 3, where m = 36q 3 + 54q 2 + 27q + 4 is an integer. = then putting r 4 in eq. (1) , we get Case 5: r 4,= a 3 = 216q 3 + 432q 2 + 288q + 64

(

)

3 a= 6 36q 3 + 72q 2 + 48q + 10 + 4 3

(

)

a =6m + 4, where m = 36q + 72q 2 + 48q + 10 is an integer. a 3 = 216q 3 + 540q 2 + 450q + 125

where 0 ≤ r < 6

(

a = 6m + 2, where m = 36q 3 + 36q 2 + 12q + 1 is an integer.

Case 6: r 5,= = then putting r 5 in eq. (1) , we get

( 216q + 108q r + 18qr ) + r .....(1) 2

)

a 3 = 216q 3 + 324q 2 + 162q + 27

2. Show that the cube of a positive integer of the form 6q + r , q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r . (NCERT Exemplar)

= then putting r 3 in eq. (1) , we get Case 4: r 3,=

Hence, one of every three consecutive positive integers is divisible by 3.

(

= a 3 6 36q 3 + 36q 2 + 12q + 1 + 2

(

)

3 a= 6 36q 3 + 90q 2 + 75q + 20 + 5 3

(

a =6m + 5, where m = 36q + 90q 2 + 75q + 20 is an integer.

)

Hence, the cube of a positive integer of the form 6q + r , q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the forms 6m, 6m, 6m +1, 6m + 2, 6m + 3, 6m + 4, 6m + 5, i.e., 6m + r .

)

= a 3 216q 3 + 108q 2 + 18q= + 1 6 36q 3 + 18q 2 + 3q + 1 3. Prove that

3 is an irrational number.

(CBSE 2019, 2023, 2024)

Topper’s Answer

Chapter 1 • Real Numbers

4. Show that there is no positive integer n for which n − 1 + n + 1 is rational. Sol. Let there be a positive integer n for which n − 1 + n + 1 be a rational number. p n − 1 + n + 1 = ,where p , q are integers ............(i) q and q ≠ 0 1 q = n −1 + n +1 p

p 2 + 2q 2 ⇒ 2 n +1 = pq p 2 + 2q 2 ⇒ n +1 = 2pq p 2 + 2q 2 is rational, n + 1 must be a perfect 2pq square of a positive integer.... (A)

Since

Subtracting (ii) from (i), we get

p 2q q n −1 + n +1 − n +1 + n −1 = − = q p p n −1 + n +1 n −1 − n +1 2 2 p − 2q ⇒ 2 n −1 = pq n −1 − n +1 q ⇒ = n − 1 − (n + 1) p p 2 − 2q 2 ⇒ n −1 = 2pq n +1 − n −1 q ⇒ = 2 p p 2 − 2q 2 Since is rational, n − 1 must be a perfect 2q 2pq ⇒ n + 1 − n − 1 = .........................................(ii) p square of a positive integer.... (B) Adding (i) and (ii), we get From (A) and (B), both n + 1 and n − 1 are p 2q perfect squares of positive integers. n −1 + n +1 + n +1 − n −1 = + q p ⇒

(

n −1 − n +1

)(

)

However, this contradicts the fact that two perfect squares differ at least by 3.

Hence, there is no positive integer n for which n − 1 + n + 1 is rational.

Competency Based Questions Multiple Choice Questions = 22 × 3x , b = 22 × 3 × 5, c = 22 × 3 × 7 and 1. If a LCM (a, b, c) = 3780, then x is equal to (CBSE 2024) (a) 1 (b) 2 (c) 3 (d) 0 2. The smallest irrational number by which 20 should be multiplied, so as to get a rational number, is

(CBSE 2024)

(a)

(b)

(d)

3. Let p be a prime number and k be a positive integer. If p divides k 2, then which of these is definitely divisible by p? k 2

(a) only k (c) only k, 7k and k 3

k3 (b) only k and 7k k (d) all , k , 7k and k 3 2

4. Which of the following is the rationalised form 5 of ? (CBSE QB) 3+ 2 3+ 2 (a) 5 (b) 5 (c) 15 − 2

(d) 15 − 10

5. If two positive integers a and b are written as a = x 3 y 2 and b = xy 3, where x , y are prime numbers, then the result obtained by dividing the product of the positive integers a, b by the LCM ( a , b ) is (a) xy (b) xy 2 (c) x 3 y 3 (d) x 2 y 2 6. If the LCM of two prime numbers p and q (p > q ) is 221, then the value of 3p − q is (a) 4 (b) 28 (c) 38 (d) 48 7. Three bells ring at intervals of 3, 5 and 18 minutes. All the three rang at 6 a.m. When will they ring together again? (a) 6:25 a.m. (b) 6:30 a.m. (c) 7:30 a.m. (d) 9:00 a.m. 8. If n is a natural number, then 92n − 42n is always divisible by (a) 5 (b) 13 (c) both 5 and 13 Chapter 1 • Real Numbers

(d) none of these

9. The remainder when the square of any prime number greater than 3 is divided by 6, is (a) 1 (b) 3 (c) 2 (d) 4

Select Response Type MCQs 1. n is a natural number such that n > 1. Which of these can definitely be expressed as a product of primes? n (i) n (ii) n (iii) 2 (a) only (i) (b) only (i) and (ii) (c) all (i), (ii) and (iii) (d) Cannot be determined without knowing n 2. The HCF of k and 93 is 31, where k is a natural number. Which of these can be true for some values of k ? (i) k is a multiple of 31 (ii) k is a multiple of 93 (iii) k is an even number

(iv) k is an odd number

(a) only (ii) and (iii) (b) only (i), (ii) and (iii) (c) only (i), (iii) and (iv) (d) all (i), (ii), (iii) and (iv)

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): The number 5n cannot end with the digit 0, where n is a natural number. Reason (R): Prime factorisation of 5 has only two factors 1 and 5. (CBSE 2023)

2. Assertion (A): The HCF of two numbers is 18 and their product is 3042, then their LCM is 169.

2. A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

Reason (R): For any two positive integers a and b HCF (a, b) + LCM (a, b) = a × b

Case Study Based Questions 1. Ajay wants to host a party on his 50th birthday in a large banquet hall having a certain number of chairs. He wants that guests should sit in different groups like in pairs, triplets, quadruplets, fives, and sixes, etc. When the banquet hall manager arranges chairs in such a pattern like in 2’s, 3’s, 4’s, 5’s and 6’s then 1, 2, 3, 4 and 5 chairs are left respectively. But, when he arranges in groups of 11’s no chair is left. (a) How many chairs are in the banquet hall? (i) 407 (ii) 209 (iii) 539 (iv) 149

(CBSE QB)

(b) If three chairs are removed, then the remaining chairs can be arranged in groups of (i) 2’s (ii) 3’s (iii) 4’s (iv) 5’s (c) If one chair is added, then the total number of chairs can be arranged in groups of (i) 2’s (ii) 3’s (iii) 4’s (iv) 11’s (d) If one chair is added to the total number of chairs, how many chairs will be left when arranged in groups of 11’s? (i) 1 (ii) 2 (iii) 3 (iv) 4

(a) Write the LCM of 60, 84 and 108. (b) If in each room the same numbers of participants are to be seated and all of them being in the same subject, what is the maximum number of participants that can be accommodated in each room. (c) (i) Write the product of HCF and LCM of 60, 84 and 108. OR

OR (e) How many chairs will be left in the original arrangement if some number of chairs is arranged in groups of 9’s? (i) 8 (ii) 1 (iii) 6 (iv) 3

(ii) W hat is the minimum number of rooms required during the event?

Answers Multiple Choice Questions

22 × 3x −1 × 5 × 7 = 3780 3780 x −1 ⇒3 = 22 × 3 × 5 × 7

1. (c) 3 Given, 2

⇒ 3x −1 = 9 ⇒ 3x −1 = 32

x −1

a = 2 ×3 = 2 ×3 ×3

On comparing the exponents of 3, we get

b = 22 × 3 × 5, c = 22 × 3 × 7

x −1 = 2 ⇒ x = 3

Here, LCM(a , b , c ) = 22 × 3 × 3x −1 × 5 × 7 (i) And given that, LCM(a, b, c) = 3780 Thus, from Eqs. (i) and (ii), we get 12

(ii)

2. (d)

We have, 20 = 2 5 From the given options, if we multiply by

i.e., 2 5 × 5, we will get a rational number.

3. (c) only k, 7k, and k 3

Remainder when 34n − 24n divided by 5 = 1 − 1 = 0

We know that by theorem, if p divides k 2, then p divides k, where p is a prime number and k is a positive integer.

This expression is always divisible by both 5 and 13.

So, p divides k, 75k, and k 3.

(6k ± 1)2= 36k 2 ± 12k + 1

We have,

3+ 2

3− 2

We have, Any prime greater than 3 is of the form 6k ± 1

4. (d) 15 − 10 5

9. (a) 1

3− 2

( 3 − 2= ) 3−2

The remainder when divided by 6 is always 1. 15 − 10

Select Response Type MCQs

5. (b) xy 2

1. (b) Only (i) and (ii)

We have,

The original number n can definitely be expressed as a product of primes.

3 2 = a x= y ,b xy 3

 taking the highest power  LCM ( a , b ) = x 3 y 3    of each prime factor x 3 y 2 × xy 3 a ×b = LCM(a , b ) x3y 3 x4y5 = = x 4 −3= y 5−3 xy 2 x3y 3 6. (c) 38 p ×q = 221

⇒ n= k Since k is a natural number, it can always be expressed as a product of prime factors.

On dividing n by 2 introduces another factor and might not result in a prime number.

221 = 13 × 17 Since p > q, let p = 17 and q = 13 3p − q = 3 (17 ) − 13 = 51 − 13 = 38 7. (c) 7:30 a.m. We have, Given intervals: 3 min, 5 min, 18 min Finding LCM(3, 5, 18): 3 = 31, 5 = 51, 18 = 2 × 32 LCM = 2 × 32 × 5 = 90 The bells rang together at 6:00 a.m. so they will ring together again after 90 minutes. 6:00 a.m. + 90 mins = 7:30 a.m. 8. (c) both 5 and 13 We have, 92n − 42n 2 2n

− (2 )

= 3 −2

Using Number Theory, Remainder when 34n divided by 5 = 1 Remainder when 24n divided by 5 = 1 Chapter 1 • Real Numbers

2. (c) only (i), (iii) and (iv) Since k must be a multiple of 31, (i) is true. Here, if k is a multiple of 93, then HCF of k and 93 cannot be 31, so (ii) is not true. The number 31 is an odd number. Thus, k cannot be even, so (iii) is not true. Since k is a multiple of 31, and 31 is odd, (iv) is true.

Prime Factorisation:

(3 )

Let: n = k 2 where k is a natural number.

The original number n can definitely be expressed as a product of primes.

We have,

2 2n

Since n is a natural number, this means that n is a perfect square of some natural number.

Assertion-Reason Based Questions 1. (a) Both A and R are true, and R is the correct explanation of A. Assertion: The expression 5n contains only the factor 5, raised to some power, and does not include the factor 2. Therefore, 5n lacks the necessary factor of 2 to be divisible by 10, and as such, it cannot end in 0. Reason: The reason given is that the prime factorisation of 5 has only two factors: 1 and 5. This statement is technically correct in describing 5 as a prime number because prime numbers only have themselves and 1 as their factors. However, this explanation is not pertinent or adequate to explain 13

why 5n does not end in the digit 0. The critical missing factor for ending in 0 is not 2, which isn’t mentioned in the reason.

●

2. (c) A is true, but R is false. We know that HCF × LCM = Product of numbers ∴18 × 169 = 3042

Case Study Based Questions 1. (a) (iii) 539 The first five conditions follow a pattern where N is 1 less than the LCM of the numbers 2, 3, 4, 5, 6. The LCM of these numbers is: LCM (2, 3, 4, 5, 6) = 60 Thus, we can rewrite N as: = N 60k − 1 k = 1: N = 60 × 1 − 1= 59

●

107 with a remainder hen divided by 5: 539 ÷ 5 = W of 4. 89 with a remainder hen divided by 6: 539 ÷ 6 = W of 5. 49 with no hen divided by 11: 539 ÷ 11 = W remainder.

All the conditions are satisfied! (b) (i) 2’s If 3 chairs are removed: N − 3= 539 − 3= 536 Checking divisibility: 536 ÷ 2 = 268 ( divisible by 2) Thus, the remaining chairs can be arranged in groups of 2’s. (c) (i) 2’s If one chair is added: N + 1= 539 + 1= 540 Checking divisibility: 540 ÷ 2 = 270 ( divisible by 2 )

55 and Is 59 divisible by 11? No, because 11 × 5 = 11 × 6 = 66, so 59 is not divisible by 11.

Thus, the new total can be arranged in groups of 2’s.

k = 2: N = 60 × 2 − 1 = 119

(d) (i) 1

110 Is 119 divisible by 11? No, because 11 × 10 = and 11 × 11 = 121, so 119 is not divisible by 11.

If one chair is added:

k = 3: N = 60 × 3 − 1 = 179 176 Is 179 divisible by 11? No, because11 × 16 = 187, so 179 is not divisible by 11. and 11 × 17 = k = 4: N = 60 × 4 − 1 = 239 231 Is 239 divisible by 11? No, because 11 × 21 = and 11 × 22 = 242, so 239 is not divisible by 11.

N + 1 = 540 49 R 1 540 ÷ 11 = (e) (i) 8 59 R 8 539 ÷ 9 = 3780 2. (a) LCM (60, 84, 108) = 2 × 2 × 3 × 5 × 7 × 9 =

k = 5: N = 60 × 5 − 1 = 299

(b) Maximum number of participants that can be accommodated in each room = HCF = (60, 84, 108)

Is 299 divisible by 11? No, because 11 × 27 = 297 308, so 299 is not divisible by 11. and 11 × 28 =

60 = 2 × 2 × 3 × 5

k = 6: N = 60 × 6 − 1 = 359

84 = 2 × 2 × 3 × 7 108 = 2 × 2 × 3 × 3 × 3

Is 359 divisible by 11? No, because 11 × 32 = 352 363, so 359 is not divisible by 11. and 11 × 33 = k = 9: N = 60 × 9 − 1 = 539 539 Is 539 divisible by 11? No, because 11 × 49 = Let’s check if N = 539 satisfies all the conditions: ● When divided by 2: 539 ÷ 2 = 269 with a remainder of 1. ●

●

179 with a remainder hen divided by 3: 539 ÷ 3 = W of 2. 134 with a remainder hen divided by 4: 539 ÷ 4 = W of 3.

So, HCF ( 60, 84, 108 ) = 2 × 2 × 3 = 12 Hence, required number of participants = 12 (c) (i) The product of HCF and LCM of 60, 84 and 108 12 3780 = 45360 = LCM × HCF =× OR (ii) Minimum number of rooms required during the event =

sum of all the participants 60 + 84 + 108 = = 21 HCF of participants 12

Additional Practice Questions 1. If n = 23 × 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, is (a) 2 (b) 3 (c) 4 (d) 7 2. If HCF(26, 169) = 13, then LCM(26, 169) = (a) 26 (b) 52 (c) 338 (d) 13 3.

5 + 3 + 2 is (a) a natural number

(b) an integer

(d) an irrational number

(b) only (i) and (iii)

(d) only (ii) and (iii)

6. The pair of positive integers, where sum is 91 and HCF is 13, is/are (i) (13, 78) (ii) (26, 65) (iii) (39, 52) (a) only (i)

(b) only (i) and (ii)

(d) all (i), (ii) and (iii)

7. Find the values of x, y, z and w in the following factor tree. Also write the prime factorisation of x. x 2 2

819

w 13

8. Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers? 9. Find the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 13 respectively.

Chapter 1 • Real Numbers

3 5 12

5+3= 8

8+5= 13

13 + 8 = 21 21 + 13 = 34 34 + 21 = 55 55 + 34 = 89 89 + 55 = 144 Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... Based on the given sequence, answer the following questions: (a) The next two numbers in the sequence are (i) 223 and 367 (ii) 233 and 367 (iii) 223 and 377

(iv) 233 and 377

(b) A rational number has a terminating decimal expansion, if the denominator is of the form 2m × 5n

(ii) 3m × 5n

(iii) 2m × 3n

(iv) 3m × 4n

where m, n are non-negative integers.

3+2 = 5

(i) y

2 +1 = 3

(c) Which of the following rational numbers (formed by taking the ratio of a term to its previous term) has a terminating decimal expansion? (i)

5 3

(ii)

21 13

(iii)

13 8

(iv)

34 21

(d) HCF of two consecutive numbers in the given sequence (except the first 3 terms) is (i) 1 (ii) 0 (iii) 2 (iv) 3

Challenge Yourself 1. P is the LCM of 2, 4, 6, 8, 10, Q is LCM of 1, 3, 5, 7, 9 and L is the LCM of P and Q. Then, which of the following is true? (a) L = 21P (b) L = 4Q (c) L = 63P (d) L = 16Q 8 8 6 4 2. If a = 6 − 8 5, b = 6 + 6 5, c = 6 + 6 5, d = 6 + 4 5, = e 4 6 + 4 5, then which of the following is a rational number? (a) abcde (b) abde (c) ab (d) cd

3. Find the correct value of 1 1 1 + + +… upto 91 terms. 9 + 10 10 + 11 11 + 12 1 ( Hint: Rationalise each term of form n + n +1

Scan me

multiply the numerator and denominator by the conjugate of the denominator, which is n − n + 1 ) 4. If X= 28 + (1 × 2 × 3 × 4 ×…× 16 × 28 ) and Y= 17 + (1 × 2 × 3 ×…× 17 ), then which of the following is/are true? (i) X is a composite number (ii) Y is a prime number (iii) X − Y is a prime number (iv) X + Y is a composite number (a) Both (i) and (iv)

(b) Both (ii) and (iii)

(d) Both (i) and (ii)

Answers Additional Practice Questions

10. (a) (iv) 233 and 377 (b) (i) 2m × 5n

1. (b) 3

2. (c) 338

3. (d) an irrational number

4. (a) 5

5. (b) only (i) and (iii)

6. (d) all (i), (ii) and (iii)

8. 30

9. 17

= = y 1639, = z 273, = w 91 7. x 3276,

13 (d) (i) 1 8

Challenge Yourself 1. (a) L = 21P

2. (b) abde

3. 7 4. (a) Both (i) and (iv)

Scan me for Exemplar Solutions 16

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. If p and q are two distinct prime numbers, then LCM (p, q) is (a) 1 (b) p (c) q

(d) pq

2. The sum of exponents of prime factors in the prime factorisation of 196 is (a) 3 (b) 4 (c) 5

(d) 2

3. The product of a non-zero rational and an irrational number is: (a) always irrational. (b) always rational. (c) rational or irrational.

(d) one.

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): (2 + 3 ) 3 is an irrational number. Reason (R): Product of two irrational numbers is always irrational. 5. Assertion (A): If HCF of 510 and 92 is 2, then the LCM of 510 and 92 is 32460. Reason (R): As HCF (a, b) × LCM (a, b) = a × b.

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. Prove that 6 − 4 5 is an irrational number, given that 5 is an irrational number. 7. Show that 11 × 19 × 23 + 3 × 11 is not a prime number.

Short Answer Questions

[4 × 3 = 12 Marks]

8. Two alarm clocks ring their alarms at regular intervals of 20 minutes and 25 minutes respectively. If they first beep together at 12 noon, at what time will they beep again together next time? 9. Find the greatest number of six digits exactly divisible by 18, 24 and 36. 10. Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm. 11. The LCM of two numbers is 9 times their HCF. The sum of LCM and HCF is 500. Find their HCF.

Long Answer Questions

[3 × 5 = 15 Marks]

12. In a teacher’s workshop, the number of teachers teaching French, Hindi, and English are 48, 80, and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are selected and all of them are of the same subject.

Self-Assessment

13. Grow more Plantations have two rectangular fields of the same width but different lengths. They are required to plant 84 trees in the smaller field and 231 trees in the larger field. In both the fields, the trees will be planted in the same number of rows but in different number of columns. (a) What is the most number of rows that can be planted in this arrangement? (b) If the trees are planted in the number of rows obtained in part (i), how many columns will each field have? 14. Prove that

2 + 3 is an irrational number.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections− section A and section B of grade X. There are 32 students in section A and 36 students in section B.

(a) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of section A or section B? (b) If the product of two positive integers is equal to the product of their HCF and LCM is true, then find the HCF (32, 36). (c) Express 36 as a product of primes. (d) If p and q are positive integers such that p = ab2 and q = a2b, where a, b are prime numbers, then find the LCM (p, q).

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks] q

1. If p q and q are natural numbers and p is the multiple of q, then what is the HCF of p q and q ? (b) p

(a) pq

(CBSE 2023)

(d) p + q

2. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is: (a) 13

(b) 65

(

(d) 1,750

(d) 2

)

3. If n is a natural number, then 2 5n + 6n always ends with (a)1

(b) 4

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

4. Assertion (A): The number 5n cannot end with the digit 0, where n is a natural number. Reason (R): Prime factorisation of 5 has only two factors 1 and 5. 5. Assertion (A):

a is a rational number, where a is a prime number.

Reason (R): The square roots of all positive integers are irrationals.

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. Given that 2 is irrational, prove that (5 + 3 2 ) is an irrational number. 7. Find the least number that is divisible by all the numbers from 1 to 10 (both inclusive).

Short Answer Questions

[4 × 3 = 12 Marks]

8. Ravi and Sikha drive around a circular sports field. Ravi takes 16 minutes to take one round while Shikha completes the round in 20 minutes. If both start at the same point at the same time and go in same direction, after how much time will they meet at the starting point? 9. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer. 10. Find the HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of two given numbers. 11. Prove that

3+ 5 is an irrational number, given that 5 is an irrational number. 5

Long Answer Questions

[3 × 5 = 15 Marks]

12. National Art convention got registrations from students from all parts of the country, of which 60 are interested in music, 84 are interested in dance and 108 students are interested in handicrafts. For optimum cultural exchange,

Self-Assessment

organisers wish to keep them in minimum number of groups such that each group consists of students interested in the same art form and the number of students in each group is the same. Find the number of students in each group. Find the number of groups in each art form. How many rooms are required if each group will be allotted a room? 13. Six bells commence tolling together and toll at interval of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together? 14. Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A garden consists of 135 rose plants planted in certain number of columns. There are another set of 225 marigold plants, which is to be planted in the same number of columns.

Read carefully the above paragraph and answer the following questions: (a) What is the maximum number of columns in which they can be planted? (i) 45 (ii) 40 (iii) 15

(iv) 35

(b) Find the total number of plants. (i) 135 (ii) 225

(iv) 45

(iii) 360

(c) Find the sum of exponents of the prime factors of the maximum number of columns in which they can be planted. (i) 5 (ii) 3 (iii) 4 (iv) 6 (d) What is the total numbers of row in which they can be planted? (i) 3 (ii) 5 (iii) 8

(iv)15

OR (e) Find the sum of exponents of the prime factors of total number of plants. (i) 2 (ii) 3 (iii) 5

(iv) 6

Polynomials

Chapter at a Glance Polynomial Definition A polynomial is an algebraic expression composed of constants, variables and non-negative integer exponents combined through addition, subtraction and multiplication.

Degree of a Polynomial The highest power of the variable in the polynomial 2 Example: In p(x ) = 3x − 2x + 7, the degree is 2.

Types of Polynomials Linear Polynomial: Expressions with degree 1 (e.g., 2x + 32) Quadratic Polynomial: Expressions with degree 2 (e.g., x 2 − 4x + 4) Cubic Polynomial: Expressions with degree 3 (e.g., x 3 − 3x 2 + 3x − 1) 1

Note: x 2 + 5 is not a polynomial, as the power of x is not a whole number.

Zeros/Roots of a Polynomial The values of x for which p ( x ) = 0. Example: Let p ( x ) = x 2 − 5x + 6

p ( 2) = 22 − 5( 2) + 6 = 4 − 10 + 6 = 10 − 10 = 0 p ( 3) = 32 − 5( 3) + 6 = 9 − 15 + 6 = 15 − 15 = 0

⇒ 2 and 3 are the zeros or roots of polynomial p ( x )

Relationship between Zeros/Roots and Coefficients

For a quadratic polynomial p ( x ) = ax 2 + bx + c , given roots are α and β : b  Coefficient of x  =−   Coefficient of x 2  a

●

Sum of zeros, α + β =−

●

Product of zeros, α × β =

c  Constant term  = a  Coefficient of x 2 

3 2 For a cubic polynomial p ( x ) = ax + bx + cx + d , Given roots are α , β and γ :

 Coefficient of x 2  b =−   a  Coefficient of x 3 

●

Sum of zeros, α + β + γ =−

●

c  Coefficient of x  Sum of product of zeros, αβ + βγ + γα = =  a  Coefficient of x 3 

●

−d  Constant term  Product of zeros, α × β × γ = =−   Coefficient of x 3  a

Chapter 2 • Polynomials

Geometrical Meaning of the Zeros of a Polynomial ●

The zeros of a polynomial are the x-coordinates where the graph of the polynomial intersects the x-axis.

●

The number of zeros of a polynomial is equal to the degree of the polynomial. Linear polynomial is a straight line. Degree = 1 Zeros = 1

Quadratic polynomial is a parabola. Degree = 2 Zeros = 2

y = ax + b

Cubic polynomial has an S-shape. Degree = 3 Zeros = 3

y = ax2 + bx + c

y = ax3 + bx2 + cx + d

NCERT Zone NCERT Exercise 2.1

(e) The number of zeros are 4 as the graph intersects x-axis at 4 points.

1. The graph of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeros of p(x), in each case. Sol. (a) The number of zeros is 0 as the graph does not intersect x-axis at any point.

Y X'

(f) The number of zeros are 3 as the graph intersects the x-axis at 3 points.

Y' Y X'

(b) The number of zeros is 1 as the graph intersect the x-axis at only 1 point.

Y X'

1. Find the zeros of following quadratic polynomials and verify the relationship between the zeros and the coefficients.

(d) The number of zeros are 2 as the graph intersects the x-axis at 2 points.

Y X'

(a) x 2 − 2x − 8

(b) 4s 2 − 4s + 1

(d) 4u 2 + 8u

(e) t 2 − 15

(f) 3x 2 − x − 4

Sol. (a) x 2 − 2x − 8

NCERT Exercise 2.2

x 2 − 4 x + 2x − 8

x (x − 4) − 2(x − 4)

( x − 4 )( x − 2 )

The value of x2 – 2x – 8 is zero when x – 4 = 0 or x +2= 0, i.e.

x = 4 or x = – 2

Therefore, the zeros of x2 – 2x – 8 are 4 and – 2.

Sum of zeros = 22

− ( Coefficient of x )

(Coefficient of x ) 2

 −2 = − =2  1 

Sum of root by adding the actual roots = 4 + ( −2 ) = 4 − 2 = 2 ( Constant term ) = − 8 Product of zeros = 1 Coefficient of x 2 = – 8 Product of zeros =4 × ( −2 ) =− 8 Hence verified.

(

)

(b) 4s 2 − 4s + 1

4s 2 − 2s − 2s + 1

2s ( 2s − 1) − 1( 2s − 1)

( 2s − 1)( 2s − 1)

( 2s − 1)

1 3 3 1 − = − Product of zeros = − × = 3 2 6 2 Hence verified.

(d) 4u2 + 8u

4u (u + 2)

The value of 4u2 + 8u is zero when 4u = 0 or

( Constant term )

0 = =  0, i.e. 2 4 Coefficient of u

(

− ( Coefficient of s )

(Coefficient of s ) 2

= −

−4 =1 4

1 1 1 × = 2 2 4

( Constant term )

Product of zeros =

(Coefficient of s ) 2

1 4

Hence verified.

u = 0 or u = – 2

Sum of zeros = − ( Coefficient of u )

(Coefficient of u ) 2

Product of zeros =

0 = =  0 4 Coefficient of u 2

(

( Constant term )

)

Product of zeros =−2 × 0 =0

Hence verified.

(e) t 2 – 15

( 15 )

t2 –

Using identity of (a2 –b2) = (a – b)(a + b)

(t − 15 )(t + 15 )

The value of t 2 – 15 is zero when t − 15 = 0 or t + 15 = 0 , i.e.

6x 2 − 7x − 3

6x 2 − 9x + 2x − 3

3x ( 2x − 3 ) + 1( 2x − 3 )

Sum of zeros =

( 2x − 3)( 3x + 1)

The value of 6x 2 − 7x − 3 is zero when 3x + 1 =0 or 2x − 3 = 0, i.e. 1 3 x = − or x = 3 2

Sum of zeros =

− ( Coefficient of x )

(Coefficient of x ) 2

Sum of zeros are =

Product of zeros =

( Constant term )

(Coefficient of x )

Chapter 2 • Polynomials

 7 7 =−  −  =  6 6

3  1 3 1 9 2 7 +−  = − = − = 2  3 2 3 6 6 6 1  3 = − −  = 6 2  

8 = −  = −2 4

Sum of zeros =−2 + 0 =−2

)

Therefore, the zeros of 4u 2 + 8u are 0 and – 2.

Product of zeros =

0 The value of 4s2 – 4s + 1 is zero when 2s − 1 =, 1 i.e. s = 2 1 1 Therefore, the zeros of 4s2 – 4s + 1 are and . 2 2 1 1 Sum of zeros = + = 1 2 2 Sum of zeros =

15 or t = − 15

− ( Coefficient of t )

( Coefficient of t ) 2

0 = −  = 0 1

0 Sum of zeros = 15 − 15 =

Product of zeros =

( Constant term )

(Coefficient of t ) 2

 −15  =  = −15  1 

(

)

Product of zeros = 15 × − 15 =−15

Hence verified.

(f) 3x 2 − x − 4

3x 2 + 3x − 4x − 4

3x ( x + 1) − 4 ( x + 1)

( x + 1)( 3x − 4 ) 23

0 The value of 3x 2 − x − 4 is zero when 3x − 4 = 0 i.e. or x + 1 = 4 when x = or x = –1 3 4 4 3 1 Sum of zeros = + ( −1) = − = 3 3 3 3 Sum of zeros =

− ( Coefficient of x )

(Coefficient of x ) 2

Product of zeros =

( Constant term )

(Coefficient of x ) 2

 −1  1 = −  =  3  3

4 4 × ( −1) =− 3 3 4  −4  =  =− 3 3  

Hence verified.

2. F ind a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively. 1 (a) , − 1 4 (d) 1, 1

1 3 1 1 (e) − , 4 4 (b)

1 , −1 4 Let the polynomial be ax2 + bx + c and its zeros be α and β. 1 b − α + β = = 4 a c αβ =−1 = a Sol. (a)

Equation can be written as

x − (sum of the roots) x + (product of the roots) = 0

x 2 − (α + β ) x + αβ = 0

1 x2 − x −1 = 0 4

(b)

4x 2 − x − 4 = 0 2,

1 3

Let the polynomial be ax2 + bx + c and its zeros be α and β . 24

b α + β =2 = − a 1 c αβ= = 3 a

Equation can be written as

1 0 x 2 − 2x + = 3 3x 2 − 3 2x + 1 = 0

Equation can be written as

x 2 − ( sum of roots ) x + ( product of roots ) = 0

x 2 − (α + β ) x + αβ = 0

x 2 − 0x + 5 = 0

x2 + 5 = 0

(d) 1, 1 Let the polynomial be ax2 + bx + c and its zeros be α and β . b 1 − α + β == a c αβ = 1= a Equation can be written as

x 2 − ( sum of roots ) x + ( product of roots ) = 0

x 2 − (α + β ) x + αβ = 0

x 2 − x + 1 =0

1 1 (e) − , 4 4 Let the polynomial be ax2 +bx+ c and its zeros be α and β . 1 b − = − α + β = 4 a 1 c αβ= = 4 a Equation can be written as

x 2 − ( sum of roots ) x + ( product of roots ) = 0

x 2 − (α + β ) x + αβ = 0

1  1 x 2 −  − x + = 0 4  4 1 1 x2 + x + = 0 4 4 4x 2 + x + 1 = 0

(f) 4, 1 x 2 − ( sum of roots ) x + ( product of roots ) = 0 Let the polynomial be ax2 + bx + c and its zeros 2 x − (α + β ) x + αβ = 0 be α and β .

b 4 − α + β == c a αβ = 1= a Equation can be written as

x 2 − ( sum of roots ) x + ( product of roots ) = 0

x 2 − (α + β ) x + αβ = 0

x 2 − 4x + 1 = 0

Multiple Choice Questions 1. Which of the following is not the graph of a quadratic polynomial? (a) (b) Y Y X'

(c)

7. If the sum of the zeros of the quadratic polynomial ax 2 + bx + c is equal to the sum of the squares of their reciprocals, then: 2 c ab 2 + bc 2 (a) 2a= (b) b 2 = 4ac

(d)

(CBSE 2011)

3. If the zeros of the quadratic polynomial x 2 + ( a + 1) x + b are 2 and -3, then: (a) a = (b) a = 5, b = −1 −7, b = −1 (d) a = 0, b = −6 (CBSE 2012)

4. The number of zeros of the polynomial x 3 − 6x 2 + 11x − 6 is: (a) 1 (b) 2 (c) 3 (d) 0 (CBSE 2013)

5. If α and β are the zeros of the polynomial x 2 − 5x + 6 , then α + β is: (a) 5 (b) 6 (c) 1 (d) 11 (CBSE 2014)

6. If α and β are the zeros of the polynomial 3x 2 − 8x + 5 then aβ is: 8 (a) 3

8 (b) − 3

5 (c) 3

5 (d) − 3

(CBSE 2016) 2

8. The zeros of the polynomial x − 4x + 4 are: (a) 2 and -2 (b) 2 and 2 (c) -2 and -2 (d) 4 and 0 (CBSE 2017)

9. If one zero of the polynomial 3x − kx + 4 is 2, then

2. A quadratic polynomial whose zeros are -3 and 4 is: (a) x 2 − x + 12 (b) x 2 + x + 12 2 x x (c) (d) 2x 2 + 2x − 24 − −6 2 2

2 = a 2c − ab 2 (d) 2bc

2 = a 2c − bc 2 (c) 2ab

the value of k is: (a) 10 (b) 8

(d) 6

(CBSE 2018)

10. The quadratic polynomial whose sum and product of zeros are -1 and -20 respectively is: 2 (a) x + x − 20

2 (b) x − x + 20

2 (c) x + x + 20

2 (d) x − x − 20

(CBSE 2019)

11. If the zeros of the polynomial x 2 + px + q are equal, then: (a) p 2 = 4q

(b) p = q

(d) p = 4q (CBSE 2020)

12. The value of k for which the polynomial

x 2 + kx + 25 has equal zeros is:

(a) 10

(b) 25

(d) -5

13. The quadratic polynomial whose zeros are 4 and -4 is:

2 (a) x − 8x − 8

2 (b) x + 8x − 8

2 (c) x + 16

2 (d) x − 16

(CBSE 2021)

14. The degree of the polynomial 4x 5 − 3x 3 + 2x − 7 is: (a) 1 (b) 3 (c) 5 (d) 7 (CBSE SQP 2023)

15. If α and β are the zeros of the polynomial x 2 − 7x + 10, then the value of

(α + β )2 − (α − β )2 − 3αβ is:

(a) 10

(b) 7

(d) -7 (NCERT Exemplar)

Chapter 2 • Polynomials

16. If one zero of the quadratic polynomial x 2 + 3x + k 2 (k + 5 − 5 ) is 2, then the value of is: 3 −1 (a) 10

(b) -10

(d) -5

(CBSE 2010)

17. If the zeros of the polynomial x 2 + px + q are reciprocal to each other, then: (a) p = q (b) q = 1

(NCERT Exemplar)

18. The sum of the zeros of the polynomial 2x 2 − 3x + 7 is: (a)

3 2

(b) −

3 2

(c)

7 2

(d) −

7 2

(NCERT Exemplar)

19. If α , β , γ are the zeros of the polynomial f ( x ) = ax 3 + bx 2 + cx + d , then α 2 + β 2 + γ 2 = (a) (c)

b 2 − ac

(b)

b + 2ac b

(d)

b 2 − 2ac a

(d) a = 0, b = −6

(CBSE Standard 2023, CBSE Basic 2022)

28. If one of the zeros of the cubic polynomial x 3 + ax 2 + bx + c is −1, then the product of the other two zeros is (a) b − a + 1 (b) b − a − 1 (d) a − b − 1

(CBSE 2018)

b − 2ac a

(CBSE 2022)

2 (d) x + 2x + 15

21. If the sum of zeros of the polynomial p (x ) = 2x 2 − k 2x + 1 is 2, then value of (c) 2 2

29. If one of the zeros of the quadratic polynomial (k − 1) x 2 + kx + 1 is −3, then the value of k is (a)

(CBSE 2023)

(b) 2

27. If the zeros of the quadratic polynomial x 2 + ( a + 1) x + b are 2 and -3, then −7, b = −1 (a) a = (b) a = 5, b = −1

100 + 10 5 + − 1 and -3 is: 4 25 2 (a) x 2 − 2x − 15 (b) x + 2x − 15

k is: (a) 2

26. What should be added to the polynomial x 2 − 5x + 4, so that 3 is the zero of the resulting polynomial? (a) 1 (b) 2 (c) 4 (d) 5

20. The quadratic polynomial whose zeros are

2 (c) x − 2x + 15

(CBSE 2022)

(CBSE 2020)

(d) p = −1

25. The graph of a polynomial P ( x ) cuts the x-axis at 3 points and touches it at 2 other points. The number of zeros of P ( x ) is (a) 1 (b) 2 (c) 3 (d) 5

4 3

(b) −

4 3

(c)

2 3

(d) −

2 3

(NCERT Exemplar)

30. If α , β are the zeros of the polynomial f ( x ) = x 2 − p ( x + 1) − c such that (α + 1)( β + 1) = 0, then c = (a) 1 (b) 0 (c) -1 (d) 2

(NCERT Exemplar)

(d)

1 2

31. Which of the following is not the graph of a quadratic polynomial? (a)

(b)

(c)

(d)

(CBSE 2024) 2

22. The zeros of a polynomial x + px + q are twice the zeros of the polynomial 4x 2 − 5x − 6 . The value of p is: 5 5 (a) − (b) (c) -5 (d) 10 2 2

(CBSE 2024)

4, then 23. If x + 2 is a factor of x 2 + ax + 2b and a + b = = a 1, = b 3 = a 3, = b 1 (a) (b)

(d) a = 5, b = −1

24. If two of the zeros of the cubic polynomial ax 3 + bx 2 + cx + d are each equal to zero, then the third zero is b c b d (a) – (b) (c) – (d) a a a a

(CBSE 2023)

(NCERT Exemplar)

32. The number of polynomials having zeros as -2 and 5 is (a) 1 (b) 2 (c) 3

(d) more than 3

Answers 1. (c)

x2 x − −6 2 2 4. (c) 3 5 6. (c) 3 8. (b) 2 and 2

15. (a) 10

16. (b) -10

17. (b) q = 1

3 2 2 20. (a) x − 2x − 15 5 22. (a) − 2 18. (a)

2. (c)

14. (c) 5

3. (d) a = 0, b = −6 5. (a) 5 2 c ab 2 + bc 2 7. (a) 2a=

9. (b) 8

2 10. (a) x + x − 20

2 11. (a) p = 4q

12. (a) 10

13. (d) x 2 − 16

b a 26. (b) 2 24. (c) −

19. (d)

b 2 − 2ac

21. (b) 2

23. (b) a = 3, b = 1 25. (d) 5 27. (d) a = 0, b = -6 4 29. (a) 3

28. (a) b − a + 1 30. (a)1

31. (d) It crosses the x-axis at three points. 32. (d) more than 3

Constructed Response Questions Very Short Answer Questions 1. Write the degree of the polynomial 5x 4 + 7x 3 + 2x + 1. (CBSE 2021) 4

Sol. The degree of the polynomial 5x + 7x + 2x + 1 is the highest power of x, which is 4. 2. If one zero of the quadratic polynomial 2x 2 + 7x + k is 3, find the value of k. (CBSE 2017)

Sol. Given 2x + 7x + k , one zero is 3. Using the zero property: 2 2 ( 3) + 7 ( 3) + k = 0 18 + 21 + k = 0 k = −39 3. Find the remainder when the polynomial x3 -3x2 + 5x -7 is divided by x - 2.

(CBSE 2019)

Sol. Substitute x = 2 into x 3 − 3x 2 + 5x − 7: f (2) = (2)3 − 3(2)2 + 5(2) − 7 f (2) = (2)3 − 3(2)2 + 5(2) − 7 =8 − 12 + 10 − 7 =−1 The remainder is -1. 2 4. If α and β are the zeros of the polynomial x − 5x + 6, find α + β .

(CBSE SQP 2020)

Chapter 2 • Polynomials

−b Sol. The sum of zeros for x 2 − 5x + 6 is given by : a − 5 α + β =− = 5 1 5. Verify whether x − 2 is a factor of x 3 − 2x 2 + x − 2. Sol. Substitute x = 2 into x 3 − 2x 2 + x − 2: ( 2)3 − 2 ( 2) 2 + ( 2) − 2 = 8 − 8 + 2 − 2 = 0 Since the remainder is 0, x − 2 is a factor.

(CBSE 2018)

6. If p ( x ) = 2x 2 − 3x + 4, find p ( 2 ). Sol. Substitute x = 2:

(NCERT Exemplar)

p ( 2) = 2 ( 2) − 3( 2) + 4 = 8−6+ 4 = 6 7. If the zeros of the quadratic polynomial x 2 + px + q are -2 and 3, find the values of p and q . Sol. Zeros are -2 and 3. b α + β =− =−p a c αβ= = q a p =− ( −2 + 3) =−1,

(CBSE 2016)

q= −6 ( −2)( 3) =

8. For the polynomial 3x 3 + 4x 2 − 7x + k , find the value of k such that x + 2 is a factor of the polynomial.

Product of new zeros =

(CBSE SQP 2022)

Sol. Substitute x = −2 into 3x 3 + 4x 2 − 7x + k

⇒k = −6

αβ

1 a = c c a

4. Find the quadratic polynomial whose zeros are 3 3, 3

Short Answer Questions

Sol. Let the zeros of the quadratic polynomial be

1. If α and β are the zeros of the polynomial x 2 + x − 2, then find the value of

α β + . β α

Sol. Here α + β =−1 and αβ =−2

= α 3= 3, β (CBSE 2024)

(α + β ) − 2αβ

= x 2 − ( sum of zeros ) x + product of zeros

= x 2 − 4 3x + 9 5. If two zeros of the polynomial 2x 3 + 3x 2 − 6x − 9 are 3 and − 3, find the third zero and verify the relationship between the sum and product of the zeros. Sol. If 3 and − 3 are zeros, their corresponding factor is:

2. Find the zeros of the polynomial 4x 2 + 4x − 3 and verify the relationship between zeros and coefficient of the polynomial. Sol. P ( x ) = 4x 2 + 4x − 3

3 1 ∴ Zeros of the polynomial are − , 2 2 Sum of zeros

Product of zeros 3 1 3 Constant term =− × =− = 2 2 4 Coefficient of x 2 3. Find the quadratic polynomial whose zeros are reciprocal of the zeros of the polynomial

f ( x ) = ax 2 + bx + c , a ≠ 0, c ≠ 0. Sol. f ( x ) = ax 2 + bx + c −b c +β ,= αβ Here, α= a a

−b 1 1 β +α b Sum of new zeros = + = =a = − c α β αβ c a

)(

(

)

x − 3 x + 3 =x 2 − 3

CBSE Marking Scheme 2024

3 1 2 = − + = − = −1 2 2 2 4 − ( Coefficient of x ) = − = 4 Coefficient of x 2

Then, sum of zeros = α + β = 3 3 + 3 = 4 3 Required polynomial

αβ

= ( 2x + 3)( 2x − 1)

Product of zeros = αβ = 3 3 × 3 = 9

( −1) − 2 ( −2 ) 5 = = − −2 2 α β 5 − + = β α 2

Thus, Required quadratic polynomial b a = x 2 + x + or cx 2 + bx + a c c

3 2 3 ( −2 ) + 4 ( −2 ) − 7 ( −2 ) + k = 0 ⇒ −24 + 16 + 14 + k = 0

α β α 2 + β2 = + = β α αβ

1 2 1 2

Thus, x − 3 is a factor of p ( x ). 2

Divide p ( x ) = 2x 3 + 3x 2 − 6x − 9 by x 2 − 3 using long division. 2x + 3 2 2x 3 + 3x 2 − 6x − 9 x −3 ( − )(2x 3 − 6x ) 3x 2 − 9 ( − )( 3x 2 − 9) 0 The third zero comes from the factor 2x + 3 : Thus, 2x + 3 = 0 3 ⇒ x =− 2 The zeros of p ( x ) are 3, − 3, and − 3 . 2 (a) Sum of zeros: 2x 3 + 3x 2 − 6x − 9 The sum of zeros of a cubic polynomial ax 3 + bx 2 + cx + d is:

Coefficient of x 2

3 = − 2 Coefficient of x 3  3 Verifying: 3 + − 3 +  −  =−  2 2 −

(

Hence verified.

)

(b) Product of zeros: The product of zeros is: −6 − Constant term = − = 3 Leading coefficient 2  3 Verifying: 3 − 3  −  = 3  2 Hence verified.

( )(

)

6. Verify that 2 and −3 are the zeros of the polynomial x 3 − 4x 2 − 11x + 30, and find the third zero. Sol. Given x 3 − 4x 2 − 11x + 30, Substitute x = 2 : ( 2 )3 − 4 ( 2 )2 − 11( 2 ) + 30

⇒ 8 − 16 − 22 + 30 = 0 Substitute x = −3 : 2 ( −3)3 − 4 ( −3 ) − 11( −3 ) + 30

⇒ −27 − 36 − 33 + 30 =0 Thus, ( x − 2 ) and ( x + 3 ) are factors of x 3 − 4x 2 − 11x + 30 ⇒ ( x − 2)( x + 3) = x 2 + x − 6 is the factor of x 3 − 4x 2 − 11x + 30 Using long division, divide the polynomial

x −5

x +x −6

( − ) ( x 3 + x 2 − 6x )

− 5x 2 − 5x + 30

x 3 − 4x 2 − 11x + 30

( −)( −5x 2 − 5x + 30) 0

x 3 − 4x 2 − 11x + 30 = ( x − 2)( x + 3)( x − 5) The third zero is 5. 7. If the squared difference of the zeros of the quadratic polynomial f ( x ) = x 2 + px + 45 is equal to 144, find the value of p. Sol. Given quadratic equation: f ( x ) = x 2 + px + 45 Sum of the roots: α + β = −p Product of the roots: αβ = 45 Given squared difference condition: (α − β )2 = 144 2 2 ( α − β ) = ( α + β ) − 4 αβ Using the identity:

144 Substituting the known values: (α + β )2 − 4 ( 45 ) = p 2 − 180 = 144 p 2 = 324 p = ±18

x 3 − 4x 2 − 11x + 30 by x 2 + x − 6: 8. If α and β are the zeros of the polynomial 8x 2 + 14x + 3, find the values of

Topper’s Answer

Chapter 2 • Polynomials

9. If α , β are the zeros of the polynomial f ( x ) = 2x 2 + 5x + k satisfying the relation 21 α 2 + β 2 + αβ = then find the value of k for this to 4 be possible. Sol. It is given that α and β are the zeros of the polynomial f ( x ) = 2x 2 + 5x + k . k 5 ∴α +β = − , and, αβ = 2 2 We have, 21 α 2 + β 2 + αβ = 4 21 ⇒ α 2 + β 2 + 2αβ − αβ = 4 21 2 ⇒ (α + β ) − αβ = 4 25 k 21 ⇒ − = 4 2 4 5 k  k − , and αβ =  2 Since α + β = ⇒ − = −1 ⇒ k = 2 2  2

(

)

For Questions 10 and 11, find the zeros of the following polynomials by factorisation method and verify the relation between the zeros and coefficients of the polynomials. 30

2 10. v + 4 3v − 15

(NCERT Exemplar)

Sol. v + 4 3v − 15 = v + 5 3v − 3v − 15 = 0

(

)

(

)

v v + 5 3 − 3 v + 5 3 = 0

(

)(

)

v − 3 v +5 3 =0 = v

3, − 5 3 are zeros of the polynomial.

(

)

Sum of zeros = 3 + −5 3 =−4 3 −b − (4 3 ) = = −4 3 1 a −15 Product of zeros = 3 × −5 3 = Also

c −15 = = −15 a 1 Hence verified. Also,

11 2 y− 3 3 2 1 2 11 Sol. 7y − y= 21y 2 − 11y = − −2 0 3 3 3 1 21y 2 − 14 y + 3 y − 2 = 0 3 1 7 y ( 3 y − 2 ) + 1( 3 y − 2 )  = 0 3 11. 7y 2 −

(

)

(NCERT Exemplar)

⇒ b = 2 − a …(i ) c Now, αβ + βγ + γα = a 3 ⇒ a ( a + b ) + ( a + b )( a + 2b ) + ( a + 2b )( a ) = 1 ⇒ a 2 + ab + a 2 + 2ab + ab + 2b 2 + a 2 + 2ab = 3

1 ( 3 y − 2 )( 7 y + 1) = 0 3 2 −1 y= , are zeros of the polynomial. 3 7 2 −1 14 − 3 11 = = Sum of zeros = + 3 7 21 21  −11  −  −b  3  11 = = Also 7 21 a 2 −1 −2 Now, product of zeros − × − 3 7 21 −2 c −2 Also, = 3 = − a 7 21

⇒ 3a 2 + 6ab + 2b 2 = 3

⇒ a 2 − 5a + a − 5 = 0 ⇒ a ( a − 5) + 1( a − 5) = 0

Then, sum of zeros = α + β =

(

)+(

3− 3

)

32 −

Required polynomial

(3 + 3 ) × (3 − 3 ) 3

( 3 ) = 9 − 3= 6= 2 9

= x 2 − ( sum of zeros ) x + product of zeros 2 = x 2 − 2x + 3 ⇒ 3x 2 − 6x + 2 2. Given that the zeros of cubic polynomial x 3 − 6x 2 + 3x + 10 are of the form a , ( a + b ) , ( a + 2b ) for some real numbers a and b, find the values of a and b as well as the zeros of the given polynomial. Sol. Given x 3 − 6x 2 + 3x + 10 , α= a , β= = γ

(a + 2b )

Coefficient of x 2 α + β +γ = Coefficient of x 3

( −6) =6 a + a + b + a + 2b =− 1 ⇒ 3a + 3b = 6 ⇒ a + b = 2 Chapter 2 • Polynomials

When a =−1, b =2 − ( −1) =3 If a = −1 and b = 3, the zeros are =−1, ( −1 + 3 ) , ( −1 + 2 ( 3 ) )

= −1, 2, 5

If a = 5 and b = −3, the zeros are = 5, ( 5 − 3 ) , ( 5 + 2 ( −3 ) )

⇒ a =−1 or a = 5 Using (i)

When a =5, b =2 − 5 =−3

3 3 3+ 3 +3− 3 6 = = = 2 3 3

Product of zeros = αβ =

⇒ ( a + 1)( a − 5) = 0

Now, b= 2 − a

Sol. Let the zeros of the quadratic polynomial be

3+ 3

)

⇒ a 2 − 4a − 5 = 0

(3 + 3 ) , (3 − 3 )

⇒ −a 2 + 4a + 5 =0

1. Find the quadratic polynomial whose zeros are

⇒ 3a + 12a − 6a + 8 + 2a − 8a − 3 = 0

Long Answer Questions

(3 + 3 ) , β (3 − 3 ) = α =

(

⇒ 3a 2 + 12a − 6a=2 3+ 2 4 + a 2 − 4a = 3

Hence verified.

Using (i) we get ⇒ 3a 2 + 6a ( 2 − a ) + 2 ( 2 − a ) = 3

(a + b ) ,

= 5, 2, − 1 3. Find the zeros of the polynomial f (= x ) 4 3x 2 + 5x − 2 3 , and verify the relationship between the zeros and its coefficients. x ) 4 3x 2 + 5x − 2 3 Sol. Given f (= Using the middle term split method we get: 4 3x 2 + 8x − 3x − 2 3 = 0

(

)

0 ( 3x + 2 ) = ( 3x + 2 )( 4x − 3 ) = 0 ⇒ 3x + 2 = 0 4x 3x + 2 − 3

3x =−2 ⇒ x =− 4x − 3 = 0 4x =

3⇒x= 2 3 3

3 4

=−

2 3 3

3 4

and x = x= −

The zeros of the polynomial are: 2 3 3

3 4

α= − and β = 31

b 5 5 3 α +β = − ⇒α + β = − = a 4 3 12 − α +β =

c −2 3 1 α ⋅β = ⇒α ⋅β = = − a 2 4 3  2 3  3 2 3⋅ 3 6 1 α ⋅ β =− − = − = −   ⋅   = 3 4 12 12 2     Hence verified.

2 3 3 8 3 3 3 −5 3 + = − + = − 3 4 12 12 12

Competency Based Questions Multiple Choice Questions

5. Shown below are the parts of graphs of two polynomials, g ( x ) and h ( x ) . When h ( x ) is divided by ( x − 3 ), the remainder is k.

1. The given linear polynomial y = f ( x ) has 5 4 3 2 1 -4 -3 -2 -1 0 -1 -2

1 2 3 4 5

(a) 2 zeros

(b) 1 zero and the zero is “3”

(d) 1 zero and the zero is “4”

5 0

-3

0 2 Y'

)(

)

2 Y'

(d) none of these

(d) (t − 2 )(t − 5 )(t − 7 )

8. If α , β , γ are the zeros of f ( x ) = ax 3 − 5x + 9 such 27, then value of a is: that α 3 + β 3 + γ 3 = − 3 (a) (b) 3 (c) 1 (d) −1

4. If zeros of the polynomial f ( x ) = x 3 − 3px 2 + qx − r are in AP, then 3 3 pq − r pq + r (a) 2p= (b) 2p= 3 pq − r (c) p=

(b) It is equal to k.

(

Which of these is true for the remainder when g ( x ) is divided by ( x − 3 ) ? (a) It is less than k.

P (t ) = t 2 + 5t − 14 t 2 − 7t + 10 t 2 + 2t − 35

-3 X

Which of these is the square root of P (t ) ? (a) (t + 2 )(t − 5 )(t + 7 ) (b) (t − 2 )(t − 5 )(t + 7 )

-3

7. P (t ) is a polynomial in t such that,

Y (c) (d)

-2

6. If f ( x ) = ax 2 + bx + c has no real zeros and a + b + c < 0, then (a) c = 0 (b) c > 0 (c) c < 0 (d) c ≥ 0

-4

(d) Cannot conclude without knowing the polynomials.

3. A certain polynomial P has a degree of 2. Polynomial P has zeros 2 and -3 and a > 0 when the function of polynomial P is written in the form of y = ax2 + bx + c. Given this information, which of the following could be the graph of polynomial P ?

-3

g(x) h(x)

2. What should be subtracted from the polynomial x 2 − 16x + 30, so that 15 is the zero of the resulting polynomial? (a) 14 (b) 15 (c) 16 (d) 30

Y (a) (b)

Select Response Type MCQs 1. Which of the following are polynomials? (i) 2x 2 + 5x + 6

(ii) x 2 +

1 x

(iii) 2x 3 + 5x x − 3

(iv) 25 −

4 x

(a) Only (i)

(b) (ii) and (iii)

(d) only (ii)

2. Which of the following polynomials has a product of roots equal to 10? (i) 2x 2 + 5x + 10 (ii) x 2 − 7x + 10 (iii) x 2 + 11x + 10

(iv) x 2 + 3x − 10

(a) Only (i)

(b) (ii) and (iii)

(d) only (ii)

3. Shown below is a part of the graph of a polynomial g(x). On dividing g(x) by which of the following will the remainder be zero?

Based on the given information answer the following questions. g(x)

-2

-1

(i) (x - 1) (ii) (x + 2) (iii) (x - 1) (a) Only (i) (b) Only (i) and (ii) (c) Only (ii), and (iii)

(d) All of these

(a) Find the value of x that satisfies the equation x 2 − 4x − 60 = 0. (b) Verify whether x = 10 is a solution of the equation. (c) Using the solution, calculate the dimensions of the rectangular plot. (CBSE 2022)

2. An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

(b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 2 1. Assertion (A): The quadratic polynomial x + 5x + 6 has real and distinct zeros.

Trikonasana

Reason (R): A quadratic polynomial has real and distinct zeros if its discriminant is positive. (CBSE 2024) 2

0 has α and β as its 2. Assertion (A): If x + px + q = −p . roots, then α + β = Reason (R): The sum of the roots of a standard 2 0 is equal quadratic polynomial of form x + px + q = to the coefficient of x with the opposite sign.

Chakrasana

(NCERT Exemplar)

Case Study Based Questions 1. A farmer uses a rectangular plot of land to plant crops. The length of the plot is x metres, and the width is ( x − 4 ) metres. The area of the plot is represented by the polynomial x 2 − 4x − 60.

Chapter 2 • Polynomials

Adho Mukha Svana

(a) The shape of the poses shown is (i) Spiral (ii) Ellipse (iii) Linear (iv) Parabola (b) The graph of parabola p ( x ) = ax 2 + bx + c opens downwards, if (i) a ≥ 0 (ii) a = 0 (iii) a < 0 (iv) a > 0

-2

-8

(i) 0

(ii) 1

(iii) 2

(iv) 3

(d) The two zeros in the above shown graph are (i) 2,4 (ii) −2,4 (iii) −8,4 (iv) 2, -8

Answers Multiple Choice Questions 1. (b) 1 zero and the zero is “3” is correct answer.

5. It is clear from the graph that both polynomial g ( x ) and h ( x ) intersect the x-axis at x = −2 and x = 2

2. Let the value to be subtracted be k. The resulting polynomial will be:

So, standard form h ( x ) = ( x − 2 )( x + 2 ) = x 2 − 4

P ( x ) = f ( x ) − k = x − 16x + 30 − k

From the graph, it is clear that opening of g ( x ) is more than that of h ( x ), so we can say h ( 3 ) > g ( 3 ).

Since 15 is a zero of the resulting polynomial, substituting x = 15 into the polynomial should satisfy: P (15 ) = 0

On dividing h ( x ) by ( x − 3 ), the remainder is k.

∴ When g ( x ) is divided by ( x − 3 ), the remainder is less than k. Thus, (a) It is less than k is correct answer.

P (15= k 0 ) (15) − 16 (15 ) + 30 −=

6. If f ( x ) = ax 2 + bx + c has no real zero,

⇒ 225 − 240 + 30 − k = 0

Then its graph does not cross x-axis.

15 − k = 0 ⇒ k = 15

Therefore, either the complete curve is above x-axis or it lies below x-axis. This means that either f ( x ) > 0 or f ( x ) < 0 for all values of x. It is given that a + b + c < 0

Thus, (b) 15 is correct answer 3. (b)

-3

Therefore, f ( x ) < 0 ⇒ f ( 0 ) < 0 ⇒ c < 0 Thus, (c) c < 0 is correct answer.

Given function of polynomial is y = ax2 + bx + c and zeros are 2 and -3x = 2, x = -3. Also a > 0, so parabola must open upwards. 3 pq − r 4. (a) 2p=

(

)(

t 2 + 5t − 14 t 2 − 7t + 10 t 2 + 2t − 35 7. Given P (t ) = By splitting the middle term we get:

(

)(

P (t ) = t 2 + 7t − 2t − 14 t 2 − 2t − 5t + 10 t 2 + 7t − 5t − 35 P (t ) = (t − 2 )(t + 7 )(t − 5 )(t − 2 )(t − 5 )(t + 7 )

f ( x ) = x 3 − 3px 2 + qx − rr . Then,

P (t ) = (t − 2 ) (t + 7 ) (t − 5 )

( a − d ) + a + ( a + d ) = 3p ⇒ 3a = 3p ⇒ a = p

Taking square root both sides we get:

But, a is a zero of f ( x ). Therefore,

P (t ) = (t − 2 ) (t + 7 ) (t − 5 )

f ( a ) = 0 ⇒ a 3 − 3pa 2 + qa − r = 0

P (t ) = )= (t − 2 )(t + 7 )(t − 5 )

)

P (t )= (t (t + 7 ) − 2 (t + 7 ) ) (t (t − 2 ) − 5 (t − 2 ) ) (t (t + 7 ) − 5 (t + 7 ) )

Let a − d , a , a + dd be the zeros of

⇒ p 3 − 3p 3 + pq − r = 0 ⇒ 2p 3 = pq − r

)

Thus, (b) (t − 2 )(t − 5 )(t + 7 ) is correct answer.

8. α + β + γ =−

Coefficient of x 2 Coefficient of x

0 and, αβγ = − = 3

9 a

3 3 3 Now, α + β + γ = 0 ⇒ α + β + γ = 3αβγ −9 ⇒ 27 =× 3 ⇒a = −1 a Thus, (d) −1 is correct answer.

Select Response Type MCQs 1. From the given options only (i) is a polynomial, because in all other cases the exponent of x is not a whole number. Thus, (a) Only (i) is correct answer. c 10 2. (i) 2x 2 + 5x + 10 , Product of roots= = = 5 a 2 c 10 2 (ii) x − 7x + 10, Product of roots= = = 10 a 1 c 10 2 (iii) x + 11x + 10 , Product of roots= = = 10 a 1 c −10 = −10 (iv) x 2 + 3x − 10 , Product of roots = = a 1 Thus, (b) (ii) and (iii) is correct answer. 3. Y g(x) X'

-2

-1 O

2 0 in given polynomial Here, x + px + q =

we have −

The curve intersecting the x axis at three points are the solution of g(x) x =−1 ⇒ ( x + 1) x =−2 ⇒ ( x + 2 ) is a factor x =1 ⇒ ( x − 1) is a factor So, on dividing g ( x ) by ( x + 1) , ( x + 2 ) and ( x − 1) the remainder will be zero. Thus, (d) All of these is correct answer.

Assertion-Reason Based Questions 2 1. Assertion (A): For x + 5x + 6, 2 Discriminant ∆= b − 4ac= 25 − 24= 1, which is positive, the quadratic polynomial x 2 + 5x + 6 has two real and distinct zeros. Reason (R): True, based on the definition and properties of the discriminant. Hence, A and R are both true, and R explains A. Correct answer: (a) Both A and R are true, and R is the correct explanation of A.

Coefficient of x

−p = = −p 1 Coefficient of x 2

Reason (R): True, based on the standard formula for the sum of the roots of a quadratic equation Correct answer: (a) Both A and R are true, and R is the correct explanation of A.

Case Study Based Questions 1. The equation for the area of the plot is x 2 − 4x − 60 = 0. 0: (a) To solve x 2 − 4x − 60 = 0 Rewrite the equation as: x 2 − 10x + 6x − 60 =

0 ( x − 10x ) + (6x − 60 ) = 2

x ( x − 10 ) + 6 ( x − 10 ) = 0

0 ( x + 6 )( x − 10 ) =

x = 10 or x = − 6 (discard as length cannot be negative).

So, x = 10

Chapter 2 • Polynomials

2. Assestion (A): The sum of the roots of a quadratic Coefficient of x polynomial is given by − . Coefficient of x 2

(b) Verify x = 10:

(10 )2 − 4 (10 ) − 60 = 100 − 40 − 60 = 0

Hence, x = 10 is a solution.

6 (c) Length = x = 10 , Width = x − 4 =

Dimensions of the plot: 10m × 6m

2. (a) (iv) Parabola (b) (iii) a < 0 (c) The graph cuts the x-axis at 2 points. Thus two zeros exist for the given graph.

(iii) 2

(d) The curve intersect the x-axis at −2 and 4.

Thus zeros are −2 and 4.

(ii) –2, 4

Additional Practice Questions 1. The number of polynomials having zeros -3 and 5 are (CBSE 2023) (a) only one (b) infinite (c) exactly two

(d) at most two

2. The zeros of the polynomial x 2 − 5x + 6 are: (a) 2 and 3 (b) -2 and -3 (c) 5 and 6

(b) a < 0, b < 0 and c < 0

(d) a < 0, b > 0 and c < 0

8. In given figure, the graph of a polynomial p(x) is given. Find the zeros of the polynomial. Y 6 5 y = p(x) 4 3 2 1

(d) -5 and -6

3. If α and β are the zeros of the polynomial x 2 − 7 x + 10 , then α + β and α × β are: (a) 7 and 10 (b) -7 and 10 (c) 7 and -10

5. The graph of y = f ( x ) is shown in the figure for some polynomial f ( x ). The number of zeros of f ( x ) are

-2 -3 -4 -5

9. Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and product of its zeros as 3, −1 and −3 respectively.

11. If f ( x ) = x 3 + x 2 − ax + b is divisible by x 2 − x write the values of a and b.

(b) 2

12. If a , β are the zeros of a polynomial such that a+β = − 6 and a β = − 4 then write the polynomial.

(d) 4

6. A quadratic polynomial whose sum and product of zeros are 2 and -1 respectively is (a) x 2 − 2x + 1

(b) x 2 + 2x − 1

2 (d) x + 2x + 1

7. If the diagram in shows the graph of the polynomial f ( x ) = ax 2 + bx + c , then f(x) = ax2 + bc + c

10. If the squared difference of the zeros of the quadratic polynomial f ( x ) = x 2 + px + 45 is equal to 144 , find the value of p .

y = f(x)

(a) 1

-6 -5 -4 -3 -2 -1 O -11 2 3 4 5 6 7

(d) -7 and -10

4. What should be subtracted from the polynomial x 2 − 16x + 30, so that 15 is the zero of the resulting polynomial? (a) 1 (b) 2 (c) 4 (d) 5

13. Find the zeros of the quadratic polynomial 3x 2 − 8x + 4 3. 14. If α and β are the zeros of the quadratic polynomial p ( s ) = 3s 2 − 6s + 4, find the value of 1 1 α β + + 2  +  + 3αβ . β α α β  15. If α and β are the zeros of the quadratic polynomial f ( x ) = x 2 − 5x + k such that α − β = 1, find the value of k. 16. If α and β are the zeros of the quadratic polynomial f ( x ) = 2x 2 − 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2 β .

O Y'

(a) a > 0, b < 0 and c > 0

−b −D , 2a 4a

17. If one zero of the polynomial f ( x ) = x 2 − 6x + b is twice the other, find the value of b. Also, find the zeros of the polynomial so obtained.

18. If one of the zero of the polynomial a 2 + 9 x 2 + 13x + 6a is reciprocal of the other, find

(

(d) The expression of the polynomial representing the wire is

)

the value of a.

(i) x 2 + 2x − 3

(ii) x 2 − 2x + 3

(iii) x 2 − 2x − 3

(iv) x 2 + 2x + 3

19. If the zeros of the polynomial x 3 − 6x 2 + 11x − 6 are α, β, and γ, find the value of α 2 + β 2 + γ 2.

23. A highway underpass is parabolic in shape as shown in the following picture.

Directions for questions 20 and 21: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A.

Shape of Cross Slope: x

20. Assertion (A): If α , β and γ are the zeros of the polynomial 6x 3 + 3x 2 − 5x + 1, then α −1 + β −1 + γ −1 = 5.

Reason (R): If α , β and γ are the zeros of the cubic b − . polynomial ax 3 + bx 2 + cx + d , then α + β + γ = a

(a) If the highway overpass is represented by x 2 − 2x − 8. Then its zeros are

21. Assertion (A): The polynomial P ( x ) = x 2 + 3x + 3 has two real zeros. Reason (R): A quadratic polynomial can have at most two real zeros. 22. Due to heavy storm an electric wire got bent as shown in the figure. It followed a mathematical shape. Answer the following questions below.

4 3 2 1 O 1 2 3 -1

-3 -2 -1

(a) Name the shape in which the wire is bent (i) Spiral (ii) Elliptical

(iii) Parabolic

(iv) Linear

(b) How many zeros are there for the polynomial representing the shape of the wire? (i) 2 (ii) 3 (iii) 1 (iv) 0 (c) The zeros of the polynomial represented by the wire are (i) –1, 5 (ii) –1, 3 (iii) 3, 5 (iv) –4, 2

Chapter 2 • Polynomials

(ii)

( 4, −2 )

(iii) ( −2, −2 )

(iv)

( −4, −4 )

(b) The highway overpass is represented graphically. Zeros of a polynomial can be expressed graphically. Number of zeros of polynomial is equal to number of points where the graph of polynomial

(i) ( 2, −4 )

(i) Intersects x -axis (ii) Intersects y -axis (iii) Intersects y-axis or x-axis (iv) None of these (c) Graph of a quadratic polynomial is a (i) straight line

(ii) circle

(iii) parabola

(iv) ellipse

(d) The representation of Highway Underpass whose one zero is 6 and sum of the zeros is 0, is (i) x 2 − 6x + 2

(ii) x 2 − 36

(iii) x 2 − 6

(iv) x 2 − 3

Challenge Yourself 1. If 1 +

Scan me

α β and 1 + are the zeros of the polynomial, β α

3x 2 − 16x + 16, then find the quadratic polynomial

whose zeros are α and β .

2. Find the quadratic polynomial whose zeros are: 3−2 5

3+2 5

3+2 5 3−2 5

(Hint: Rationalise the denominator. Multiply the numerator and the denominator by the

conjugate of the denominator. Conjugate of (a + b) is (a - b)) 3. For what value of p will the equation

p + 1 ax − c = p − 1 x 2 − bx

have roots equal in magnitude but opposite in sign? 4. Prove that the ( −r + p + q ) x 2 + 2px + ( −p + q + r ) =0 equation has rational roots. 5. Find the value of c for which one root of ax 2 + bx + c =shall 0 be n times the other.

Answers Additional Practice Questions 1. (b) infinite

2. (b) −2 and −3

3. (a) 7 and 10

4. (b) 2

5. (d) 4

6. (c) x 2 − 2x − 1

7. (a) a > 0, b < 0 and c > 0

− 3 and x = −1 8. x =

{

}

9. f ( x= ) k x 3 − 3x 2 − x + 3 , where k is any non-zero real number. 10. p = ±18 12. x 2 + 6x − 4 14. 8 25   16. g ( x ) = k  x 2 − x + 41  2   18. 3

20. (b) Both A and R are true, but R is not the correct explanation of A. 21. (d) A is false, but R is true. 22. (a) (iii) Parabolic

(b) (i) 2

(d) (iii) x 2 − 2x − 3

23. (a) (ii) ( 4, −2 )

(b) (i) Intersects x-axis

(d) (ii) x 2 − 36

11. a = 2 and b = 0 2 13. and 2 3 3 15. 6

Challenge Yourself

17. 2 and 4

3. p =

14 19. α + β + γ =

1. x 2 + 4x + 3 a −b a +b

2 2. 11x + 58x + 11

5. c =

nb 2

a (n + 1)

Scan me for Exemplar Solutions 38

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

2 1. If one zero of the polynomial x − 4x + k is 2, then the value of k is:

(a) 4

(b) 0

(d) 2

(d) 4

2. The degree of the polynomial 2x 3 − 3x 2 + x − 52 is: (a) 1

(b) 2

2 3. The sum and product of the zeros of the polynomial x + 5x + 6 are:

(a) -5 and 6

(b) 5 and -6

(d) 5 and 6

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(

)

(

)

4. Assertion (A): 2 − 3 is one zero of the quadratic polynomial then other zero will be 2 + 3 . Reason (R): Irrational zeros always occurs in pairs.

5. Assertion (A): The polynomial p ( x ) =x 3 − 6x 2 + 11x − 6 has three distinct real roots.

Reason (R): The factorisation of p ( x ) =x 3 − 6x 2 + 11x − 6 gives ( x − 1)( x − 2)( x − 3).

Very Short Answer Questions

[2 × 2 = 4 Marks] 2

6. For what value of k does the polynomial x + kx + 16 have the sum of zeros as 0? 7. If one zero of the polynomial 6x 2 − 13x + 6 is

Short Answer Questions

2 , find the other zero. 3 [4 × 3 = 12 Marks]

8. Find a quadratic polynomial whose zeros are 4 and -5. 9. If one zero of the polynomial x 2 + px + q is 3, and the other zero is -2, then find the values of p and q. 10. Find the zeros of the quadratic polynomial x 2 − 15 and verify the relationship between the zeros and the coefficient of the polynomial. (CBSE 2024) 11. If α and β are the zeros of the quadratic polynomial x 2 − 6x + 8, find the quadratic polynomial whose zeros are α + 2 and β + 2.

Long Answer Questions

[3 × 5 = 15 Marks] 2

12. Find the zeros of the polynomial 3x − 11x + 6 3 and verify the relationship between the zeros and the coefficients of the polynomials. 13. If the sum of squares of the zeros of the quadratic polynomial x 2 − 16x + k is 226, find k. 14. If α and β are zeros of quadratic polynomial f ( x ) = x 2 − 4x + 3, find the value of α 4 β 2 + α 2 β 4. (CBSE 2019)

Self-Assessment

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A student wants to design an open box by cutting out squares of side x from each corner of a rectangular sheet of metal. The dimensions of the sheet are 20 cm × 16 cm. The volume of the box formed is represented by the polynomial 4x 3 − 72x 2 + 320x .

(a) Write the expression for the volume of the box in terms of x. (b) Factorise 4x 3 − 72x 2 + 320x . (c) If the volume of the box is 320cm 3 , find the possible values of x. (CBSE 2021)

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. If one root of the polynomial f ( x ) = 5x 2 + 13x + k is reciprocal of the other, then the value of k is 1 (a) 0 (b) 5 (c) (d) 6 6 2. If the product of two zeros of the polynomial f ( x ) = 2x 3 + 6x 2 − 4x + 9 is 3, then its third zero is (a)

3 2

(b) −

3 2

(c)

9 2

(d) −

3. If α and β are the zeros of the polynomial f ( x ) = ax 2 + bx + c , then (a)

b 2 − 2ac a2

(b)

b 2 − 2ac c2

(c)

β2

9 2

b 2 + 2ac

(d)

b 2 + 2ac c2

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): If the graph of the polynomial f ( x ) = ax 2 + bx + c , a > 0 is equal to touches x-axis at its lowest point, b then its zeros are equal to − . 2a Reason (R): If the graph of the quadratic polynomial touches x-axis, then its zeros are equal. 5. Assertion (A): The polynomial x 3 − 2x 2 − x + 2 is divisible by x −1.

Reason (R): A polynomial p ( x ) is divisible by x − k if p (k ) = 0

(CBSE 2022)

Very Short Answer Questions 6. If α and β are the roots of the quadratic equation x 2 − 5x + 6, find the value of

[2 × 2 = 4 Marks]

α β + . β α

7. Determine the value of k for which one root of the quadratic equation 2x 2 − 9x + k is twice the other.

Short Answer Questions

[4 × 3 = 12 Marks]

8. Find the quadratic polynomial whose roots are the reciprocals of the roots of the polynomial x 2 − 3x + 2. 9. If the sum of the squares of the roots of the quadratic equation x 2 − px + 16 is 40, find the value of p. 10. If a - b, a and a + b are zeros of the polynomial f(x) = 2x3 - 6x2 + 5x - 7, write the value of a. 2 11. The quadratic polynomial x − kx + 24 has roots in the ratio 2:3. What is the value of k ?

Self-Assessment

Long Answer Questions

[3 × 5 = 15 Marks] 2

12. If α and β be the zeros of the polynomial x + 13x + 30, then find the quadratic polynomial whose zeros are 3α + β and α + 3β . 13. If

2 and -3 are zeros of quadratic polynomial ax 2 + 7x + b , find the value of a and b. 3

(CBSE 2019)

14. If α and β are zeros of quadratic polynomial 5x 2 + 5x + 1, find the value of (a) α 2 + β 2

(b) α

−1

+ β −1 (CBSE 2024)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A farmer plans to build a rectangular fenced garden such that the area of the garden is 300 square metres. He wants the length of the garden to be 5 metres more than twice its width. Let the width of the garden be x metres.

Using the information provided: (a) Form a quadratic equation in terms of x to represent the problem. (b) Find the width and length of the garden by solving the quadratic equation. (c) Verify that the product of the roots of the quadratic equation equals the constant term divided by the coefficient of x2. (d) Explain why a negative value of x is not valid for this situation.

Pair of Linear Equations in Two Variables

Chapter at a Glance Equation: An equation is a mathematical statement where an ‘equal to’ (=) sign establishes a relationship of equality between two algebraic expressions. It is a form of algebraic expression that demonstrates equivalence. Example: 2= x 3, 2x += 3 0, 2x + 3= y 5 etc. Linear Equation: A linear equation is defined as one where the highest power of any variable is one. This equation also represent a straight line. 5 0, 3x −= 7 8 etc. Example: 2x += Linear Equation in One Variable: An equation of the form ax + b =, 0 where a and b are real numbers and a ≠ 0 is called linear equation in one variable. Example: 2x + 5 = 0 etc.

Linear Equations in Two Variables A linear equation in two variables takes the form ax + by + c =, 0 where: ● x and y are variables. ●

a, b, and c are real numbers.

●

a and b are both non zero.

The solution to such an equation is a pair of values ( x , y ) that satisfy the equation. Graphically, this equation represents a straight line on the Cartesian plane. Example: ( 0, 2 ) is the solution of the equation 6x − y = −2 because when we substitute x = 0 and y = 2 into 6x − y = −2 we get: 6 × 0 − 2 =−2 ⇒ 0 − 2 =−2 ⇒ − 2 =−2

Pair of Linear Equations When two linear equations involve the same set of variables, they are termed as a pair of linear equations in two variables. The general representation is: 1. a1x + b1 y + c = 0 2. a 2 x + b2 y + c 2 = 0

Algebraic Methods of Solving Linear Equations ●

●

Substitution Method: In this method, we express one variable in terms of the other and substitute it into the second equation. Elimination Method: This method eliminates one variable by adding or subtracting the given equations.

Graphical Method of Solving Linear Equations ●

Convert both equations into y = mx + c form.

●

Plot both lines on a graph.

●

Find the intersection point.

●

The intersection point gives the solution.

Chapter 3 • Pair of Linear Equations in Two Variables

Case I: Intersecting Lines: If the lines intersect at a single point, the system has a unique solution, and the equations are consistent. y-axis

(a, b) x-axis

= x a= , y b is the solution of above lines.

Case II: Parallel Lines: If the lines are parallel and distinct, there is no solution, rendering the system inconsistent. y-axis

x-axis

Case III: Coincident Lines: If the lines coincide, there are infinitely many solutions, and the equations are dependent and consistent. y-axis

Solutions

(a1, b1) (a2, b2) (a3, b3)

x-axis

Conditions for Solvability The solvability of the system depends on the coefficients of the equations: In a1x + b= y + c 2 0, 1 y + c1 0 and a 2 x + b2= ●

a  b  Unique Solution: If  1  ≠  1  , the system has a unique solution and the equation is consistent.  a 2   b2 

●

 a1   b1   c1  No Solution: If  =    ≠   , the system has no solution and the equation is inconsistent.  a 2   b2   c 2 

●

 a1   b1   c1  I nfinitely Many Solutions: If  =   =   , the system has infinitely many solutions and the equation is a 2   b2   c 2   dependent and consistent.

NCERT Zone NCERT Exercise 3.1 1. Form the pair of linear equations in the following problems, and find their solutions graphically. (a) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. 44

(b) 5 pencils and 7 pens together cost `50, whereas 7 pencils and 5 pens together cost `46. Find the cost of one pencil and that of one pen. Sol. (a) Let the number of girls be x and the number of boys be y . According to the question the algebraic representation is

x+y = 10 (1)

x − y = (2) 4

For x + y = 10 or x = 10 – y , y = 10 − x x y

12 11 10 9 8

5x + 7y = 50

7 6

5 3

x + y = 10

7 6

(4, 6) (5, 5) (6, 4)

(4, 0)

5 4 2

1 2 −1

−2 −1

−2 −3

(3, −1)

(b) Let the cost of one pencil be x And the cost of one pen = y

According to the question, the algebraic representation is 7x + 5 y = 46

for 5x + 7 y = 50

For

7x + 5 y = 46

( 46 − 5 y ) 7

,y=

−2

7x − 5y = 46

9 10 11

(8, −2)

From the figure it can be observed that these lines intersect each other at. (3, 5).

2. On comparing the ratios a1 , b1 and c1 . Find out c2 a 2 b2

whether the lines representing the following pairs of linear equations at a point are parallel or coincident?

(a) 5x − 4y + 8 = 0 7x + 6y − 9 = 0 (b) 0 (ii ) 9x + 3y + 12 =

10 0

18x + 6 y + 24 = 0

Sol. (a) Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

( 50 − 7 y ), y = 50 − 5x x

Therefore, the cost of a pencil and pen are rupees 3 and rupees 5 respectively.

Therefore, the number of girls and boys in a class of 7 and 3 respectively.

1 2 −1

−5

From the figure it can be observed that these lines intersect each other at point (7, 3).

5x + 7 y = 50

−9 −8 −7 −6 −5 −4 −3 −2 −1

−4

(10, 0)

−3

(5, 1)

(7, 3)

x−y=4

−4

(3, 5)

4 9

Hence the graphic representation is as follows.

−2

(−4, 10)

−1

−2

(−2, 12)

For x − y = 4 or x = 4 + y, y = x − 10 x

−4 10

46 − 7 x 5

Chapter 3 • Pair of Linear Equations in Two Variables

We obtain

a1 = 5

b1 = − 4

c1 = 8

a2 = 7

b2 = 6

c2 = − 9

a1 5 b1 4 2 = , =− =− 6 3 a 2 7 b2

Here, a1 b1 ≠ a 2 b2

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(c)

(b) 9x + 3 y + 12 = 0

(d) 5x − 3 y = 11

18x + 6 y + 24 = 0

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

We obtain

a1 = 9

b1 = 3

c1 = 12

= a 2 18 = b2 6 = c 2 24

a1 9 1 b1 3 1 c1 12 1 = = , = = , = = a 2 18 2 b2 6 2 c 2 24 2

Here,

a1 b1 c1 = = a 2 b2 c 2

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations. (c) 6x − 3y +10 = 0

2x − y + 9 = 0

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

We obtain a1 = 6 b1 = −3

a 2 == 2 b2 −1 c2 = 9

a1 6 3 c1 10 b1 −3 = = = 3, = = 3, = c2 b2 −1 9 a2 2 1

c1 = 10

Here, a1 b1 c = ≠ 1 c2 a 2 b2

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

(e)

3 5 7 x+ y= 2 3 9x − 10 y = 14 −10x + 6 y = −22 4 8 x + 2y = 3 2x + 3 y = 12

Sol. (a) 3x + 2 y = 5

2x − 3 y = 7

Above equation can also be written as 3x + 2 y − 5 = 0

2x − 3 y − 7 = 0

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

We obtain a1 = 3 b1 = 2

a2 = 2 b2 = −3 c2 = −7

a1 3 b1 2 = , =− a 2 2 b2 3

c1 = − 5

Here, a1 b1 ≠ a 2 b2

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. (b) 2x − 3 y = 8

4x − 6 y = 9

Above equation can also be written as

4x − 6 y − 9 = 0

2x − 3 y − 8 = 0

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

We obtain a1 = 2 b1 = −3

a 2 == 4 b2 −6 c2 = −9

a1 2 1 b1 −3 1 c1 8 = = , = = , = a 2 4 2 b2 −6 2 c 2 9

2x − 3 y = 7

Here,

(b) 2x − 3 y = 8

a1 b1 c = ≠ 1 c2 a 2 b2

a b c 3. On comparing the ratios 1 , 1 and 1 , find out a 2 b2 c2 whether the following pair of linear equations is consistent, or inconsistent?

(a) 3x + 2 y = 5 46

4x − 6 y = 9

c1 = −8

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent. (c)

3 5 7 x+ y= 2 3

(e)

9x −10 y = 14

Above equation can also be written as 3 5 x + y −7 = 0 2 3

9x −10 y − 14 = 0

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0

We obtain

3 a1 = 2

a 2 == 9 b2 −10 c2 = −14

c1 = −7

5 3 a1 2 3 1 b1 5 1 = = = , = 3 = = − a 2 9 18 6 b2 −10 −30 6

c1 7 1 = = c 2 14 2

Here,

5 b1 = 3

4 8 x + 2y = 3

2x + 3 y = 12

Above equation can also be written as 4 0 x + 2 y − 8 = 3 2x + 3 y − 12 = 0 Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0 We obtain, 4 a1 = b1 = 2 3

c1 = −8

a 2 = 2 b2 = 3 c 2 = −12 4 a1 3 4 2 b 2 c −8 2 = = = , 1 = , 1 = = 6 3 b2 3 c 2 −12 3 a2 2 Here, a b c 1 = 1 = 1 a 2 b2 c 2

a1 b1 ≠ a 2 b2

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. (d) 5x − 3 y = 11

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent. 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (a) x + y = 5

− 1 0x + 6 y = − 22

Above equation can also be written as

5x − 3 y − 11 = 0

−10x + 6 y + 22 = 0

c1 = −11

a 2 = −10 b2 = 6 c2 = 22

2x + 2 y = 10

(b) x − y = 8

Comparing these equations with a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0 We obtain a1 = 5 b1 = −3

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

3x − 3 y = 16

4x − 2 y − 4 = 0

(d) 2x − 2 y − 2 = 0

4x − 4 y − 5 = 0

Sol. (a) x + y = 5

a 5 1 b −3 1 c −11 1 1 = =− , 1 = = − , 1 = =− 6 a 2 −10 2 b2 2 c 2 22 2

2x + 2 y = 10

Here, a b c 1 = 1 = 1 a 2 b2 c 2

Above equation can also be written as

x + y −5= 0

2x + 2 y − 10 = 0

Chapter 3 • Pair of Linear Equations in Two Variables

Here, a1 = 1 a2 = 2

And −5 1 a1 1 b1 1 c1 = = = , = , a 2 2 b2 2 c 2 −10 2

Since,

a1 b1 c1 = = a 2 b2 c 2

b1 = 1 b2 = 2

c1 = −5 c2 = −10

a1 1 b1 −1 1 c1 −8 1 = , = , = = = a 2 3 b2 −3 3 c 2 −16 2

Since, a1 b1 c = ≠ 1 c2 a 2 b2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent. (c) 2x + y − 6 = 0

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

4x − 2 y − 4 = 0

Here, a1 = 2

Now, For x + y = 5 or x=5−y

a2 = 4 b2 = −2 c2 = −4

And a1 2 1 b1 1 1 = = , = = − a 2 4 2 b2 −2 2

And for 2x + 2 y = 10 or

10 − 2 y x= 2

Hence, the graphic representation is as follows.

2x +

(0, 5)

2y =

And 4x − 2y − 4 = 0 Or 4x − 4 y= 2

Above equation can also be written as

x − y −8 = 0

3x − 3 y − 16 = 0 Here,

a1 = 1

a2 = 3 b2 = −3 c2 = −16

And

b1 = −1

c1 = −8

3 4

11 10 9

4x − 2y = 4

7 6

5 (1, 4) 4 3

(0, 6)

3x − 3 y = 16

2x + y = 6

Hence, the graphic representation is as follows.

(b) x − y = 8

y= 6 − 2x

Since, a1 b1 ≠ a 2 b2

(5, 0)

c1 = −6

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. For, x 0 2 3 2x + y − 6 = 0 Or

1 −1 O −1

b1 = 1

(2, 2)

(1, 0)

1 2 −1 −2 −3 −4 −5

−7 −6 −5 −4 −3 −2 −1 −7 −6 −5 −4 −3 −2 −1

(3, 4)

9 10 11

From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.

6. Given the linear equation 2x + 3 y − 8 =, 0 write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(d) 2x − 2y − 2 = 0 4x − 4y − 5 = 0 Here,

a1 = 2

a2 = 4 b2 = −4 c2 = −5

And a1 2 1 b1 −2 1 c1 −2 2 = , = = , = = = a 2 4 2 b2 −4 2 c 2 −5 5

b1 = −2

c1 = −2

(a) Intersecting lines (b) Parallel lines (c) Coincident lines Sol. (a) Intersecting lines:

Since,

a1 b1 c1 = ≠ a 2 b2 c 2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent. 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. find the dimensions of the garden. Sol. Let the width of the garden be x and length be y. According to the question, y −x = 4 (1) 1 (2) 36 ⇒ y + x = 36 × 2 (y + x) = 2 For equation (1) y − x = 4 or y = x + 4 x y

0 4

8 12

12 16

And for y + x = 36 or = y 36 − x x y

0 36

16 20

36 0

44 36

For second line, we can take a 2 3x + 4 y − 2 =, 0 As 1 = , a2 3 b1 3 a b = and 1 ≠ 1 b2 4 a 2 b2

(b) Parallel lines:

For this condition,

a1 b1 c1 = ≠ a 2 b2 c 2

Hence, the second line can be

6x + 9 y − 8 = 0

We can see here a1 b1 c1 = ≠ a 2 b2 c 2

a1 2 1 b1 3 1 c1 −8 = = , = = , = =1 a 2 6 3 b2 9 3 c 2 −8

For coincident lines,

a1 b1 c1 = = a 2 b2 c 2

y−x=4

Hence, the second line can be

(16, 20)

4x + 6y − 16 = 0

We can see here

a1 b1 c1 = = a 2 b2 c 2

−3 −2 −1

(0, 4) O

For this condition, two equations must satisfy the relation. a1 b1 ≠ a 2 b2

Hence, the graphic representation is as follows.

(0, 36)

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

4 −4 −12

(12, 16) (8, 12) 12

(7, 3) (36, 0) 28

y + x = 36

a1 2 1 b1 3 1 c1 −8 1 = = , = = , = = a 2 4 2 b2 6 2 c 2 −16 2

Chapter 3 • Pair of Linear Equations in Two Variables

7. Draw the graphs of the equations x − y + 1 = 0 and 3x + 2 y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. Sol. Here, x − y + 1 = 0 or y = x + 1 0

–5

3 5 x− y= −2 2 3 x y 13 + = 3 2 6

From (1), we obtain

x = 14 – y (3)

Substituting this value in equation (2), we obtain

14 − y − y = 4 ⇒ 2= y 14 − 4 ⇒ y = 5

Substituting this in equation (3), we obtain

= x 14 − 5 ⇒ x = 9

thus = x 9,= y 5

–3

(f)

14 (1) (1) Sol. (a) x + y = 2

Hence, the graphic representation is as follows.

3x − 8 y = 0

x − y = (2) 4 ( 2)

12 − 3x y= 2

(0, 1) (–1, 0)

3x + 2 y − 12 = 0

(0, 6)

2x + 3 y = 0

(e)

x–y 1 0

(b) s − t = 3 (1) (1)

5 (2, 3) (1, 2) 1 (4, 0) 3

–1 O1 –1 –3 –5

X 7

3x + 2y – 12 0

–7 Y'

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

NCERT Exercise 3.2

s t + = (2) 6 3 2 From (1), we obtain s = t + 3 (3) Substituting this value in equation (2), we obtain t +3 t + = 6 3 2

2 (t + 3 ) + 3t = 36 ⇒ 2t + 6 + 3t = 36

5t + 6 = 36 ⇒ 5= t 36 − 6 ⇒ t = 6

Substituting this in equation (3), we obtain

s= 6 + 3 ⇒ s = 9

thus= s 9,= t 6

9x − 3 y = 9 (2)

From (1), we obtain

(a) x + y = 14

1. Solve the following pair of linear equations by the substitution method. x−y = 4

(b) s − t = 3 s t + = 6 3 2 (c) 3x − y = 3

9x − 3 y = 9

(d) 0.2x + 0.3 y = 1.3 50

0.4x + 0.5 y = 2.3

( 2)

(3 + y )

(3) 3 Substituting this value in equation (2), we obtain 3( 3 + y ) −y = 3 ⇒ 3+ y − y = 3 ⇒3=3 3 This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by = y 3x − 3 .

(1) ( 2)

Therefore, one of its possible solutions is x = 1, y = 0. (d) 0.2x + 0.3 y = 1.3 (1)

13 − 3 y 2 Substituting this value in equation (3), we obtain

0.4x + 0.5 y = 2.3 (2)

From (1), we obtain (1.3 − 0.3y ) x= (3) 0.2

Substituting this value in equation (2), we obtain 0.4 (1.3 − 0.3 y ) 0.2

2.3 + 0.5 y =

2 (1.3 − 0.3 y ) + 0.5 y = 2.3

2.6 − 0.6 y + 0.5 y = 2.3 ⇒ −0.1y =2.3 − 2.6

−0.1y = −0.3 ⇒ y = 3 (4)

Substituting this in equation (3), we obtain (1.3 − 0.9 ) 0.4 x= ⇒ x= 0.2 0.2

x =2

Thus x = 2 and y = 3

(e)

2x + 3 y = 0 (1)

3x − 8 y = 0 (2)

From equation (1), we obtain

− 3y

(3) 2 Substituting this value in equation (2), we obtain  3y  −3 y − 3 − 8y = 0 ⇒ − 8y = 0  2  2   3y + 2 2y = 0 (Hint: 8 = 2 2 ) 2

3y + 4 y = 0 ⇒ y = 0 (4)

Substituting the value of y in equation (3), we obtain x = 0 Thus = x 0= and y 0 3 5 (f) x − y = −2 (1) 2 3 x y 13 + = (2) 3 2 6 From equation (1), we obtain

− 12 (3) 9x − 10 y = From equation (2), we obtain 2x + 3 y = (4) 13

Chapter 3 • Pair of Linear Equations in Two Variables

x =

(13 − 3y ) 2

− 10 y = −12

9 (13 − 3 y ) − 20 y = −24

117 − 27 y − 20 y = −24 ⇒ −47 y = −141 ⇒ y = 3 (5)

Substituting this value in equation (4), we obtain 2x + 9 = 13 ⇒ 2x = 4 ⇒ x = 2 Thus = x 2= and y 3 2. Solve 2x + 3 y = 11 and 2x − 4 y = − 24 and hence find the value of ‘m’ for which= y mx + 3. Sol. 2x + 3y = 11

(1)

2x − 4y = −24 (2) From equation (1), we obtain 11 − 3 y x= (3) 2 Substituting this value in equation (2), we obtain  11 − 3 y  2  − 24 ⇒ 11 − 3 y − 4 y = −24  − 4y =  2  −7 y = −24 − 11 ⇒ −7 y = −35 ⇒ y = 5 (4) Putting this value in equation (3), we obtain 11 − 3 × 5 4 x= ⇒ x= − = −2 2 2 Thus, x = −2, y = 5 Also= y mx + 3 5 = m ( −2 ) + 3 ⇒ −2m = 2 ⇒ m = −1 3. Form the pair of linear equations for the following problems and find their solution by substitution method. (a) The difference between two numbers is 26 and one number is three times the other. Find them. (b) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (c) The coach of a cricket team buys 7 bats and 6 balls for `3800. Later, she buys 3 bats and 5 balls for `1750. Find the cost of each bat and each ball. (d) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is `105 and for a journey of 15 km, the charge paid is `155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km. 51

8 , if 2 is added to both the 11 numerator and the denominator. If, 3 is added to both the numerator and the denominator it 5 becomes . Find the fraction. 6

(e) A fraction becomes

(f) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Sol. (a) Let the first number be x and the other number be y such that y > x.

Substituting this value in equation (2), we obtain

( 3800 − 6 y )

+ 5y = 1750 7 11400 − 18 y + 35 y = 12250 = 17 y 12250 − 11400 17 y= 850 ⇒ y= 50

Substituting this in equation (3), we obtain = x

3800 − 300 3500 ⇒= ⇒= x x 500 7 7

Hence, the cost of a bat is `500 and that of a ball is `50.

According to the given information,

y = 3x (1) (1)

(d) Let the fixed charge be `x and per km charge be `y.

y −x = 26 (2)

According to the given information,

On substituting the value of y from equation (1) into equation (2), we obtain

x + 10y = 105

(1)

3x − x = 26 ⇒ 2x = 26 ⇒ x = 13

x + 15y = 155

(2)

From (1), we obtain

Substituting this in equation (1), we obtain y = 39

Hence, the numbers are 13 and 39.

(b) Let the larger angle be x and smaller angle be y. We know that the sum of the measures of angles of a supplementary pair is always 180º.

According to the question,

x+y = 180º (1)

x−y = 18º (2)

From (1), we obtain x 180º − y (3) =

= x 105 − 10 y (3) ( 3)

Substituting this in equation (2), we obtain

105 − 10 y + 15 y = 155

5 y = 50 ⇒ y = 10 (4)

Putting this in equation (3), we obtain

= x 105 − 100 ⇒ x = 5

Hence, fixed charge = `5

And per km charge = `10

Charge for 25 km= x + 25 y

= 5 + 250 = `255

Substituting this in equation (2), we obtain

180°− y − y = 18° ⇒ 180°− 2 y= 18°

(e) Let the fraction be

−2 y = − 180°+ 18° ⇒ −2 y = −162° ⇒ y= 81°

x . y According to the given information,

Putting this in equation (3), we obtain x= 180° − 81°= 99°

x +2 9 = y + 2 11

11x + 22 = 9 y + 18

11x − 9 y = −4 (1)

x +3 5 = y +3 6

6x + 18 = 5 y + 15

6x − 5 y = −3 (2)

From equation (1), we obtain 9y − 4 x= (3) 11

Hence, the angles are 99º and 81º.

7x + 6 y = 3800

(1)

3x + 5y = 1750

(2)

From (1), we obtain

3800 − 6y 7

(3)

Substituting this in equation (2), we obtain

54 y − 24 − 55 y = −33

− y =−33 + 24 ⇒ − y =−9 ⇒ y = 9

Putting this in equation (3), we obtain

Thus fraction is

( 9y − 4 ) 11

− 5y = −3

81 − 4 77 ⇒ x= ⇒ x =7 11 11

(b) 3x + 4y = 10

2x − 2y = 2

9x = 2y + 7 x 2y (d) + = −1 2 3 y x − = 3 3 Sol. (a) By elimination method

x 7 = y 9

x + y = 5

(1)

(f) Let the age of Jacob be x and the age of his son be y.

2x − 3y = 4

(2)

Multiplying equation (1) by 2, we obtain

2x + 2y = 10

5 year hence

Age of Jacob = x + 5

Age of his son = y + 5

According to the given question,

( x + 5) = 3( y + 5)

x + 5 = 3 y + 15

x − 3 y = (1) 10

Five years ago

Age of Jacob= x − 5

Age of his son= y − 5

According to the given question,

( x − 5 )= 7 ( y − 5 )

x − 5 = 7 y − 35

x − 7y = −30 (2)

From (1), we obtain

x = 10 + 3 y (3)

Substituting this value in equation (2), we obtain 10 + 3 y − 7 y = −30 ⇒ 10 − 4 y = −30

−4 y = −30 − 10 ⇒ 4 y = 40 ⇒ y = 10

Putting this value in equation (3), we obtain

x = 10 + 30 ⇒ x = 40

(3)

Subtracting equation (2) from equation (3), we obtain 6 5 y = 6 ⇒ y = 5 Substituting the value in equation (1), we obtain

6 6 19 5 ⇒ x= 5 − ⇒ x = x+ = 5 5 5

Thus x =

By substitution method

x + y = 5

(1)

2x − 3y = 4

(2)

From equation (1), we obtain

x = 5 − y

Putting this value in equation (2), we obtain

2 ( 5 − y ) – 3y = 4

19 6 , y= 5 5

(3)

6 5 Substituting the value in equation (3), we obtain

x= 5 −

10 − 2 y − 3 y = 4 ⇒ −5 y = −6 ⇒ y =

19 6 25 − 6 ⇒ x= ⇒x= 5 5 5 6 19 Thus x = and y = 5 5

(b) By elimination method

Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

3x + 4 y = 10 (1)

NCERT Exercise 3.3

2x – 2 y = 2

Multiplying equation (2) by 2, we obtain

4x – 4 y = 4 (3)

Adding equation (1) and (3), we obtain

7x = 14 ⇒ x =

1. Solve the following pair of linear equations by the elimination method and the substitution method: (a) x + y = 5 2x – 3 y = 4 Chapter 3 • Pair of Linear Equations in Two Variables

(2)

14 ⇒ x =2 7

Substituting in equation (1), we obtain

Putting this value in equation (2), we obtain

6 + 4y = 10 ⇒ 4 y = 4 ⇒ y = 1

Thus x = 2, y = 1

 4 + 5y  0 9 0 ⇒ 3( 4 + 5y ) − 2 y − 7 =  − 2y − 7 =  3 

By substitution method

3x + 4 y = 10 (1)

2x – 2 y = 2 (2)

From equation (2), we obtain

2x= 2 + 2 y

x = 1 + y (3)

Putting this value in equation (1), we obtain

3 (1 + y ) + 4 y = 10 ⇒ 3 + 3 y + 4 y = 10

7y + 3 = 10 ⇒ 7 y = 7 ⇒ y = 1

Substituting the value in equation (3), we obtain

(d) By elimination method

x= 1+1 ⇒ x = 2

12 + 15 y − 2 y − 7 = 0 ⇒ 13 y + 5 = 0 5 13 y = −5 ⇒ y = − (1) 13

( 2)Substituting the value in equation (2), we obtain −10 + 91  5  9x =2  −  + 7 ⇒ 9x = 13 13   9x =

Thus 9 5 and y = − x= 13 13

81 9 ⇒ x= 13 13

x 2y + = −1 (1) 2 3

3x − 5 y − 4 = 0 (1)

9= x 2 y + 7 (2)

y x − (1= 3 3 ) 2) This (equation (1) can be written as

Equation 2 can also be written as

3x + 4 y = −6

9x − 2 y − 7 = 0

Multiplying equation (1) by 3, we obtain

And Equation (2) can be written as

9x − 15 y − 12 = 0

3x − y = 9

(3)

Subtracting equation (4) from equation (3), we obtain 5 y = −15 ⇒ y = −3

Subtracting equation (3) from equation (2), we obtain 5 13 y + 5 = 0 ⇒ y= − 13

Substituting in equation (1) , we obtain

5 25 0 ⇒ 3x + 0 −4= −4= 13 13 39x + 25 − 52 = 0 ⇒ ( 39x − 27 ) = 0 13 27 9 39x = 27 ⇒ x = ⇒ x= 39 13 9 5 Thus x = and y = − 13 13 3x + 5

3x − 5 y − 4 = 0 (1)

9= x 2 y + 7 (2)

Equation 2 can also be written as

9x − 2 y − 7 = 0

From equation (1), we obtain 4 + 5y x= 3

(3)

Substituting this value in equation (4), we obtain

3x − ( −3 ) =9 ⇒ 3x + 3 = 9

3x = 6 ⇒ x = 2

Thus x = 2, y = −3

By substitution method x 2y + = −1 (1) 2 3

By substitution method

(2)

y = 3 (2) 3

x−

This equation (1) can be written as 3x +(41y) = −6 (3)

And ( 2) Equation (2) can be written as

3x − y = (4) 9

From equation (4), we obtain

9+ y (5) 3

Putting this value in equation (3), we obtain

 9+ y  3 −6  + 4y =  3 

9 + y + 4y = −6 ⇒ 5 y = −15 ⇒ y = −3

Substituting the value in equation (5), we obtain 9−3 x= ⇒ x =2 3 Thus x = 2 and y = −3

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (a) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction? (b) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? (c) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. (d) Meena went to bank to withdraw `2000. She asked the cashier to give her `50 and `100 notes only. Meena got 25 notes in all. Find how many notes of `50 and `100 she received. (e) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid `27 for a book kept for seven days, while Susy paid `21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Sol. (a) Let the numerator = x

Let the denominator = y

Therefore, the fraction =

According to the first condition,

x +1 = 1 ⇒ x +1 = y −1 y −1

x y

x − y =−2… (1) (1)

According to the second condition,

x 1 = y +1 2

2x= y + 1

2x − y =1… (2) (2)

Chapter 3 • Pair of Linear Equations in Two Variables

Multiply equation (1) by 2 and subtract from equation (2), we get: 2x − y − 2x + 2y = 1 + 4 y = 5 Putting the value of y = 5 in equation (1), we get:

x − 5 =−2 ⇒ x = 3

Hence, the fraction=

x 3 = y 5

(b) Let the age of Nuri = x years

Let the age of Sonu = y years

5 years ago,

Age of Nuri = x − 5 years

Age of Sonu = y − 5 years

According to the question,

x − 5= 3 ( y − 5 ) ⇒ x − 5 = 3 y − 15

x − 3y = −10… (1) (1)

After 10 years,

Age of Nuri = x + 10 years

Age of Sonu = y + 10 years

According to the question,

x + 10 = 2 ( y + 10 ) ⇒ x + 10 = 2 y + 20

x − 2y = 10 (2)

Subtracting equation (2) from equation (1), we get: x − 3y − x + 2y = −10 − 10 − y =−20 ⇒ y = 20 Putting the value of y = 20 in equation (1), we get:

x − 3 ( 20 ) = −10 ⇒ x − 60 = −10 ⇒ x = 50

Hence, the age of Nuri is 50 years, and the age of Sonu is 20 years. (c) Let the one’s place = x

Let the ten’s place = y

Therefore, the number = 10y + x

Sum of the digits is 9, therefore

x + y =9… (1) (1)

Number obtained by reversing the digits = 10x + y

According to the question,

9 (10 y + x= ) 2 (10x + y )

90 y + 9x = 20x + 2 y 55

10 y + 8 y = 0

x + y − x + 3y = 25 − 40

x − 8y = 0 (2)

− y =−15

Putting the value of y in equation (1), we get

x + 15 = 25 ⇒ x = 10

Subtracting equation (2) from equation (1), we get: x + y − x + 8y = 9 − 0 9 y = 9 ⇒ y = 1

Putting the value of y = 1 in equation (1), we get:

x +1 = 9 ⇒ x = 8

Therefore, the number = 10 y += x 10 (1) += 8 18

Hence, the two-digit number is 18.

(d) Let the number of notes of `50 = x

Let the number of notes of `100 = y

Total number of notes is 25, therefore

x+y = 25 (1)

The total amount of `50 and `100 is `2000, therefore

50x + 100 y = 2000

x + 2y = 40 (2)

Subtracting equation (2) from equation (1), we get:

Hence, the number of `50 notes is 10 and the number of `100 notes is 15. (e) Let the charge for the first three days = `x

Let the additional charge for each day = ` y

Charge for seven days = `27, therefore

x + 4 y =27… (1) (1)

Charge for 5 days = `21, therefore

x + 2y = 21 (2)

Subtracting equation (2) from equation (1), we get

x + 4y − x − 2y = 27 − 21 2y = 6 ⇒ y = 3

Putting the value of y = 3 in equation (1), we get

x + 4 ( 3) = 27 ⇒ x = 15

Hence, the fixed charge is `15 and the additional charge for each day is `3.

Multiple Choice Questions 1. The difference between two numbers is 15. If the larger number is represented as x and the smaller number as y, which of the following equations correctly represents this situation?

Y 4

(a) x − y = 15

(d) x + y = 30

–5 –4 –3 –2 –1 0 –1

(a) parallel

–3 –4

a1 b1 c1 (a) = = a 2 b2 c 2

(b) always coincident (c) intersecting or coincident

3. The given pair of linear equations is non-intersecting. Which of the following statement is true?

y b2

0 2 c

–2

2. If a pair of linear equations is consistent, then the lines will be

(d) always intersecting

x 2

x a1

(b) x + y = 15 (c) x − y = −15

y 1 b

3 0 1 c 2

a1 b1 c1 ≠ = a 2 b2 c 2

a1 b1 c1 (b) = ≠ a 2 b2 c 2 (d)

a1 b1 c1 + + a 2 b2 c 2

(CBSE SQP 2023)

4. The value of k for which the pair of equations kx= y + 2 and 6= x 2 y + 4 has infinitely many solutions,

(a) is k = 3

(b) does not exist

(d) is k = 4

5. The lines represented by linear equations x = a and = y b ( a ≠ b ) are

(a) Intersecting at ( a , b )

(b) Intersecting at (b , a )

(d) coincident

(a) consistent with unique solution. (b) consistent with infinitely many solutions. (c) inconsistent (d) inconsistent but can be made consistent by extending these lines.

(CBSE Basic 2024)

6. 3 chairs and 1 table cost `900; whereas 5 chairs and 3 tables cost `2,100. If the cost of 1 chair is `x and the cost of 1 table is `y, then situation can be represented algebraically as (a) 3x= + y 900, 3x = + 5 y 2100

(CBSE QB)

9. The sum of the digits of a two-digit number is 9. If 27 is subtracted from the number, its digits are interchanged. Which of these is the product of the digits of the number?

(b) x += 3 y 900, 3x += 5 y 2100

(a) 8

7. In the given figure, graphs of two linear equations are shown. The pair of these linear equations is Y

(b) 14

(d) x += 3 y 900, 5x += 3 y 2100

(CBSE 2024)

p 8. What is the value of q if + 3q = 6 and 2p − 2q = 10? 2 (a) 1 (b) 4 (c) 6 (d) 16

(d) 20 (CBSE QB)

10. The value of c for which the pair of equations cx − y = 2 and 6x − 2 y = 3 will have infinitely many solutions is (a) 3 (b) −3 (c) −12 (d) no value 11. Sum of numerator and denominator of a fraction is 12. If the denomimator is increased by 3, the 1 fraction becomes , then the fraction is 2 (a)

8 7

(b)

6 7

(c)

5 7

(d)

4 7

Answers 1. (a) x − y = 15

6. (c) 3x= + y 900, 5x += 3 y 2100

2. (c) intersecting or coincident

7. (a) consistent with unique solution

a1 b1 c1 3. (b) = ≠ a 2 b2 c 2

8. (a) 1 9. (c) 18

4. (a) k = 3

10. (d) no value 5 11. (c) 7

5. (a) Intersecting at ( a , b )

Constructed Response Questions Very Short Answer Questions 1. Show that x = 2, y = 1 is not a solution of the system of simultaneous linear equations 2x + 7 y = 11 and x − 3y = 5.

Chapter 3 • Pair of Linear Equations in Two Variables

11 Sol. The given system of equations is 2x + 7 y =

(i)

x − 3y = 5 (ii) Putting x = 2 , y = 1 in equation (i), we obtain L.H.S = 2 × 2 + 7 × 1 = 11 = R.H.S 57

So, x = 2 and y = 1 satisfy equation (i) Putting x = 2, y = 1 in equation (ii), we obtain L.H.S =2 × 1 − 3 × 1 =−1 ≠ R.H.S Hence, x = 2, y = 1 is not a solution of the given system of equations. 2. If 2x + y = 13 and 4x − y = 17, then find the value of x − y . Sol. We can add these two equations to eliminate y ( 2x + y ) + ( 4x − y ) = 13 + 17 Simplifying this, we get: 6x = 30 ⇒ x = 5 Now, substitute x = 5 into the first equation

8 30 8 22 x+ = 6 ⇒ x= − = 5 5 5 5 5. Find out whether the following pair of linear equations are consistent or inconsistent: 5x − 3 y = 11, −10x + 6 y = 22 Sol. Since,

5 −3 11 1 1 1 = ≠ Or, − = − ≠ −10 6 22 2 2 2

Thus, both lines are parallel. So, the given system is inconsistent. 6. If x = −a and y = b is the solution of the pair of equations x − y = 2 and x + y = 4, find the values of a and b.

2 ( 5) + y = 13 ⇒ 10 + y = 13 ⇒ y = 3

Sol. Let x − y = 2

So, x − y = 5 − 3 = 2

4, x + y = (ii)

3. Using graphical method, find whether following system of linear equations is consistent or not: 10 x = 5 and y =3

(i)

First, substitute x = −a and y = b into the equation (i) 2 ⇒ −a − b = 2 ⇒ a +b = −2 ( −a ) − b = Now, substitute x = −a and y = b into the equation (ii)

Sol.

4 ( −a ) + b = 5

−a + b = 4 Now we have two new equations: a + b = −2 −a + b = 4

–5

First, add these two equations together: (a + b ) + ( −a + b ) =−2 + 4

Both lines are intersection at a point.

2b = 2 ⇒ b = 1

Thus, –5 the given equations are consistent.

Now, substitute b = 1 back into one of the equations to find a.

4. Solve for x and y : x + y = 6 ; 2x − 3 y = 4 Sol. We have x+y = 6 ; (i) 2x − 3 y = 4 (ii)

a + 1 =−2 ⇒ a =−2 − 1 =−3 7. In fig. ABCD is a rectangle. Find the values of x and y . D

First, we can solve equation 6 x= − y Now substitute in equation (ii) 2 ( 6 − y ) − 3y = 4 ⇒ 12 − 2 y − 3 y = 4 8 12 − 5 y = 4 ⇒ −5 y = −8 ⇒ y = 5 8 Substitute y = into equation (i) 5

x+y

14 cm

x–y A

30 cm

(CBSE 2018)

Sol. Since, ABCD is a rectangle, so opposite sides are equal.

Thus, x + y = 30 ⇒ and x − y = 14

Number = of girls Number of boys + 4

On adding two equations, we get

i.e., x = y + 4 or, x − y = 4

x + y + x − y = 30 + 14 ⇒ 2x = 44 ⇒ x = 22 Hence, 22 + y = 30 ⇒ y = 30 − 22 = 8 8. For what value of k , the pair of linear equations 3x + y = 3 and 6x + ky = 8 does not have a solution? 3 1 3 Sol. For no solution, = ≠ 6 k 8

(CBSE SQP 2021)

Therefore, k = 2 9. For the pair of equations λ x + 3 y + 7 = 0 and 2x + 6 y − 14 = 0. To have infinitely many solutions, the value of λ should be 1. Is the statement true? Give reasons. Sol. λ x + 3 y + 7 = 0 and 2x + 6 y − 14 = 0 a1 λ= , b1 3,= c1 7 Here,= a 2 = 2, b2 = 6, c 2 = −14 a1 b1 c1 If = = then system has infinitely many a 2 b2 c 2 solutions. λ 3 7 ⇒ = = 2 6 −14 3 = ⇒λ =1 2 6

and

7 1 = = − ⇒λ = −1 2 −14 2

Hence λ , does not have a unique value. So, for no value of λ for the given pair of linear equations has infinitely many solutions. 10. Ten students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys. Represent this situation algebraically. Sol. Let the number of girls be x and the number of boys be y. It is given that total ten students took part in the quiz. Number of girls + Number of boys = 10 i.e., x + y = 10 It is also given that the number of girls is 4 more than the number of boys.

Chapter 3 • Pair of Linear Equations in Two Variables

Thus, the algebraic representation of the given situation is x + y = 10, x − y = 4 11. 37 pens and 53 pencils together cost `320, while 53 pens and 37 pencils together cost `400. Find the cost of a pen and that of a pencil. Sol. Let the cost of a pen be `x and that of a pencil be `y. 320 (i) Then, 37x + 53 y = 53x + 37 y = 400 (ii) Adding equations (i) and (ii), we get: 90x + 90 y= 720 ⇒ x + y= 8 (iii) Subtracting equation (i) and (ii), we get: 16x − 16 y = 80 ⇒ x − y = 5 (iv) Adding equations (iii) and (iv), we get: 2x = 13 ⇒ x = 6.5 Substituting x = 6.5 in equation (iii), we get: y =− 8 6.5 = 1.5 Hence, cost of one pen = `6.50 and cost of one pencil = `1.50

Short Answer Questions 1. Solve graphically the following pair of linear equations: 2 y − 3x = 14 2x + 3 y = 8 Hence, shade the region enclosed by these lines and y-axis. (CBSE 2017) Sol. Two solutions of linear equations. −3x + 2 y = (i) 14 ( i ) 2x + 3 y = 8 (ii) ( ii ) The solutions for each equation are given below: For equation (i): x

−2

For equation (ii): x

The graphical representation of the given pair of linear equations is as follows:

3. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present? (CBSE 2019)

Y 2x +

3y =

(0, 7) B 7 6

(−2, 4)A 3x =

4 3 2 1

2y −

Sol. Let Sumit’s age = s and his son’s age = x. s = 3x

−5 −4 −3 −2 −1

According to question:

C (1, 2)

s += 5 2.5 ( x + 5 ) Substitute s = 3x into the second equation:

1 2 3 4

3x + 5= 2.5x + 12.5 ⇒ 0.5x = 7.5 ⇒ x = 15 Sumit's age = = s 3= x 45

The two lines intersect at the point ( −2, 4 ).

4. If 2x + y = 23 and 4x − y = 19, find the value of

So, x = −2, y = 4 is the required solution. 2. Three lines x + 3 y= 6, 2x − 3 y= 12 and x = 0 are enclosing a beautiful triangular park. Find the points of intersection of the lines graphically and the area of the park, if all measurements are in km. (CBSE 2015)

Sol. x + 3 y = 6 (i) (i )

And 2x − 3 y = (ii) 12 ii ) 6−x For equation (i): y = 3 x

y  − 2 . x 

( 5y − 2x ) and  Sol. We have 2x + y = 23 4x − y = 19

Adding equations: 6x = 42 ⇒ x = 7 Substituting x in one of the equations: 2 ( 7 ) + y = 23 ⇒ y = 9 Hence, 5 y − 2x = 5 ( 9 ) − 2 ( 7 ) = 45 − 14 = 31 and,

y 9 − 2 = − 2 =−0.71428 x 7

5. Find c if the system of equations cx + 3 y + ( 3 −= c ) 0; 12x + cy = − c 0 has infinitely many solutions. (CBSE 2019)

A ( 6, 0 ) , B ( 0, 2 )

2x − 12 Equation (ii): y = 3 x

Sol. We have

−4

−2

cx + 3 y + ( 3 − c ) = 0

A ( 6, 0 ) , C ( 0, − 4 )

12x + cy − c = 0

Vertices A ( 6, 0 ) , B ( 0, 2 ) and C ( 0, − 4 )

For Infinite Solutions: We need:

1  ∴ Area of ∆ABC =  × 6 × 6  km 2 2 

c 3 ⇒ 2 c =36 ⇒ c =6 or c =−6 = 12 c

= 18 km 2

Verifying c = 6 3 3 −c = −c c

B(0, 2) 2 x+3 y=6 1 A(6, 0) −3 −2 −1 −1 −2 −3 −4 −5 Y'

c 3 3 −c = = −c 12 c

1 2

3 4 5 12 y= 3 −

2x C(0, −4)

3 −3 ⇒ = 6 −6

1 1 = vaild 2 2 3 9 = c = −6, not vaild −6 −6 Only c = 6 vaild.

6. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay `1,500 as hostel charges while another student B, who takes food for 26 days, pays `1,800. Find the fixed charges and the cost of food. (CBSE 2023) Sol. Let the monthly fixed charges of hostel be `x and cost of food be `y per day. A.T.Q 20 x+ y= 1500 (i) 26 x+ y= 1800 (ii) Subtracting equations (i) & (ii) 26 x + y − x − 20= y 1800 − 1500

Substituting N = 16 back into either equation for R: R = 3 (16 ) − 6 = 48 − 6 = 42 Therefore, Rashmi is 42 years old and Nazma is 16 years old. 8. In a 2-digit number, the digit at the unit’s place is 5 less than the digit at the ten’s place. The product of the digits is 36. Find the number. (CBSE Standard 2024) Sol. Let’s denote the digit at the tens place as t and the digit at the units place as u. The digit at the unit’s place is 5 less than the digit at the ten’s place: u = t − 5 The product of the digits is 36: t ⋅u = 36 Substitute u = t − 5 into the second equation:

6 y= 300 ⇒ y= 50

t ⋅ (t − 5= ) 36t 2 − 5t − 36= 0

Substitute in eq (i) we get

This quadratic equation can be factored as:

x + 20 ( 50= = 1500 ) 1500 ⇒ x + 1000

0 (t − 9)(t + 4 ) =

1500 x= − 1000 = 500 500, = x = y 50

Therefore, t = 9 or t = −4 (but a digit cannot be negative), so t = 9.

Hence, the monthly fixed charges of hostel be `500 and cost of food be `50 per day.

Using u = t − 5:

7. Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now? (CBSE Standard 2024)

Sol. Let’s denote Rashmi’s current age as R and Nazma’s current age as N . Three years ago, Rashmi’s age was R − 3 and Nazma’s age was N − 3. Since Rashmi was thrice as old as Nazma at that time, the equation is: R −= 3 3( N − 3)

Ten years later, Rashmi’s age will be= R + 10 Nazma’s age will be= N + 10.

9 u = −5= 4 Therefore, the two-digit number is 94. 9. Rehana went to a bank to withdraw `2,000. She asked the cashier to give her `50 and `100 notes only. Rehana got 25 notes in all. Find how many notes of `50 and `100 did she receive. (CBSE Standard 2024)

Sol. Let’s denote the number of `50 notes as x and the number of `100 notes as y . Rehana received a total of 25 notes: x+y = 25 The total amount withdrawn is `2,000, using `50 and `100 notes:

Since Rashmi will be twice as old as Nazma then, the equation is:

50x + 100 y = 2000

R + 10= 2 ( N + 10 )

50x + 100= y 2000 ⇒ x + 2= y 40

The first equation simplifies to:

Simplify the second equation:

3 R − = 3N − 9 ⇒ R= 3N − 6

Now we have two simultaneous equations: x+y = 25

The second equation simplifies to:

2 x+ y= 40

10 R + = 2N + 20 ⇒ R = 2N + 10

Subtract the first equation from the second equation to solve for y :

By setting these two expressions for R equal to each other: 3N − 6= 2N + 10N= 16 Chapter 3 • Pair of Linear Equations in Two Variables

x + 2 y − ( x + y ) = 40 − 25 15 y=

Substitute y = 15 back into the first equation to find x : x + 15 = 25 ⇒ x = 10 Therefore, Rehana received 10 notes of `50 and 15 notes of `100. 1 10. A fraction becomes when 2 is subtracted 3 1 from the numerator and it becomes when 1 is 2 subtracted from the denominator. Find the fraction. Sol. Let the fraction be

x y

(CBSE 2019)

According to the first condition, x −2 1 = y 3 ⇒ 3x − 6 = (i) y According to the second condition, x 1 = y −1 2 ⇒ 2x =y − 1 ⇒ y = 2x + 1 (ii) From eq’s. (i) and (ii), we get 3x − 6 = 2x + 1 ⇒x = 7 Substituting value of x in eq (i), we get = y 3(7 ) − 6

21 y= − 6 = 15

7 Hence, fraction is . 15 11. In a painting competition of a school a child made Indian national flag whose perimeter was 50 cm. Its area will be decreased by 6 square cm, if length is decreased by 3 cm and breadth is increased by 2 cm then find the dimension of flag. (CBSE 2015)

Substitute Equation (3) into Equation (2): 2 ( 25 − y ) − 3 y = 0 50 − 2 y − 3 y = 0 50 − 5 y = 0 10 y= Substitute y = 10 back into Equation (3) 25 = x − 10 15 x= = ⇒ x 15 cm = and y 10 cm Length of the flag = 15 cm and Breadth of the flag = 10 cm 12. Draw the graphs of the pair of linear equations: x + 2y = 5 and 2x − 3 y = −4. Also, find the points where the lines meet the x-axis. (CBSE 2014, 2015)

Sol. Given: x + 2 y = 5 5−x ⇒y= For line to meet x − axis y = 0 2 x

And 2x − 3 y = −4 2x + 4 ⇒y= 3 x

−2

Thus, the lines meet x-axis at ( 5, 0 ) and ( −2, 0 ) respectively. Y

Sol. Let length of the flag be x cm and breadth of the flag be y cm 2(x + y ) = 50

x + 2y = 5

or x + y = 25 (1) (1)

From (1), we get = x 25 − y (3) ( 3) 62

(−2, 0) −4 −3 −2 −1

(1, 2) (3, 1)

1 0 −1 −2 −3

4 =−

(4, 4)

( x − 3)( y + 2 ) = xy − 6

2x − 3 y = 0 (2) ( 2)

According to the condition xy + 2x − 3 y − 6 = xy − 6

y −3

(5, 0) 1

13. Half of the difference between two numbers is 2. Sum of greater number and twice the smaller number is 13. Find the numbers. (CBSE 2023) Sol. Let x be the greater number and y be the smaller number. According to question, 1 2 (x − y ) = 2 ⇒ x − y =2 × 2

∠ C =(x + y ) =20 + 30 =50 ∠D = (3 y − 10) =× 3 30 − 10 = 80 Hence, the required values of x and y are 20° and 30° respectively and the values of the four angels i.e., ∠A, ∠B, ∠C and ∠D are 130°, 100°, 50° and 80°, respectively.

Long Answer Questions 1. Determine graphically the coordinates of the vertices of triangle, the equations of whose sides are given by 2 y − = x 8, 5 y − = x 14 and y − 2x = 1.

⇒ x − y = (i) 4 Also, x + 2 y = (ii) 13 On solving equation (i) and (ii), we have: x − y − ( x + 2 y ) = 4 − 13

(CBSE 2020)

Sol. Given,

−3 y = −9 3 y= Putting the value of y = 3 in (i), we get: 3 x− = 4 ⇒x = 7 Numbers are 7 and 3. 14. The angles of a cyclic quadrilateral ABCD are ∠A= (6x + 10) , ∠ B = (5x ) , ∠ C = (x + y ) and

∠= D (3 y − 10) . Find x and y and hence the values

of the four angles.

2y − x = 8

⇒ x = 2y − 8

x = 2y − 8

−8

5y − x = 14

⇒ x = 5 y − 14

x = 5y − 14

−4

and y − 2x = 1

y = 1 − 2x

⇒ y =1 + 2x

Plotting the above points and drawing lines joining them, we get the graphical representation: Y

(NCERT Exemplar)

Sol. We know that, by property of cyclic quadrilateral,

Sum of opposite angles = 180 



∠ A + ∠ C = (6x + 10) + (x + y ) = 180



[Given ∠ A = (6x + 10) , ∠ C = (x + y ) ] 



C(−4, 2)

⇒ 7x + y = 170 (i) And ∠ B + ∠ D = (5x ) + (3 y − 10) = 180 



[Given ∠ = B (5x ) , ∠ D = 3 y − 10) ] ⇒ 5x + 3 y = 190 (ii) On multiplying Eq. (i) by 3 and then subtracting, we get 3 × ( 7x + y ) − ( 5x + 3 y ) = 510 − 190 ⇒ 21x + 3 y − 5x − 3 y= 320° ⇒ x= 20° On putting x= 20° in Eq. ( i ) , we get 7 × 20° + y= 170° ⇒ y= 30° ∠A= (6x + 10) =× 6 20 + 10 = 130 ∠B = (5x ) =× 5 20 = 100

Chapter 3 • Pair of Linear Equations in Two Variables

(−8, 0)

(0,

, 5) B(2 ) A(1, 3

− 2y

(6, 4)

1 5y − x =

2 1 (0, 1)

−8 −7 −6 −5 −4 −3 −2 −1 −1 −2 y − 2x = 1

−3 Y'

Hence, the coordinates of the vertices of the triangle ABC are A (1, 3 ) , B ( 2, 5 ) and C ( −4, 2 ) . 2. Points A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If they travel in same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? What steps do you suggest to save petrol? (CBSE 2015) 63

Sol. Let the speed of car at A be u km/h and car at B be v km/h.

∴ Speed of walking is 3 km/h and speed of cycling is 18 km/h.

Case 1: When both cars travel in the same direction

When Rajiv walked and cycled for equal time, distance covered =3 × 1.5 + 18 × 1.5

100 km

Let both the cars meet at point C in 5 hours. Car at A travels distance AC, whereas car at B travels distance BC. ∴ AC =× 5 u and BC =× 5 v Now, AC − BC = 100 ⇒ 5u − 5v = 100 ⇒u −v = 20 (i) Case 2: When both cars travel in opposite directions Let both cars meet at point D. u

100 km

Car at A will travel distance AD, whereas car at B will travel distance BD. ∴ AD = 1 × u and BD = 1× v 100 (ii) Now, AD + BD = 100 ⇒ u + v = On adding (i) and (ii), we get 2u = 120 ⇒ u = 60 From ( ii ) , we get 60 + v = 100 ⇒ v = 40 3. Rajiv walks and cycles at uniform speeds. When he walks for 2 hrs and cycles for 1 hr , distance travelled is 24 km. When he walks for 1 hr and cycles for 2 hrs, distance travelled is 39 km. Find his speed of walking and cycling. If he walked and cycled for equal time in 3 hrs how much distance does he cover? (CBSE 2013) Sol. Let the speed of walking be x km/h and the speed of cycling be y km/h. According to question, 2x + y = 24 (i) (i ) x + 2y = 39 (ii) ( ii ) Multiply (i) by 2 and subtracting from (ii), we get x + 2 y − ( 4x + 2 y ) = 39 − 48 ⇒ 3x = 9 ⇒x = 3 Putting the value of x in (i), we get, 2×3 + y = 24 ⇒ y = 24 − 6 = 18 64

4. Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. If the difference of the digits is 3, determine the number. Sol. Let the ten’s and unit digit be y and x respectively. So, the number is 10y + x The number, when its digits are reversed, becomes 10x + y. So, 7 (10 y + x= ) 4 (10x + y ) ⇒ 70 y + 7x = 40x + 4 y ⇒ 70 y − 4 y = 40x − 7x ⇒ 2y = x (i)

= 4.5 + 27 = 31.5 km

and x−y = 3 (ii) From (i) and (ii), we get 2y − y = 3 = y 3= and x 2(3)= 6

Hence, the number is 36. 5. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water. (CBSE 2019) Sol. Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. Then speed of boat in upstream =

( x − y ) km hr

and speed of boat in downstream = (x + y) km/h According to question,

30 44 + = 10 (i) (i ) x−y x+y And 40 55 = 13 (ii) + x−y x+y 1 1 Putting = u= and v in ( i ) and ( ii ) , we get x−y x+y 30u + 44v = 10 (iii) 40u + 55v = 13 ( (iv) iv ) Multiplying ( iii ) by 40 and ( iv ) by 30, we get 1200u + 1760v = 400 (v) (v )

1200u + 1650v = 390 ( (vi) vi ) Subtracting ( vi ) from ( v ) , we get 1 110v = 10 ⇒ v = 11 From ( v ) ,

1760 400 − 160 400 ⇒ 1200u = = 11 240 1 ⇒= u = 1200 5 1200u +

1 1 ∴ =⇒ x − y = (vii) 5 x−y 5 1 1 and = ⇒ x + y = 11 (viii) x + y 11 Adding ( vii ) and ( viii ) , we get x − y + x + y = 5 + 11 ⇒ 2x = 16 ⇒ x = 8 Using (viii) 8 + y = 11 ⇒ y = 3

0 = 6s − 6t − 36 (2) Solving equations (1) and (2) simultaneously, starting with (2): 6s − 6t = 36 ⇒ s − t = 6(3)

Substituting s = t + 6 into equation (1): −4 (t + 6 ) + 6t − 24 = 0 −4t − 24 + 6t − 24 = 0 2t − 48 = 0 ⇒ t = 24 s =t + 6 = 24 + 6 = 30km/h Now, finding d : d = s × t = 30 × 24 = 720km Therefore, the length of the journey is 720 km.

7. Solve the following pair of equations: 2 3 4 9 2 and = − = −1 + x y x y ∴ Speed of the boat in still water = 8 km/h Sol. The pair of equations given as: And speed of the stream = 3 km/h 2 3 4 9 2 and = − = −1 + x y x y 6. A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time and if the train were slower by 6 km/h, it would have taken 6 hours more than the scheduled time. Find the length of the journey.

Sol. Let s be the original speed of the train in km/h, t be the scheduled time in hours, and d be the distance of the journey in km. The distance equation with the original speed is: d = s ×t If the train was 6 km/h faster, the speed would be s + 6 km/h, and the time would be t − 4 hours: d = ( s + 6 ) ⋅ (t − 4 )

(CBSE 2015)

Let’s solve these equations by substituting: 1 1 Let u = and v = . x y Thus, the equations become: 2u + 3v = 2 (1)

4u − 9v = −1 (2) Multiplying the first equation by 2: 4u + 6v = 4 (3) Now subtract the second equation from this: 4u + 6v − ( 4u − 9v ) = 4 − ( −1) 1 15v = 5 ⇒ v = 3 v Substitute in equation 3, we get,

s × t = ( s + 6 )(t − 4 )

1 4u + 6   = 4 3 1 4u= 4 − 2 ⇒ u = 2 1 1 = ⇒ x =2⇒x =4 Therefore: x 2 1 1 = ⇒ y = 3 ⇒ y = 9 y 3

s × t = ( s − 6 )(t + 6 )

The solutions for x and y are:= x 4,= y 9

If the train was 6 km/h slower, the speed would be s − 6 km/h, and the time would be t + 6 hours: d = ( s − 6 ) ⋅ (t + 6 )

Since d = s ⋅t , we substitute this into the other equations:

Expanding both equations and simplifying: s × t = s × t − 4s + 6t − 24 0= −4s + 6t − 24 (1) s × t = s × t + 6s − 6t − 36

Chapter 3 • Pair of Linear Equations in Two Variables

8. For what values of m and n the following system of linear equations has infinitely many solutions. 3x + 4 y = 12 and (m + n ) x + 2 (m − n ) y =5m − 1 (CBSE 2015, 2018)

Sol. Let 3x + 4 y = 12 (m + n ) x + 2 (m − n ) y =5m − 1

(1) (2)

For infinitely many solutions: a1 b1 c1 = = a 2 b2 c 2

3 4 12 Thus, = = m + n 2 (m − n ) 5m − 1 3 4 3 2 = ⇒ = m + n 2 (m − n ) m + n m − n 3 (m − n ) = 2 (m + n ) 3m − 3n = 2m + 2n ⇒ m = 5n (3) Using other ratio, 3 12 = ⇒ 3 ( 5m − 1 = ) 12 (m + n ) m + n 5m − 1 Substitute m = 5n , 3 ( 5 ( 5n ) −= 1) 12 ( 5n + n ) ⇒ 3 ( 25n= − 1) 12 ( 6n ) ⇒ 75n= − 3 72n ⇒ 3n =3 ⇒ n =1 ⇒ m =5n =5 Thus= m 5,= n 1

9. Two people are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other with different speeds, they will meet in 2 hours. Had they walked in the same direction with same speeds as before, they would have met in 8 hours. Find their walking speeds. (CBSE 2023)

Sol. Let the speed of person 1 be x km/h and the speed of person 2 be y km/h. Distance = Speed × Time Given, Distance = 16 km

Case 1: Towards each other, Time = 2h x km/h A

y km/h B

16 km

( x + y ) km/h Distance = 2 ( x + y ) km ⇒ 16 = 2x + 2 y Speed=

⇒x+y = 8 (1) Case 2: Towards each other, Time = 8h x km/h

y km/h

( x − y )km / h Distance (AB) = 8 ( x − y ) km ⇒ 16 = 8x − 8 y Speed =

⇒x−y = 2 (2) Adding 1 and 2, we get x + y + x − y = 8 + 2 ⇒ 2x = 10 ⇒ x = 5 Substituting in 1, we get: 5 + y =8 ⇒ y =3 Speed of person 1 = 5 km/h, Speed of person 2 = 3 km/h

Competency Based Questions Multiple Choice Questions 1. Consider the equations as shown: b ( x − a )( y − b ) = ( x − 2a )  y −  2  1  a 2   + y  y +  − 2xy =5 + ( x − y ) xx +  2b  2   On comparing the coefficients, a student says these pairs of equations is consistent. Is he/she correct? Which of these explains why? (a) Yes; because they are parallel lines (b) Yes; because they are intersecting lines (c) No; because they are parallel lines (d) No; because they are intersecting lines 2. The sum of the digits of two digits number is 9. Also 9 times the number is twice the number obtain by reversing the order of digits. Find the number. 66

(a) 18

(b) 15

(d) 21

3. In a classroom, there are 30 beings altogether, including students and their pet dogs. If the total count of legs in the room is 90, and knowing that each student has two legs and each dog has four legs, how many students and dogs are present? (a) 20 students, 10 pets (b) 25 students, 5 pets (c) 15 students, 15 pets (d) 28 students, 2 pets 4. A landscaper charges `120 for 4 small plants and 2 large plants. Another set of 3 small plants and 5 large plants costs `230. How much does each small and large plant cost?

(a) Small = `20, Large = `30

(i) ∠ A = 25

(b) Small = `10, Large = `40

(ii) ∠ C = 82

(iii) ∠ B = 82, ∠ C = 73

(d) Small = `50, Large = `60 5. A total of 24 fruits, consisting of apples and oranges, cost `60. If each apple costs `1 and each orange costs `3, how many of each fruit were bought? (a) 12 apples, 12 oranges

(iv) ∠B = 73° (a) Only (i)

(b) (i) and (iii)

(d) (iii) only

Assertion-Reason Based Questions

(b) 18 apples, 6 oranges

irections: Each Question consists of two statements: D an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

(c) 6 apples, 18 oranges (d) 15 apples, 9 oranges 6. A theater has 200 seats. One ticket for adult costs `25 and for children `10. If the total revenue from full capacity is `3500, how many adults and child tickets were sold?

(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A.

(a) 100 adults, 100 children

(b) 120 adults, 80 children

(d) A is false, but R is true.

(c) 140 adults, 60 children (d) 80 adults, 120 children 7. If the lines given by 3x + 2ky = 2 and 2x + 5 y = 1 are parallel, then the value of k is 5 4

(b)

2 5

15 4

(d)

3 2

(a) − (c)

a circle with centre O ( 2, 3 ) , then the value of x is 2.

Reason (R): Centre of the circle s the mid-point of each chord of the circle. (CBSE 2023)

A (1, 3)

Case Study Based Questions

C 4 5

(a) 15

(b) 10

a1 b1 ≠ a 2 b2

2. Assertion (A): If the point A ( 4, 3 ) and B ( x , 5 ) lie on

Reason (R): The system of linear equations a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0 has a unique solution if

8. In the figure, the area of ∆ABC is

B –4 –3 –2 –1

1. Assertion (A): The number of common solutions for the system of linear equations 5x + 4 y + 6 = 0 and 10x + 8 y = 12 is zero.

(d) 2.5

Select Response Type MCQs 1. In ∆ABC , ∠A = x °, ∠B = ( 3x − 2 ) ° and ∠C = y °. Also ∠c − ∠B = 9°.

Chapter 3 • Pair of Linear Equations in Two Variables

1. Essel World is one of India’s largest amusement parks that offers a diverse range of thrilling rides, water attractions, and entertainment options for visitors of all ages. The park is known for its iconic “Water Kingdom” section, making it a popular destination for family outings and fun-filled adventure. The ticket charges for the park are `150 per child and `250 per adult. (CBSE 2024) On a day, the cashier of the park found that 300 tickets were sold and an amount of `55,000 was collected. Based on the above information, answer the following questions:

(a) If the number of children visited be x and the number of adults visited be y, then write the given situation algebraically. (b) (i) How many children visited the amusement park that day?

OR (ii) How many adults visited the amusement park that day? (c) How much amount will be collected if 250 children and 100 adults visit the amusement park?

Answers Multiple Choice Questions

9 × (10 A + B ) =2 × (10B + A ) .

1. (b) Yes; because they are intersecting lines b ( x − a )(b − y ) = ( x − 2a )  y −  2 

90 A + 9B − 20B = 2 A ⇒ 90 A − 2 A = 20B − 9B

⇒ xy − bx − ay + ab = xy −

bx − 2ay + ab 2

88 A = 11B ⇒ 8A = B Now, substitute B = 8 A into the first condition:

bx ⇒ − ay =… 0 (i) (i ) 2 2

1   a  +  y +  − 2xy =5 + (x − y )2 Also,  x +  2b   2  ay x + y2 + − 2xy =5 + x 2 + y 2 − 2xy 2b 2 x ay 5… (ii) ⇒ + = ( ii ) 2b 2 From equations (i) and (ii), we have ⇒ x2 +

bx − ay = 0 2

x + aby =10b … ( iii ) (iii) a b2y − − = −10b 2 2 2

(

90 A + 9B = 20B + 2 A

8 A+ A= 9 ⇒ 9A = 9 ⇒ A = 1 So, B = 8 A = 8 × 1 = 8 Therefore, the number is 18. 3. (C): Let s be the number of students and p the number of pets. We know: s + p = 30 2s + 4p = 90 Solving this system, = s 30 − p Substituting in the second equation: 2 ( 30 − p ) + 4p = 90 60 − 2p + 4p = 90 ⇒ 2p = 30 ⇒ p = 15 So, s = 30 − 15 = 15 4. (B): Let x be the cost of a small plant and y the cost of a large plant.

)

⇒ −a 2 + b 2 y = −10b 2

From the problem:

10b 2 ⇒y= a b2 + 2

4x + 2 y = 120

(

)

3x + 5 y = 230

10b 2 20b 2 From (i), we have bx = 2ay = 2a × = 2 2 a b2 + 2 b + 2 2 20b ⇒ x =2 b +2 Thus, the point of intersection is

(

20b 2 10b 2 = ,y x = 2 b +2 a b2 + 2

(

)

Hence, these two lines are intersecting lines. 2. (A): Let’s denote the two-digit number as AB , where A is the tens digit and B is the units digit. The actual number can be represented as 10A + B . From the problem, we have two conditions: A +B = 9. 68

Multiplying the first equation by 5: 20x + 10y = 600 Multiplying the second equation by 2: 6x + 10 y = 460 On subtracting: ( 20x + 10 y ) − ( 6x + 10 y ) =600 − 460 14x = 140 ⇒ x = 10 4(10) + 2y = 120 ⇒ 2y = 80 ⇒ y = 40 5. (C): Let a be the number of apples, and o the number of oranges So, a + o = 24 (1) 3 a + o = (2) 60

Subtracting equation 1 from 2, we get, 2o = 36 ⇒ o = 18 So, a = 24 − 18 = 6 6. (A): Let x be the number of adult tickets and y be the number of child tickets: x + y = 200 25x + 10 y = 3500 Solving:= y 200 − x

 Therefore, ∠ A = x= 25

∠ B =(3x − 2) =(3 × 25 − 2) =73

 ∠C = y= 82

Assertion-Reason Based Questions 1. (b) Both A and R are true, but R is not the correct explanation of A.

25x + 10 ( 200 − x ) = 3500

We have, 5x + 4 y + 6 = 0 and 10x + 8 y = 12

25x + 2000 − 10x = 3500

a1 5 1 Here, = = a 2 10 2

b1 4 1 = = b2 8 2

c1 6 1 = = − c 2 −12 2

15x = 1500 100 x= So, y = 100

a1 b1 c1 ≠ (i) 7. (c) Condition for parallel lines is = a 2 b2 c 2

Given lines, 3x + 2ky −= 2 0, 2x + 5 y − = 1 0

Here, a1 = 3, b1 = 2k , c1 = −2, a 2 = 2, b2 = 5, c 2 = −1

a1 b1 c1 Since, = ≠ a 2 b2 c 2

From Eq. (i),

So, the pair of linear equation has no solution.

3 2k = 2 5 15 k= 4

15 Therefore, k = 4 8. (c) 7.5

Base = 5

Height = 3

1 Area of Triangle = × 5 × 3 = 7.5 2

Select Response Type MCQs 1. (c) (i), (ii) and (iv)

The total sum of the angles in a triangle is180°.

∠ A + ∠ B + ∠C = 180

Given, ∠= A x  , ∠= B (3x − 2) and ∠ C = y  .

x + 3x − 2 + y = 180

4x + y = 182 ⇒ y = 182 − 4x

∠C − ∠ B = 9

y − 3x + 2 =9… (i) (i )

On putting value of y from Eq. ( i ) , we get

182 − 4x − 3x + 2 = 9 ⇒ 184 − 7x = 9

7x = 175 ⇒ x = 25

On putting x = 25 in Eq. ( i ) , we get

y= 182 − 4 × 25 ⇒= y 182 − 100 ⇒ y = 82

Chapter 3 • Pair of Linear Equations in Two Variables

Given reason is true but that’s not the correct explanation for A. 2. (c) A is true, but R is false.

OA = OB (radius)

⇒ OA 2 = OB 2

⇒ ( 4 − 2 ) + ( 3 − 3 ) =( x − 2 ) + ( 5 − 3 )

⇒ 4 = (x − 2) + 4 ⇒ (x − 2) = 0 ⇒ x = 2

So, Assertion is true.

Clearly Reason is not correct as the centre of circle is the mid-point of chords passing through it.

Case Study Based Questions 1. (a) Number of children visited = x

Number of adults visited = y

According to question,

The algebraic equation is x + y = 300

(b) (i) The amount collected = `55000

The algebraic equation is 150x + 250 y = 55000

or 15x + 25 y = 5500

From (i)= y 300 − x

Put in (ii), we get

15x + 25 × ( 300 − x ) = 5500

⇒ −10x + 7500 =5500 ⇒ 2000 = 10x ⇒ x = 200

So the number of children visited = 200 69

(ii) Put x = 200 in (i), we get

200 + y = 300 ⇒ y = 100

So the number of adults visited = 100

= `(37500 + 25000) = `62500 Hence the amount collected = `62,500

Additional Practice Questions 1. The pair of equations ax + 2 y = 9 and 3x + by = 18 represent parallel lines, where a , b are integers, if (a) a = b

(b) 3a = 2b

(d) ab = 6 (CBSE 2023)

2. The system of equations 2x + 3 y = 5 and 4x + ky = 10 has infinitely many solutions. What is the value of k? (a)

3 2

(b) 6

(Board Term I 2015)

10. The sum of two numbers is 20, and their difference is 4. What are the two numbers? 1 11. A fraction becomes when 1 is subtracted from 3 1 the numerator and it becomes when 8 is added 4 to its denominator. Find the fraction. (CBSE 2020) (Board Term I 2017)

(d) Infinitely many

4. What will be the solution for the equations x − 2y = 4 and x − 2 y = 4? (a) (4, 0)

13. Sum of two numbers is 105 and their difference is 45. Find the numbers. (CBSE 2024)

14. Two chairs and three tables cost `5650 whereas three chairs and two tables cost `7100. Find the cost of a chair and a table separately. (Board Term I 2016)

15. For what value of a, the pair of linear equations: ax + 3y = a − 3; 12x + ay = a has no solution? (CBSE 2013)

(b) No solution (c) Infinitely many solutions (d) (2, −1) 5. Mallesh has some cows and some hens in his shed. The total number of legs is 92 and the total number of heads is 29 . Find the number of cows in his shed. (a) 12 (b) 14 (c) 17 (d) 19 6. Check whether the following system of equation is consistent or not. If consistent, solve graphically x − 2 y += 4 0, 2x − y −= 4 0 (CBSE 2025) 7. Find the value of k for which the following pair of linear equations have infinitely many solutions. 2x + 3 y = 7, (k + 1) x + ( 2k –1) y = 4k + 1 (CBSE 2019)

8. If the total cost of 7 pens and 4 pencils is `310, and the total cost of 4 pens and 7 pencils is `250, find the cost of one pen and one pencil. 70

3x + y = 1; ( 2k − 1) x + (k –1) y = 2k + 1

12. Solve by elimination 3x = y + 5 and 5x − y = 11.

(d) 3

3. The ratio of a two-digit number and the sum of its digits is 7 : 1. How many such two-digit numbers are possible? (a) 1 (b) 4 (c) 9

9. F ind k so that the following pair of linear equations has no solution.

16. The ratio of income of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3, if each of them manage to save `2000/month. Find their monthly incomes. (CBSE 2017)

17. The sum of a two-digit number and number obtained by reversing the order of digits is 99. If the digits of the number differ by 3, then find the numbers. 18. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present? (CBSE 2019)

19. Solve the following pair of linear equations by substitution method: 3x + 2y − 7 = 0 and 4x + y − 6 = 0 (Board Term I 2015)

20. Places A and B are 80 km apart from each other on a highway. A car starts from A and another from B at the same time. If they move in same direction they meet in 8 hours and if they move towards each other they meet in 1 hour 20 minutes. Find the speed of cars. (CBSE SQP 2018) Directions: Question 21 consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

(b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 21. Assertion (A): The pair of equations x + 2 y = 7 and 2x + 4 y = 14 has exactly one solution. Reason (R): The second equation is a multiple of the first, indicating that the lines are coincident.

(a) Both A and R are true, and R is the correct explanation of A.

Challenge Yourself 1. What is the solution set of and

Scan me

12 5 + = −7 2x + 3 y 3x − 2 y

8 6 + = −10? 2x + 3 y 3x − 2 y

2. An examination consists of 100 questions. Two marks are awarded for every correct option. If one mark is deducted for every wrong option and half mark is deducted for every question left, then a person scores 135 . Instead, if half mark is deducted for every wrong option and one mark is deducted for every question left, then the person scores 133. Find the number of questions left unattempted by the person. (a) 14

(b) 16

(d) 12

3. Find the value of x and y. x +y=7

4. Sanjana travels 660 km, partly by train and partly by car. If she covers 300 km by train and the rest by car, it takes 13.5 hours. But, if she travels 360 km by train and the rest by car, she takes 30 minutes longer. Find the time taken by Sanjana if she travels 660 km by car. (in hours) (a) 13

(b) 14

(d) 11

5. The ratio of monthly incomes of Mr X and Mr Y is 3 : 4 and the ratio of their monthly expenditures is 5 : 7. If the ratio of their monthly savings is 3 : 2 and Mr X saves `500 more than Mr Y per month, then find the monthly income of Mr Y. (a) `35000 (b) `32000 (c) `26000

x + y = 11

(d) `22000

Answers Additional Practice Questions 1. (d) ab = 6

2. (b) 6

3. (b) 4 4. (c) Infinitely many solutions 5. (c) 17

6. Consistant

7. k = 5 8. Pen = `35.46, Pencil = `15.45 9. k = 2 11.

10. 12 and 8

5 12.= x 3,= y 4 12

13. 75 and 30

14. `2000, `550

15. a = −6

16. `18000, `14000

17. 63, 36

18. 45 years

19. = x 1,= y 2 20. 35 km/h; 25 km/h 21. (d). A is false, but R is true.

Challenge Yourself 1  1.  ,1  2 

2. (a) 14

3. x = 9, y = 4

4. (d) 11

5. (d) `22000 Scan me for Exemplar Solutions

Chapter 3 • Pair of Linear Equations in Two Variables

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. The pair of linear equations y = 0 and y = −6 has (a) Unique solution

(b) No solution

2. The value of k for which the system of equations kx + 2 y = 5 and 3x + 4 y = 1 have no solution is 3 3 2 (a) k = (b) k ≠ (c) k ≠ (d) k = 15 2 2 3 2

(CBSE 2023)

3. The solution of the pair of equations x + y = a + b and ax − by = a − b is (a)= x b= , y a

(b) x = −a , y = b

(c)= x a= , y b (d) x = a , y = −b (CBSE 2023)

Assertion-Reasoning Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): A system of equations having two variables can have more than one solution. Reason (R): Two lines can intersect at more than one point in a plane. 5. Assertion (A): The equations 4x − 5 y = 2 and −8x + 10 y = −4 have no solution. Reason (R): These equations are equivalent, hence they overlap each other on the graph.

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was `1500 after 4 year of service and `1800 after 10 years of service, what was his starting salary and what is the annual increment? 7. Using graphical method, find whether following system of linear equations is consistent or not: x = 0 and y = −7 (CBSE 2023)

Short Answer Questions

[4 × 3 = 12 Marks]

8. For what value of k , will the following pair of equations have infinitely many solutions 2x + 3 y = 7 and (k + 2 ) x − 3 (1 − k ) y =5k + 1 (CBSE 2019)

9. The present age of a father is three more than three times the age of his son. Three years hence the father’s age will be 10 years more than twice the age of the son. Determine their present ages. (CBSE 2020)

913 and131x + 217 y = 827, then solve the equations for the values of x and y . 10. If 217x + 131y = (CBSE 2023)

11. Sum of two numbers is 105 and their difference is 45. Find the numbers. (CBSE Standard 2024)

Long Answer Questions

[3 × 5 = 15 Marks]

12. Using graphical method, solve the following system of equations: 3x + y + 4 = 0 and 3x − y + 2 = 0 (CBSE Standard 2024)

13. A motorboat covers a distance of 16 km upstream and 24 km downstream in 6 hours. In the same time it covers a distance of 12 km upstream and 36 km downstream. Find the speed of the boat in still water and that of the stream. (CBSE 2019)

14. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A coaching institute conducts Mathematics classes in two batches I and II for rich and poor children whose fees are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is `9000 and from batch II is `26000. Assume that each poor child pays `x per month and each rich child pays `y per month.

Based on the above information, answer the following questions: (a) Represents the information given above in terms of x and y. (b) Find the monthly fee paid by a poor child. (c) Find the difference in the monthly fee paid by a poor child and a rich child. (d) If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II? (CBSE Standard 2023)

Self-Assessment

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. For what value of k do the equations 3 (k − 1) x + 4 y = 24 and 15x + 20 y = 8 (k + 13 ) have infinite solutions? (a) 1 (b) 4 (c) 3 (d) 2 2. Swathi starts her job with certain monthly salary and earns a fixed increament every year. If her salary was 22,500 per month after 6 years of service and `30,000 per month after 11 years of service. Find her salary after 8 years of service. (a) 24000 (b) 25500 (c) 26000 (d) 24500 3. The value of k for which the system of equations x + y − 4 = 0 and 2x + ky = 3, has no solution, is (a) −2 (b) ≠ 2 (c) 3 (d) 2 (CBSE 2020)

Assertion-Reasoning Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): The pair of linear equations −2x − 3 y = 1 and 6 y + 4x = −2 represents coincident lines. a1 b1 c1 Reason (R): Two lines a1x + b1 y + c1 = 0 and a 2 x + b2 y + c 2 = 0 are coincident lines if = = . a2 b2 c2 5. Assertion (A): The graph of the equation y = 3x + 7 is a straight line parallel to the y-axis. Reason (R): Every linear equation in two variables represents a straight line when plotted on a graph.

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. For which value(s) of p , will the lines represented by the following pair of linear equations be parallel? 3x − y − 5 = 0 6x − 2 y − p = 0 7. For what value of k , does the system of linear equations

(CBSE SQP 2020)

2x + 3 y = 7 3k (k − 1) x + (k + 2 ) y = have an infinite number of solutions?

Short Answer Questions 8. Solve for x and y: 23x + 24y = 23 24x + 23y = 24

(CBSE 2019) [4 × 3 = 12 Marks]

9. Solve the following system of linear equations graphically x − y +1 = 0 x+y = 5 10. If the system of linear equations 2x + 3 y = 7 and 2ax + ( a + b ) y = 28 have infinite number of solutions, then find the values of ‘a ’ and ‘b’. (CBSE Standard 2024)

11. The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.

Long Answer Questions

[3 × 5 = 15 Marks]

12. Determine graphically the number of solutions of the following pair of linear equations: 3x − y = 7 and 2x + 5 y + 1 = 0. 13. Students of a class are made to stand in rows. If one student is extra in each row, there would be 2 rows less. If one student is less in each row there would be 3 rows more. Find the number of students in the class. 14. Tara scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each wrong answer, then Tara would have scored 50 marks. Assuming that Tara attempted all questions, find the total number of questions in the test. (CBSE Standard 2024)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Lokesh is a production manager in Mumbai, hires a taxi everyday to go to his office. The taxi charges in Mumbai consists of a fixed charges together with the charges or the distance covered. His office is at a distance of 10 km from his home. For a distance of 10 km to his office. Lokesh paid `105. While coming back home, he took another route. He covered a distance of 15 km and the charges paid by him were `155.

Based on the above information, answer the following questions. (a) What are the fixed charges? (b) What are the charges per km? (c) If fixed charges are `20 and charges per km are `10, then how much Lokesh have to pay for travelling a distance of 10 km?

Find the total amount paid by Lokesh for travelling 10 km from home to office and 25 km from office to home. (Fixed charges and charges per km are same as in parts (a) and (c)). (CBSE 2023)

Self-Assessment

Quadratic Equations

Chapter at a Glance Quadratic Equation A quadratic equation is a polynomial equation of degree 2, represented as: ax 2 + bx + c = 0 Where x is the variable and a, b, c are real numbers, and a ≠ 0. 0 is called standard form. The form ax 2 + bx + c = 2 0 is the standard form. Example: 2x + 3x + 5 = x 2 − 5 + 4x= 0, x + x 2 + 9= 0 are not in their standard form.

Roots of a Quadratic Equation 0 are the values of x that satisfy the equation. There are at The roots (solutions) of a quadratic equation ax 2 + bx + c = most two roots.

Methods to Solve Quadratic Equations 1. Solution by Factorisation Method: Split the middle term and factorise. 0 Step I: Obtain given quadratic equation in the form ax 2 + bx + c =. Step II: Split the mid term in two parts such that their sum is equal to mid term and product is equal to the product of the term containing x 2 and constant term. Step III: Factorise the quadratic equation obtained in step II by grouping method. Step IV: Factors obtained in step III are then equated with zero and the values of x are found. 0 Example: x 2 − 5x + 6 = 0 Compare equation with ax 2 + bx + c =. a = 1, b = –5, c = 6 Product of ac =1 × 6 =6 Factor pairs of 6 are 6 × 1and 2 × 3, 5 which is same as per middle term. As, 2 + 3 = ∴ x 2 − 5x + 6 = x 2 − 3x − 2x + 6 = x (x – 3) – 2 (x –3) = ( x − 2 )( x − 3 ) = 0

So, the roots are x = 2 and x = 3.

2. Solution by Quadratic Formula (Sridharacharya’s Formula): Use the formula: x = 0. equation ax 2 + bx + c = 0. Example: Find the roots of 2x 2 − 8x + 6 = • Identify coefficients: In this equation, a = 2, b = −8 and c = 6.

• Apply the quadratic formula: x =

= x

− ( −8 ) ±

( −8 )2 − 4 × 2 × 6 2×2

− ( −8 ) ± 64 − 48 − ( −8 ) ± 16 8 ± 4 = = 2×2 2×2 4

−b ± b 2 − 4ac for the quadratic 2a

8 + 4 12 = = 3 4 4 8−4 4 = = 1 x= 4 4

x • Solve for x : =

0. Thus, x = 3 and x = 1 are the roots of 2x 2 − 8x + 6 =

Nature of Roots The discriminant ( ∆ or D ) (D = b 2 − 4ac ) determines the nature of roots: • D > 0: Two distinct real roots. 0 Example: Consider the quadratic equation x 2 − 5x + 6 = Here, a = 1, b = –5, c = 6 D =( −5)2 − 4 × 1 × 6 = 25 − 24 =1 Since D > 0, the equation has two distinct real roots. • D = 0: Two equal real roots. 0 Example: Consider the quadratic equation x 2 − 4x + 4 = Here, a = 1, b = -4, c = 4 D =( −4)2 − 4 × 1 × 4 =16 − 16 = 0 Since D = 0, the equation has a repeated root, hence two equal real roots. • D < 0: No real roots (complex roots). 0 Example: Consider the quadratic equation x 2 + x + 1 = a 1,= b 1,= c 1 Here,= D =12 − 4 × 1 × 1 =1 − 4 =−3 Since D < 0, the equation has no real roots but complex roots.

NCERT Zone NCERT Exercise 4.1 1. Check whether the following are quadratic equations:

( x + 1)2 = 2 ( x − 3) 2 ( −2 )( 3 − x ) (b) x − 2x = (c) ( x − 2 )( x + 1) = ( x − 1)( x + 3 ) (d) ( x − 3 )( 2x + 1)= x ( x + 5 ) (e) ( 2x − 1)( x − 3 ) = ( x + 5 )( x − 1) 2 (f) x 2 + 3x + 1 = ( x − 2 ) 3 (g) ( x + 2 )= 2x ( x 2 − 1) 3 3 2 (h) x − 4x − x + 1 = ( x − 2 ) 2 Sol. (a) ( x + 1) = 2 ( x − 3 ) 2 x + 2x + 1= 2 ( x − 3 ) (a)

x 2 + 2x + 1= 2x − 6 0 x 2 + 2x + 1 − 2x + 6 = 0 x 2 + 7 = 0. It is of the form ax 2 + bx + c = Hence, given equation is a quadratic equation. Chapter 4 • Quadratic Equations

(b) x 2 − 2x = ( −2 )( 3 − x ) x 2 − 2x =−6 + 2x x 2 − 2x − 2x + 6 = 0 x 2 − 4x + 6 = 0 0. It is of the form ax 2 + bx + c = Hence, given equation is a quadratic equation. (c)

( x − 2 )( x + 1) = ( x − 1)( x + 3 )

x 2 − 2x + x − 2 = x 2 + 3x − x − 3 x 2 − x − 2 = x 2 + 2x − 3 x 2 − x − 2 − x 2 − 2x + 3 = 0 −3x + 1 = 0 3x − 1 = 0 0. It is not of the form ax 2 + bx + c = Hence, given equation is not a quadratic equation. (d)

( x − 3)( 2x + 1)= x ( x + 5)

2x 2 − 6x + x − 3 = x 2 + 5x 2x 2 − 5x − 3 − x 2 − 5x = 0 77

from now will be 360. We would like to find Rohan’s present age.

x 2 − 10x − 3 = 0 2 0. It is of the form ax + bx + c =

Hence, given equation is a quadratic equation. (e)

( 2x − 1)( x − 3) = ( x + 5)( x − 1)

2x 2 − 7x + 3 = x 2 + 4x − 5 2x 2 − x 2 − 7x − 4x + 3 + 5 = 0 x 2 − 11x + 8 = 0 0. It is of the form ax 2 + bx + c = Hence, given equation is a quadratic equation. (f) x 2 + 3x + 1 =

( x − 2 )2

Sol. (a) Let the breadth of the plot be x m. Hence, the length of the plot is (2x + 1) m. Area of a rectangle = Length × Breadth ∴ 528= x ( 2x + 1) ⇒ 528 = 2x 2 + x

x 2 + 3x + 1= x 2 − 4x + 4 x 2 − x 2 + 3x + 4x + 1 − 4 = 0 7x − 3 = 0 0. It is not of the form ax 2 + bx + c = Hence, given equation is not a quadratic equation. (g)

(d) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

3 2x ( x 2 − 1) ( x + 2 )=

2x 2 + x − 528 = 0 (b) Let the consecutive integers be x and x + 1. It is given the product is 306. ∴ x ( x + 1) = 306 0 x 2 + x − 306 = (c) Let Rohan’s age be x.

x 3 + 8 + 6x 2 + 12x= 2x 3 − 2x

Hence, his mother’s age = x + 26

x − 2x + 6x + 12x + 2x + 8 = 0

After 3 years, Rohan’s age = x + 3

−x 3 + 6x 2 + 14x + 8 = 0

Mother’s age = x + 26 + 3 = x + 29

0. It is not of the form ax + bx + c = Hence, given equation is not a quadratic equation. 3

(h) x − 4x − x + 1 =

(x − 2)

x 3 − 4x 2 − x + 1= x 3 − 8 − 6x 2 + 12x 3

x − x − 4x + 6x − x − 12x + 1 + 8 = 0 0 2x 2 − 13x + 9 = 0. It is of the form ax 2 + bx + c = Hence, given equation is a quadratic equation. 2. R epresent the following situations in the form of quadratic equations. (a) The area of a rectangular plot is 528 m 2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. (b) The product of two consecutive positive integers is 306. We need to find the integers. (c) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years

According to question, product of their ages after 3 years is 360. ∴ ( x + 3 )( x + 29 ) = 360 x 2 + 32x − 273 = 0 (d) Let the speed of train be x km/h. 480 Time taken to travel 480 km = h. x In second condition, let the speed of train = (x - 8) km/h. It is also given that the train will take 3 hours more to cover the same distance.  480  ∴Time taken to travel 480 km = h + 3  hrs   x distance speed × time =

480  + 3 = 480  x 

( x − 8 ) 

480 + 3x − 3x −

3840 − 24 = 480 x

3840 24 = x

3x 2 − 24x − 3840 = 0 x 2 − 8x − 1280 = 0

NCERT Exercise 4.2 1. Find the roots of following quadratic equations by factorisation:

1 ( 4x − 1)2 8 Roots of this equation are the values for which

0 (a) x − 3x − 10 =

(b) 2x 2 + x − 6 = 0

0 ( 4x − 1)2 = ∴ ( 4x= − 1) 0 or ( 4x= − 1) 0

1 1 or x = 4 4 2 (e) 100x − 20x + 1 = ⇒x

2x 2 + 7 x + 5 2 = 0 1 (d) 2x 2 − x + = 0 8 (e) 100x 2 − 20x + 1 = 0 (c)

100x 2 − 10x − 10x + 1

0 Sol. (a) x 2 − 3x − 10 =

10x (10x − 1) − 1(10x − 1)

x 2 − 5x + 2x − 10

x ( x − 5) + 2 ( x − 5)

(10x − 1)(10x − 1)

( x − 5)( x + 2 )

Roots of this equation are the values for which 0 (10x − 1)2 =

Roots of this equation are the values for which ( x − 5)( x + 2 ) = 0

∴ x= − 5 0 or x= +2 0

x = 5 or x = −2 2x 2 + 4x − 3x − 6

2x ( x + 2 ) − 3 ( x + 2 )

( x + 2 )( 2x − 3)

Roots of this equation are the values for which ( x + 2 )( 2x − 3) = 0 (c)

∴ x= + 2 0 or 2 x= −3 0 3 ⇒x = −2 or x = 2

2 x 2 + 5x + 2x + 5 2

) ( ( 2 x + 5)( x + 2 ) x

(

2x +5 + 2

2x +5

)

Roots of this equation are the values for which

( 2 x + 5)( x + 2 ) = 0

∴ 2 x += 5 0 or x + = 2 0

5 or x = x= − − 2 2 1 (d) 2x 2 − x + = 0 8 1 16x 2 − 8x + 1 8

(

)

1 16x 2 − 4x − 4x + 1 8 1 (4x ( 4x − 1) − 1( 4x − 1) 8

(

Chapter 4 • Quadratic Equations

x =

1 1 or x = 10 10

)

(b) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. Find out the number of toys produced on that day. Sol. (a) Let the number of John’s marbles be x.

2x 2 + 7 x + 5 2 = 0

∴ (10= x − 1) 0 or (10= x − 1) 0

2. (a) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.

(b) 2x 2 + x − 6 = 0

Therefore, number of Jivanti’s marble = 45 – x

After losing 5 marbles,

Number of John’s marbles = x – 5

Number of Jivanti’s marbles = 45 − x − 5 = 40 − x

It is given that the product of their marbles is 124.

∴ ( x − 5 )( 40 − x ) = 124 ⇒ 40x - 200 - x2 + 5x =124

x 2 − 45x + 324 = 0

x 2 − 36x − 9x + 324

x ( x − 36 ) − 9 ( x − 36 )

0 ( x − 36 )( x − 9) =

∴ x −= 36 0 or x= −9 0

= x 36 = or x 9

If the number of John’s marble = 36,

9 Then, number of Jivanti’s marbles = 45 − 36 = 79

If number of John’s marble = 9

36 Then, number of Jivanti’s marble = 45 − 9 = (b) Let the number of toys produced be x. ∴ Cost of production of each toy = Rs (55 − x )

It is given that, total production cost of the toys = Rs 750

∴ x ( 55 − x ) = 750

x 2 − 55x + 750 = 0

x 2 − 25x − 30x + 750 = 0

x ( x − 25 ) − 30 ( x − 25 ) = 0

0 ( x − 25 )( x − 30 ) =

x − 25 = 0 or x − 30 = 0

= x 25 = or x 30 Hence, the number of toys will be either 25 or 30. 3. Find two numbers whose sum is 27 and product is 182.

x 2 + x − 182 = 0 x 2 + 14x − 13x − 182 = 0

0 x ( x + 14 ) − 13 ( x + 14 ) = ( x + 14 )( x − 13 ) = 0 x + 14 = 0 or x − 13 = 0 −14 or x = 13 x =

Since considering positive integers, x can only be 13. x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14. 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Sol. Let the base of right angle triangle be x cm. By Pythagoras theorem, ( Base)2 + ( Altitude)2 = ( Hypotenuse)2 2

Sol. Let the first number be x then the second number is 27 - x.

x 2 + ( x − 7 ) = 132

Therefore, their product = x (27 − x)

2x 2 − 14x − 120 = 0

It is given that the product of these numbers is 182. 182 Therefore, x ( 27 − x ) = 2

x − 27x + 182 = 0 2

x − 13x − 14x + 182 = 0 x ( x − 13 ) − 14 ( x − 13 ) = 0 0 ( x − 13 )( x − 14 ) = x − 13 = 0 or x − 14 = 0 = x 13 = or x 14 If first number = 13 Other number = 27 − 13 = 14 If first number = 14 13 Other number = 27 − 14 = Therefore, the numbers are 13 and 14. 4. Find two consecutive positive integers, sum of whose squares is 365. Sol. Let the consecutive positive integers be x and x + 1. 2

x 2 + ( x + 1) = 365 x 2 + x 2 + 1 + 2x = 365 2x 2 + 2x − 364 = 0

x 2 + x 2 + 49 − 14x = 169 x 2 − 7x − 60 = 0 x 2 − 12x + 5x − 60 = 0 0 x ( x − 12 ) + 5 ( x − 12 ) = 0 ( x − 12 )( x + 5 ) = 12 0 or x = +5 0 x −=

x = 12 or x = −5 Since the sides are positive, x can only be 12. Therefore, the base of given triangle is 12 cm and altitude of this triangle will be (12 - 7) = 5 cm. 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article. Sol. Let the number of articles produced be x. Therefore, cost of production of each article = Rs (2x + 3) It is given that the total production is = 90.

Therefore, real roots exist for the given equation and they are equal to each other. b b The roots will be – and − 2a 2a

x ( 2x + 3 ) = 90 2x 2 + 3x − 90 = 0 2x 2 − 12x + 15x − 90 = 0 2x ( x − 6 ) + 15 ( x − 6 ) = 0

b −4 3 4 3 2 3 2 = − = = = 2a 2×3 6 3 3 2 2 ∴ The roots are and . 3 3 −

( x − 6 )( 2x + 15 ) = 0 ( x= − 6 ) 0 or ( 2x += 15 ) 0 15 2 As the number of articles produced can only be a positive integer, therefore, x is 6.

x = 6 or x = −

Hence, number of articles produced = 6

a= 2, b = −6, c = 3

Cost of each article = 2 × 6 + 3 = Rs 15

NCERT Exercise 4.3 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them 0 (a) 2x 2 − 3x + 5 = (b) 3x 2 − 4 3 x + 4 = 0

Comparing with standard equation ax 2 + bx + c = 0 Discriminant = b 2 − 4ac 2

= ( −6 ) − 4 ( 2 )( 3)

= 36 − 24

= 12

As b 2 − 4ac > 0 Therefore, distinct and real roots exist for the given equation as follows. By using quadratic formula, we get

0 (c) 2x 2 − 6x + 3 = Sol. (a) 2x 2 − 3x + 5 = 0

Comparing with standard equation ax 2 + bx + c = 0 a = 2, b = −3, c = 5 Discriminant = b 2 − 4ac

= ( −3 ) − 4 ( 2 )( 5 )

= 9 − 40

= −31 2

As b − 4ac < 0

6±

( −6)2 − 4 ( 2)( 3) 4

6 ± 36 − 24 4

6 ± 12 4

6±2 3 4

−b ± b 2 − 4ac 2a

3± 3 2

Therefore, no real root is possible for given equation.

0 (b) 3x 2 − 4 3 x + 4 =

Therefore, roots are x =

Comparing with standard equation ax 2 + bx + c = 0 a = 3, b = −4 3, c = 4 Discriminant = b 2 − 4ac

(

) − 4 ( 3)( 4 )

= −4 3

= 48 − 48

As b 2 − 4ac = 0

Chapter 4 • Quadratic Equations

3+ 3 3− 3 = or x 2 2

2. F ind the values of k for each of the following quadratic equations, so that they have two equal roots. 0 (a) 2x 2 + kx + 3 = 0 (b) kx ( x − 2 ) + 6 = 0 has Sol. (a) We know that if an equation ax 2 + bx + c = 2 0. two equal roots, its discriminant b − 4ac = Comparing with standard equation ax 2 + bx + c = 0

= a 2,= b k= ,c 3 Discriminant = b 2 − 4ac

(k )2 − 4 ( 2 )( 3 )

= k 2 − 24

For equal roots, discriminant = 0 ∴ k 2 − 24 = 0

k = ±2 6 0 (b) kx ( x − 2 ) + 6 = Comparing to standard equation with ax 2 + bx + c = 0 k , b −2k , c = 6 a ==

Discriminant = b 2 − 4ac 2

= ( −2k ) − 4 (k )( 6 ) = 4k 2 − 24k

For equal roots, discriminant = 0 0 ∴ 4k 2 − 24k = 4k 2 = 24 k k 2 = 6k 2 0 k − 6k =

0 k (k − 6 ) = = k 0 or k −= 6 0 = k 0= or k 6

However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’. Therefore, if this equation has two equal roots, k should be 6 only. 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. Sol. Let the breadth of mango grove be l. Length of mango grove will be 2l. Area of mango grove = ( 2l )(l ) = 2l 2 l 2 = 400 2

l − 400 = 0

= ( 0 ) − 4 (1)( −400 )

= 1600

l = ±20

kx 2 − 2kx + 6 = 0

Therefore, 2l 2 = 800

Discriminant = b 2 − 4ac

Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

k = ± 24

a = 1, b = 0, c = −400

Here, b 2 − 4ac > 0

k 2 = 24

Comparing with standard equation al 2 + bl + c = 0, we get

However, breadth cannot be negative. Therefore, breadth of mango grove = 20 m 40m Length of mango grove = 2 × 20 = 40 m 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Sol. Let the age of one friend be x years Age of other friend will be ( 20 − x ) years 4 years ago, age of 1st friend = ( x − 4 ) years Age of 2nd friend = ( 20 − x − 4 )=

(16 − x ) years

According to question, 48 ( x − 4 )(16 − x ) = 48 16x − x 2 − 64 + 4x = −x 2 + 20x − 64 − 48 = 0 0 x 2 − 20x + 112 = Comparing to standard equation ax 2 + bx + c = 0, we get 1, b −20, c = 112 a ==

Discriminant = b 2 − 4ac 2

= ( −20 ) − 4 (1)(112 )

= 400 − 448

= −48 2

As b − 4ac < 0 Therefore, no real root is possible for this equation and hence, this situation is not possible. 5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Sol. Let the length and breadth of the park be l and b .

a == 1, b −40, c = 400

Perimeter = 2 (l + b )

Discriminant = b 2 − 4ac 2

2 (l + b ) = 80

= ( −40 ) − 4 (1)( 400 )

40 (l + b ) =

= 1600 − 1600

= b 40 − l

= 0

Area = l × b

Therefore, this equation has equal roots. And hence this situation is possible.

= l ( 40 − l )

Roots of this equation, b l = − 2a −40 l = − 2 l = 20

= 40l − l 2 Therefore, 40l − l 2 = 400

40l − l 2 − 400 = 0

l 2 − 40l + 400 = 0

Comparing with standard equation ax 2 + bx + c = 0, we get

Therefore, length of park, l = 20 m Breadth of park, b = 40 − l = 40 − 20= 20 m

Multiple Choice Questions 1. Which of the following is not a quadratic equation? (a) 2(x − 1)2 = 4x 2 − 2x + 1 (b) 2x − x 2 = x 2 + 5

(a) 0

(b)

25 4

(c)

4 25

(d) 7 (CBSE SQP 2024)

7. The roots of the quadratic equation 3x 2 − 4 3x + 4 = 0 are:

2. Which of the following equations has two distinct real roots? 9 0 0 (a) 2x 2 − 3 2x + = (b) x 2 + x − 5 = 4 2 2 0 (c) 3x + 2x = (d) 5x − 3x + 1 = 0 (NCERT Exemplar)

(a) not real

(b) real and equal

(d) irrational and distinct (CBSE SQP 2024)

8. If one root of the quadratic equation x 2 − 5x + k = 0 is twice the other, what is the value of k ? 5 50 (a) 5 (b) (c) (d) 1 3 9

3. If x = 2 is a solution of the equation x 2 + 7x − k = 0 the value of k is: (a) 2 (b) 9 (c) 18 (d) 21

0 are −6 and −2, what 9. If the roots of x 2 + px + 12 = is the value of p ? (a) 8 (b) 6 (c) 5 (d) 4

0 are: 4. The roots of the equation x 2 + 3x − 10 = (a) 2, −5 (b) −2, 5 (c) 2, 5 (d) −2, −5

0 differ by 2, what is the 10. If the roots of x 2 − kx + 24 = value of k ? (a) 4 (b) 6 (c) 8 (d) 10

(CBSE 2023)

5. The quadratic equation x 2 + x + 1 = 0 has roots. (a) real and equal

(b) irrational

(d) not-real (CBSE 2024)

6. The value of k for which the quadratic equation kx 2 − 5x + 1 = 0 does not have a real solution, is

Chapter 4 • Quadratic Equations

11. If one root of 2x 2 + 3x + c = 0 is 2, what is the value of c ? (a) −14 (b) −12 (c) −10 (d) −8 12. Which of the following quadratic equations has the sum of its roots as 4? 2 0 (a) 2x − 4x + 8 = 4 (c) 2x 2 − x +1 = 0 2

0 (b) −x 2 + 4x + 4 = (d) 4x 2 − 4x + 4 = 0 (CBSE 2023) 83

13. Which constant must be added and subtracted to 3 solve the quadratic equation 9x 2 + x − 2 = 0 by 4 the method of completing the square? (a)

1 8

(b)

1 64

(c)

1 4

(d)

9 64

14. (x 2 + 1)2 − x 2 = 0 has (a) four real roots (c) no real roots

(b) two real roots (d) one real root

8. (c) 50 9 10. (d) 10

11. (a) -14

0 12. (b) −x 2 + 4x + 4 =

13.

(NCERT Exemplar)

Answers 1. (d) (x 2 + 2x )2 = x 4 + 3 + 4x 2 2

0 (b) x + x − 5 =

(a) 2, −5

(d) not-real

(d) 7

(b) real and equal

(a) 8

(b)

14. (c) no real roots

1 64

Constructed Response Questions Very Short Answer Questions 1. The product of two consecutive positive integers is 240. Formulate the quadratic equation whose roots are these integers. Sol. Let two consecutive positive integers be x and x + 1. Then, their product is x ( x + 1). It is given that the product is 240. ∴ x ( x + 1)= 240 ⇒ x 2 + x − 240= 0, which is the required quadratic equation. 2. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, formulate the quadratic equation to find the sides of the two squares. Sol. Let the length of each side of a square be x metres. Then, its perimeter is 4x. It is given that the difference of the perimeters of two squares is 24 m.

0 3. Find the roots of the equation x 2 + 7x + 10 = (CBSE 2021)

0 Sol. x 2 + 7x + 10 = x 2 + 2x + 5x + 10 = 0 x ( x + 2) + 5( x + 2 ) = 0 0 ( x + 2 )( x + 5) = 0 x +2= 0 or x + 5 = x = −2 or x = −5 4. Solve the quadratic equation

a 2b 2 x 2 + b 2 x − a 2 x − 1 = 0 by factorisation method. Sol. We have, 0 ⇒ a 2b 2 x 2 + b 2 x − a 2 x − 1 =

( ) ( ) ⇒ ( a x + 1)b x − ( a x + 1) = 0 ⇒ ( a x + 1)(b x − 1) = 0

⇒ a 2b 2 x 2 + b 2 x − a 2 x + 1 = 0 2

⇒ a 2 x= + 1 0, b 2 x= −1 0

1 1 ⇒ a 2 x =−1, b 2 x =⇒ 1 x =− 2 , x = 2 ∴ Perimeter of second square = 24 + 4x metres a b ⇒ Length of each side of second square 24 + 4x = = 6+x 4

∴ x + (6 + x ) = 468 2

(

)

⇒ x 2 + 36 + 12x + x 2 = 468 2

⇒ 2x + 12x − 432 = 0 2

⇒ x + 6x − 216 = 0, which is the required equation.

5. Find the roots of the quadratic equation 3x 2 − 2x − 3 = 0 2

Sol.

3x − 2x − 3 = 0

3x 2 − 3x + x − 3 = 0 3x x − 3 + 1 x − 3 = 0

(

x− 3

) (

0 )( 3x + 1) =

x− 3= 0 or

)

3x + 1 = 0 1 x = 3 or x = − 3

(CBSE 2015)

6. Write the discriminant of the quadratic equation ( x + 5)2 = 2 ( 5x − 3). (CBSE 2019) 2

Sol. ( x + 5 ) = 2 ( 5x − 3 ) x 2 + 25 + 10x = 10x − 6 x 2 + 31 = 0 −124 D = 02 − 4 (1)( 31) = 7. For what value of k, the roots of the equation (CBSE 2019) x 2 + 4x + k = 0 are real? Sol. For real solution, D ≥ 0 4 – 4 (1)(k ) ≥ 0

16 – 4k ≥ 0 16 ≥ 4k k ≤4 8. Find the values(s) of k for which the quadratic 0 has equal roots. equation x 2 + 2 2kx + 18 = Sol. For equal roots, D = 0

(CBSE SQP 2020)

8k 2 − 72 = 0 8k 2 = 72 k2 = 9 k = ±3

0 then find the value of k . 9. If x = 3 is one root of the equation x 2 − 2kx − 6 =, Topper's Answer

0 10. Determine the nature of the roots of x 2 − x + 1 =. 2 0. Sol. The given equation is x + x + 1 =

a 1,= b 1 and, c = 1 We have:=

∴D =b 2 − 4ac =12 − 4 × 1 × 1 =−3 D < 0 thus, roots of the given equation are not real. 11. If x = −2 is the common solution of quadratic x 2 + bx + b = 0, then 0 equations ax 2 + x − 3a =and 2 a b . find the value of (CBSE 2022) 0 Sol. x = −2 is the solution of ax 2 + x − 3a =and 2 x + bx + b = 0

⇒ a( −2) + ( −2 ) − 3a =0 2

⇒ 4a − 2 − 3a = 0 ⇒ a = 2 0 and ( −2)2 + b ( −2 ) + b = 4 − 2b + b = 0 ⇒ b = 4 2

∴a b= (2) × 4= 16 Chapter 4 • Quadratic Equations

12. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. (NCERT Exemplar) Sol. Let n be a required natural number. Square of a natural number diminished by 84 = n 2 − 84 and thrice of 8 more than the natural number = 3 (n + 8 ) Now, by given condition, n 2 − 84 = 3 (n + 8 ) ⇒ n 2 − 84 = 3n + 24 ⇒ n 2 − 3n − 108 = 0 ⇒ n 2 − 12n + 9n − 108 = 0 ⇒ n (n − 12 ) + 9 (n − 12 ) = 0 ⇒ (n − 12 )(n + 9 ) = 0 = ⇒ n 12

( n ≠ −9 )

Hence, the required number is 12. 85

13. Find the discriminant of the quadratic equation 4x 2 − 5 = 0 and hence comment on the nature of roots of the equation. 2

0 Sol. 4x − 5 =

Hence, required quadratic equation is x 2 + 4x − 12 = 0.

2. Solve for x:

Here, a = 4, b = 0 and c = −5

2x + 9 + x = 13 (CBSE 2016) Sol. Squaring both sides to eliminate the square root:

D = b 2 − 4ac

( 2x + 9)2 = (13 − x )2

= 0 − 4 × 4 × ( −5 ) = 80 > 0 Since, D > 0, therefore roots are real and distinct. 14. Find the value of x using factorisation method. 1 2 6 + = x − 2 x −1 x 1 2 6 Sol. We have, + = x − 2 x −1 x

( x − 1) + 2 ( x − 2 ) 6 ⇒ = ( x − 2 )( x − 1) x 3x − 5 6 ⇒ 2 = x − 3x + 2 x ⇒ 3x 2 − 5x = 6x 2 − 18x + 12

⇒ 3x 2 − 13x + 12 =0 ⇒ 3x 2 − 9x − 4x + 12 =0 ⇒ 3x ( x − 3 ) − 4 ( x − 3 ) = 0 ⇒ ( x − 3)( 3x − 4 ) = 0 ⇒ x − 3 = 0 or 3x − 4 = 0 4 ⇒= x 3 or= x 3

Short Answer Questions

1. A two-digit number is such that the product of the digits is 12. When 36 is added to the number the digits interchange their places. Formulate the quadratic equation whose root(s) is (are) digit(s) of the number. Sol. Let the ten’s digit of the number be x. It is given that the product of digits is 12. 12 12 Unit’s digit , Number = = 10x + x x Number obtained by reversing the digits is 12 120 10 + x= +x x x If 36 is added to the number the digits interchange their places. 12 120 10x + + 36 = +x x x 108 0 ⇒ 9x − + 36 = x 0 ⇒ 9x 2 − 108 + 36x = ⇒ x 2 + 4x − 12 = 0

2x + 9= 169 − 26x + x 2 x 2 − 28x + 160 = 0 0 ( x − 20 )( x − 8 ) = x − = 0 ⇒ x = 20 20 x − =0⇒x =8 8

3. The sum of two numbers is 15. If the sum of their 3 reciprocals is , find the two numbers. 10

(CBSE 2023)

Sol. We are given: 15 The sum of two numbers is 15: x + y = 3 1 1 3 The sum of their reciprocals is : + = 10 x y 10 Now, 1 1 x+y + = x y xy 15: Substituting x + y = 15 3 = xy 10 15 × 10 =3 × xy 150 = 3xy xy = 50 Using the quadratic equation identity: t 2 − ( x + y )t + xy = 0

15 and xy = 50: Substituting x + y = t 2 − 15t + 50 = 0 Solve for t using the quadratic formula: − ( −15 ) ± t=

( −15)2 − 4 (1)( 50 ) 2 (1)

15 ± 225 − 200 t= 2 15 ± 25 t= 2 15 ± 5 t= 2

15 + 5 20 = = 10 2 2 15 − 5 10 Second root:= = = 5 t 2 2 The two numbers are 10 and 5. First root:= t

4. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new 29 fraction and original fraction is . Find the 20 original fraction. Sol. Let the numerator and the denominator of the fraction be x and x + 3, respectively. Original fraction=

x x +3

Now, 2 is added to both the numerator and the denominator. New fraction =

x +2 x +5

x 2 + 5x + x 2 + 3x + 2x + 6 29 ⇒ = 20 x 2 + 3x + 5x + 15

2x 2 + 10x + 6 29 ⇒ 2 = x + 8x + 15 20 2

⇒ 40x + 200x + 120 = 29x + 232x + 435 ⇒ 11x 2 − 32x − 315 = 0 ⇒ 11x 2 − 77x + 45x − 315 = 0 ⇒ 11x ( x − 7 ) + 45 ( x − 7 ) = 0 ⇒ (11x + 45 )( x − 7 ) = 0 45 x= − or x = 7 11

⇒ x 2 − 41x + 180 = 0 ⇒ x 2 − 36x − 5x + 180 = 0 ⇒ x ( x − 36 ) − 5 ( x − 36 ) = 0 ⇒ ( x − 36 )( x − 5 ) = 0 So, either x − = 0 ⇒ x = 36 Or 36 x −5= 0⇒ x = 5

Hence, man’s present age is 36 years and his son’s age is 9 years. 6. At t min past 2 pm, the time needed by the minute t2 hand of a clock to show 3 pm was found to be 4 min. Find t. (NCERT Exemplar) Sol. We know that, the time between 2 pm to 3 pm = 1 h = 60 min Given that, at t min past 2 pm, the time needed by the min hand of a clock to show 3 pm was t found to t2 be 3 min less than min i.e. 4 t 2  t +  − 3  = 60 4  ⇒ 4t + t 2 − 12 = 240

Since x represents the numerator of a fraction and 45 should be a positive integer, we discard x = − 11 and accept x = 7. x 7 7 Thus, = = x + 3 7 + 3 10 7 So the original fraction is . 10 5. The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at the time. Find their present ages. Sol. Let the present age of the man be x years. Then present age of his son is ( 45 − x ) years. Five years ago, man’s age = ( x − 5 ) years

Chapter 4 • Quadratic Equations

⇒ 40x − x 2 + 5x − 200 = 4x − 20

45 − x = 45 − 36 = 9

x ( x + 5 ) + ( x + 2 )( x + 3 ) 29 ⇒ = 20 ( x + 3)( x + 5)

And his son’s age ( 45 − x − 5 )=

( x − 5)( 40 − x ) = 4 ( x − 5)

But, the father’s age can never be 5 years. Therefore, when x = 36 then

According to the question: x x + 2 29 + = x + 3 x + 5 20

Then according to the question,

( 40 − x ) years.

⇒ t 2 + 4t − 252 = 0 ⇒ t 2 + 18t − 14t − 252 = 0 ⇒ t (t + 18 ) − 14 (t + 18 ) = 0 ⇒ (t + 18 )(t − 14 ) = 0 ∴t = 14 min Hence, the required value of t is 14 min. 7. In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of his marks would have been 180. Find his marks in two subjects. Sol. Let marks obtained by P in Mathematics be x , then in science = ( 28 − x ) It is given that,

⇒ 24x − x 2 + 72 − 3x = 180

( )( ) ⇒ 4m c − 4 (1 + m )(c − a ) = 0 ⇒ m c − (1 + m )(c − a ) = 0

⇒ x 2 − 21x + 108 = 0

⇒ x 2 − 12x − 9x + 108 = 0

⇒ m 2 c 2 − c 2 + a 2 − m 2c 2 + m 2 a 2 = 0

180 ( x + 3)( 28 − x − 4 ) =

⇒ (2mc )2 − 4 1 + m 2 c 2 − a 2 = 0

⇒ ( x + 3 )( 24 − x ) = 180

2 2

⇒ x ( x − 12 ) − 9 ( x − 12 ) = 0

[Dividing throughout by 4]

(

)

⇒ −c 2 + 1 + m 2 a 2 =0

(

⇒ x − 12 = 0 Or x − 9 = 0

⇒ c = a 1 + m2

Therefore, when x = 12 then

Hence proved.

Hence, marks in Mathematics 12 and marks in Science 16. Or, when x = 9 then 28 − x = 28 − 9 = 19 Hence, marks in Mathematics 9 and marks in Science 19.

)

10. A chess board contains 64 equal squares and the area of each square is 6.25 cm2. A border round the board is 2 cm wide. Find the length of the side of the chess board. Sol. Let the length of the side of the chess board be cm. Then, 2 cm 2 cm

8. Find the value of k for which the equation x 2 + k ( 2x + k − 1) = 0 has real and equal roots. (CBSE 2017)

Chess board

Sol. We are given the quadratic equation:

2 cm

x 2 + k ( 2x + k − 1) = 0

(

x cm

)

x + 2kx + k − k = 0 For a quadratic equation to have real and equal roots, the discriminant must be zero: D =b 2 − 4ac =0 Substituting the values:

(

)

4k − 4k + 4k = 0

⇒ x = 24 cm

4k = 0

(

0 has ) x + 2mcx + c − a = c a (1 + m ). equal roots, show that=

9. If the equation 1 + m

(CBSE 2017)

Sol. The given equation is

0. (1 + m ) x + ( 2mc ) x + (c − a ) = 2

Here, A = 1 + m 2 , B = 2mc and C= c 2 − a 2. Since the given equation has equal roots, therefore 2

0. D = 0 ⇒ B − 4 AC =

⇒ x 2 - 8x + 16 = 400

⇒ (x - 24) (x + 16) = 0

k =0

∴ ( x - 4)2 = 64 × 6.25

⇒ x 2 - 24x + 16x - 384 = 0

4k − 4 k − k = 0 2

Area of 64 squares = (x - 4)2

⇒ x 2 - 8x - 384 = 0

0 ( 2k ) − 4 (1) k 2 − k = 2

2 cm

28 − x = 28 − 12 = 16

⇒ −c 2 + a 2 + m 2a 2 =0

⇒ ( x − 12 )( x − 9 ) = 0

11. Find the value of k for which the roots of the equation 3x 2 − 10x + k = 0 are reciprocal of each other. (CBSE 2019) Sol. We are given the quadratic equation: 3x 2 − 10x + k = 0 If the roots of a quadratic equation are reciprocal of each other, then: Product of the roots = 1 (i) 0 For a quadratic equation of the form: ax 2 + bx + c = c the product of the roots is given by: α ⋅ β = a

k 3 k From (i) and (ii), we get: = 1 3 k= 3 ×1 So, αβ =

(ii)

12. If p, q, r and s are real numbers such that = pr 2 (q + s ), then show that at least one of the equations x 2 + px + q = 0 has 0 and x 2 + rx + s = real roots. Sol. Given equations are x 2 + px + q = 0 (i) 2

0 (ii) and, x + rx + s =

Let D1 and D2 be the discriminants of equations (i) and (ii) respectively. Then, 2

(

⇒ D1 + D2 = p − 4q + r − 4s = p + r

) − 4 (q + s )

 pr  ⇒ D1 + D2 = p 2 + r 2 − 4    2 

0 has real roots, then Sol. If the equation x 2 + kx + 64 = D ≥ 0.

To find the nature of its roots, we find the discriminant:

⇒ k 2 ≥ 256 ⇒ k ≥ 16 [since k > 0] (i) 0 has real roots, then If the equation x 2 − 8x + k = ⇒ 64 − 4k ≥ 0

(

)

(

⇒ k ≤ 16 (ii) From (i) and (ii), we get k =16. 0 has no (a + b ) x + 2 (ac + bd ) x + (c + d ) = 2

real roots. Sol. The given equation is

(CBSE 2017)

0 ) x + 2 (ac + bd ) x + (c + d ) =

Chapter 4 • Quadratic Equations

)

= −4a 2d 2 − 4b 2c 2 + 8abcd

(

= −4 a 2d 2 + b 2c 2 − 2abcd

)

= −4(ad − bc )2

⇒D <0 Hence, the given equation has no real roots. 15. Solve: x =

1 2−

Sol. We have, x =

= ⇒x 2 −

2−

,x ≠2

1 2−x

1 2−

2−

1 2−x

1 1 = ⇒x 1 2−x 2− 2(2 − x ) − 1 4 − 2x − 1 2−x

1 3 − 2x 3 − 2x = ⇒x = ⇒x 2−x 2 ( 3 − 2x ) − ( 2 − x ) 4 − 3x 2− 3 − 2x

⇒ x ( 4 − 3x ) =( 3 − 2x ) ⇒ 4x − 3x 2 =3 − 2x ⇒ 3x 2 − 6x + 3 =0 ⇒ x 2 − 2x + 1 =0

14. If ad ≠ bc , then prove that the equation

(

)

= 4a 2c 2 + 4b 2d 2 + 8abcd − 4a 2c 2 − 4a 2d 2 − 4b 2c 2 − 4b 2d 2

= ⇒ x

⇒ 4k ≤ 64

a +b

)(

= 4 a 2c 2 + b 2d 2 + 2abcd − 4 a 2c 2 + a 2d 2 + b 2c 2 + b 2d 2

D ≥0

(

= [2 ( ac + bd )]2 − 4 a 2 + b 2 c 2 + d 2

⇒ k 2 − 4 × 64 ≥ 0

⇒ −4(ad − bc )2 < 0

13. Find the positive value of k, for which the equation x 2 + kx + 64 = 0 and x 2 − 8x + k = 0 will both have real roots. (CBSE 2016)

⇒ (ad − bc )2 < 0 for all values of a , b , c and d

⇒ At least one of the two equations has real roots.

2 ( ac + bd ) , C = ), B = (c + d )

⇒ ad − bc ≠ 0

⇒ At least one of D1 and D2 is greater than or equal to zero.

It is given that ad ≠ bc

pr    pr= 2 (q + s ) ∴q + s=  2   ⇒ D1 + D2 = p 2 + r 2 − 2pr = (p − r )2 ≥ 0

2 D= B 2 − 4 AC

k =3

D1 = p 2 − 4q and D2 = r 2 − 4s

(

a +b Here, A =

⇒ ( x − 1) = 0 ⇒ x = 1, 1

16. Solve for x: 2x + 3 x +1 x − 2 + = 4− ; x ≠ 1, −2, 2 x −1 x + 2 x −2 89

Sol. We have, ⇒

17. Solve for x:

x +1 x − 2 2x + 3 + = 4− x −1 x + 2 x −2

(

)

x 2 + 5x − a 2 + a − 6 = 0

( x + 1)( x + 2 ) + ( x − 2 )( x − 1) 4 ( x − 2 ) − ( 2x + 3) = Sol. We are given the quadratic equation: x −2 ( x − 1)( x + 2 )

( ) x + 5x + ( 6 − a − a ) = 0 x 2 + 5x − a 2 + a − 6 = 0

x 2 + 3x + 2 + x 2 − 3x + 2 4x − 8 − 2x − 3 ⇒ = x −2 x2 + x − 2

(

)

−b ± b 2 − 4ac 2a Substituting the values: x=

( 2x − 11) ( x 2 + x − 2 )

⇒ 2x 3 − 4x 2 + 4x − 8= 2x 3 + 2x 2 − 4x − 11x 2 − 11x + 22 2

⇒ −4x + 4x − 8 = −9x − 15x + 22 ⇒ 5x 2 + 19x − 30 =0 ⇒ 5x 2 + 25x − 6x − 30 =0 ⇒ 5x ( x + 5 ) − 6 ( x + 5 ) =0 ⇒ ( 5x − 6 )( x + 5 ) =0 ⇒x =

Using the quadratic formula:

2x 2 + 4 2x − 11 ⇒ 2 = x −2 x +x −2 ⇒ 2x 2 + 4 ( x − 2 )=

(CBSE OD Set-II 2019) (CBSE Board 2015)

6 , −5 5

(

)

(

−5 ± 52 − 4 (1) 6 − a 2 − a

)

2 (1)

−5 ± 25 − 24 + 4a 2 + 4a 2

−5 ± 1 + 4a 2 + 4a 2

)

18. If the roots of the equation a 2 + b 2 x 2 − 2 ( ac + bd ) x + c 2 + d 2 = 0 are equal, prove that

a c = . b d

Topper's Answer

Long Answer Questions 1. If x = 2 and x = 3 are roots of the equation 3x 2 − 2kx + 2m = 0, find the values of k and m. Sol. Given x = 2 and x = 3 are roots of the equation.

⇒ 4k − 2m = 12 (i) 3 ⋅ 32 − 2 ⋅ 3 ⋅ k + 2m = 0 ⇒ 27 − 6k + 2m = 0 ⇒ 6k − 2m = (ii) 27

∴ 3 ⋅ 22 − 2 ⋅ 2 ⋅ k + 2m = 0

6k − 27 = 2m

⇒ 12 − 4k + 2m = 0

Substitute in equation (i), we get:

(CBSE 2017)

4k − ( 6k − 27 ) = 12 ⇒ 4k − 6k + 27 = 12 ⇒ − k = −15 2 15 ⇒k = 2 Substituting in equation (i) we get:  15  4   − 2m = 12  2 ⇒ 30 − 2m = 12

(NCERT Exemplar)

⇒ − m = −18 2 ⇒m = 9 Thus, we get: k =

15 and m = 9. 2

2. Divide 27 into two parts such that the sum of their 3 reciprocals is . (CBSE 2018) 20 Sol. Let the two numbers be x and 27 − x .

Sol. Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 m × 40 m. So, the distance between the pond and the lawn would be the same around the pond. Say x m. x x x

The given condition states: 1 1 3 + = x 27 − x 20

( 27 − x ) + x 3 = x ( 27 − x ) 20

x 50 m Now, length of rectangular lawn ( l1) = 50 m

27 3 = x ( 27 − x ) 20 = 540 3x ( 27 − x )

and breadth of rectangular lawn (b1) = 40 m ∴ Length of rectangular pond (l2 ) = 50 − ( x + x ) = 50 − 2x and breadth of rectangular pond (b2)

= 180 x ( 27 − x )

= 40 − ( x + x ) = 40 − 2x

27 × 20 =3 × x ( 27 − x )

Also, area of the grass surrounding the pond = 1184 m²

x ( 27 − x ) = 180 27x − x 2 = 180 x 2 − 27x + 180 = 0 Solve using the quadratic formula: x=

− ( −27 ) ±

40 m

( −27 ) − 4 (1)(180 ) 2 (1)

27 ± 729 − 720 x= 2

27 ± 9 27 ± 3 = x = = 15, 12 2 2

The two parts are 15 and 12. 3. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be 1184 m² (see figure). Find the length and breadth of the pond. Chapter 4 • Quadratic Equations

∴ Area of rectangular lawn - Area of rectangular pond = Area of grass surrounding l1 × b1 − l2 × b2 = 1184

⇒ 50 × 40 − ( 50 − 2x )( 40 − 2x ) = 1184

(

)

⇒ 2000 − 2000 − 80x − 100x + 4x 2 = 1184 2

⇒ 80x + 100x − 4x = 1184

⇒ 4x 2 − 180x + 1184 = 0 ⇒ x 2 − 45x + 296 = 0 ⇒ x 2 − 37x − 8x + 296 = 0 ⇒ x ( x − 37 ) − 8 ( x − 37 ) = 0 ⇒ ( x − 37 )( x − 8 ) = 0 ⇒x = 8 At x = 37, length and breadth of the pond are −24 and −34, respectively, but length and breadth cannot be negative. So, x = 37 is not possible. 91

Length of pond = 50 − 2x = 50 − 2 ( 8) = 50 − 16 = 34 m

and breadth of pond = 40 - 2x = 40 - 2(8) = 40 - 16 = 24 m Hence, the required length and breadth of the pond are 34 m and 24 m, respectively. 4. The diagonal of a rectangular field is 16 metre more than the shorter side. If the longer side is 14 metre more than the shorter side, then find the length of the sides of the field. (CBSE 2015) Sol. Let the length of shorter side be x m. ∴ Length of diagonal = ( x + 16 ) m and length of longer side= x + 14 m Using Pythagoras Theorem, 2

x + (x + 14) = (x + 16)

⇒ x 2 + x 2 + 28x + 196 =x 2 + 32x + 256 ⇒ x 2 − 4x − 60 = 0 ⇒ x 2 + 6x − 10x − 60 = 0 ⇒ ( x + 6 )( x − 10 ) = 0 ⇒x = −6 or x = 10 ∴ Length of sides are 10 m and 24 m. 5. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/h and time of flight increased by 30 minutes. Find the original duration of flight. (CBSE 2020) Sol. Let the original speed of the aircraft be x km/h. 600 hours. x After the speed is reduced by 200 km/h, the new The original time of flight is

speed becomes x − 200 km/h.

600 The new time of flight is hours. x − 200

According to the problem, 600 600 1 − = x − 200 x 2

1 x ( x − 200 ) 2 1 600x − 600x + 120000 = x ( x − 200 ) 2 1 120000 = x ( x − 200 ) 2 240000 = x ( x − 200 )

240000 = x 2 − 200x

600x − 600 ( x − 200 ) =

Rearrange into a quadratic equation

x 2 − 200x − 240000 = 0

Substitute the values into the quadratic formula:

− ( −200 ) ±

( −200 )2 − 4 (1)( −240000 ) 2 (1)

200 ± 40000 + 960000 2 200 ± 1000000 x = 2 200 ± 1000 x = 2 Calculate the two possible values of x

200 + 1000 1200 = = 600 2 2 200 − 1000 − 800 x = = = −400 2 2 Since speed cannot be negative, the original speed of the aircraft must be x = 600 km/h. = x

The original time of flight is: 600 600 Time = = = 1 hour x 600

The original duration of the flight was 1 hour.

6. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it? Topper's Answer

7. A pole has to be erected at a point on the boundary of a circular park with a diameter of 17 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Find the distances from the two gates where the pole is to be erected. (CBSE 2016)

Sol.

x+7

Let P be the location of the pole such that its distance from gate B, x metres. ∴ AP =x + 7 AB is diameter, ∠ APB = 90 and AB = 17 m. Using Pythagoras Theorem, x 2 + (x + 7)2 = (17)2 x 2 + x 2 + 14x − 240 = 0 Or 2

x + 7x − 120 = 0

−7 ± 49 + 480 2 −7 ± 23 x= 2 x= 8, − 15 ( rejected ) x=

8. Two pipes running together can fill a cistern in 3 1 13 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern. Sol. Suppose the faster pipe takes x minutes to fill the cistern. Therefore, the slower pipe will take ( x + 3 ) minutes to fill the cistern. Since the faster pipe takes x minutes to fill the cistern. Therefore, Portion of the cistern filled by the faster pipe in one 1 minute = x ⇒ Portion of the cistern filled by the faster pipe in 40 1 40 40 minutes = × = 13 x 13 13x Similarly, Portion of the cistern filled by the slower pipe in 40 13 1 40 40 minutes= × = x + 3 13 13 ( x + 3 ) 40 It is given that the cistern is filled in minutes. 13 40 40 ∴ + = 1 13x 13 ( x + 3 ) ⇒

1 1 13 + = x x + 3 40

⇒

x + 3 + x 13 = x ( x + 3 ) 40

⇒ 40 ( 2x + = 3 ) 13x ( x + 3 ) ⇒ 80x + 120 = 13x 2 + 39x ⇒ 13x 2 − 41x − 120 = 0 ⇒ 13x 2 − 65x + 24x − 120 = 0

= ∴ x 8 m and = x + 7 15 m

⇒ 13x ( x − 5 ) + 24 ( x − 5 ) = 0

Hence, distance between two gates = 8 m and 15 m.

⇒ ( x − 5 )(13x + 24 ) = 0

Chapter 4 • Quadratic Equations

[ x > 0 ∴13x + 24 ≠ 0]

−B ± D we get x= x= , 2A

⇒ x −5= 0⇒ x = 5 Hence, the faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 minutes to fill the cistern. 9. Solve for x: 2

 2x   2x    + 5   − 24 = 0, x ≠ 5 x −5 x − 5 (CBSE 2016)

Sol.  2x  + 5  2x  − 24 = 0  x −5  x −5

9(a + b ) ± 3(a − b ) ⇒x = 2×9

3(a + b ) ± (a − b ) ⇒x = 6 3 a + 3 b ( 3a + 3b ) − (a − b ) ( ) + (a − b ) and ⇒ x = ⇒x = 6 6

( 4a + 2b ) and ( 2a + 4b ) ⇒x = ⇒x = 6 6

a + 2b 2a + b and ⇒ x = are required solutions. ⇒x = 3 3

2x =y x −5

Let

9 ( a + b ) ± 9(a − b )2

34 11. We have, Solve: ( x − 3 )( x − 4 ) = 2 ( 33 ) 34 Sol. We have, ( x − 3 )( x − 4 ) =2 33 34 ⇒ x 2 − 7x + 12 − 2 = 0 33

0 Then, y + 5 y − 24 = ⇒ ( y + 8 )( y − 3 ) = 0 = y or − 8 3 Putting y = 3, we get, 2x = 3 x −5

⇒ x 2 − 7x +

2= x 3x − 15

⇒ x 2 − 7x +

x = 15

13034 332

0 =

98 133 0 × = 33 33

231 98 × 133 x+ = 0 33 33 × 33

Again for y = −8,

⇒ x2 −

2x = −8x + 40

98 × 133  98 133  ⇒ x2 −  + = 0 x + 33 × 33  33 33 

10x = 40

⇒ x2 −

2x = −8 x −5

x =4

98   133 98 × 133   ⇒ x2 − x  −  x− 0 = 33   33 33 × 33  

Hence, x = 15 or 4. 10. Solve the following quadratic equation: 9x 2 − 9 ( a + b ) x + 2a 2 + 5ab + 2b 2  = 0

(CBSE 2016)

Sol. Given: 9x 2 − 9 ( a + b ) x + 2a 2 + 5ab + 2b 2 = 0. 0 Now comparing with Ax 2 + Bx + C = We get, A = 9, B = −9 ( a + b ) , C = 2a 2 + 5ab + 2b 2 Now discriminant, D = B 2 − 4 AC

(

2 =( −9( a + b ) ) − 4 × 9 × 2a 2 + 5ab + 2b 2

(

= 9 (a + b ) − 4 × 9 2a + 5ab + 2b

(

= 9 9(a + b ) − 4 2a + 5ab + 2b

( = 9 ( a + b − 2ab ) = 9(a − b ) 2

)

))

= 9 9a + 9b + 18ab − 8a − 20ab − 8b 2

Now using the quadratic formula,

98 133 98 × 133 x− x+ = 0 33 33 33 × 33

98  133  98   ⇒ xx −  − 0 x −  = 33 33 33     98  133   ⇒  x −  x − 0 = 33  33   98 133 = or x 33 33

= ⇒x

12. Using quadratic formula, solve the

)

(

)

abx 2 + b 2 − ac x − bc = 0 for x. Sol. Given quadratic equation is

)

(

)

abx 2 + b 2 − ac x − bc = 0 x=

(

)

− b 2 − ac ± (b 2 − ac )2 − 4 ( ab )( −bc ) 2ab

(CBSE 2005)

(

)

∆= (C + 1) 36 (C + 1) − 12 (C + 9 ) 

− b 2 − ac ± b 4 − 2ab 2c + a 2c 2 + 4ab 2c ⇒x = 2ab

(

)

∆= (C + 1)[ 36C + 36 − 12C − 108] ∆=

− b 2 − ac ± (b 2 + ac )2 ⇒x = 2ab

(

)

For the quadratic to have real and equal roots, the discriminant must be zero:

(

) (

− b 2 − ac ± (b 2 − ac )2 + 4ab 2c ⇒x = 2ab

− b 2 − ac + b 2 + ac ⇒x = 2ab −b 2 + ac + b 2 + ac ⇒x = 2ab 2ac ⇒x = 2ab

(

) (

− b 2 − ac − b 2 + ac ⇒x = 2ab 2 −b + ac − b 2 − ac ⇒x = 2ab −2b 2 ⇒x = 2ab ⇒= x

)

(C + 1)[24C − 72]

= ∆ 24 (C + 1)(C − 3 )

24 (C + 1)(C − 3 ) = 0 This equation is satisfied when either C + 1 = 0 or C −3= 0: C + 1 = 0 gives C = −1 C− = 0 gives C = 3 3

7 hours. 8 The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

14. Two water taps together can fill a tank in 1

)

−b c ,= x b a

13. Find the value of C for which the quadratic 0 has real equation: (C + 1) x 2 − 6 (C + 1) x + 3 (C + 9 ) = and equal roots. (CBSE 2024) Sol. For the quadratic to have real and equal roots, the discriminant must be zero: ∆ =0 The given quadratic equation is: 0 (C + 1) x 2 − 6 (C + 1) x + 3 (C + 9) = From this, we can identify the coefficients: a= C + 1 b = −6 (C + 1) = c 3 (C + 9 ) Now, apply the discriminant formula: ∆= b 2 − 4ac −6 (C + 1), and= c 3 (C + 9 ): Substitute a= C + 1, b = 2 ∆ =  −6 (C + 1)  − 4 (C + 1) 3 (C + 9 ) 

First, expand the terms: 2

= ∆ 36 (C + 1)  − 4 (C + 1) 3 (C + 9 )    2 = ∆ 36 (C + 1) − 12 (C + 1)(C + 9 )

Now, factor out (C + 1) from both terms: Chapter 4 • Quadratic Equations

Expand 36 (C + 1) and 12 (C + 9 ) :

(CBSE 2016, 2017, 2019)

Sol. Let the time taken by the smaller diameter tap = x h. Therefore, time for larger diameter tap = ( x − 2 ) h. 7 15 Total time taken = 1 = h. 8 8 Portion filled in one hour by smaller diameter 1 1 and by larger diameter tap = x −2 x According to the problem, tap =

1 1 8 + = x x − 2 15 8 x −2+x = × x (x − 2) 15 8x 2 − 16x 2x − 2 = 15 15 ( 2x − 2 )= 8x 2 − 16x 30x − 30 = 8x 2 − 16x 8x 2 − 46x + 30 = 0 4x 2 − 23x + 15 = 0 4x 2 − 20x − 3x + 15 = 0 0 ( 4x − 3)( x − 5) = 3 = or x 5 x = 4 3 3 5 If x = , then x − 2 = − 2 =− . 4 4 4 3 Since time cannot be negative, we neglect x = . 4 Therefore, x = 5 and x − 2 = 5 − 2 = 3 Hence, time taken by larger diameter tap = 3 hours and time taken by smaller diameter tap = 5 hours. 95

15. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. (CBSE 2016, 2017, 2018)

Sol. Let the speed of stream be x km/h. Then, the speed of boat upstream (18 − x ) km/h and speed of boat downstream = (18 + x ) km/h According to the question, 24 24 − 1 = 18 − x 18 + x ⇒ 24 (18 + x ) − 24 (18 − x )= 182 − x 2

(

⇒ 432 + 24x − 432 + 24x = 324 − x

)

⇒ 48x = 324 − x 2

16. Solve Sol.

1 1 1 1 = + + , where a ≠ 0, b ≠ 0. a +b + x a b x

1 1 1 1 = + + a +b + x a b x

1 1 1 1 ⇒ = + + a +b + x a b x 1 ab + ax + bx ⇒ = a +b + x abx ab ⇒ x − (a + b + x ) = ab + ax + bx ab ⇒ x − a −b − x = a +b + x ⇒ − (a + b ) =

⇒ x + 48x − 324 = 0

ab a +b + x

⇒ x ( a + b + x ) =−ab

⇒ x 2 + 54x − 6x − 324 = 0 ⇒ ( x + 54 )( x − 6 ) = 0

⇒ x 2 + ( a + b ) x + ab = 0

⇒ x += 54 0 or x = −6 0 ⇒x = −54 or x = 6 Since, speed cannot be negative. Hence, the speed of stream = 6 km/h.

⇒ ( x + a )( x + b ) = 0 ⇒x = −a or x = −b

Competency Based Questions Multiple Choice Questions

5. The value of

1. Meera and Sarita solve an equation. In solving Meera commits a mistake in constant term and find the roots 8 and 2. Sarita commits a mistake in the coefficient of x and find the roots -9 and -1. The correct roots are (a) 9, 1 (b) −9, 1 (c) 9, − 1 (d) −9, − 1 2. Which among the following equations can be categorised as a quadratic equation? (CBSE QB) 50 (a) x(x − 2)2 =

2 ( 2 + x )( 4x − 7 ) (b) 4x =

(d) x 2 +

1 0 = x

3. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number. (a) 3 (b) 8 (c) 4 (d) 7 4. The number of real roots of the equation x2 + 3 x + 2 = 0, is (a) 2

(b) 3

(d) 4

6 + 6 + 6 +… is

(a) 3.5

(b) 4

(d) -3

6. A student is trying to find the roots of 3x 2 − 10x − 8 = 0 by splitting the middle term as follows: 0 Step 1: 3x 2 − 10x − 8 = Step 2: 3x 2 − mx + nx − 8 = 0 What could be the values of m and n? (a) m = 8 and n = 2 (b) m = −8 and n = −2 (c) m = 12 and n = 2

(

(d) m = −12 and n = −2

)

0 has no real 7. If x 2 a 2 + b 2 + 2x ( ac + bd ) + c 2 + d 2 = roots, then (a) ad ≠ bc (b) ad < bc (c) ad > bc

(d) All of these

8. If α and β are the zeros of the polynomial p ( x ) = kx 2 − 30x + 45k and α + β = αβ , then the value of k is 3 2 2 3 (a) − (b) (c) (d) − 2 3 3 2

Select Response Type MCQs

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

1. Which of the following equations represents a quadratic equation that has no real roots? 0 (i) x 2 − 6x + 9 =

2 0 (ii) x + 4x + 5 =

(iii) x 2 − 2x − 3 = 0

0 (iv) x 2 + 7x + 12 =

(a) Only (ii)

(b) Only (i) and (ii)

(d) Only (i), (ii) and (iv)

(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.

2. Aman solved a quadratic equation and found it roots to be real.

1. Assertion (A): If 5 + 7 is a root of a quadratic equation with rational coefficients, then its other root is 5 − 7 .

Which of these could represent the graph of the equation Aman solved? (i)

(ii)

2 0

4 2

-2

-4

-2

(iii) 8

-2

Reason (R): Irrational roots of a quadratic equation with rational coefficients occur in conjugate pairs. 1 1 1 2. Assertion (A): In the expression , + = x −3 x +5 6 x can’t have values 3 and −5. Reason (R): If discriminant D =b 2 − 4ac > 0, then the roots of the quadratic equation ax 2 + bx + c = 0 are real and unequal.

Case Study Based Questions

(iv)

-2

1. A company manufactures and sells a product. The company has observed that the total profit from selling x number of units of the product can be represented by the quadratic equation:

P (x ) = −10x 2 + 600x − 5000

(CBSE 2023)

(a) Only (iii)

(b) Only (i) and (ii)

(d) Only (i), (ii) and (iv)

3. Rahul follows the below steps to find the roots of the 0, by splitting the middle equation 3x 2 − 11x − 20 = term. Step 1: 3x 2 − 11x − 20 = 0 0 Step 2: 3x 2 − 15x + 4x − 20 = 0 Step 3: 3x ( x − 5 ) − 4 ( x − 5 ) = Step 4: ( 3x − 4 )( x − 5 ) = 0 4 Step 5: x = − and −5 3 In which step did Rahul make the first error? (a) Step 1 and 2 (b) Step 2 and 3 (c) Step 3 and 5 Chapter 4 • Quadratic Equations

(d) Step 4 and 5

Where: P ( x ) represents the profit in rupees when x units of the product are sold. x is the number of units sold. Based on the above information, answer the following questions: (a) For which value of x (the number of units sold) will the company make zero profit? (b) If the company wants to earn at least `3000 in profit, what is the minimum number of units it should sell? 97

(c) Calculate the discriminant of the quadratic equation. What does the discriminant tell you about the number of real solutions for the profit to be zero? 2. The motorboat travels at a speed of 20 km/h. To travel a distance of 15 km, it required an additional hour going upstream compared to the time it took to travel downstream.

(a) If the speed of stream is x km/h, then the speed of the motorboat in upstream is (i) 20 km/h (ii) ( 20 + x ) km/h

(iii) ( 20 − x ) km/h

(iv) ( −20 ) km/h

(b) If the speed of stream is x km/h, then the speed of the motorboat in downstream is (i) ( 20 + x ) km/h (ii) ( x − 20 ) km/h

(iii) 20 km/h

(iv)

20 km/h x

0 (iv) x 2 − 20x − 400 = 0 (iii) x 2 + 30x − 400 =

(d) The speed of the current is (i) 20 km/h (ii) 10 km/h

(iii) 15 km/h

(iv) 25 km/h

OR (e) Time taken by the motorboat to cover 15 km upstream is 1 (i) 1 hour (ii) 1 hours 2 (iii) 2 hours (iv) 3 hours

Answers Multiple Choice Questions

0 with a ≠ 0. Since, it is not of the form ax 2 + bx + c =

1. (a) 9, 1

Hence, it is not a quadratic equation. From option (c), we have 3x 2 + 5x = 9 ( x − 3 )

From Meera’s solution,

⇒ 3x 2 + 5x = 9x − 27

Sum of roots = 8 + 2 = 10

⇒ 9x − 27 − 3x 2 − 5x = 0

From Sarita’s solution,

⇒ 3x 2 + 4x − 27 = 0

Product of roots = (-9)(-1) = 9

0 with a ≠ 0. Since, it is of the form ax 2 + bx + c = Where a = −3 , b = 4 and c = −27

0 ∴ Equation is x 2 − 10x + 9 = 0 ⇒ x 2 − 9x − x + 9 =

Hence, it is a quadratic equation. 1 3 0 0 ⇒ x +1 = From option (d), we have x 2 + = x

⇒ ( x − 9 )( x − 1) = 0 ⇒ x = 9, 1 2. (c) 3x 2 + 5x = 9 ( x − 3 )

(

From option (a), we have x x 2 − 2

50 )= 2

x(x 4 − 4x 2 + 4) = x 5 − 4x 3 + 4x x 5 − 4x 3 + 4x − 50 = 0 Since, it is not of the form ax 2 + bx + c= 0, a ≠ 0 . Hence, it is not a quadratic equation. From option (b), we have 4x 2 = ( 2 + x )( 4x − 7 ) ⇒ 4x 2 = 8x − 14 + 4x 2 − 7x 0 ⇒ x − 14 = 98

0 a ≠ 0. Since, it is not of the form ax 2 + bx + c =, Hence, it is not a quadratic equation. 3. (b) 8 Let the number be x. Then, according to the question, 160 x + 12 = x ⇒ x 2 + 12x − 160 = 0 ⇒ x 2 + 20x − 8x − 160 = 0 0 ⇒ ( x + 20 )( x − 8 ) =

⇒ x =−20, 8 Since, the number is natural, so x = 8.

Select Response Type MCQs

4. (c) 0 2

0 We have, x + 3 x + 2 =

1. a. Only (ii)

x 2 > 0, x > 0 and + 2 > 0

∴x + 3 x + 2 ≠ 0 for any real x.

0 (i) x 2 − 6x + 9 =

5. (c) 3

D = ( −6)2 − 4 (1)( 9 ) = 36 − 36 = 0 (This equation has a discriminant of 0, implying a repeated real root.)

Let y = 6 + 6 + 6 +…

Hence no real solution exist.

⇒ y =

D = (4)2 − 4 (1)( 5 ) = 16 − 20 = −4 (This equation has a negative discriminant, implying no real roots.)

6+y

On squaring both sides, we get y 2= 6 + y

(iii) x 2 − 2x − 3 = 0

0 ⇒ y 2 − 3 y + 2 y − 6 =

D =− ( 2)2 − 4 (1)( −3 ) =4 + 12 =16 (This equation has a positive discriminant, implying two distinct real roots.)

0 ⇒ ( y − 3 )( y + 2 ) = ⇒ y = 3, − 2 3[since y cannot be negative] ⇒ y = 6. (c) m = 12 and n = 2 0 We have, 3x 2 − 10x − 8 = By splitting the middle term, we have 3x 2 − 12x + 2x − 8 = 0

0 Given equation in step 2 is 3x 2 − mx + nx − 8 =. On comparing, we get m = 12 and n = 2. 7. (d) All of these We have,

(

)

0 x 2 a 2 + b 2 + 2x ( ac + bd ) + c 2 + d 2 = Roots are not real. 2

(

⇒ 2 ( ac + bd )  − 4 a + b

2 0 (iv) x + 7x + 12 =

D = (7)2 − 4 (1)(12 ) = 49 − 48 = 1 (This equation has a positive discriminant, implying two distinct real roots.) 2. d. Only (i), (ii) and (iv) For non-real roots, the graph of the quadratic equation lies completely above the x-axis. So, graphs of (i), (ii), and (iv) indicate that the quadratic equation has real roots and roots of the quadratic equation for graph (iii) are non-real. ∴ Option (d) is correct. 3. (c) Step 3 and 5 0 Step 1: 3x 2 − 11x − 20 =

⇒ B − 4 AC < 0

0 (ii) x 2 + 4x + 5 =

)(c + d ) < 0 2

⇒ 4a 2c 2 + 4b 2d 2 + 8abcd < 4a 2c 2 + 4b 2d 2 +

4a 2d 2 + 4b 2c 2 ⇒ 2abcd < a 2d 2 + b 2c 2 2

⇒ ( ad − bc ) > 0 ⇒ ad ≠ bc and ad < bc or ad > bc 2 8. (c) 3 45k  −30  − We have, α + β =  and αβ = k  k  30 ⇒ α + β = and αβ = 45 k 30 2 ∴α + β = αβ ⇒ = 45 ⇒ k = k 3

0 Step 2: 3x 2 − 15x + 4x − 20 = 0 Step 3: 3x ( x − 5 ) + 4 ( x − 5 ) = 0 Step 4: ( 3x + 4 )( x − 5 ) = −4 Step 5: x = and 5 3

Assertion-Reason Based Questions 1. (a) Both A and R are true, and R is the correct explanation of A. 0 be the quadratic equation. Let ax 2 + bx + c = Then, roots of this equation are given by x =

−b ± b 2 − 4ac 2a

⇒ −

b b 2 − 4ac ± 2a 2a

This shows that irrational roots can exist only as conjugate pairs. Chapter 4 • Quadratic Equations

2. (b) Both A and R are true, but R is not the correct explanation of A. 1 1 1 + = We have, x −3 x +5 6 ⇒ 6 ( x + 5 + x − 3 ) = ( x − 3 )( x + 5 )

(b) −10x 2 + 600x − 5000 ≥ 3000 −10x 2 + 600x − 8000 ≥ 0 −x 2 + 60x − 800 ≥ 0 x 2 − 60x + 800 ≤ 0

⇒ 6 ( 2x + 2 ) = x 2 + 2x − 15

x =

⇒ 12x + 12 = x + 2x − 15 2

0 ⇒ x − 10x − 27 = Using the quadratic formula, x =

− ( −10 ) ±

= x

( −10 )2 − 4 ⋅1 ⋅ ( −27 )

2 ⋅1 10 ± 100 + 108 ⇒ x = 2 10 ± 208 ⇒ x = 2 10 ± 4 13 ⇒ x = 2 ⇒ x = 5 ± 2 13 So, the roots of the given equation are 5 + 2 13 and 5 − 2 13. Hence, x cannot have values 3 and −5. So, assertion is true. Also, reason is true but reason is not the correct explanation of assertion.

Case Study Based Questions 1. (a) Given −10x 2 + 600x − 5000 = 0 x=

−600 ± 6002 − 4 ( −10 )( −5000 ) 2 ( −10 )

x =

−600 ± 360000 − 200000 −20

x =

−600 ± 160000 −20

x =

−600 ± 400 −20

Now, solving for the two possible values of x: −600 + 400 −200 x = = 10 = −20 −20 = x

−600 − 400 −1000 = = 50 −20 −20

The company makes zero profit when they sell 10 or 50 units.

100

− ( −60 ) ± ( −60)2 − 4 (1)( 800 ) 2 (1)

60 ± 3600 − 3200 60 ± 400 = 2 2

⇒ = x

60 + 20 60 − 20 = 40 or = = 20 x 2 2

20 ≤ x ≤ 40 To achieve a profit of ` 3000 or more, Company should sell between 20 to 40 units (c) ∆ = b 2 - 4ac ∆ = 6002 - 4(-10)(-5000) ∆ = 360000-200000 ∆ = 160000 Since the discriminant is positive, there are two real solutions for x, meaning there are two possible values for the number of units sold where the profit is zero. 2. (a) (iii) Speed of motorboat = 20 km/h Speed of stream = x km/h Speed upstream of the motoboat = (20 - x) km/h (b) (i) Speed of motorboat = 20 km/h Speed of stream = x km/h Speed downstream of the motorboat = (20 + x) km/h (c) (iii) Time taken by the motorboat to cover 15 hours 20 − x Time taken by the motorboat to cover 15 km 15 downstream = hours 20 + x 15 15 1 − = 20 − x 20 + x 15 km upstream =

⇒

15 ( 20 + x ) − 15 ( 20 − x )

= 1

( 20 − x )( 20 + x ) ⇒ 15 ( 20 + x ) − 15 ( 20 − x ) = ( 20 − x )( 20 + x ) ⇒ 15 × 2x = 400 − x 2 ⇒ 30x = 400 − x 2 ⇒ x 2 + 30x − 400 = 0

(d) (ii) The speed of the current is given by

(e) (ii) Time taken by the motorboat to cover 15 15 1 15 km upstream = = = 1 hours 20 − x 10 2

x + 30x − 400 = 0

⇒ ( x + 40 )( x − 10 ) = 0 ⇒x = 10 km/h

Additional Practice Questions 0 , what is the 1. For the equation x 2 − 8x + 15 = product of its roots? (a) 15 (b) -15 (c) 23 (d)-23 2. Which quadratic equation has a discriminant of 64? 2

0 (a) 2x − 16x + 32 = 2

0 (c) x − 16x + 32 =

0 (b) x − 8x + 16 = 0 (d) x 2 − 16 x + 48 =

0 then 3. If x = 0.3 is a root of the equation x 2 − 0.9k = k is equal to (a) 1 (b) 10 (c) 0.1 (d) 100 (CBSE 2023)

4. Which of the following represents a quadratic equation that has no real roots? 0 (a) x 2 − 4x + 4 =

2 0 (b) x − 4x + 1 =

0 (d) x 2 + x + 4 =

5. Find the value of x which satisfy the equation 3x + 7 − 2x + 3 = 1. (a) 2 and −2

(b) 3 and 4

(d) 3 and −1

6. A plane left 30 minutes later than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km/h. Find its usual speed. (CBSE 2018) 7. Find the value of k for which the roots of the 0 equation 3x 2 − 10x + k =are reciprocal of each other. (CBSE 2019) 8. Find the value of k for which the quadratic equation 0, k ≠ −1 has equal (k + 1) x 2 − 6 (k + 1) x + 3 (k + 9) = roots. (CBSE 2019) 1 9. If x = − is a solution of the quadratic equation 2 3x 2 + 2kx − 3 =, 0 find the value of k. (CBSE 2015, 2017)

14 5 −1 = 10. Solve the equation: ; x ≠ −3, −1 for x. x +3 x +1

(CBSE 2014)

11. Find the value of

30 + 30 + 30 +……∞

Chapter 4 • Quadratic Equations

12. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers. (CBSE 2016) 13. Solve the following quadratic equation for x: a +b   a x2 +  + 0 x +1 = a + b a   (CBSE 2016) 14. A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number. (CBSE 2016)

(

)

0 has 15. If the equation 1 + m 2 x 2 + 2mcx + c 2 − a 2 =

(

)

c 2 a 2 1 + m2 . equal roots, show that= (CBSE 2017)

16. Sum of the areas of two squares is 544 m². If the difference of their perimeters is 32 m, find the sides of the two squares. (CBSE 2020) 17. Find x in terms of a ,b and c : a b 2c + = ;x ≠ a, b, c x −a x −b x −c

(CBSE 2016)

18. A motorboat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream. (CBSE 2016) 19. The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and 16 its reciprocal is 2 , find the fraction. 21

(CBSE 2016)

20. To fill a swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. Find how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. (CBSE 2015) 21. A shopkeeper buys some books for ` 80. If he had bought 4 more books for the same amount, each book would have cost ` 1 less. Find the number of books he bought. (CBSE 2012) 101

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

The shortest distance between gates 1 and 2 is 29 m, find the width of the park. Show your work.

(a) Both A and R are true, and R is the correct explanation of A.

(b) The caretaker of the park is attempting to plant saplings in the form of a square. That is, number of rows of saplings is the same as the number of columns of saplings. On arranging the saplings, he found that 24 saplings were still left with him. When he increased the number of rows and columns by 1, he found that he was short of 25 saplings. Find the number of saplings available with him. Show your work.

(b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 22. Assertion (A): The sum of the roots of the quadratic 0 is 5. equation 2x 2 − 10x + 12 = Reason (R): The formula for the sum of the roots of b the quadratic equation ax 2 + bx + c = 0 is − . a 23. Assertion (A): The roots of the equation x 2 − 6x + 9 =are 0 real and equal.

(c) The city corporation is planning to install a rectangular sand table for the kids in the play area. The carpenter is asked to design a rectangular table whose length is thrice its width and the area is 108 sq. m. Determine whether such a table can be made. Show your steps. (CBSE QB)

Reason (R): The discriminant b 2 − 4ac of the equation is greater than zero. 24. Assertion (A): For the quadratic equation x 2 + 2x + 5 = 0 the roots are non-real. Reason (R): If the discriminant of a quadratic is negative, then the roots are not real.

26. A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.

25. Bangalore city corporation is building parks in residential areas across the city. Shown below is one such park. The rectangular park consists of various components like walking track, kids play area, open gym, pond etc. Gate 1 WALKING TRACK Lawn

Width

Lawn

Open gym Pond

Shelter

Lawn

Kids play area

(a) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.

Gate 3

(b) Write the corresponding quadratic equation in standard form.

WALKING TRACK Gate 2

Length

(Note the image is not to scale)

Gate 1 Based on the above information, answer the following questions. (a) Gate 3 has been placed exactly opposite to gate 1 on the boundary of the park. The distance between gate 3 and gate 2 is 1 m more than the distance between gate 3 and gate 2. 102

(c) (i) Find the value of x, the length of side of tile by factorisation. OR (ii) Solve the quadratic equation for x, using quadratic formula. (CBSE 2024)

Scan me

Challenge Yourself 1. Find the equation whose roots are x 2 + 5x + 3 2. Solve <x x +2 3

a a − a ±b

4. If the roots of the equation 3x 2 + 9x + 2 = 0 are in the ratio m : n , then find

m n . + n m

3. Solve 8x 2n − 8x − 2n = 63. (Hint: Substitute

3 2n y =x

)

Answers Additional Practice Questions

22. (a) Both A and R are true, and R is the correct explanation of A.

1. (a) 15

2. (d) x2 -16x + 48 = 0

3. (c) 0.1

4. (d) x + x + 4 = 0

5. (d) 3 and -1

6. 500 km/h

7. k = 3

8. k = 3

9. k = −

10. 1, 4

11. 6

12. 9, 10, 11

a +b a − ,− 13. x = 14. 24 a a +b

15. Proved

23. (c) A is true, but R is false.

16. 20 cm, 12 cm

17. x = −

18. 8 km/h

19.

bc + ac − 2ab ,0 a + b − 2c

3 7

20. 20 hours, 30 hours

24. (a) Both A and R are true, and R is the correct explanation of A. 9 4

25. (a) 20 m

26. (a) 200 = x 2 128(x + 1)2 0 (b) 9x 2 − 32x − 16 = (c) (i) x = 4

Challenge Yourself 0 1. bx 2 − 2ax + a = 3. x = 2

21. 16

(b) 600

2. −2 < x < −1 4.

−3 3 2

Scan me for Exemplar Solutions Chapter 4 • Quadratic Equations

103

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

0? 1. Is -8 a solution of 3x 2 + 8x + 2 = (a) Yes

(b) No

(d) None of these (NCERT Exemplar)

0 are real, is 2. The values of k for which the roots of the equation x 2 + 4x + k = (a) k ≥ 4

(b) k ≤ 4

(d) k ≤ −4 (CBSE 2023)

3. Which of the following equations has two distinct real roots? 9 0 0 0 (a) 2x 2 − 3 2x + = (b) x 2 + x − 5 = (c) 3x 2 + 2x = 4

0 (d) 5x 2 − 3x + 1 =

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

0 are real and Reason (R): If discriminant D = b 2 − 4 > 0 , then the roots of the quadratic equation ax 2 + bx + c = unequal. 0 has equal roots. 5. Assertion (A): Quadratic equation 3x 2 − 6x + 3 = 0 have equal roots if discriminant D > 0 . Reason (R): The quadratic equation ax 2 + bx + c =

Very Short Answer Questions

[2 × 2 = 4 Marks] 2

0 6. Find the nature of roots of the equation 2x − 4x + 3 =. (CBSE 2019) 2

0 and x + bx + b = 0, then find the value 7. If x = −2 is the common solution of the quadratic equation ax + x − 3a = 2 of a b. (CBSE 2022)

Short Answer Questions

[4 × 3 = 12 Marks] 2

8. Solve the quadratic equation: ( x − 1) − 5 ( x − 1) − 6 =. 0 (CBSE 2015)

9. A rectangular garden has an area of 96 square metres. If its length is 4 metres more than its width, form a quadratic equation to determine its dimensions. 10. Solve for x: 1 1 11 − = , x ≠ −4, 7. x + 4 x + 7 30 11. The product of two consecutive positive integers is 306. Find the integers. 104

(CBSE 2020)

Long Answer Questions

[3 × 5 = 15 Marks] 2

+ 64 0 and x − 8= x + k 0 both will have the 12. Find the positive values of k for which quadratic equations x + kx= real roots. (CBSE 2016)

13. The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is 16 2 , find the fraction. 21

(CBSE 2016)

14. The difference between the radii of the smaller circle and the larger circle is 7 cm, and the difference between the areas of the two circles is 1078 sq. cm. Find the radius of the smaller circle. (CBSE Comptt. 2017)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/hrs while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km. (CBSE QB 2021)

(a) What will be the distance covered by Ajay’s car in two hours?

(i) 2(x + 5) km

(ii) (x - 5) km

(iii) 2(x + 10) km

(iv) (2x + 5) km

(b) Which of the following quadratic equations describe the speed of Raj’s car? 0 (i) x 2 − 5x − 500 =

0 (ii) x 2 + 4x − 400 =

(iii) x 2 + 5x − 500 = 0

0 (iv) x 2 − 4x + 400 =

(iii) 25 km/h

(iv) 10 km/h

(iii) 25 hours

(iv) 16 hours

(i) 20 km/h

(ii) 15 km/h

(d) How much time did Ajay take to travel 400 km?

(i) 20 hours

Self-Assessment

(ii) 40 hours

105

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. If one root of the equation 3x2 - 8x - (2k+1) = 0 is seven times the other, then the value of k is (a) 7 (b) 5 (c) − 5 (d) − 7 3 3 3 3 2. Which of the following equations has no real roots? 0 (a) x 2 − 4x + 3 2 =

0 (b) x 2 + 4x − 3 2 =

0 (c) x 2 − 4x − 3 2 =

0 (d) 3x 2 + 4 3x + 4 = (NCERT Exemplar)

α2 + β2 0, then find the value of 3. If α and β are the roots of the equation x − 12x + 32 = . α +β 2

(a)

−8 3

(b)

8 3

(c)

−20 3

(d)

20 3

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

a 0. for a quadratic equation 2x 2 + ax − a 2 = 2 −b ± b 2 − 4ac 0 x= Reason (R): For quadratic equation ax 2 + bx + c =, . 2a 4. Assertion (A): The values of x are

0 has real and equal roots. 5. Assertion (A): The quadratic equation 2x 2 − 4x + 2 = Reason (R): A quadratic equation has equal roots if its discriminant is zero.

Very Short Answer Questions

[2 × 2 = 4 Marks]

0 is the reciprocal of the other, then find the value of k. 6. If one root of the equation (k − 1) x 2 − 10x + 3 = (CBSE SQP 2020)

0 and hence comment on the nature of roots of the 7. Find the discriminant of the quadratic equation 4x 2 − 5 = equation. (CBSE 2023)

Short Answer Questions

(

[4 × 3 = 12 Marks]

)

(

)

0 has equal roots, then show that= c 2 a 2 1 + m2 . 8. If the equation 1 + m 2 x 2 + 2mcx + c 2 − a 2 = (CBSE 2017)

9. Find the values of k, for which the quadratic equation (k + 4 ) x + (k + 1) x + 1 = 0 has equal roots. 2

(CBSE 2020)

106

10. A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number. 11. If x =

2 0, find the values of a and b. and x = −3 are roots of the quadratic equation ax 2 + 7x + b = 3

Long Answer Questions

(CBSE 2016) (CBSE 2016)

[3 × 5 = 15 Marks]

12. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed? (CBSE 2018) 13. Divide 27 into two parts such that the sum of their reciprocals is

3 . 20

(CBSE 2018)

0 are equal, then show that 14. If the roots of the quadratic equation ( x − a )( x − b ) + ( x − b )( x − c ) + ( x − c )( x − a ) = a= b= c . (CBSE 2015)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. The speed of a motorboat is 20 km/hrs. For covering the distance of 15 km, the boat took 1 hour more for upstream than downstream.

(CBSE QB)

(a) Let the speed of the stream be x km/h, then the speed of the motorboat in upstream will be:

(i) 20 km/h

(ii) (20 + x) km/h

(iii) (20 - x) km/h

(iv) 2 km/h

(b) What is the relation between speed, distance, and time? Distance Time

(ii) Distance = Speed × Time

(iii) Time = Speed × Distance

(iv) Speed = Distance × Time

(i) Speed=

(ii) x² + 20x - 400 = 0

(iii) x² + 30x + 400 = 0

(iv) x² - 20x - 400 = 0

(iii) 15 km/h

(iv) 25 km/h

(d) What is the speed of the current?

(i) 20 km/h

Self-Assessment

(ii) 10 km/h

107

Arithmetic Progressions

Chapter at a Glance Sequence A sequence is an ordered list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term. Sequences can be finite (with a fixed number of terms) or infinite (extending indefinitely). Example: ●

2, 4, 6, 8, 10 (Each term increases by 2).

●

50, 45, 40, 35, 30 (Each term decreases by 5).

Arithmetic Progression An Arithmetic Progression (AP) is a type of sequence where the difference between any two consecutive terms is constant. This constant is called the common difference (d). AP series: a, a + d, a + 2d, a + 3d,............a + (n – 1)d Example: ●

3, 7, 11, 15, 19 (The common difference is 4).

●

100, 90, 80, 70 (Here, the common difference is −10 ).

Common Difference The common difference (d) is the fixed value that is added (or subtracted) to each term of an AP to get the next term. It can be calculated by subtracting any term from the term that follows it. Formula:= d an +1 − an Where an and an +1 are consecutive terms of the AP. Example: Find the common difference in the given AP 5, 9, 13, 17. d =9−5= 4

nth Term (an) of an AP

The nth term of an AP represents the term’s value at the nth position. Formula: an =a + (n − 1)d Where: a = First term of the AP, d = Common difference, n = Term position Example: Find the 10th term of the AP: 2, 5, 8, 11. Here, a = 2, d = 5 − 2 = 3, n = 10 a10 =2 + (10 − 1) 3 =2 + 27 =29 So, the 10th term is 29.

nth Term (an) of an AP from the End

To find the nth term from the end of a finite AP, we use a formula that considers the last term and the common difference. Formula: an =− l (n − 1)d 108

Where: l = Last term of the AP, d = Common difference, n = Term position from the end Example: For the AP: 3, 7, 11, ..., 19, find the 2nd term from the end. = l 19,= d 4,= n 2 Here, an = 19 − ( 2 − 1) 4 = 19 − 4 = 15 So, the 2nd term from the end is 15.

The Sum of n Terms of an AP The sum of the first n terms of an AP is the total of the terms from the first to the nth position. Formula: n Sn = ⋅ (2a + (n − 1) ⋅ d ) 2 Or, if the last term (l) is known: n Sn = ⋅ ( a + l ) 2 Example: Find the sum of the first five terms of the AP: 2, 4, 6, 8, 10. Here, a = 2, d = 4 − 2 = 2, n = 5 S5=

5 5 2 × 2 + ( 5 − 1) 2= ( 4 + 8 ) = 30 ( ) 2 2

NCERT Zone NCERT Exercise 5.1

air remaining in the cylinder at a time. In other 1 3 words, after every stroke, only 1 – = tℎ part 4 4 of air will remain. Therefore, volumes will be V,  3   3 2  3 3  4 V,   V,   V…   4 4

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why? (a) The taxi fare after each km when the fare is `15 for the first km and `8 for each additional km. (b) The amount of air present in a cylinder when a vacuum pump removes 1 of the air remaining in 4 the cylinder at a time. (c) The cost of digging a well after every metre of digging, when it costs `150 for the first metre and rises by `50 for each subsequent metre. (d) The amount of money in the account every year, when `10000 is deposited at compound interest at 8% per annum. Sol. (a) It is observed that

Taxi fare for 1st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 + 8 = 31

Taxi fare for first 4 km = 31 + 8 = 39 Clearly 15, 23, 31, 39 … forms an AP because every term is 8 more than the preceding term. (b) Let initial volume of air in a cylinder be V lit. In 1 each stroke, the vacuum pump removes of 4 Chapter 5 • Arithmetic Progressions

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an AP (c) Cost of digging for first metre = 150

Cost of digging for first 2 metres = 150 + 50 = 200

Cost of digging for first 3 metres = 200 + 50 = 250

Cost of digging for first 4 metres = 250 + 50 = 300

Clearly, 150, 200, 250, 300 … forms an AP because every term is 50 more than the preceding term. (d) We know that if `P is deposited at r % compound interest per annum for n years, our n r   money will be P  1 +  after n years.  100  Therefore, after every year, our money will be 2

8  8    8   10000  1 + , 10000  1 +  , 10000  1 + 100  ...  100      100 

It can be obsered that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an AP. 109

2. Write first four terms of the AP when the first term a and the common difference d are given as follows (a) = a 10, = d 10 (b) a = −2, d = 0 (c) a = 4, d = −3 1 (d) a = −1, d = 2 −1.25, d = − 0.25 (e) a = = a 10, = d 10 Sol. (a)

Let the series be a1 , a 2 , a 3 , a 4 , a5 , ...

a1= a= 10

a 2 = a1 + d = 10 + 10 = 20

a 3 = a 2 + d = 20 + 10 = 30

a 4 = a 3 + d = 30 + 10 = 40

First four terms of this AP are 10, 20, 30, and 40. (b) a = −2, d = 0

Let the series be a1 , a 2 , a 3 , a 4, ... a1 = a = −2

a 2 =a1 + d =−2 + 0 =−2

a 3 =a 2 + d =−2 + 0 =−2

a 4 =a 3 + d =− 2 + 0 =−2

Clearly, the series will be 1 1 –1, − , 0, ... 2 2

First four terms of this AP are –1, − −1.25, d = −0.25 (e) a =

Let the series be a1 , a 2 , a 3 , a 4, ...

a1= a= −1.25

a 2 = a1 + d = −1.25 − 0.25 = −1.50

−1.50 − 0.25 = −1.75 a3 = a2 + d =

a 4 = a3 + d =

1 1 , 0, ... 2 2

−1.75 − 0.25 = −2.00

First four terms of this AP are −1.25, −1.50, −1.75 and −2.00. 3. For the following APs, write the first term and the common difference. (a) 3, 1, −1, −3, ... (b) −5, −1, 3, 7, ... (c)

1 5 9 19 , , , , ... 3 3 3 3

(d) 0.6, 1.7, 2.8, 3.9, ... Sol. (a) 3, 1, −1, −3, …

Here, first term, a = 3

Common difference, d =− 1 3 =−2

First four terms of this AP are −2, −2, −2 and −2.

(b) −5, −1, 3, 7, …

Common difference, d = ( −1) − ( −5 ) =−1 + 5 =4

Let the series be a1 , a 2 , a 3 , a 4, ...

a1= a= 4

a 2 = a1 + d = 4 − 3 = 1

a 3 =a 2 + d =1 − 3 =−2

a 4 =a 3 + d =− 2 − 3 =−5

First four terms of this AP are 4, 1, −2 and −5. 1 −1, d = (d) a = 2 Let the series be a1 , a 2 , a 3 , a 4, ...

a1 = a = −1

a 2 = a1 + d =−1 +

1 1 =− 2 2 1 1 − + = 0 a3 = a2 + d = 2 2 1 1 a 4 =a3 + d = 0 + = 2 2

Here, first term, a = −5

(c)

1 5 9 19 , , , ,… 3 3 3 3

Here, first term, a =

Common difference, d =

5 1 4 − = 3 3 3

(d) 0.6, 1.7, 2.8, 3.9, …

Here, first term, a = 0.6

Common difference, d = 1.7 − 0.6 = 1.1 4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms. (a) 2, 4, 8, 16… (b) 2,

5 7 , 3, … 2 2

110

1 3

(e) 3, 3 + 2, 3 + 2 2, 3 + 2 2, ... (f) 0.2, 0.22, 0.222, 0.2222, ... (g) 0, − 4, − 8, − 12, ...

It is an AP because difference between consecutive terms is equal.

⇒ −6 − ( −10 ) =−2 − ( −6 ) =2 – ( −2 ) =4

1 1 1 1 (h) − , − , − , − , ... 2 2 2 2 (i) 1, 3, 9, 27, ...

d =4

5th term = 2 + 4 = 6

6th term = 6 + 4 = 10

7th term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14.

(j) a , 2a , 3a , 4a , ... (k) a , a2, a3, a4, …. (l)

8 , 18,

(m)

32 , ….

9 , 12 ,…..

(n) 12 , 32 , 52 , 72 (o) 12 , 52 , 72 , 73 Sol. (a) 2, 4, 8, 16… It is not an AP because difference between consecutive terms is not equal.

(e) 3, 3 + 2, 3 + 2 2, 3 + 2 2 , … It is an AP because difference between consecutive terms is equal. ⇒ 3 + 2 − 3 = 2 , 3+ 2 2 ( 3 + 2 ) = 3+ 2 2 − 3 − 2 =

d= 2

5th term = 3 + 3 2 +

2 = 3 +4 2

5 7 , 3, … 2 2 It is an AP because difference between consecutive terms is equal. 5 5 1 ⇒ −= 2 3 –= 2 2 2 1 d = 2 7 1 5th term = + = 4 2 2 1 9 6th term = 4+ = 2 2 9 1 7th term = + = 5 2 2 9 Therefore, next three terms are 4, and 5. 2 (c) −1.2, − 3.2, − 5.2, − 7.2…

6 term = 3 + 4 2 +

2 = 3 +5 2

7th term = 3 + 5 2 +

2 = 3 +6 2

It is an AP because difference between consecutive terms is equal.

⇒ 4−2≠8−4

(b) 2,

⇒ −3.2 – ( −1.2 ) =− 5.2 ( −3.2 ) = −7.2 − ( −5.2 ) = −2

d = −2

5 term =−7.2 − 2 =−9.2

6th term =−9.2 − 2 =−11.2

7th term =−11.2 − 2 =−13.2

Therefore, next three terms are −9.2, − 11.2, and −13.2 (d) −10, − 6, − 2, 2 ,…

Chapter 5 • Arithmetic Progressions

Therefore, next three terms are (3 + 4 2 ), (3 + 5 2 ), (3 + 6 2 ) (f) 0.2, 0.22, 0.222, 0.2222,... It is not an AP because difference between consecutive terms is not equal.

⇒ 0.22 − 0.2 ≠ 0.222 − 0.22

(g) 0, −4, −8, −12,... It is an AP because difference between consecutive terms is equal.

⇒ −4 − 0 =−8 − ( −4 ) =−12 − ( −8 ) =−4

d = −4

5th term =−12 − 4 =−16

6th term =−16 − 4 =−20

7th term = −20 − 4 = −24

Therefore, next three terms are −16, −20 and −24 .

1 1 1 1 − , − , − , − ,… 2 2 2 2 It is an AP because difference between consecutive terms is equal. (h)

1  1 1  1 ⇒ − −  −  =− –  −  = 0 2  2 2  2 d=0 1 1 5th term = − + 0 = − 2 2

111

1 1 6 term = − + 0 = − 2 2

7th term = −

(o) 12, 52, 72, 73

It is an AP because difference between consecutive terms is equal.

1 1 +0= − 2 2

1 1 1 Therefore, next three terms are − , − , − . 2 2 2 (i) 1, 3, 9, 27,… It is not an AP because difference between consecutive terms is not equal.

⇒ 3–1 ≠ 9–3

(j) a, 2a, 3a, 4a … It is an AP because difference between consecutive terms is equal.

⇒ 2a – a = 3a – 2a = 4a – 3a = a

d=a

5th term = 4a + a = 5a

6th term = 5a + a = 6a

7th term = 6a + a = 7a

Therefore, next three terms are 5a, 6a, and 7a.

⇒ 52 − 12 = 72 − 52 = 73 − 72 = 24

d = 24

5th term = 73 + 24 = 97

6th term = 97 + 24 = 121

7th term = 121 + 24 = 145

Therefore, next three terms are 97, 121, 145.

NCERT Exercise 5.2 1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP (a) (b) (c) (d) (e)

(k) a , a2, a3, a4, ... It is not an AP because difference between consecutive terms is not equal. ⇒ a2 – a ≠ a3 – a2 (l)

8 , 18,

32 , ...

It is an AP because difference between consecutive terms is equal.

⇒ 2 2 –

d= 2

5th term = 4 2 +

2 =5 2

6th term = 5 2 +

2 =6 2

7th term = 6 2 +

2 =7 2

Therefore, next three terms are 5 2 , 6 2, 7 2.

(m)

2 = 3 2 –2 2 =

9 , 12 ,…

It is not an AP because difference between consecutive terms is not equal.

7 –18 ... –18.9 3.5

3 ... –3 2.5 0

8 10 18 ... 105

... 0 –5 3.6 ...

Sol. (a) a = 7, d = 3, n = 8, an = ?

For an AP

an =a + (n − 1)d

=7 + ( 8 − 1) 3 =7 + ( 7 ) 3 =7 + 21 =28

Hence, an = 28

(b) Given that a = −18, n = 10, an == 0, d ?

0= −18 + (10 − 1)d ⇒ 18 = 9d ⇒ d = 2

Hence, common difference, d = 2

−3, n = 18, an = −5 (c) Given d =

−5 = a + (18 − 1)( −3 )

−5 = a − 51 ⇒ a = 46

−18.9, d = 2.5, an == 3.6, n ? (d) a =

3.6 = −18.9 + (n − 1) 2.5

22.5 =

(n) 12, 32, 52, 72

It is not an AP because difference between consecutive terms is not equal.

n −1 = 9 ⇒ n = 10

a 3.5,= d 0,= n 105, = an ? (e)=

112

⇒ 6− 3≠ 9− 6

⇒ 32 − 12 ≠ 52 − 32

(n − 1) 2.5

an =3.5 + (105 − 1) 0 = 3.5

2. Choose the correct choice in the following and justify (a) 30th term of the AP: 10, 7, 4,..., is (i) 97

(ii) 77

(iii) −77

(iv) − 87

1 (b) 11th term of the AP: −3, − , 2,... is 2 (i) 28 (ii) 22 1 (iii) −38 (iv) −48 2 Sol. (a) Given

a 4 =a + ( 4 − 1)d

3= a + 3d

On subtracting (I) from (II),

We obtain −10 = 2d ⇒ d = −5

From equation (I), we obtain 13 = a + ( −5 ) ⇒ a = 18

a 3 = 18 + ( 3 − 1)( −5 ) = 18 + 2 ( −5 ) = 18 − 10 = 8

Therefore, the missing terms are 18 and 8 respectively.

AP 10, 7, 4, ...

7 − 10 = −3 a = 10, d =

a 30 = 10 + ( 30 − 1)( −3 )

a 30 = 10 + ( 29 )( −3 ) = 10 − 87 = −77

Hence, the correct answer is (iii).

(b) Given

1 AP −3, − , 2,... 2 1 5 1 a = −3 , d =− − ( − 3 ) =− + 3 = 2 2 2 5 5 a11 =−3 + (11 − 1)   =−3 + (10 )   =−3 + 25 = 22 2 2 Hence, the answer is (ii).

3. In the following APs find the missing term in the boxes (a) 2,

, 26

(b)

, 13,

(d) –4, (e)

, , 38,

Sol. (a) 2,

, –22

, 26

For this AP, a = 2 and a 3 = 26

Thus, a 3 =2 + ( 3 − 1)d

26= 2 + 2d ⇒ 24 = 2d ⇒ d = 12

a 2 =2 + ( 2 − 1)12 =14

Therefore, 14 is the missing term.

(b)

, 13,

For this AP, a 2 = 13 and a 4 = 3

a 2 =a + ( 2 − 1)d

13= a + d (I)

Chapter 5 • Arithmetic Progressions

For this AP,

1 2

1 19 9= a = 5 , a= 4 2 2 a 4 =a + ( 4 − 1)d

19 19 3d = 5 + 3d ⇒ − 5 = 2 2 9 3 = 3d ⇒ d = 2 2 a 2 =a + d = 5 +

3 13 = 2 2

 3 5 + 2  = 8 a3 = a + 2d =  2

13 Therefore, the missing terms are and 2 8 respectively. (d) –4,

,3 1 ,9 2 , ,

... (II)

For this AP,

a = −4 and a6 = 6

a6 =a + ( 6 − 1)d

6 =− 4 + 5d ⇒ 10 = 5d ⇒ d = 2

a 2 =a + d =−4 + 2 =−2

a 3 =a + 2d =−4 +2 ( 2 ) =0

a 4 =a + 3d =−4 + 3 ( 2 ) =2

a5 =a + 4d =− 4 + 4 ( 2 ) =4

Therefore, the missing terms are −2, 0, 2, and 4 respectively. (e)

, 38,

For this AP,

a 2 = 38 and a6 = −22

, –22

a 2 =a + ( 2 − 1)d

113

38= a + d (1)

a6 =a + ( 6 − 1)d

−22 =a + 5d

On subtracting equation (1) from (2), we obtain

−22 − 38 = 4d ⇒ − 60 = 4d ⇒ d = −15

6. Check whether −150 is a term of the AP 11, 8, 5, 2 ... 11 and d =− 8 11 = −3 Sol. For this AP, a =

(2)

a= a 2 − d= 38 − ( −15 )= 53

a 3 =a + 2d =53 + 2 ( −15 ) =23

a 4 =a + 3d =53 + 3 ( −15 ) =8

a5 =+ a 4d = 53 + 4 ( −15 ) = −7

Therefore, the missing terms are 53, 23, 8 and –7 respectively. 4. Which term of the AP 3, 8, 13, 18… is 78? Sol. 3, 8, 13, 18 … For this AP, a = 3 and d = a 2 − a1 = 8 − 3 = 5

Let −150 be the n th term of this AP −150 = 11 + (n − 1)( −3 ) ⇒ −150 =11 − 3n + 3 −164 = −3n ⇒ n =

164 3

As, n is not an integer. Therefore, −150 is not a term of this AP. th

7. Find the 31st term of an AP whose 11 and the 16th term is 73. Sol. Given that, a11 = 38 and a16 = 73

a11 = a + (11 − 1)d ⇒ 38= a + 10d

a16 = a + (16 − 1)d ⇒ 73= a + 15d

75 = (n − 1) 5 ⇒ (n − 1) = 15 ⇒ n = 16

38 − 70 = a ⇒ a = −32

5. Find the number of terms in each of the following AP (a) 7, 13, 19, …., 205 1 (b) 18, 15 , 13, …., −47 2 Sol. (a) 7, 13, 19, …., 205

For this AP, a = 7 and d = a 2 − a1 = 13 − 7 = 6

Let n terms be there in this AP

an = 205

Therefore, 205 =7 + (n − 1) 6

198 =

Therefore, the given series has 34 terms in it.

(n − 1) 6 ⇒ 33= (n − 1) ⇒ n = 34

1 (b) 18, 15 , 13, …., −47 2

1 5 For this AP , a 18,= = d 15 − 18 = − 2 2 Let n terms be there in this AP.

Therefore, an = −47

 5 −47 = 18 + (n − 1)  −   2

114

 5 −65 = (n − 1)  −  ⇒ (n − 1) = 26 ⇒ n = 27  2 Therefore, the given AP has 27 terms in it.

(2)

On subtracting (1) from (2), we obtain

78 =3 + (n − 1) 5 Hence, 16th term of this AP is 78.

(1)

Similarly,

35 = 5d ⇒ d = 7

Let nth term of this AP is 78.

term is 38

From equation (1), 38 =a + 10 × ( 7 ) a 31 = a + ( 31 − 1)d = −32 + 30 ( 7 ) = −32 + 210 = 178 Hence, 31st term is 178. 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. Sol. Given, a 3 = 1 2 and a50 = 106

a 3 =a + ( 3 − 1)d ⇒12= a + 2d (I) Similarly,

a50 = a + ( 50 − 1)d ⇒106= a + 49d

(II)

On subtracting (I) from (II), we obtain 94 = 47d ⇒ d = 2

From equation (I), we obtain 12= a + 2 ( 2 ) ⇒ a = 12 − 4 = 8

a 29 = a + ( 29 − 1)d= 8 + ( 28 ) 2 = 64 Therefore, 29th term is 64.

9. If the 3rd and the 9th term of an AP are 4 and −8 respectively. Which term of this AP is zero? Sol. Given that, a 3 = 4 and a9 = −8 a 3 = a + ( 3 − 1)d ⇒ 4= a + 2d (I)

a9 =a + ( 9 − 1)d ⇒ −8 = a + 8d (II)

Subtracting equation (I) from (II), we obtain −12 = 6d ⇒ d = −2

Given, difference between

From equation (I), we get 4 =a + 2 ( −2 ) ⇒ 4= a − 4 ⇒ a = 8 Let n th term of this AP be zero.

0 =8 + (n − 1)( −2 ) ⇒ 0 =8 − 2n + 2 2n = 10 ⇒ n = 5

Hence, 5th term of this AP is 0.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. Sol. For an AP, an =a + (n − 1)d a17 = a + (17 − 1)d ⇒ a17= a + 16d Similarly, a10= a + 9d It is given that a17 − a10 =7 ⇒ ( a + 16d ) − ( a + 9d ) =7 7d = 7 ⇒ d = 1 Therefore, the common difference is 1. 11. Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term? Sol. Given: 3, 15, 27, 39, … a = 3, d = a 2 − a1 = 15 − 3 = 12 a54 = a + ( 54 − 1)d = 3 + ( 53 )(12 ) = 3 + 636 = 639 Now a54 +132 = 639 + 132 = 771 Now finding the term of this AP which is 771. Let n th term be 771. 771 =3 + (n − 1)12 ⇒ 768 =

a1000 = a 2 + (1000 − 1) d = a 2 + 999d

(n − 1)12

64 ⇒ n = 65 (n − 1) =

Therefore, 65th term was 132 more than 54th term. 12. Two APs have the same common difference. The difference between their 100th term is 100, What is the difference between their 1000th terms? Sol. Let first term of both these APs be a1 and a 2 respectively and common difference of these APs be d . For 1st AP, a100 = a1 + (100 − 1) d = a1 + 99d a1000 = a1 + (1000 − 1) d = a1 + 999d For 2nd AP, a100 = a 2 +(100 − 1) d = a 2 + 99 d

Chapter 5 • Arithmetic Progressions

100th term of these APs = 100 Therefore, (a1 + 99d ) − (a 2 + 99d ) = 100 ⇒ a1 − a 2 = 100

(1)

Difference between 1000th terms of these APs (a1 + 999d ) − (a 2 + 999d ) = a1 − a 2 From equation (1), the difference, a1 − a 2 = 100 Hence, difference between 1000th terms of these AP will be 100. 13. How many three digits numbers are divisible by 7? Sol. 3-digit number divisible by 7 are 105, 112, 119, … All these are terms of an AP, a = 105. The maximum possible 3-digit number is 999. When we divide it by 7, remainder will be 5. d=7 So, 999 − 5 = 994 is the maximum possible 3-digit number divisible by 7. The series follows. 105, 112, 119, …, 994 Let 994 be the n

term of AP

= a 105, = d 7 and = an 994. = n ? 994 = 105 + (n − 1) 7 ⇒ 889 =

(n − 1) 7

127 ⇒ n = 128 (n − 1) =

Therefore, 128 three-digit numbers are divisible by 7. 14. How many multiples of 4 lie between 10 and 250? Sol. First multiple of 4 that is greater than 10 is 12. Therefore, 12, 16, 20, 24, … All are divisible by 4 and thus, are terms of an AP with a = 12 and d = 4. When we divide 250 by 4, remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4. The series is as follows. 12, 16, 20, 24, …, 248 Let 248 be the n th term of this AP. = a 12, = d 4,= an 248 236 248 = 12 + (n − 1) 4 ⇒ = n −1 4 59= n − 1 ⇒ n = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

115

15. For what value of n, are the n th terms of two APs: 63, 65, 67, ... and 3, 10, 17,… equal? Sol. a = 63 and d = a 2 − a1 = 65 − 63 = 2 an = 63 + (n − 1) 2 = 63 + 2n − = 2 61 + 2n

(1)

For AP: 3, 10, 17, …

Given, a 4 + a8 = 24 3 a + d + a + 7d = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 (1) a6 + a10 = 44

5 a + d + a + 9d = 44

a = 3 and d = a 2 − a1 = 10 − 3 = 7

7 7n − 4 (2) an =3 + (n − 1) 7 =3 + 7n −= It is given, n th term of these APs are equal to each other.

2a + 14d = 44 ⇒ a + 7d = 22

(2)

Subtracting equation (1) from (2), we get 2d = 22 − 12 ⇒ 2d = 10 ⇒ d = 5 From equation (1), we get

Equating both equations,

5 a+ d= 12

We get, 61 + 2n = 7n − 4 61 + 4 = 5n ⇒ 5n = 65 ⇒ n = 13

Therefore, 13th terms of both these APs are equal to each other.

a + 5( 5) = 12 ⇒ a + 25 =⇒ 12 a = −13 a 2 =a + d =−13 + 5 =−8

a 3 = a 2 + d =− 8 + 5 =−3 Therefore, the first three terms of this AP are −13, −8, and −3.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. Sol. a 3 = 16

a + ( 3 − 1)d = 16 ⇒ a + 2d = 16

a10= a + 9d

(1)

a7 − a5 = 12

a + ( 7 − 1)d ]−[ a + ( 5 − 1)d  = 12 12 ⇒ 2d = 12 ⇒ d = 6 (a + 6d ) − (a + 4d ) =

From equation (1), we get, a + 2 ( 6 ) = 16 a + 12 =⇒ 16 a = 4 Therefore, AP will be 4, 10, 16, 22, …

19. Subba rao started work in 1995 at an annual salary of `5000 and received an increment of `200 each year. In which year did his income reach `7000? Sol. It is observed the incomes that Subba Rao has obtained in various years are in AP as every year, his salary is increased by `200. Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, … Here, = a 5000 = and d 200

17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Let after nth year, his salary be `7000.

Sol. Given AP can be written in reverse order as

200 (n − 1) = 2000 ⇒ (n − 1) = 10 ⇒ n = 11

253, 248, 243, …, 13, 8, 5 Where, a = 253, d = 248 − 253 = −5, n = 20

Thus, 7000= 5000 + (n − 1) 200

Therefore, in 11th year, his salary will be `7000.

a 20 = a + ( 20 − 1)d

20. Ramkali saved `5 in the first week if a year and then increased her weekly savings by `1.75. If in the nth week, her weekly savings become `20.75, find n.

Therefore, 20th term from the last term is 158.

a 5,= d 1.75 and = an 20.75.= n ? Sol. Given,=

a 20 =253 + (19 )( −5 ) = 253 − 95 = 158 18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Thus,

Sol. a 4 =a + ( 4 − 1)d = a + 3d

1) (n −=

Similarly, a8 = a + 7d a6 = a + 5d

116

20.75 =5 + (n − 1)1.75 15.75 1575 63 = = = 9 1.75 175 7

n −1 = 9 ⇒ n = 10 Hence, n = 10.

NCERT Exercise 5.3 1. Find the sum of the following APs. (a) 2, 7, 12, …, to 10 terms (b) −37, −33, − 29, …, to 12 terms (c) 0.6, 1.7, 2.8,..., to 100 terms (d)

1 1 1 , , ,…,to 11 terms 15 12 10

So, Finding nth term of this AP by using formula,

[7 + (n – 1)(3.5)] = 84 ⇒ 7 + ( 3.5 )n − 3.5 = 84 ⇒ 3.5n =84 + 3.5 − 7 ⇒ 3.5n = 80.5 ⇒ n = 23 Therefore, there are 23 terms in the given AP.

We know, S= n

We know, n Sn = (a + l ) to find sum of n terms of AP, 2 23 = S23 ( 7 + 84 ) = 1046.5 2 (b) 34 + 32 + 30 +…+ 10

a = 34, d = 32 – 34 = –2, l = 10

Sol. (a) 2, 7, 12, …, to 10 terms

a = 2 , d = 7 − 2 = 5, n = 10 n 2a + (n − 1)d  2 10  4 + (10 − 1) 5 ⇒ Sn= 2  ⇒ 5 ( 4 + 45 ) =5 × 49 =245

(b) −37, − 33, − 29,…, to 12 terms

a = −37, d =−33 − ( −37 ) =4, n = 12

⇒ Sn =

12  −74 + ( 12 –1 ) 4  2  ⇒ 6(–74 + 44) = 6 × (–30) = –180

a = 0.6, d = 1.7 − 0.6 = 1.1, n = 100

= ⇒ Sn (d)

100 1.2 + ( 100 –1 )1.1 2 

⇒ 50 (1.2 + 108.9 ) =× 50 109.5 = 5505 1 1 1 , , , …, to 11 terms 15 12 10 1 1 1 1 = a = ,d = − = , n = 11 15 12 15 60 11  2 1  + ( 11 − 1)  2  15 60 

⇒ S= n

11  2 1  11  4 + 5  11 9 33 ⇒ + ⇒ ⇒ × ⇒ 2  15 6  2  30  2 30 20

2. Find the sums given below: 1 (a) 7 + 10 + 14 +…+84 2 (b) 34 + 32 + 30 +…+ 10 (c) −5 + ( − 8 ) + ( −11) +…+ ( −230 ) 1 + 14 +…+ 84 2 21 21 − 14 7 a = 7, d = −7 = = = 3.5, l = 84 2 2 2 We are not aware of the number terms in the given AP. Sol. (a) 7 + 10

Chapter 5 • Arithmetic Progressions

We are not aware of the number of terms in the given AP. So, Finding nth term of this AP by using formula, 34 + (n − 1)( −2 )  = 10 ⇒ 34 − 2n + 2 = 10 ⇒ −2n = –26 ⇒ n = 13 Therefore, there are 13 terms in the given AP. We know, n Sn = (a + l ) to find sum of n terms of AP, 2 13 = S13 ( 34 + 10 ) = 286 2 (c) −5 + ( − 8 ) + ( −11) +…+ ( −230 ) a = −5, d =−8 − ( −5 ) =−8 + 5 =−3, l = −230 We are not aware of the number terms in the given AP. So, Finding nth term of this AP by using formula,  −5 + (n − 1)( −3 )  =−230 ⇒ −5 − 3n + 3 =−230 ⇒ −3n = −228 ⇒ n = 76 Therefore, there are 76 terms in the given AP. We know, n (a + l ) to find sum of n terms of AP, 2 76 S76= ( −5 − 230 ) = −8930 2 3. In an AP: Sn =

(a) Given a = 5, d = 3, an = 50, Find n and Sn (b) Given a =7, a13 = 35, Find d and S13 (c) Given a12 = 37, d = 3 , Find a and S12 117

(d) Given a 3 = 15, S10 = 125, Find d and a10.

(e) Given d = 5, S9 = 75, Find a and a9.

(e) Given d = 5, S9 = 75, Find a and a9. (f) Given a = 2 , d = 8, Sn = 90, Find n and an

9 9 ⇒ S9 =[2a + (9 – 1)5] ⇒ 75 = [ 2a + 40] 2 2

(g) Given a = 8, an = 62, Sn = 210, Find n and d .

⇒ 150 = 18a + 360 ⇒ −210 = 18a ⇒ a = −

(h) Given an = 4, d = 2, Sn = –14, Find n and a. a 3,= n 8,= S 192 , Find d . (i) Given=

(j) Given l = 28, S = 144, and there are total 9 terms. Find a.

Sol. (a) Given= a 5,= d 3, an = 50, Find n and Sn .

an =5 + (n − 1) 3 ⇒ 50 =5 + 3n − 3

⇒ 48 = 3n ⇒ n = 16

16 [10 + (16 – 1 )3] ⇒ 8 × 55 = 440 2 Therefore, n = 16 and Sn = 440 Now, S16 =

(b) Given a = 7, a13 = 35, Find d and S13

a13 = 7 + (13 – 1) d ⇒ 35= 7 + 12d 7 ⇒ 28 = 12d ⇒ d = 3 13 7 13 ⇒ S13 =[14 + (13 – 1) ] ⇒ × 42 = 273 2 3 2 7 Therefore, d = and S13 = 273 3

a12 = a + (12 − 1) 3 ⇒ 37= a + 33 ⇒ a = 4

(f) Given= a 2,= d 8 ,= Sn 90, Find n and an n n [4 + (n − 1)8] ⇒ 90 = [ 4 + 8n − 8] 2 2 n ⇒ 90= [8n − 4 ] ⇒ 8n 2 − 4n − 180 = 0 2 ⇒ 2n 2 − n − 45 = 0 ⇒ 2n 2 − 10n + 9n − 45 = 0

90 =

⇒ 2n (n − 5 ) + 9 (n − 5 ) = 0 ⇒ (n − 5 )( 2n + 9 ) = 0 9 2 We discard negative value of n because n cannot be in negative or fraction. It must be a positive integer.

⇒ n = 5 ,−

Therefore, n = 5 and an = 34

(g) Given a = 8, an = 62, Sn = 210, Find n and d

62 = 8 + (n − 1)d = 8 + nd − d

= 62 = 8 + nd − d = nd − d = 54 = nd = 54 + d (1) n n 210 = [16 + ( n – 1) d ] = [16 + nd − d ] 2 2 Putting equation (1) in above equation, n n 210 = [16 + 54 + d − d ] = × 70 2 2 210 × 2 = ⇒ n = 6 70 Putting value of n in equation (1) 54 6d = 54 + d ⇒ d = 5 54 Therefore, n = 6 and d = 5

12 ⇒ S12 =[8 + (12 – 1 )3] ⇒ 6(8 + 33) 2 ⇒ 6 × 41 = 246

Therefore, a = 4 and S12 = 246

(d) Given a 3 = 15, S10 = 125, Find d and a10 .

35 2 35 35 −35 + 120 85 a9 = − + ( 9 − 1)( 5 ) = − + 40 = = 3 3 3 3 35 85 Therefore, a = − and a9 = 3 3

a 3 =a + ( 3 − 1)d ⇒ 15= a + 2d ⇒ a = 15 − 2d (1)

10 ⇒ S10 =[ 2a + (10 − 1)d ] 2

⇒ 125 = 5 ( 2a + 9d ) = 10a + 45d

Put (1) in above equation,

125 = 5 2 (15 − 2d ) + 9d 

⇒ 125 = 150 + 25d ⇒ 125 – 150 = 25d

(h)

Given an = 4, d = 2, Sn = −14 , Find n and a.

⇒ −25 = 25d ⇒ d = −1

4 = a + (n − 1) 2 = a + 2n − 2

⇒ 4 =a + 2n − 2 ⇒ 6= a + 2n ⇒ a= 6 − 2n (1) n n −14 =[2a + (n – 1)2] = [ 2a + 2n − 2] 2 2

a10 = a + (10 − 1)d

Put value of d and equation (1) in above equation, a10 = 15 − 2d + 9d = 15 + 7d = 15 + 7(−1) = 15 − 7 = 8

118

Therefore, d = −1 and a10 = 8

n [2a + 2n − 2] 2 Putting equation (1) in above equation,

−28 = n 2 ( 6 − 2n ) + 2n − 2 

⇒ −14 =

⇒ −28= n (12 − 4n + 2n − 2 ) ⇒ −28= n (10 − 2n )

As we know,

⇒ 2n 2 − 10n − 28 = 0 0 ⇒ n 2 − 5n − 14 =

Sn =

⇒ n 2 − 7n + 2n − 14 = 0 0 ⇒ n (n − 7 ) + 2 (n − 7 ) =

⇒ (n + 2 )(n − 7 ) = 0 ⇒ n = −2, 7

We discard negative value of n because n cannot be in negative or fraction. It must be a positive integer.

Which means n = 7.

Putting value of n in equation (1),

a ⇒ 6 − 2n = 6 − 2 ( 7 ) = 6 − 14 = −8

Therefore, n = 7 and a = −8

a 3,= n 8,= S 192 , Find d . (i) Given=

8 6 + ( 8 − 1)d  = 4 ( 6 + 7d ) 192= 2 ⇒ 192 = 24 + 28d ⇒ 168 = 28d ⇒ d = 6

(j) Given l = 28 , S = 144 , and there are total 9 terms. Find a . 9 [a + 28] 2

144 =

⇒ 288 = 9[a + 28] ⇒ 32= a + 28 ⇒ a = 4

4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636? Sol. a = 9, d = 17 − 9 = 8, Sn = 636 n 18 + (n − 1)( 8 )  ⇒ 1272= n (18 + 8n − 8 ) 636 = 2 2 0 ⇒ 1272 = 18n + 8n 2 − 8n ⇒ 8n + 10n − 1272 = 2

⇒ 4n + 5n − 636 = 0 Using quadratic formula we have, −5 ± 5 − 4 ( 4 )( −636 ) −5 ± 25 + 10176 n= ⇒n = 8 8 2

96 106 106 −5 ± 101 ⇒n = ,− = 12, − ⇒n = 8 8 8 8 As, we discard negative values of n, because it cannot be negative it can only be positive. Therefore, n = 12. 12 terms of the given sequence make sum equal to 636. 5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. Sol. a = 5, Last term = l 45, = = Sn 400 Chapter 5 • Arithmetic Progressions

n [a + l ] is used to find sum of nth terms of AP. 2 n 400 n = 400 = 5 + 45] ⇒ ⇒n = 16 [ 2 50 2 16 10 + (16 − 1)d  ⇒ 400 = 8 (10 + 15d ) 400 = 2  320 8 ⇒ 400 =80 + 12d ⇒ 320 = 120d ⇒ d= = 120 3 6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Sol. a = 17, l = 350 and d = 9 350 = 17 + (n − 1)( 9 ) ⇒ 350 = 17n + 9n − 9 ⇒ 342 = 9n ⇒ n = 38 S38 =

38 34 + ( 34 − 1) 9 2 

⇒ S38= 19 ( 34 + 333 )= 6973 Therefore, there are 38 terms and their sum is equal to 6973. 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149? Sol. a 22 = 149 149 = a + ( 22 − 1)( 7 ) ⇒ 149 =+ a 147 ⇒ a = 2 22  4 + ( 22 − 1) 7  Now, S22= 2  ⇒ S22 = 11( 4 + 147 ) ⇒ S22 = 1661

Therefore, Sum of first 22nd terms of AP is equal to 1661. 8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively? Sol. It is given 2nd and 3rd terms of AP are 14 and 18 respectively. 14 =a + ( 2 − 1)d ⇒ 14 =a + d

(1)

And, 18 =a + ( 3 − 1)d ⇒ 18 =a + 2d

(2)

These equations consist of two variables. Using (1), we get, a = 14 − d Putting value of a in equation (2), 18 = 14 − d + 2d ⇒d = 4 119

Therefore, common difference, d = 4 Put value of d in (1), 18= a + 2 ( 4 ) ⇒a = 10 = S51

51 51 20 + ( 51 − 1)d=  ( 20 + 200 ) 2  2

51 = × 200 =51 × 110 =5610 2

Let us find difference between consecutive terms of sequence.

11 = − 7 4,

1= 5 − 11 4,

19= − 15 4

Therefore, difference between consecutive term is constant which means terms a1, a 2, … an from an AP.

We have sequence in form 7, 11, 15, 19…..

a = 7 and d = 4

Therefore, sum of first 51 terms of an AP is equal to 5610.

15 15 15 S15 = 14 + (15 − 1) 4  = (14 + 56 ) = × 70 2 2 2

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, Find the sum of first n numbers?

=15 × 35 = 525

Sol. It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289. 7 2a + ( 7 − 1)d  49 ⇒ 98= 7 ( 2a + 6d ) 49 = 2 ⇒ 7 = a + 3d ⇒ a = 7 − 3d (1) 17 2a + (17 − 1)d  ⇒ 578= 17 ( 2a + 16d ) = And, 289 2  ⇒ 34 = 2a + 16d ⇒ 17 =a + 8d Put (1) in the above equation, we get 17 = 7 − 3d + 8d ⇒ 10 = 5d ⇒ d = 2 Put d in (1), we get, a = 7 − 3d = 7 − 3 ( 2 ) = 7 − 6 = 1 n Sn = [ 2 (1) + (n − 1) 2 ] 2 n n ⇒ Sn =[ 2 + 2n − 2 ] ⇒ Sn = × 2n ⇒ Sn =n 2 2 2 Therefore, sum of n terms of AP is equal to n 2 . 10. Show that a1 , a 2, …. an , … from an AP where an is

Therefore, Sum of first 15 terms of an AP is equal to 525. (b) We need to show that a1 , a 2, … an from an AP where an = 9 − 5n Let us find values of a1, a 2 a1, a 3, … using an = 9 − 5n

a1 = 9 − 5 (1) = 9 − 5 = 4

a 2 =9 − 5 ( 2 ) =9 − 10 =−1

a 3 =9 − 5 ( 3 ) =9 − 15 =−6

a 4 =9 − 5 ( 4 ) =9 − 20 =−11

So, the sequence is of form 4, − 1, − 6, − 11…..

Let us find difference between consecutive terms of sequence.

−1 − ( 4 ) =−5,

−6 − (1) =−6 + 1 =−5,

11 − ( −6 ) =−5

Therefore, difference between consecutive term is constant which means terms a1 , a 2 , … an from an AP.

We have sequence in form 4, − 1, − 6, − 11…..

(a) an = 3 + 4n

a = 4 and d = −5

(b) an = 9 − 5n

Now, S15= 15 8 + (15 – 1)( −5 ) = 15 ( 8 − 70 )= 15 × ( −62 )

defined as below:

Also find the sum of first 15 terms in each case. Sol. (a) We need to show that a1, a 2, … an from an AP where an = 3 + 4n Let us find values of a1 , a 2 a1, a 3, … using an = 3 + 4n

a1 = 3 + 4 (1) = 3 + 4 = 7

a 2 = 3 + 4 ( 2 ) = 3 + 8 = 11

a 3 =3 + 4 ( 3 ) =3 + 12 =15

a 4 =3 + 4 ( 4 ) =3 + 16 =19

So, the sequence is of form 7, 11, 15, 19…..

120

⇒ 15x ( −31) = −465

Therefore, Sum of first 15 terms of an AP is equal to −465. 1 1. If the sum of the first n terms of an AP is 4n – n2, What is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, Find the 3rd, the 10th and the nth terms? Sol. It is given that sum of n terms of an AP is equal to (4n − n 2)

It means S= 4n − n 2 n Let us calculate S1 and S2 using S= 4n − n 2 n

a = 1 and d = 3 − 1 = 2, l = 49 We need to find n first.

S1 = 4 (1) − (1) = 4 − 1 = 3

49 =1 + (n − 1) 2 ⇒ 49 =+ 1 2n − 2

S2 = 4 ( 2 ) − ( 2 ) = 8 − 4 = 4

⇒ 50 = 2n ⇒ n = 25

First term= a= S1= 3 … (1)

Applying formula, Sn =

Let’s find common difference now. Any AP can be written in form of general terms like a , a + d , a + 2d … We have sum of first two terms equal to 4 which means S2 = 4 Therefore, We can say that a + ( a + d ) = 4 Put Value of a from (1), we get 2a + d = 4 ⇒ 2 ( 3) + d = 4 ⇒ 6 +d = 4 ⇒d = −2 2nd term of AP = a2 =a + ( 2 − 1)d =3 + ( 2 − 1)( −2 ) =3 − 2 =1 3rd term of AP = a3 =a + ( 3 − 1)d =3 + ( 3 − 1)( −2 ) =3 − 4 =−1 10 term of an AP = a10 = a + (10 − 1)d =+ 3 (10 − 1)( −2 ) th

= 3 − 18 = −15 nth term of an AP = an =a + (n − 1)d =3 + (n − 1)( −2 ) =3 − 2n + 2 ⇒ 5 − 2n 12. Find the sum of the first 40 positive integers divisible by 6. Sol. The positive integers that are divisible by 6 are 6, 12, 18, 24 … It is observed that these are making an AP a = 6 and d = 6 40 S 40 = [12 + (40– 1) 6] = 20 (12 + 39 × 6= ) 20 (12 + 234 ) 2 =20 × 246 =4920 13. Find the sum of the first 15 multiples of 8. Sol. The first 15 multiples of 8 are 8, 16, 24, 32… 15 times These are in an AP, a = 8 and d = 16 − 8 = 8, n = 15 S15 =

15 15 15 [16 + (15– 1) 8] = 8) (16 + 14 ×= (16 + 112 ) 2 2 2

15 = × 128 =15 × 64 = 960 2 14. Find the sum of the odd numbers between 0 and 50. Sol. The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49 Therefore, it can be observed that these odd numbers are in an AP as difference between consecutive terms is constant. Chapter 5 • Arithmetic Progressions

S25 =

n [(a + l ) 2

25 25 (1 + 49) = × 50 = 25 × 25 = 625 2 2

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: `200 for the first day, `250 for the second day, `300 for the third day, etc, the penalty for each succeeding day being `50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? Sol. It is observed penalties are in AP. a = 200 and d = 50 Penalty has to be paid if he delay the work by 30 days = S30 = S30

30 2 ( 200 ) + ( 30 − 1 = ) 50 15[ 400 + 1450] 2 

S30 = 15 (1850 ) = 27750

Therefore, Contractor has to pay `27750 as penalty. 16. A sum of `700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is `20 less than its preceding prize, find the value of each of the prizes. Sol. Let value of 1st prize = ` a Let value of 2nd prize = `(a − 20 ) Let value of 3rd prize = `(a − 40 ) We have sequence of the form: a , a − 20, a − 40, a − 60,… It is an AP. First term = a , common difference d= (a − 20 ) − a =−20

n = 7 (Because of total 7 prizes) S7 = `700 7 7 2a + ( 7 − 1)( −20 )  ⇒ 700 = S= [2a − 120] 7 2 2 ⇒ 200 =2a − 120 ⇒ 320 =2a ⇒ a =160 Value of 1st prize = `160 Value of 2nd prize = `160 – 20 = `140 Value of 3rd prize = `140 – 20 = `120 Value of 4th prize = `120 – 20 = `100 121

Value of 5th prize = `100 – 20 = `80

Sol. Length of semi circle

Value of 6th prize = `80 – 20 = `60

Circumference of circle 2π r = = = πr 2 2

Value of 7 prize = `60 – 20 = `40 th

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII . There are three sections of each class. How many trees will be planted by the students? Sol. The number of trees planted by class I = number of sections × 1 = 3 × 1 = 3 The number of trees planted by class II = number of sections × 2 = 3 × 2 = 6 The number of trees planted by class III = number of sections × 3 = 3 × 3 = 9

Length of semi circle of radii 0.5 cm = π ( 0.5 ) cm Length of semi circle of radii 1.0 cm = π (1.0 ) cm Length of semi circle of radii 1.5 cm = π (1.5 ) cm So, we have a sequence of the form: π ( 0.5 ) , π (1.0 ) , π (1.5 )...13 terms (There are total of 13 semi circles) To find total length of spiral, find sum of sequence. Total length of spiral = π ( 0.5 ) + π (1.0 ) + π (1.5 )….13 terms ⇒ Total length of spiral= π ( 0.5 + 1.0 + 1.5 )….13 terms (I) = a= 0.5, d = 1.0 − 0.5 = 0.5 , n = 13 S13 =

13 1 + (13 − 1) 0.5 = 6.5 (1 + 6 )= 6.5 × 7 = 45.5 2 

Therefore, We have sequence of the form 3, 6, 9, … 12 terms

Therefore, 0.5 + 1.0 + 1.5 + 2.0….13 terms = 45.5

To find total number of trees planted by all students, We need to find sum of sequence 3, 6, 9, 12...12 terms.

Total length of spiral = π ( 0.5 + 1.5 + 2.0 +…..13 terms )

First term= a= 3 Common difference = 9 = 6 − 3 = 3 12 n= 12 S12 = 6 + (12 − 1) 3 =6 ( 6 + 33 ) =6 × 39 =234 2 18. A spiral is made up of successive semi circles, with centres alternately at A and B, Starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5cm, 2.0 cm,… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? 22 (Take π = ) 7

[Hint: Length of successive semicircles is l1 , l2 , l3 , l 4 ,… With centres at A, B, A, B, …, respectively.]

= π= ( 45.5) 143 cm 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row,18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Sol. Number of logs in bottom row = 20 Number of log sin next row = 19 Number of logs in next to next row = 18 Sequenced form 20, 19, 18….

First term= a= 20

We need to find how many rows make total of 200 logs. n ⇒ 200= [40 + (n − 1)( −1) ⇒ 400 = n ( 40 − n + 1) 2 ⇒ 400 = 40n − n 2 + n ⇒ n 2 − 41n + 400 = 0

Common difference == d 19 − 20 = −1

l1 B l2 l4

122

Put this in (I), we get

It is a quadratic equation,

0 ⇒ n 2 − 25n − 16 n + 400 = 0 ⇒ n (n − 25 ) − 16 (n − 25 ) = ⇒ (n − 25 )(n − 16 ) ⇒ n = 25, 16

We cannot take n = 25 because we cannot have more than 20 rows in the sequence.

Sol. Distance of first potato from the starting point = 5 metres

At most, we can have 20 or less number of rows.

∴ Distance covered by competitor to pick 1st potato and put in bucket = 5 × 2 = 10 m

Therefore n = 16 , Which means 16 rows make total number of logs equal to 200. n ( a +l ) 2 = 200 8 ( 20 + l ) ⇒ 200 =160 + 8l Applying formula, Sn =

Distance of 2nd potato from the starting point = 5 + 3 = 8 metres ∴ Distance covered by competitor to pick 2nd potato and put in bucket = 8 × 2 = 16 metres

40 = 8l ⇒ l = 5

Therefore, There are 5 logs in top most row and there are total of 16 rows. 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).

Distance of 3rd potato from the starting point = 8 + 3 = 11 metres ∴ Distance covered by competitor to pick 3rd potato and put in bucket = 11 × 2 = 22 metres ∴ we have a sequence of form 10, 16, 22, …10 terms To calculate total distance covered, we need to find 10 + 16 + 22 +….10 terms First term= a= 10 Common difference =d =16 − 10 =6

10 n=

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

10 S10 = 20 + (10 − 1) 6  =5 ( 20 + 54 ) =5 × 74 =370 2

Therefore, total distance covered by competitor is equal to 370 metres.

Multiple Choice Questions 1. If the first three terms of an AP are 3p − 1,3p + 5 and 5p + 1 respectively, then the value of p is (CBSE 2024)

(a) 2

(b) –3

(d) 5

(b) −3b

(d) −3

3. The next 4th term of the AP 18, 50, 98 is (a) 128

(b) 140

(d)

200

4. If k , 2k − 1 and 2k + 1 are three consecutive terms of an AP, the value of k is (a) −2 (b) 3 (c) −3 (d) 6 5. The common difference of the AP whose nth term is given by a= 3n + 7 , is n (a) 7 (b) 3 (c) 3n (d) 1 Chapter 5 • Arithmetic Progressions

(a)

(b)

(c)

(d) 112

7. If the sum of the first n terms of an AP be 3n 2 + n and its common difference is 6, then its first term is (a) 2 (b) 3 (c) 1 (d) 4

2. The common difference of the AP 1 1 − 6b 1 − 12b , , , ... is 2b 2b 2b (a) 3b

6. The next term of the AP 7 , 28, 63,…

1 1 1 are in AP. Then, x = , , x +2 x +3 x +5 (a) 5 (b) 3 (c) 1

8. If

(d) 2

9. The first term of an AP is p and the common difference is q , then its 10th term is

(CBSE 2020)

(a) q + 9p

(b) p − 9q

(d) 2p + 9q

10. The sum of first 20 odd natural numbers is (a) 100 (b) 210 (c) 400 (d) 420

123

11. Which term of the AP 21, 42, 63, 84,… is 210?

(c)

(NCERT Exemplar)

(a) 9th

(b) 10th

(d) 12th

12. Two APs have same common difference. The first term of one of these is –1 and that of the other is –8. Then the difference between their 4th terms is (NCERT Exemplar, CBSE SQP 2024)

n (n + 1)

17. If 7 times the 7 th term of an AP is equal to 11 times i ts 11th term, then its 18 th term will be (a) 7

(b) 11

(b) –8

(a) 501th

(d) –9

(b) 502th

(d) none of these 19. The sum of first 16 terms of the AP 10, 6, 2,… is (NCERT Exemplar)

14. The 21st term of the AP whose first two terms are –3 and 4, is (NCERT Exemplar) (a) 17 (b) 137

(a) –320 20. If

(b) 320

(b) 4, 10, 16, 22

(d) –400

5 + 9 + 13 +  to n terms 17 = , then n = 7 + 9 + 11 +  to (n + 1) terms 16

(a) 8

(d) –143

15. If four numbers in AP are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are (a) 5, 10, 15, 20

(b) 7

(d) 11

21. Three numbers in AP have the sum 30. What is the middle term? (CBSE 2024) (a) 4 (b) 10 (c) 16 (d) 8

( )

22. The next 4th term of the AP 18, 50, 98,…. is

(CBSE 2022)

(a) 128

(d) none of these

(b) 140

16. Sum of n terms of the series

2 + 8 + 18 + 32 + is (a)

(d) 0

(d) 8

18. The 9 th term of an AP is 449 and 449 th term is 9. The term which is equal to zero is

(a) –1

13. The first and last terms of an AP are 1 and 11. If the sum of its terms is 36, then the number of terms will be (a) 5 (b) 6

(d) 1

n (n + 1)

(d)

(b) 2n (n + 1)

200

Answers 1. (d) 5

2. (d) −3

3. (c) 162

4. (b) 3

5. (b) 3

6. (d) 112

7. (d) 4

8. (c) 1

9. (c) p + 9q

10. (c) 400

11. (b) 10th

12. (c) 7

13. (b) 6

14. (b) 137

15. (a) 5, 10, 15, 20

124

16. (c)

n (n + 1)

2 19. (a) −320 22. (c) 162

17. (d) 0

18. (c) 458th

20. (b) 7

21. (b) 10

Constructed Response Questions Very Short Answer Questions 1. In an AP, if the common difference d is −4 and the seventh term ( a7 ) is 44, then find the first term. Sol. Given: a7 = 44 Common difference d = −4

6. Find the n th term of the AP a, 3a, 5a, ...

(CBSE SQP 2021)

Sol. Given: a, 3a, 5a,… First term: a1 = a Common difference: d = 3a − a = 2a The formula for the nth term of an AP is:

Using the formula for the 7th term:

an =a + (n − 1)d

44 =a + ( 7 − 1)( −4 )

an = a + 2a (n − 1) = a + 2an − 2a = 2an − a

44 =a + ( 7 − 1)( −4 )

So, an = a + (n − 1) ⋅ 2a

44= a − 24

= an a ( 2n − 1)

68 a=

7. Determine k so that k 2 + 4k + 8, 2k 2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP. Sol. For three terms to be in AP, the second term must be the average of the first and third terms:

2. Find the value of a 25 − a15 for the AP: 6, 9, 12, 15, ... Sol. Given: First term ( a ) = 6 Common difference ( d ) = 3 a 25 = a + 24d = 6 + 24 ( 3 ) = 78 a15 =+ a 14d = 6 + 14 ( 3 ) = 48 a 25 − a15 = 78 − 48 = 30 3. In an Arithmetic Progression, if d = −4, n = 7, an = 4, then find a. Sol. Given: d = −4 , d = −4, a7 = 4 Using the formula:

4 =a + ( 7 − 1)( −4 ) ⇒ 4 =a + 6 ( −4 ) 4= a − 24 ⇒ a = 28

4. How many natural numbers are there between 1 and 1000 which are divisible by 5 but not by 2?

Sol. a1 = 5, a2 = 15, a3 = 25,...., these numbers form an AP where a = 5 and d = 10. an = 995 995 = 5 + (n − 1)10 → n − 1 = 99 → n = 100 5. If the common difference of an AP is 5, then what is a18 − a13. (NCERT Exemplar) Sol. We need to find a18 − a13

a18 − a13 = a + (18 − 1)d ]−[ a + (13 − 1)d  = ( a + 17d ) − ( a + 12d ) = 17d − 12d = 5d ⇒ = 5 × 5 = 25

Chapter 5 • Arithmetic Progressions

(

) (

2 2k 2 + 3k + 6 = k 2 + 4k + 8 + 3k 2 + 4k + 4 2

)

4k + 6k + 12 = 4k + 8k + 12 0 = 2k → k = 0 8. Which term of the following AP 27, 24, 21,… is zero?

Sol. Given 27, 24, 21,...

(CBSE SQP 2020)

First term: a = 27 Common difference: d = 24 − 27 = −3 So, an = 0 0 = 27 + (n − 1)( −3 ) 0 =27 − 3 (n − 1) n − 1=

27 = 9 ⇒ n = 10 3

9. If the ratio of sum of the first m and n terms of an AP is m 2 : n 2, show that the ratio of its mth and nth terms is ( 2m − 1) : ( 2n − 1) . m 2a + (m − 1)d  m 2a + (m − 1)d 2 ⇒ = n n 2a + (n − 1)d 2a + (n − 1)d  2 ⇒ 2an + mnd − dn = 2am + mnd − md Sm m 2 Sol. = = Sn n 2

⇒ 2a(m − n) − d(m − n) = 0 ⇒ (2a − d)(m − n) = 0 ⇒ 2a = d

...(i)

am a + (m − 1)d a + 2 (m − 1) a 2m − 1 = = = ( From ( i ) ) an a + (n − 1)d a + 2 (n − 1) a 2n − 1 Hence proved.

125

10. If Sn , the sum of the first n terms of an AP is given 2

by = Sn 2n + n , then find its n term. th

(CBSE 2019)

Sol. Substituting n − 1 in the given formula:

(

)

Sn −1 = 2(n − 1) + (n − 1= ) 2 n − 2n + 1 + n − 1 2

Sn −1 = 2n − 3n + 1 Now, a= Sn − Sn −1= n 2

( 2n + n ) − ( 2n − 3n + 1) 2

an= 2n + n − 2n + 3n −= 1 4n − 1 11. Find the common difference of the AP 1 1 − p 1 − 2p , , , ... p p p Sol. Given

1 1 − p 1 − 2p , ... , , p p p

(CBSE 2020, 2013)

∴n th term, an =a + (n − 1)d =3 + (n − 1) × 2 =2n + 1 ⇒ an = 2n + 1

Short Answer Questions 1. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. Sol. Given, a 3 = 12 and a50 = 106

a 3 =a + ( 3 − 1) d ⇒ 12= a + 2d (I) Similarly, a50 = a + ( 50 − 1) d ⇒ 106= a + 49d (II)

1−p 1 = − d p p

On subtracting (I) from (II), we obtain 106 − 12 = a + 49d − a − 2d ⇒ 94 = 47d ⇒ d = 2

(1 − p ) − 1 −b d= = = −1 p p

From equation (I), we obtain

12. Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12. (CBSE 2019)

16 (1) Sol. a 3 = 16 ⇒ a + ( 3 − 1)d = 16 ⇒ a + 2d =

Using question statement a7 − a5 = 12

12= a + 2 ( 2 ) ⇒ a = 12 − 4 = 8 a 29 = a + ( 29 − 1)d= 8 + ( 28 ) 2 = 64 Therefore, 29th term is 64. 2. If p th term of an AP is q and q th term is p, then prove that its n th term is ( p + q − n ) . (CBSE 2023)

a + ( 7 − 1)d  − a + ( 5 − 1)d  = 12

Sol. Let the first term of AP be a and common difference be d.

6d − 4d = 12 ⇒ d = 6

Substituting d = 6 into Equation 1:

AP = a, a + d, a + 2d, ... ⇒ an = a + (n − 1) d

a + 2(6 ) = 16 ⇒ a = 4

ap = a + (p − 1)d = q ⇒ a + pd − d = q (1)

So, the AP is 4, 10, 16, 22, 28, 34, 40, … 13. If the 17 term of an AP exceeds its 10 term by 7, find the common difference. (CBSE 2019) th

3 + 9d = 21 ⇒ 9d = 18 ⇒ d = 2

Sol. a17 = a + (17 − 1)d = a + 16d a10 = a + (10 − 1)d = a + 9d Given: a17 − a10 = 7 ⇒ a + 16d − a − 9d = 7 7d = 7 ⇒ d = 1

14. If the 10th term of an AP is 21 and the sum of its first ten terms is 120, find its n th term.

Sol. Given: a10 = 21 ⇒ a + 9d = 21 (i)

aq = a + (q − 1)d = q ⇒ a + qd − d = p (2) Subtracting 1 from 2, we get, a + qd − d − a − pd + d = p − q ⇒ d(q − p) = −1(q − p) ⇒ d = −1

Using equation 1 and value of d, we get, a + p(−1) + 1 = q ⇒ a + 1 = p + q (3) an = a + (n − 1)d = a + (n − 1)(−1) an = (a + 1) − n ⇒ p + q − n (Using 3) Hence proved.

10 24 ( 2a + 9d ) = 120 ⇒ 2a + 9d = 2 ⇒ a + ( a + 9d ) = 24 ⇒ a + 21 = 24 ( From ( i ) )

Sol. The first 8 multiples of 3 are:

⇒a = 3

3, 6, 9, 12, 15, 18, 21, 24

Also, S10 = 120 ⇒

From (i), we get,

126

3. Find the sum of first 8 multiples of 3. (CBSE 2018)

This forms an Arithmetic Progression (AP) where:

3 a = , d = 3, n = 8 Sum of 8 terms: 8 S8 = × 2 ( 3 ) + ( 8 − 1)( 3 )  =4 × [6 + 7 × 3] =4 × [6 + 21] 2 =4 × 27 =108

4. If the 2nd term of an AP is 13 and 5th term is 25, then what is its 7th term? (NCERT Exemplar)

Sol. Using a 2 = 13: a + ( 2 − 1)d = 13 ⇒ a + d = 13 (1)

Using a5 = 25: a + ( 5 − 1)d = 25 ⇒ a + 4d = 25 (2) Subtract Equation 1 from Equation 2:

(a + 4d ) − (a + d ) = 25 − 13

a + 4 = 13 ⇒ a = 9 Using the formula: 4. Which term of the arithmetic progression 3, 15, 27, 39 .... will be 120 more than its 21st term? (CBSE 2019)

Sol. The AP is: 3, 15, 27, 39, … Where, a = 3 , d = 15 − 3 = 12 a 21 = 3 + ( 21 − 1) × 12 =3 + 20 × 12 = 3 + 240 = 243

360 = (n − 1) × 12

6. How many terms are there in an AP whose first and fifth terms are -14 and 2, respectively, and the last term is 62? Sol. Given: a = −14 , a5 = 2 , l = 62

(a5 = 2 ): 2 = −14 + ( 5 − 1)d 16 = 4 4 62 = −14 + (n − 1) × 4 2= −14 + 4d ⇒= d

Sol. Given:

a7 =a + ( 7 − 1)d = 9 + 6 ( 4 ) = 33

363 = 3 + (n − 1) × 12

a11 = a + (11 − 1)d = 6 + 10 ( 4 ) = 46

76 76 = (n − 1) × 4 ⇒ n − 1 = ⇒ n = 20 4 7. If m times the mᵗʰ term of an AP is equal to n times its nᵗʰ term, show that the (m + n)ᵗʰ term of the AP is zero.

Now, substitute d = 4 into Equation 1:

Using the AP formula:

a + 3( 4 ) = 18 ⇒ a = 6

62 + 14 = (n − 1) × 4

a + 4d − a − d = 12 ⇒ 3d = 12 ⇒ d = 4

= 243 + 120 = 363 So, a= a 21 + 120 n

92 = 4 23 Substituting d = 4 in Equation 2: 23d = 92 ⇒= d

m ⋅ a + (m − 1)d  =n ⋅ a + (n − 1)d  ma + m (m − 1)d =na + n (n − 1)d ma − na = n (n − 1)d − m (m − 1)d a (m −= n ) d n (n − 1) − m (m − 1)  The (m + n)ᵗʰ term is: am +n = a + (m + n − 1)d From the previous equation: d n (n − 1) − m (m − 1)  a=  m −n Substituting this in am +n :

360 d n (n − 1) − m (m − 1)  n − 1 = ⇒ n = 31 = am +n + (m + n − 1)d 12 m −n 5. The sum of the first 30 terms of an AP is 1920. If the  n (n − 1) − m (m − 1) + (m + n − 1)(m − n )  fourth term is 18, find its 11th term. am +n = d    m −n (CBSE 2020)   30 2 2 2 2  n − n − m + m + m + mn − m − mn − n + n  Sol. For S30 = 1920 : × 2a + ( 30 − 1)d  = 1920 am +n = d   2   m −n   15 × [ 2a + 29d ] = 1920 am +n = 0 1920 2a + 29d = ⇒ 2a + 29d = (1) 128 Thus, the (m + n)ᵗʰ term of the AP is zero. 15 For a 4 = 18 : a + ( 4 − 1)d = 18 ⇒ a + 3d = 18 (2) 8. The first term of an AP is 5, the last term is 45, and the sum of all the terms is 400. Find the number of Multiply Equation 2 by 2: terms and the common difference of the AP. 2a + 6d = 36 Sol. Given: Subtract from Equation 1: n n 400 = × ( 5 + 45 ) ⇒ 400= × 50 ⇒ 800 = 50n ( 2a + 29d ) − ( 2a + 6d ) = 128 − 36 2 2 Chapter 5 • Arithmetic Progressions

127

800 = n = 16 50

Now, First term + Last term

So, the number of terms n = 16.

Thus, twice the middle term is equal to the sum of the first and last term.

=+ 2 302 = 304 =× 2 152 = 2a51

The formula for the last term in an AP is: l =a + (n − 1)d

11. The sum of three numbers in AP is -3, and their product is 8. Find the numbers.

Substituting the known values: 45 =+ 5 (16 − 1)d ⇒ 45= 5 + 15d ⇒ 40 = 15d 40 8 = d = 15 3

9. Show that the sum of all terms of an AP whose first term is a , the second term is b and the last term is c (a + c )(b + c − 2a ) is equal to . 2 (b − a )

Sol. Let the numbers be ( a − d ) , a , ( a + d ). Given: ( a − d ) + a + ( a + d ) =−3 ⇒ 3a =−3 ⇒ a =−1 It is also given that the product of the numbers is 8. ∴ ( a − d )( a )( a + d ) = 8

(

)

⇒ a a2 −d2 = 8

(

⇒ ( −1) 1 − d

) =8 ⇒ d =9 ⇒ d =±3 2

Sol. Given, first term = a , second term = b .

If d = 3 , the numbers are −4, − 1, 2.

d= b − a and an = c

If d = −3, the numbers are 2, − 1, − 4 .

⇒ c = a + (n − 1)(b − a )

Long Answer Questions

⇒c −a =

(n − 1)(b − a )

1. The ratio of the 11th term to 17th term of an AP is 3 : 4. Find the ratio of 5th term to 21st term of the same AP Also, find the ratio of the sum of first 5 terms to that of first 21 terms. (CBSE 2023)

c −a ⇒ n −1 = b −a ⇒= n

c −a +1 b −a

= ⇒n

c − a + b − a b + c − 2a = b −a b −a

CBSE Marking Scheme 2023 Sol. Given,

∴ Sum of all terms of an AP.

a + 10d 3 = 1 a + 16d 4

⇒ 4a + 40d = 3a + 48d ⇒ a = 8d

 b + c − 2a  n Sn=  ( a + c ) (a + l =)  2  2 (b − a )  Hence,

(i) 1

a5 a + 4d 8d + 4d Therefore = = (Using (i)) a 21 a + 20d 8d + 20d

∴a5 : a 21 = 3:7

(a + c )(b + c − 2a ) Sn = 2 (b − a )

5 ( 2a + 4d ) 5 (16d + 4d ) S5 2 = 1 = 10. Consider the AP 2, 5, 8, 11, …, 302 . Show that twice of S21 21 2a + 20d ( ) 21 (16d + 20d ) 2 the middle term of the above AP is equal to the sum 5 × 20d 25 of its first and last term. = = ( From ( i ) ) 1 × d 21 36 189 Sol. Clearly, 2, 5, 8, 11, …, 302 is an AP with a = 2 and Therefore, S5 : S21 = 25 :189 d = 3.

Let there be n terms in the given AP. Then, n th term= 302 ⇒ a + (n − 1)d= 302 ⇒ 2 + 3 (n − 1)= 302 ⇒ 3n= 303 ⇒ n= 101 th

 n +1  st Clearly, n is odd. Therefore,   i.e. 51 term 2   is the middle term and is given by a51 = a + 50d = 2 + 50 × 3 = 152

128

2. The angles of a triangle are in AP The greatest angle is twice the least. Find all the angles. (NCERT Exemplar)

Sol. Let the angles be x, x + d, x + 2d We are given that the greatest angle is twice the least, so: x + 2d = 2x (1)

Also, the sum of all angles of a triangle is 180°, so x + ( x + d ) + ( x + 2d ) = 180 (2) From equation 1: x + d = 2x ⇒ 2d = x ⇒ d = 2 Now substitute d =

x 2

x into equation 2: 2

x 8x + x 9x = 180 ⇒ = 180 = 180 ⇒ 2 2 2

9x= 360 ⇒ x= 40

T10

–3

–9

Here it is clear that T1 + T2 +…+ T6 = 180 and T1 + T2 +…+ T10 = 180. 4. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X . Find value of X . Sol. The numbers of houses are 1, 2, 3, 4...49. They are in AP, where a = 1 and d = 1 .

Least angle x = 40 

Greatest angle: x + 2d = 40 + 40 = 80 3. How many terms of the arithmetic progression 45, 39, 33, must be taken so that their sum is 180? Explain the double answer. Sol. Let n terms of the AP 45, 39, 33... must be taken so that their sum is 180. 45, d = 39 − 45 = −6 We have, a = Sn = 180 n ⇒ 2 × 45 + (n − 1) × ( −6 )  =180 2 180 ⇒ n  45 − (n − 1) × 3 = ⇒ n ( 45 − 3n + 3 ) = 180 ⇒ n ( 48 − 3n ) = 180 ⇒ n (16 − n ) = 60 ⇒ 16n − n 2 = 60 ⇒ n 2 − 16n + 60 = 0 ⇒ n 2 − 10n − 6n + 60 = 0 ⇒ n (n − 10 ) − 6 (n − 10 ) = 0 ⇒ (n − 10 )(n − 6 ) = 0 ⇒ = n 10 or = n 6 Hence, 10 terms or 6 terms should be taken to get sum 180. Now, the explanation of double answer.

Chapter 5 • Arithmetic Progressions

Hence both answers are possible.

x  x +  x +  + (x + x ) = 180 2 

Middle angle x + d = 40 + 20 = 60

Sum = 0

x  x  x + x +  + x + 2⋅  = 180 2  2 

4x +

T6 = a + 5d = 45 + 5 × ( −6 ) = 15

Let the X th number house be the required house. Sum of number of houses preceding X th house is equal to S X −1 , i.e., X −1 [2a + ( X − 2)d ] 2

S = X −1

a 1,= d 1, so: Given= X ( X − 1) X −1 X −1 ⋅X = ( 2 + X − 2=) 2 2 2

S X −= 1

Sum of number of houses following X th house is equal to S 49 − S X : S 49 = S X=

49 49 49 2a + ( 49 − 1)d  = ( 2 + 48 ) = ⋅ 50 2  2 2

X X X 2a + ( X − 1)d = ) ( X + 1)  2 ( 2 + X − 1= 2 2

2450 − X ( X + 1) 49 X S 49 − S X = ⋅ 50 − ( X + 1) = 2 2 2 Now, given that sum of number of houses before X = sum of number of houses after X : X ( X − 1) 2450 − X ( X + 1) = 2 2 Multiply both sides by 2: X ( X − 1= ) 2450 − X ( X + 1) 2 X 2 + X 2 + X − X= 2450 ⇒ 2 X = 2450

X2 = 1225 ⇒ X =1225 = ±35 X = 35

129

5. Prerna saves `32 during the first month, `36 in the second month and `40 in the third month. If she continues to save in this manner, in how many months will she save `2000? (CBSE 2023)

32 + 36 + 40 + ... = 2,000

1 2

⇒= a 32, = d 4

1 2

Let ‘n’ be the number of months then

Sn = 2000

n ⇒ ( 2 × 32 + (n − 1) × 4 ) = 2000 2 ⇒ n 2 + 15n − 1000 = 0 1 ⇒ n 2 + 40n − 25n − 1000 = 0 ⇒ (n + 40 )(n − 25 ) = 0 1 ⇒n = −40,n = 25 ⇒n = 25 [Rejecting negative value of n ] 1 So, Prerna will save `2,000, in 25 months. 6. Find the sum of n terms of the series 1  2  3   4 − n  +  4 − n  +  4 − n  + ...       Sol. Let the given series be 1  2  3  S = 4 −  +  4 −  +  4 −  + ... n  n  n  1 2 3  = [ 4 + 4 + 4 +…] −  + + +… n n n   = 4 [1 + 1 + 1 +…] −

1 [1 + 2 + 3 +…] n

= S1 − S2 S1= 4 [1 + 1 + 1 +…] 1, = a = d 0 S1 = 4 ×

n n   2 × 1 + (n − 1) × 0   Sn = (2a + (n − 1)d )  2 2  

⇒ S1 = 4n = S2

130

1 [1 + 2 + 3 +…] n

S2 =

1 n × 2 × 1 + (n − 1) × 1 n 2

1 = [2 + n − 1] 2

CBSE Marking Scheme 2023 Sol. According to question

1, a = d = 2 −1 = 1

1 = [1 + n ] 2 Thus, S = S1 − S2 = 4n −

1 [1 + n ] 2

8n − 1 − n 7n − 1 = S = 2 2

7n − 1 . 2 7. The sum of the first p , q , r terms of an AP are a , b , c respectively. Show that

Hence the sum of the n terms of the series is

a b c 0 (q − r ) + (r − p ) + (p − q ) = p q r Sol. Let A be the first term and D be the common difference of the given AP It is given that a = Sum of p terms ⇒a=

p {2 A + (p − 1) D} ⇒ 2pa = {2 A + (p − 1) D} (i) 2

b = Sum of q terms ⇒b=

q {2 A + (q − 1) D} ⇒ 2qb = {2 A + (q − 1) D} (ii) 2

c = Sum of r terms ⇒ c= r {2 A + (r − 1) D} ⇒ 2c = {2 A + (r − 1) D} (iii) r 2 Multiplying (i), (ii) and (iii) by (q − r ) , (r − p ) and

(p − q ) respectively and adding, we get

2a 2b 2c (q − r ) + (r − p ) + (p − q ) p q r = {2A + (p - 1)D}(a - r) + {2A + (q - 1)D}(r - p) + {2A + (r - 1)D}(p - q) = 2A(q - r + r - p + p - q) + D{(p - 1)(a - r) + (q - 1) (r - p) + (r - 1)(p - q)} = 2 A × 0 + D × 0= 0 ⇒

a p c 0 (q − r ) + (r − p ) + (p − q ) = p q r

Competency Based Questions Multiple Choice Questions 1 1− 4x 1− 8x 1. The common difference of the AP , , 2x 2x 2x is (CBSE 2024)

(a) -2x

(b) -2

(d) 2x

2. Let a1 , a 2 , a 3 , ... be the terms of an AP. If a1 + a 2 + a 3 + ... + ap p 2 a =≠ (p q ) , then 6 = 2 a1 + a 2 + a 3 + ... + aq q a 21 6 41 (a) (b) 21 11 (c)

11 41

(d)

(d) 8

4. Given below is an arithmetic progression. X and Y are unknown. 4 1 , 6 3 , X , 11 3 , Y , 16 3 4 4 4 4 Which of these are X and Y? (CBSE QB) 3 3 (a) = X 8= , Y 13 4 4 3 1 (b) = X 8= , Y 14 4 4 1 1 (c) = X 9= , Y 14 4 4 1 3 (d) = X 9= , Y 13 4 4 5. In an AP, if = Sn n ( 5 + 3n ) and t n = 32 , then the value of n is (a) 4 (b) 5 (c) 6 (d) 7 6. In a game, a player must gather 20 flags positioned 5 m apart in a straight line. The starting point is 10 m away from the first flag. The player starts from the starting point, collects the 20 flags and comes back to the starting point to complete one round. What will be the total distance covered by a player upon completing one round? (CBSE QB) (a) 105 m (b) 210 m (c) 220 m (d) 1150 m

Chapter 5 • Arithmetic Progressions

Pragun has built a solid inverted lego pyramid as shown below. The number of studs in successive floors forms an arithmetic progression. Pragun figures out that the sum of the number of studs used in the first p floors is given by 6p 2 − 2p .

(

)

11 21

3. If a1 , a 2 , a 3 , a 4 , a5 are the first five terms of an AP such that a1 + a 3 + a5 = −12 and a1 .a 2 ⋅ a 3 = 8. Find the first term and common difference. (a) 2 (b) 4 (c) 6

7. The cylindrical bumps on top of lego blocks are called studs.

How many studs are there in the 5th floor? (a) 140 (b) 88 (c) 64 (d) 52 8. Mr. Sharma buys a property every year for 12 years. Every year he buys x acres more than the previous year. If in the 8th year he bought 45 acres of land and in the 5 th year he bought 30 acres of land, how many acres did he buy in the last year? (CBSE QB) (a) 5 + 11( 5 )  acres

(b) 10 + 11( 5 )  acres

(d) 5 + 12 (10 )  acres

9. Rakesh loves to travel and he travels every year. He has seen 8 different cities in his first year. Thereafter every year he has seen 2 cities. If he had followed this pattern, how many cities did he see by the end of 10 years of travel? (CBSE QB) (a) {8 + 9 ( 2 )} cities

(b) {8 + 10 ( 2 )} cities

(d) {16 + 10 ( 2 )} cities

10. The sum of n terms of the series 3 + 12 + 27 + 48 + ..., is (a) (c)

2n (n + 1) 3 3n (n + 1) 2

(b) (d)

3n (n − 1) 2

2n (n − 1) 3

131

Select Response Type MCQs 1. Shown below are some squares whose sides form an arithmetic progression (AP).

6 cm 9 cm

15 cm

12 cm

(CBSE QB)

Which of these are also in AP? (i) The areas of these squares. (ii) The perimeters of these squares. (iii) The length of the diagonals of these squares. (a) only (ii) (b) only (i) and (ii)

2. In a class 4 students asked to write unique arithmetic progression (AP). Their responses are: Student Rahul

AP 14, 7, 0, − 7, − 14….

Somya

−6, 6, − 6, 6, − 6, 6….

Jatin

4. Assertion (A): The sum of 20 terms of the AP 1, 3, 5, 7, ….. is 400. Reason (R): The sum of first n odd natural numbers is n2

Case Study Based Questions

Ria

3. Assertion (A): Sum of all 11 terms of an AP whose middle most term is 30, is 330. Reason (R): Sum of first n terms of an AP is given by n Sn = .l is the middle term. 2 [a + l ]

1. In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below:

330’ . 300’ / 100 m =110 m

3, 5.3, 7.6, 9.9, 12.2…. −36, − 31, − 26, − 21, − 16

210’ - 240’ 64m-73m

2. Assertion (A): The sequence 7, 10, 13, 16, ... is an arithmetic progression. Reason (R): In an arithmetic progression, the difference between consecutive terms is constant.

Which of these students correctly came up with AP? (i) Rahul (ii) Somya (iii) Ria

(iv) Jatin

(a) Only (i) and (iv) (b) Only (i), (ii) and (iv) (c) Only (i), (iii) and (iv) (d) Only (iii) and (iv)

Assertion-Reason Based Questions Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): The 10th term of the arithmetic progression 5, 8, 11, 14, ... is 32. Reason (R): In an arithmetic progression, the nth term is given by 132

L1 L1 L1 L1 L1 L1 L1 L1

The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane. Based on given information, answer the following questions, using concept of Arithmetic Progression. (a) What is the length of the 6th lane? (b) How long is the 8th lane than that of 4th lane? (c) (i) While practicing for a race, a student took one round each in first six lanes. Find the total distance covered by the student. OR (c) (ii) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student. (CBSE 2025) 2. A factory produces a certain number of units in the first year. Each subsequent year, the production

increases by a fixed number of units. In the 6th year, the factory produced 16,000 units, and in the 9th year, it produced 22,600 units.

(a) Determine the annual increase in production. (b) Calculate the production in the first year. (c) Find the total production over the first 10 years. 3. The cannon fires every 2 min, with the first shot occurring 10 min after the start of the fight. Additionally, the weight of each cannonball

increases by 0.5 kg with each successive shot, starting with the first ball weighing 0.5 kg.

Based on the above information, answer the following questions. (a) How many balls are fired after the first 30 min of fight? (b) W hat is the ball’s weight when the 12th ball is fired? (c) After how much time will the ball of 10 kg be fired?

Answers Multiple Choice Questions 1. (b) In an AP, the common difference is given by: d = a 2 − a1 = a 3 − a 2 d So, =

1 − 4x 1 − 2x 2x

(1 − 4x ) − 1

1 − 4x − 1 −4x = = −2 2x 2x

2x 11 2. (c) 41 Let d be the common difference for the AP:

a1 + 5 × 2a1 11a1 11 = = = a1 + 20 × 2a1 41a1 41 3. (a) 2 Let a1 = a 3 − 2d , a 2 = a3 − d , a3 = a3 , a4 = a 3 + d , a5 = a 3 + 2d

Then a1 + a 3 + a5 = −12, ( given ) ⇒ a 3 − 2d + a 3 + a 3 + 2d = −12 ⇒ 3a 3 = −12 ⇒ a 3 = −4

Also, a1 ⋅ a 2 ⋅ a 3 = 8 ( given )

a1 , a 2 , a 3 , ... p Sp 2 2a1 + ( p − 1)d  p 2 = Then, = Sq q 2a + q − 1 d  q 2 )  1 ( 2 2a1 + ( p − 1)d p ⇒ = 2a1 + (q − 1)d q

⇒ a1 ⋅ a 2 = −2 ( a 3 = −4 )

 p 2a1 + (q − 1)d  ⇒ q 2a1 + ( p − 1)d= ⇒ 2a1q + pqd − qd = 2a1p + pqd − pd

1 1 4. = (c) X 9= , Y 14 4 4 1 3 Given, 4 , 6 , X , 11 3 , Y , 16 3 are in AP. 4 4 4 4 Then, the common difference will be equal.

⇒ pd − qd= 2a1p − 2a1q

⇒ d ( p −= q ) 2a1 ( p − q ) ⇒= d 2a1 Now

a1 + ( 6 − 1)d Term 6 a6 = = Term 21 a 21 a1 + ( 21 − 1)d

Chapter 5 • Arithmetic Progressions

⇒ ( a 3 − 2d )( a 3 − d ) = −2

⇒ ( −4 − 2d )( −4 − d ) =−2 ⇒ ( 2 + d )( 4 + d ) = 1 ⇒ d 2 + 6d + 9 = 0 ⇒ (d + 3)2 = 0 ⇒ d = −3 ∴a1 =a 3 − 2d =−4 − 2 ( −3 ) =2

3 1 3 6 − 4 =X − 6 4 4 4

133

27 17 27 37 1 − =X − ⇒= X = 9 4 4 4 4 4 3 1 3 Also, 6 − 4 = Y − 11 4 4 4 27 17 47 57 1 − =Y − ⇒ = Y ⇒ Y = 14 4 4 4 4 4

Also, in the 5th year he buys =a + ( 5 − 1) x

Given, Sn =n ( 5 + 3n ) =5n + 3n

∴ Now, in the last year, i.e., in 12 th year Mr. Sharma buys

⇒ 30 = a + 4x ⇒ a + 4x = 30 (ii) Subtracting (ii) from (i), we have ⇒ 3x = 15 ⇒ x = 5 From equation (i), we get

5. (b) 5

⇒ a = 45 − 7x = 45 − 7 × 5 = 10 ⇒ a = 10

and t n = 32

10 + 11( 5 )  acres = a + (12 − 1) x =

We know that t= Sn − Sn −1 n

(

32= 5n + 3n

9. (a) [{8 + 9(2)}] cities

) − {5(n − 1) + 3(n − 1) } 2

a 8,= d 2 We have,=

32 = 5n + 3n 2 − 5n + 5 − 3n 2 − 3 + 6n

∴Total number of cities travelled in 10 years =

32= 2 + 6n ⇒ 30 = 6n ⇒ n = 5

an = a + (n − 1)d

6. (b) 210 m

a10 = 8 + (10 − 1)2 = 8 + 9(2)

Given, number of flags = 20 Distance between the flag = 5 m

10. (c)

The position of flags is shown in figure

3n (n + 1) 2

Sol. Let Sn = 3 + 12 + 27 + 48 + ... to n terms.

10 m

Then Sn = 3 + 2 3 + 3 3 + 4 3 + ... to n terms

S starting point F1

F20

∴ Total distance covered by player upon completing one round = 2 SF1 + ( F1F2 + F2 F3 +…+ F19F20 )    = = 2 10 + 5 + 5  +  + 5 2[10 + 95] = 210 m 19   7. (d) 52 Sum of studs used in the first p floors S= 6p 2 − 2p p

( )

⇒ S= n

3 (1 + 2 + 3 + 4 + ... to n terms= )

3n (n + 1)

 n (n + 1)  1 + 2 + 3 + ... + n =  2  

Select Response Type MCQs 1. (c) only (ii) and (iii) Given squares are shown below.

We know that a= Sp − Sp −1 p

(

)

2 = 6p 2 − 2p − 6 ( p − 1) − 2 ( p − 1)    = 6p 2 − 2p −  6p 2 + 6 − 12p − 2p + 2   

( i ) The area of the squares are 62 , 92 , 122 , 152

= 6p 2 − 2p − 6p 2 + 14p − 8= 12p − 8

Now, number of studs in the 5th floor = 12 × 5 − 8 = 60 − 8 = 52

( ii ) The perimeters of the squares are

(

)

8. (b) 10 + 11( 5 )  acres Let Mr Sharma buys a acres in the first year. In the 8th year he buys =a + ( 8 − 1) x acres (here d = x from question) ⇒ 45 = a + 7x ⇒ a + 7x = 45 (i)

134

6 cm 9 cm

12 cm

15 cm

i.e., 36, 81,144, 225 are not in AP.

6 × 4, 9 × 4, 12 × 4, 15 × 4

i.e., 24, 36, 48, 60

Here, 36 − 24 = 48 − 36 = 60 − 48 = 12

So, the perimeters of the squares are in AP. ( iii ) The length of the diagonals = Side 2

Thus, the length of diagonals are 6 2, 9 2,12 2, and15 2 respectively.

and S11 =

11 2a + (11 − 1)d  2

= 15 2 − 12 2 = 3 2

11 = [2a + 10d ] 2 =11( a + 5d ) =11 × 30 =330

So, the length of the diagonals are in AP.

Reason is false, since l is the last term.

Hence, statements ( ii ) and ( iii ) are correct.

4. (a) Both A and R are true, and R is the correct explanation of A.

Here, 9 2 − 6 2 = 12 2 − 9 2

2. (C) Only (i), (iii) and (iv) For Rahul a 2 − a1 = 7 − 14 = −7, a 3 − a 2 =0 − 7 =−7, so it is an AP For Somya a 2 − a1 =−6 − 6 =−12,

a 3 − a 2 = 6 − ( −6 ) = 12, so it is not an AP For Ria a 2 − a1 = 5.3 − 3= 2.3, a 3 − a 2 = 7.6 − 5.3 = 2.3, so it is an AP For Jatin a 2 − a1 =−31 − ( −36 ) =5,

a 3 − a 2 =−26 − ( −31) =5, so it is an AP

Assertion-Reason Based Questions 1. (a) Both A and R are true, and R is the correct explanation of A. To find the 10th term of the AP: an = a + (n − 1) × d a10 = 5 + (10 − 1) × 3 = 5 + 27 = 32

1, 3, 5, 7, 9,…, ( 2n − 1) are first n odd natural numbers. Let S be their sum. Then, S = 1 + 3 + 5 + 7 + 9 +…+ (2n − 1) n = ⇒ Sn = n2 2 (1 + 2n − 1) n  Sn = (a1 + an )  Using 2   So, Reason is true. Using this statement, we find that 2 1 + 3 + 5 + 7 +… upto 20 terms = 20 = 400

So, Assertion is also true and Reason is a correct explanation for Assertion. Hence, option (a) is correct.

Case Study Based Questions 1. (a) For 6th lane; n = 6

Assertion (A) is true.

an = a + (n − 1) × d

Reason (R) is true and correctly explains the assertion.

a6 = 400 + ( 6 − 1) × 7.6 = 400 + 5 × 7.6 = 400 + 38

2. (a) Both A and R are true, and R is the correct explanation of A. The difference between each consecutive term is consistent: • 10 − 7 = 3

= 438 metres (b) For the 4th and 8th lanes: a 4 = 400 + ( 4 − 1) × 7.6 = 400 + 3 × 7.6 = 400 + 22.8 = 422.8 metres

a8 = 400 + ( 8 − 1) × 7.6 = 400 + 7 × 7.6 = 400 + 53.2

• 13 − 10 = 3

= 453.2 metres

• 16 − 13 = 3

Difference between 8th and 4th lanes:

Thus, Assertion (A) is true.

a8 − a= 453.2 − 422.8 = 30.4 metres 4

Reason (R): In an arithmetic progression, the difference between consecutive terms is constant. Reason (R) is true and it correctly explains why the sequence is an AP.

3. (c) A is true, but R is false Middle most term =

n +1 2

11 + 1 = = 6th term 2 Then, a6 = 30 ⇒ a + 5d = 30

Chapter 5 • Arithmetic Progressions

Using the sum formula for the first n terms of an AP n ( 2a + (n − 1)d ) 2 6 S6= ( 2 × 400 + (6 − 1)× 7.6 ) 2 S= n

= S6 3 ( 800 + 5 × 7.6 = = ) 3 ( 800 + 38 ) 3 × 838 = 2514 metres

135

(d) (ii) Total distance from lane 4 to 8: S 4 to 8 = a 4 + a5 + a6 + a7 + a8 S 4 to 8 = 422.8 + 430.4 + 438 + 445.6 + 453. 2 S 4 to 8 = 2190 metres 2. (a) Annual Increase in Production: • Let the production in the first year be a units, and the annual increase be d units. • The production in the 6th year (a6 ) is given by: a6 =a + 5d =16,000 (i) • The production in the 9th year (a9) is: a9 =a + 8d =22,600 (ii) • Subtracting the first equation from the second: d ) 22,600 − 16,000 (a + 8d ) − (a + 5= 3= d 6600 ⇒ = d 2200 • Therefore, the annual increase in production is 2,200 units. (b) Production in the First Year: • Using the value of d in the equation for the 6th year: a + 5 × 2,200 = 16,000 a + 11,000 = 16,000 a = 5,000 Therefore, the production in the first year is 5,000 units. (c) Total Production Over the First 10 Years: • The total production over the first n years is given by the sum of the first n terms of an AP: n Sn = × ( 2a + (n −1)d ) 2 • For n = 10:

The first cannonball is fired 10 minutes after the start, and each subsequent shot occurs every 2 minutes. Therefore: • Time of first shot = 10 minutes • Interval between shots = 2 minutes To find out how many balls are fired in the first 30 minutes, we need to consider the shots from 10 minutes up to 30 minutes. Number of shots = 1+

Last shot time − First shot time Interval

Last shot time = 30 minutes Number of shots =1 +

30 − 10 =1 + 10 =11 shots 2

(b) W hat is the ball’s weight when the 12th ball is fired? The weight of the cannonballs starts at 0.5 kg and increases by 0.5 kg with each successive shot. • Weight of nth ball = Initial weight + (n − 1) × Increment • Weight of 12th ball = 0.5 + (12 − 1) × 0.5 = 0.5 + 11 × 0.5 = 0.5 + 5.5 = 6.0 kg (c) After how much time will the ball of 10 kg be fired? The weight of the ball reaches 10 kg and increases at a rate of 0.5 kg per shot. We set up the equation to find the nth shot where the weight equals 10 kg:

10 × ( 2 × 5,000 + 9 × 2,200 ) S10 = 2 S10 = 5 × (10,000 + 19,800 )

Weight of nth ball = 0.5 + (n − 1) × 0.5 = 10

S10 = 149,000

The 20th ball weighs 10 kg. Now, we find the time when the 20th ball is fired:

• Therefore, the total production over the first 10 years is 149,000 units. 3. (a) How many balls are fired after the first 30 min of fight?

0.5 + 0.5n − 0.5 = 10 0.5n = 10 = n

10 = 20 0.5

= Time First shot time + (n − 1) × Interval Time = 10 + ( 20 − 1) × 2 = 10 + 38 = 48 minutes

Additional Practice Questions 1. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. (CBSE 2009) (a) 1 (b) 2 (c) 3 (d) 4

136

2. The 11th term from the end of the AP: 10, 7, 4, …, − 62 is (a) 25

(b) 16

(d) 0

3. The value of p for which ( 2p + 1) ,10 and ( 5p + 5 ) are three consecutive terms of an AP is (a) -1 (b) -2 (c) 1 (d) 2 4. Find the common difference of the AP: 1 p +1 p + 2 , , , ... p p p

(CBSE 2011)

5. Determine whether -150 is a term of the AP: 17, 12, 7, 2, ... (CBSE 2011) 6. The sum of the first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3. Find the first term of the AP. (CBSE SQP 2017) 7. 250 logs are stacked in the following manner:

Reason (R): nth term of AP is given by an =a + (n − 1)d 1 13. The 26th , 11th and last term of an AP are 0, 3 and − , 5 respectively. Find the common difference and the number of terms.

(NCERT Exemplar)

14. The eighth term of an AP is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term. 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown alongside). In how many rows, are the 250 logs placed and how many logs are there in the top row? 8. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. 9. If the number of terms of an AP is ( 2n + 1) , then what is the ratio of the sum of the odd terms to the sum of even terms? 1 1 and nth term be , n m th then show that its (mn ) term is 1.

10. If the mth term of an AP be

11. Which term of the AP: 65, 61, 57, 53,… is the first negative term? Directions: Question 15 consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

15. Find the middle term of the AP 213, 205, 197,…, 37 . (CBSE 2015)

16. Find the sum of two middle terms of the AP: 4 −2 1 1 (NCERT Exemplar) − , − 1, , − ,…, 4 . 3 3 3 3 1 1 17. If the m th term of an AP is and the nth term is , n m 1 show that the sum of mn terms is (mn + 1) . 2

(CBSE 2015, 2017)

18. Solve the equation: 1 + 4 + 7 + 10 +……x = 287. (NCERT Exemplar)

19. A man repays a loan of `3250 by paying `20 in the first month and then increases the payment by `15 every month. How long will it take him to clear the loan?

Challenge Yourself 1. The sum of an AP’s 3rd and 7th terms is 30, and the sum of the 5th and 8th terms is 50. Find the first three terms of the AP. 2. If jth, kth and lth terms of an AP are p, q and r, respectively. Show that 0 (k – l )p + (l – j )q + ( j – k )r =

Chapter 5 • Arithmetic Progressions

Scan me th

3. If the ( p + 1) Term of an AP is three times the th

(q + 1)th the term prove that the ( 4p + 1) Term is th three times the ( p + q + 1) Term. 4. If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of the squares of

137

their reciprocals, then show that ab 2 , ca 2 , bc 2 are in AP. 5. If a , b , c are in AP, show that 1

b+ c

c+ a

a+ b

6. If

a n +1 + b n +1

is the A.M. between a and b . Then, an + bn find the value of n .

(Hint: A.M. between a and b is

are in AP

a +b ) 2

(Hint: 2b= a + c as a , b , c are in AP)

Answers Additional Practice Questions

12. (d) A is false, but R is true.

1. (b) 1

2. (c) −32

3. (d) 2

5. No

1 p

6. First term (a) = 2 7. 20 rows, 3 logs 8. 89 numbers 11. 18th term

1 13. d = − ,n= 27 5 15. 125

14. 3

18. 40

19. 20 months

16. 3

Challenge Yourself 9. n + 1 n

1. −35 , −5, 5 3 3

6. 0

Scan me for Exemplar Solutions 138

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. Which term of the AP: 5, 10, 15, 20, ... is 100? (a) 10th (b) 15th

(d) 25th

2. The 4th term from the end of the AP −11, − 8, − 5, ……., 49 is (a) 37 (b) 40 (c) 43

(d) 58

3. If the sum of the first 10 terms of an AP is 150, what is the average of these terms? (a) 10 (b) 15 (c) 20

(d) 25

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): a , b , c are in AP if 2b= a + c . Reason (R): In an AP the sum of the terms equidistant from the beginning and the end is always same and is equal to the sum of first and least term. 5. Assertion (A): The sum of the first n terms of an AP is given by = Sn 3n 2 − 4n. Then its nth term a= 6n − 7. n Reason (R): nth term of an AP, whose sum to n terms is Sn , is given by a= Sn − Sn −1 . n

Very Short Answer Questions th

6. If 9

[2 × 2 = 4 Marks] th

term of an AP is zero, prove that its 29

term is double the 19

term.

(NCERT Exemplar)

7. Find the sum of all two digit natural number which when divided by 3 yield 1 as remainder.

Short Answer Questions

[4 × 3 = 12 Marks]

8. The 3rd term of an AP is 15 and the 7th term is 27. Find the first term and the common difference. 9. If the sum of the first n terms of an AP is given by = Sn 3n 2 + 5n , find the AP. 1 2 10. Find the number of terms in the series 20 + 19 + 18 +… of which the sum is 300, explain the double answer. 3 3 11. The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 :15 . Find the number. (NCERT Exemplar, CBSE 2018, 2020)

Long Answer Questions

[3 × 5 = 15 Marks]

12. The ratio of the 11 th term to the 18 th term of an AP is 2 : 3. Find the ratio of the 5 th term to the 21 st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms. 13. A thief runs with a uniform speed of 100 m / minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m / minute in the first minute and increases his speed by 10 m / minute every succeeding minute. After how many minutes the policeman will catch the thief? 14. The angles of a quadrilateral are in AP. If the smallest angle is 75°, find the other three angles. Self-Assessment

139

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. 1 1 Each step rises of m and a tread of m (see Figure). Let V1 , V 2V 3 ,…, V15 denote respectively the volumes of 4 2 concrete required to build the first, second, third, ..., fifteenth step. Based on the above information answer the following questions:

1 2

m 50 m

1 4

1 2

3 m

(a) Heights of first, second, third, …,15th steps form an AP with common difference (i)

1 m 4

(b) The value of V 2 is (i) 25 m 3

(ii)

1 m 2

(iii)

(ii) 50 m 3

3 m 4

(iv) −

1 m. 4

(iii) 12.5 m 3

(iv) 6.25 m 3

(iii) 75 m 3

(iv) 6.25 m 3

(d) The sum of the surface areas of 15 treads is (i) 350 m 2 (ii) 400 m 2

(iii) 375 m 2

(iv) 475 m 2

(e) The total volume of the concrete required to build the terrace is (i) 800 m 3 (ii) 375 m 3 (iii) 650 m 3

(iv) 750 m 3

140

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. Two AP’s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th terms is (a) 11 (b) 3 (c) 8 (d) 5 (CBSE SQP 2024)

2. If

5 + 9 + 13 +  to n terms 17 = , then n = 7 + 9 + 11 +  to (n + 1) terms 16

(a) 8

(b) 7

(d) 11

3. Mr. Prasad is collecting donations for a new school building in a village. His wife starts by donating `1,25,000. His brother also contributes by donating `54,000. Thereafter every person who contributes in the donation pays `5,250 more than the previous donor. How much amount was Mr. Prasad able to collect after 23 donations? (a) `1,64,250

(b) ` 24,00,750

(d) ` 25,70,250

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

n 2a + (n − 1)d  2

5 5 5. Assertion (A): −5, − , 0, ,... is an AP 2 2 Reason (R): The terms of an AP cannot have both positive and negative rational numbers.

(CBSE SQP 2024)

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. Find the common difference of an AP in which the 8th term is 17 and the 4th term is 9. 7. The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Find the AP.

Short Answer Questions

(CBSE 2020) [4 × 3 = 12 Marks]

8. The 14th term of an AP is twice its 8th term. If its 6th term is -8, find the first term and the common difference. 9. If the sum of the first 7 terms of an AP is 49 and that of its first 17 terms is 289, find the sum of its first n terms 10. Solve the equation: −4 + ( −1) + 2 + ... + x =437 . 11. Which term of the Arithmetic Progression (AP) −7, − 12, − 17, − 22,… will be −82? Is -100 any term of the AP? Give reason for your answer.

Self-Assessment

(CBSE 2019)

141

Long Answer Questions

[3 × 5 = 15 Marks]

12. The sum of the first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1:3. Find the first term of the AP. 13. Sums of the first p , q , r terms of an AP are a , b , c respectively. Prove that a b c + + = 0 p (q − r ) q (r − p ) r ( p − q ) 14. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623. (NCERT Exemplar)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are n potatoes in the line (See Figure). Each competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in the bucket, and she continues in the same way until all the potatoes are in the bucket. Based on the above information answer the following questions:

(a) Distance run by the competitor to pickup and drop first potato in the bucket, is

(i) 5 cm

(ii) 8 m

(iii) 10 m

(iv) 7 m

(b) Distance run by the competitor to pick up and drop nth potato in the bucket, is (i) ( 3n + 2 ) m

(ii) 2 ( 3n + 2 ) m

(iii) 2 ( 3n − 1) m

(iv) 6 (n − 1) m

(i) 36 metres

(ii) 40 metres

(iii) 86 metres

(iv) 76 metres

(d) Total distance run by the competitor to pick up and drop n potatoes in the bucket is (i) n ( 3n + 2 ) m

(ii) 2n ( 3n + 2 ) m

(iii) n ( 3n + 7 ) m

(

)

(iv) 3n 2 + 7 m

(e) If d1 , d2 , d3 , …, dn denote distances run by the competitor to pick up first, second, third, ..., nth potato respectively, then d1 , d2 , d3 , …, dn form an AP with common difference

142

(i) 3

(ii) 6

(iii) 7

(iv) 5

Triangles

Chapter at a Glance A

Congruent Figures

Two figures are said to be similar if they have the same shape and the same size.

Similar Figures

Two figures are said to be similar if they have the same shape but not necessarily the same size. ●

C F

A P

All congruent figures are similar but the similar figures need not to be congruent

Conditions of Similarity of Figures

The conditions of similarity of different figures are as follows: ●

Circle: Two circles are always similar.

●

Square: Two squares are always similar.

●

Equilateral triangle: Two equilateral triangles are always similar.

●

Polygon: Two polygons of the same number of sides are similar if:

●

◆

All the corresponding angles are equal and

◆

All the corresponding sides are in the same ratio (or proportion)

Triangle: Two triangles are said to be similar if: ◆

Their corresponding angles are equal, and

◆

Their corresponding sides are in the same ratio (or proportional).

Basic Proportionality Theorem or Thales Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given PQ || BC ⇒

AP AQ = PB QC

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. AP AQ Given = PB QC ⇒ PQ || BC

A N P

M Q

Criteria for Similarity of Triangles ●

ngle–Angle–Angle (AAA) criterion Or A B’ C Angle–Angle (AA) criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then their third angle and corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Chapter 6 • Triangles

143

●

ide–Side–Side (SSS) criterion: If in two triangles, sides of one triangle are S proportional to (i.e., in the same ratio of)the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. ide–Angle–Side (SAS) criterion: If one angle of a triangle is equal to one angle S of the other triangle and the sides including these angles are proportional, then the two triangles are similar. If ∆ABC  ∆DEF then:

Area of ∆ABC  AB   BC   AC  = = = Area of ∆DEF  DE   EF   DF 

NCERT Zone NCERT Exercise 6.1

Two squares with different sides 2 cm

1. Fill in the blanks using correct word given in the brackets:

1 cm

(a) All circles are __________. (congruent, similar)

(d) Two polygons of the same number of sides are similar, if (i) their corresponding angles are __________________ and (ii) their corresponding sides are __________________. (equal, proportional) Sol. (a) Similar

2 cm

1 cm

(b) All squares are __________. (similar, congruent) (c) All __________ triangles are similar. (isosceles, equilateral)

1 cm

2 cm

(b) Trapezium and square P

Triangle and parallelogram A

(b) Similar

(ii) Proportional

2. Give two different examples of pair of (a) Similar figures

B S

3. State whether the following quadrilaterals are similar or not: D

(b) Non-similar figures Sol. (a) Two equilateral triangles with different sides.

S 1.5 cm R 1.5 cm

2 cm 1 cm 60° 1 cm 60° 60° 1 cm

144

60°

2 cm

1.5 cm

1.5 cm Q

3 cm

3 cm A

C 3 cm

3 cm

2 cm

60°

Sol. Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1: 2, but their corresponding angles are not equal.

NCERT Exercise 6.2 1. In Fig. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Hence,

Therefore, EF is parallel to QR.

(ii)

Sol. (a) Let EC = x cm It is given that DE || BC .

1 cm E

Chapter 6 • Triangles

Therefore, EF is not parallel to QR.

2.5

3. In Fig., if LM || CB and LN || CD , prove that AM AN = AB AD

(NCERT Exemplar)

Sol. Given, LM || CB By using Proportionality theorem, AM AL = AB AC

(i)

Similarly, LN || CD AN AL ∴ = AD AC

M N

(ii)

C D

AM AN = AB AD 4. In Fig, DE || AC and DF || AE. Prove that BF BE = FE EC

3 cm

0.18 18 9 = = 1.28 128 64 0.36 9 = Q 2.56 64 PE PF Hence, = PQ PR Therefore, EF is parallel to QR.

From (i) and (ii), we obtain,

(b) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

PE = PQ PF = PR

(a) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

PF 3.6 = = 1.5 FR 2.4 PE PF Hence, ≠ EQ FR

PF = 0.36 cm

2. E and F are points on the sides PQ and PR respectively of a ∆PQR . For each of the following cases, state whether EF ||QR:

A D C

F E Sol. In ∆ABC , DE  AC BD BE (i) (Basic proportionality theorem) ∴ = DA EC

3.6 cm F

9 cm

(b) Let AD = x cm It is given that DE || BC. By Basic proportionality theorem, we get AD AC A = DB EC D 1.8 cm 7.2 cm E x 1.8 = 7.2 5.4 B 5.4 cm 1.8 × 7.2 x = = 2.4 5.4 C ∴ AD = 2.4 cm

Sol. (a) Given that,PE = 3.9 cm, EQ = 3 cm, P PF = 3.6 cm, FR = 2.4 cm PE 3.9 = = 1.3 3.9 cm EQ 3

PE PF = EQ FR

By Basic proportionality 3 cm theorem, we get AD AC B = DB EC 1.5 1 = 3 x 3 ×1 ∴ EC = 2 cm x = =2 1.5

1.5 cm D

0.3

5.4 cm

4.5 cm

B (i)

PF 8 = FR 9

8 cm

cm 0. 18

1.8 cm E

4 cm E

3 cm

7.2 cm

PE 4 8 = = EQ 4.5 9

A 1.5 cm 1cm E D

(b) Given, PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

In ∆BAE , DF  AE

2.4 cm R

BD BF ∴ = DA FE

(ii) (Basic proportionality theorem)

From, (i) and (ii), we get BE BF = EC FE 145

5. In Fig, DE || OQ and DF || OR . Show that EF || QR .

(NCERT Exemplar)

Sol. In ∆POQ , DE  OQ PE PD ∴ = EQ DO

A D

O Q

(i) (Basic proportionality theorem)

In ∆POR , DF  OR ∴

PF PD = FR DO

(ii) (Basic proportionality theorem)

From (i) and (ii), we get PE PF = EQ FR ∴ EF  QR

(Converse of basic)

6. In Fig, A, B and C are points on OP , OQ and OR respectively such that AB || PQ and AC || PR. Show that BC ||QR. Q

(i) (Basic proportionality theorem)

(ii) (Basic proportionality theorem)

From (i) and (ii), we get OB OC = BQ CR ∴ BC  QR

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. AO CO Show that = BO DO

(Converse of basic)

Sol. The given figure in which PQ is a line segment drawn through mid point P of line B, Such that PQ  BC By using basic proportionality theorem, we get

A P

AQ AP = QC PB AQ 1 = (P is the mid-point of AB , ∴ AP = PB QC 1 ⇒ AQ = QC Or, Q is the mid-point of C .

In ∆ADC , EO  CD

By using Basic proportionality theorem, we get AE AO = (i) ED OC

B F

In ∆ABD , OE  AB

7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

146

Hence, by using basic proportionality theorem, we get

Sol. Draw a line EF through point O, such that EF  CD

In ∆POR , AC  PR OA OC ∴ = AP CR

By using basic proportionality theorem, we get A AP 1 = PB 1 P AQ 1 And = Q QC 1 B AP AQ ∴ = PB QC

PQ  BC .

Sol. In ∆POQ , AB  PQ OA OB ∴ = AP BQ

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). Sol. The given figure in which PQ is a line segment drawn through mid point P and Q of line AB and C , Such = that AP PB = And AQ QC

So, by using basic proportionality theorem, ED OD = AE BO AE BO (ii) ⇒ = ED OD From (i) and (ii), we get AO BO = OC OD AO OC ⇒ = BO OD

10. The diagonals of a quadrilateral ABCD intersect AO CO each other at the point O such that = . BO DO Show that ABCD is a trapezium. Sol. Draw a line OE  AB In ∆ABD , OE  AB By using Basic of proportionality, AE BO = ED OD

A E

(i)

B O

However, it is given that AO OB = (ii) OC OD From (i) and (ii), AE AO = ED OC ⇒ EO  DC [By the converse of basic ⇒ AB  OE  DC proportionality theorem] ⇒ AB  CD ∴ ABCD is a trapezium.

NCERT Exercise 6.3 1. State which pairs of triangles in Fig. are similar. Write the similarity criterion used for answering the question and also write the pairs of similar triangles in the symbolic form. P (a) 60°

A 60° 80° 40°

80°

(b)

A 2

2.5

(c)

6 C

2.7

M 2.5 70° 5

∠C = ∠R = 40°

80°

(b) In ∆ABC and ∆PQR BC 2.5 1 = = PR 5 2 AB 2 1 A = = QR 4 2 3 2 AC 3 1 = = B 2.5 PQ 6 2

(e)

C Q

2.5 B

(f)

80° 80° 3

70° E

80°

Chapter 6 • Triangles

80°

30°

M 2.5 70° N

6 F

∴∆MNL ~ ∆PQR

[SAS similarity criterion]

BC 3 1 = = EF 6 2 ∠ A ≠ ∠ F ≠ 80°

70°

10 D

2.5 80°

C E

∴∆ABC is not similar to ∆DEF A

Since the three ratios are not the same.

40°

Hence, ∆ABC ∼ ∆QRP (SSS criterion)

70°

80°

40°

C Q B Therefore, ∆ABC ∼ ∆PQR (AAA similarity)

(e) In ∆ABC and ∆DEF AB 2.5 1 = = DF 5 2

60°

P 6 L

(d)

∠B = ∠Q = 80°,

(d) In ∆MNL and ∆PQR MN 2.5 1 = = PQ 5 2 ML 5 1 = = QR 10 2 ∠ M= ∠ Q= 70°

40°

Sol. (a) ∠A = ∠P = 60°,

80°

Since the angles between two sides are not the same. (f) In ∆DEF , ∠D + ∠E + ∠= F 180° ⇒ 70° + 80° + ∠= F 180° ⇒ ∠F = 30°

Similarly, in ∆PQR ,

∠P + ∠Q + ∠= R 180°

⇒ 80° + 30° + ∠= P 180°

70° 80°

80° 30°

R 147

⇒ ∠P = 70°

In ∆DEF and ∆PQR ,

∠D = ∠P [= 70°]

∠E = ∠Q [= 80°]

∠F = ∠R [= 30°]

Thus, ∆DEF ∼ ∆PQR (AAA similarity)

In ∆PQS and ∆TQR , QR QT From equation ( 2 )  = QS QP  ∠Q = ∠Q [Common ] Thus, ∆PQS ~ ∆TQR [SAS similarity] 5. S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS . Show that ∆RPQ ~ ∆RTS .

= 125° and 2. In Figure, ∆ODC ∼ ∆OBA , ∠BOC ∠CDO =°. 70 Find ∠DOC , ∠DCO , and ∠OAB . Sol. Since DOB is a straight line: ∠DOC + ∠COB = 180°

∠DOC= 180° − 125°= 55° Since ∆DOC is a triangle:

∠DCO + ∠CDO + ∠DOC =180

6. In the figure, if ∆ABE ≅ ∆ACD , show that ∆ADE ~ ∆ABC .

[Corresponding angles

∠ OAB= 55°

of similar triangles]

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O . Using similarity criterion for two triangles, show that: OA OB = OC OD

Sol. In ∆DOC and ∆BOA , ∠DOC = ∠BOA [Vertically Opposite Angles] D ∠CDO = ∠ABO [Alternate Angles] ∠DCO = ∠BAO [Alternate Angles] Thus, ∆DOC ~ ∆BOA [AAA similarity] Since ∆DOC ~ ∆BOA , we get

O C

T P

∠ APE = ∠ CPD [ Vertically Opposite Angles]

∠AEP = ∠CDP [ = 90°]

Thus, ∆AEP ~ ∆CDP [AA similarity].

(b) In ∆ABD and ∆CBE ,

= ∠ ADB ∠ CEB [= 90°]

∠PQR = ∠PRQ 1

QR QT ⇒ = From equation (1)  (2) QS QP  148

Sol. (a) In ∆AEP and ∆CDP ,

(b) ∆ABD ~ ∆CBE (d) ∆PDC ~ ∆BEC

Sol. In ∆PQR ,

Thus, ∆ADE ~ ∆ABC [SAS similarity].

QR QT 4. In the figure, = and ∠ 1 = ∠ 2. QS PR Show that ∆PQS ~ ∆TQR .

Given that, QR QT = QS PR

AB = AC [Corresponding sides of similar triangles] (1) AD = AE [Corresponding sides of similar triangles] (2) B In ∆ADE and ∆ABC , AD AE = AB AC AB AC From equation (1) and ( 2 )  = AC AB  ∠ A = ∠ A [Common ]

7. In the figure, altitudes AD and CE of ∆ABC intersect each other at the point P . Show that: C D (a) ∆AEP ~ ∆CDP

DO CO = BO AO BO AO ⇒ = DO CO

Thus, PQ = PR (1)

Sol. Given that ∆ABE ≅ ∆ACD ,

Given that ∆ODC ∼ ∆OBA , ∠ OAB = ∠ OCD

Thus, ∆RPQ ~ ∆RTS . [AA similarity]

125°

T Q

[Given ] ∠R = ∠R [Common ] ∠RTS = ∠QPS

∠ DCO + 70° + 55° = 180° ∠ DCO= 55°

Sol. In ∆RPQ and ∆RST ,

70°

∠ ABD = ∠ CBE [Common ]

Thus, ∆ABD ~ ∆CBE [AA similarity].

∠ PAE = ∠ DAB [Common ]

Thus, ∆AEP ~ ∆ADB [AA similarity].

∠DCB = ∠HGE [CD and GH are the bisectors of equal angels]

(d) In ∆PDC and ∆BEC ,

∠PDC = ∠BEC [Each 90°]

∠PDC = ∠BEC [Common]

Thus, ∆PDC ~ ∆BEC [AA similarity].

∠ A ∠= = F , and∠ ACD ∠ FGH

8. E is a point on the side AD produced of a parallelogram ABCD , and BE intersects CD at F . Show that ∆ABE ~ ∆CFB . A

CD AC = GH FG

∠DCB = ∠HGE ∠B = ∠E Thus, ∆DCB ~ ∆HGE [AA similarity].

Sol. In ∆ABE and ∆CFB , ∠A = ∠C [Opposite angles of parallelogram] ∠AEB = ∠CBF [Alternate angles as AE || BC ] Thus, ∆ABE ~ ∆CFB [AA similarity]. 9. In the figure, ABC and AMP are two right triangles, right-angled at B and M , respectively. Prove that: C

(a) ∆ABC ~ ∆AMP AB AD (b) = PQ PM

Thus, ∆ACD ~ ∆FGH [AA similarity].

(b) In ∆DCB and ∆HGE ,

In ∆ACD and ∆FGH ,

11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC . If AD ⊥ BC and EF ⊥ AC , prove that ∆ABD ~ ∆ECF .

F E

Sol. Given that ABC is an isosceles triangle, AB = AC ⇒ ∠ ABD = ∠ ECF

Sol. (a) In ∆ABC and ∆AMP ,

∠ ADB= ∠ EFC=[ 90°]

= ∠ ABC ∠ AMP [= 90°]

Thus, ∆ABD ~ ∆ECF [AA similarity].

∠ A = ∠ A [Common ]

Thus, ∆ABC ~ ∆AMP [AA similarity].

(b) Since ∆ABC ~ ∆AMP , AB AD = PQ PM

In ∆ABD and ∆ECF ,

[Corresponding parts of similar triangles].

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF , such that D and H lie on sides AB and FE of ∆ABC and ∆EFG , respectively. If ∆ABC ~ ∆EFG , show that: F

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of PQR (see Figure). Show that ∆ABC ~ ∆PQR . P

Sol. AD and PM are the medians of the triangle. Therefore, CD AC A (a) = BC QR H GH FG BD = = and QM D 2 2 (b) ∆DCB ~ ∆HGE Given that, (c) ∆DCA ~ ∆HGF B C G E AB BC AD = = Sol. (a) Given that ∆ABC ~ ∆EFG, PQ QR PM ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE, AB 1 BC AD = = ∠ACB = ∠FGE PQ 2 QR PM AB BD AD ∠ACD = ∠FGH [CD and GH are the bisectors of = = equal angles] PQ QM PM Chapter 6 • Triangles

149

In ∆ABD and ∆PQM,

AD = DE [by Construction] PM = ML So, the diagonals of ABEC bisecting each other at D, therefore ABEC is a parallelogram.

AB BD AD = = PQ QM PM

∴∆ABD ∆PQM [SSS similarity ]

Similarly, PQLR is also a parallelogram.

∠ABD = ∠PQM [Corresponding angles of similar triangles] In ∆ABC and ∆PQR, ∠ABD = ∠PQM [ Proved above]

∴AC = BE , AB = EC and PR = QL , PQ = LR Given that,

AB BC = PQ QR ∴∆ABC ∆PQR [SAS similarity ] 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC . Show that CA= ² CB × CD .

A B D

We know that the corresponding angles of similar triangles are equal. Therefore, ∠ BAE = ∠ QPL (1) B

∠ ACD = ∠ BCA [Common ]

Similarly, ∆AEC ~ ∆PLR and ∠CAE = ∠RPL (2)

∴∆ADC ∆BAC [ AA similarity ] We know that the corresponding sides of similar triangles are proportional. Therefore, CA CD = CB CA ∴CA 2 =CB × CD

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR . Show that (CBSE 2023, 2024) ∆ABC ~ ∆PQR . P

Adding the equations (1) and (2), we get ∠BAE + ∠CAE = ∠QPL + ∠RPL ⇒ ∠ CAB = ∠ RPQ (3) In ∆ABC and ∆PQR , AB AC = PQ PR

∠ CAB = ∠ RPQ [Given ] ∴∆ABC ∼ ∆PQR [SAS similarity ]

15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. A

Sol. Given that, AB AC AD = = PQ PR PM Produce AD and PM to E and L such that AD = DE and PM = DE . Now join B to E , C to E , Q to L and R to L. Since BD and PM are medians of the triangle, therefore: = BD DC = and QM MR

150

∴∆ABE ∼ ∆AQL [SSS similarity ]

Sol. In ∆ADC and ∆BAC ,

∠ ADC = ∠ BAC [Given ]

AB AC AD = = PQ PR PM AB BE 2 AD ⇒ = = PQ QL 2PM AB BE AE ⇒ = = PQ QL PL

6m D 4m F

28 m

Sol. Let CD be the pole and AB be the tower. Therefore, DF and BE are the shadows of the pole and tower. In ∆ABE and ∆CDF ,

∠ CDF = ∠ ABE [ Angle of sun at same place] = ∠ CDF ∠ ABE [ Each 90°]

∴∆ABE ∼ ∆CDF [ AA similarity ]

AB BE = CD DF

∠A = ∠P , ∠B = ∠Q , ∠C = ∠R (2)

AB 28 = ⇒ AB = 42 m 6 4 Hence, the height of the tower is 42 m.

AD and PM are medians of the triangle. Therefore, BC QR BD = = and QM (3) 2 2

16. If AD and PM are medians of triangles ABC and PQR, respectively, where P ∆ABC ~ ∆PQR, prove that:

From equation (1) and (3), we have BD AB = (4) QM PQ

AB AD = PQ PM

In ∆ABD and ∆PQM ,

Sol. Given that: ∆ABC ~ ∆PQR

We know that the corresponding sides of similar triangles are proportional. Therefore, AB AC BC = = (1) PQ PR QR and,

∠ABD = ∠PQM From equation ( 2 )  BD AB From equation ( 4 )  = QM PQ 

∴∆ABD ∼ ∆PQM [SAS similarity ]

AB AD ⇒ = PQ PM

Multiple Choice Questions 1. In the given figure,

AD AE = and ∠ADE = 70°, BD EC

∠BAC = 50°. Then, find angle ∠BCA. A

6 cm

5 cm B

3 cm

(b) 2.5 cm

(d) 3.2 cm

(b) similar but not congruent (c) neither congruent nor similar

3. In ABC and DEF , ∠B = ∠E , ∠ F = ∠ C and AB = 3DE . Then, the two triangles are (a) congruent but not similar

4 cm

x cm

(a) BD · CD = BC²

(a) 3.6 cm

Chapter 6 • Triangles

(d) 60°

(d) congruent as well as similar

D 3 cm

5. In the figure, if, ∠BAC = 90° and AD ⊥ BC, then,

(CBSE 2023)

2 cm

(d) 10

2. In the given figure, ∆ABC ∼ ∆QPR . If AC = 6 cm, BC = 5 cm, QR = 3 cm, and PR = x, then the value of x is A

(a) 6 (c) 8

E C

(b) 50°

(b) 12.5

(a) 70°

4. In the given figure, DE  BC . The value of x is

(b) AB · AC = BC² (d) AB · AC = AD²

6. If the lengths of the diagonals of a rhombus are 16 cm and 12 cm, then the length of the sides of the rhombus is: (a) 9 cm (b) 10 cm (c) 8 cm

(d) 20 cm

AB BC CA 7. If in two ∆ABC and ∆PQR, = = then QR PR PQ (a) ∆PQR ~ ∆CAB

(b) ∆PQR ~ ∆ABC

(d) ∆BCA ~ ∆PQR

151

EF DE (c) = RP QR

(a) 20.25 cm

(b) 27 cm

(d) 64 cm

11. In the given figure, QR || AB, RP || BD,

The value of x is: (a) 1 (b) 6

D 3

12. In ∆ABC, DE || AB. If AB = a, DE = x, BE = b, and EC = c, then x in terms of a, b, and c is: B ac ac (a) (b) b b +c ab ab (c) (d) c b +c

+ 5x

(d) 1 cm

m 7cD

(b) 2.25 cm

3 cm

18. In the given figure, DE || AC. Find the value of x (a) 1 (b) 2

2x + 10 B

x–7

19. In the given figure (not to scale), AM : MC = 3 : 4, BP : PM = 3 : 2 and BN = 12 cm. Find AN. (a) 10 cm

M P

(b) 12 cm (c) 14 cm (d) 16 cm

A E

14. In the given figure, given that quad. ABCD ∼ quad. PQRS. Determine the values of x, y, z.

152

17. In the given figure, DE || BC. If AD = 3 cm, AB = 7 cm, and EC = 3 cm, then the length of AE is (CBSE 2023) cm (a) 2 cm 3

28 7 34 = ,y = ,z 5 4 6

(b) 3 cm

(d) Alternate segment theorem

(b) SAS similarity criterion

(a) = x

Which of these criteria can be used to prove that ∠RSQ is similar to ∠RTP?

(d) RHS similarity criterion T

(a) 9 cm

(b) Thales theorem

13. In the following figure, Q is a point on PR and S is a point on TR. QS is drawn and ∠RPT = ∠RQS.

(a) AAA similarity criterion

(a) Pythagoras theorem

(d) 9

2 3

This theorem is known as

16. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.”

(CBSE 2023)

(a) ∠B = ∠E (b) ∠A = ∠D

CP = 5x + 4, PD = 3x.

28 8 35 = ,y = ,z 5 4 6

15. In the given figure, AB || PQ. If AB = 6 cm, PQ = 2 cm, and OB = 3 cm, then the length of OP is

AB BC = 10. If in ∆ABC and ∆DEF, then they will be DE FD similar when:

CQ = x + 2, QA = x,

3 cm

9. If ∆PQR ∼∆ABC, PQ = 6 cm, AB = 8 cm, and the perimeter of ∆ABC is 36 cm, then the perimeter of ∆PQR is:

27 7 35 (c) = = ,y = ,z x 5 4 6 (d) = x

B Q

6 cm

DE DF (c) = QR PQ

28 7 35 (b) = = ,y = ,z x 5 4 6

2 cm

8. If in two ∆DEF and= ∆PQR, ∠ D ∠= Q and ∠ R ∠ E , then which of the following is not true? EF DF DE EF (a) = (b) = PR PQ PQ RP

20. Which pair of the given quadrilaterals is similar?

3 cm 2 cm

2 cm

4 cm

(II)

110°

5 cm

3 cm 70°

(III) 100°

7 cm

70°

80°

40°

5 cm

80°

3 cm

70°

6 cm

6 cm 100°

80°

70°

40°

5 cm 120°

3 cm

3 cm 3 cm

6 cm

(I)

3 cm

70°

3 cm

80°

70°

110°

4 cm

110°

4 cm

3 cm

5 cm

2.5 c m

80°

100°

5 cm

70°

2.5 cm

110°

140°

4 cm

5 cm

110°

4 cm

2 cm

5 cm 140°

60° 7 cm

(IV)

(a) I

(b) II

(d) IV

Answers 1. (d) 60

2. 2.5 cm

13. (a) AAA similarity criterion

3. (b) similar but not congruent

4. (d) x = 10 cm

14. (b) = x

5. (c) BD· CD = AD²

6. (b) 10 cm.

9. (b) 27 cm

EF DF 8. (a) = PR PQ 10. (c) ∠B = ∠D

11. (a) x = 1

12. (b) x =

7. (a) ∆PQR ~ ∆CAB

28 7 35 = ,y = ,z 5 4 6 16. (b) Thales Theorem

15. (d) 1 cm 17. (b) 2.25 cm

18. (a) 1 cm

19. (c) 14 cm

20. (c) III

ac b +c

Constructed Response Questions Very Short Answer Questions

2. If the area of two similar triangles is equal, then prove that they are congruent. Sol. Given, ∆ABC ~ ∆PQR, and ar(∆ABC) = ar(∆PQR) To prove: ∆ABC ≅ ∆PQR

1. In ∆ABC , D and E are points on sides AB and AC respectively such that DE  BC and AD : DB = 3 : 1. If EA = 6.6cm, then AC is equal to? Sol. Given, AD : DB = 3 : 1 In ∆ABC , since DE  BC , we apply the Basic Proportionality Theorem (BPT): AD AE = DB EC

A D B

3 6.6 = 1 EC 6.6 EC = = 2.2 cm 3 AC = AE + EC = 6.6 + 2.2 = 8.8 cm Chapter 6 • Triangles

E C

Proof: Since ∆ABC ~ ∆PQR, we have:

ar ( ∆ABC ) AB 2 BC 2 CA 2 = = = (i) ar ( ∆PQR ) PQ 2 QR 2 RP 2 Also, given that ar(∆ABC) = ar(∆PQR), we get:

ar ( ∆ABC )

ar ( ∆PQR )

= 1 From equation (i), we have:

153

Sol. Let the side of the other triangle be x cm. ∴ Ratio of perimeters of two similar triangles is equal to the ratio of their corresponding sides. 9 30 = x 20 ⇒x = 6 cm

AB 2 BC 2 CA 2 = = = 1 PQ 2 QR 2 RP 2 AB BC CA = = = 1 PQ QR RP which implies, = AB PQ = , BC QR = , CA RP Thus, by the SSS (Side-Side-Side) Congruence Theorem,

8. The ratio of the corresponding altitudes of two 3 similar triangles is . Is it correct to say that the 5 6 ratio of their areas is ? Why? 5 Sol. False

∆ABC ≅ ∆PQR Hence proved. 3. In the given figure, AB = AC. E is a point on CB produced. If AD is perpendicular to BC and EF perpendicular to AC, prove that ∆ABD is similar A to ∆ECF. (CBSE 2019) Sol. AB = AC ⇒ ∠ C = ∠ B (i) F

By the property of the area of two similar triangles,  Area1   Altitude1   =   Area 2   Altitude2 

In ∆ABD & ∆ECF , ∠ADB = ∠EFC (Each 90°) ∠ ABD = ∠ ECF (By (i))

∆ABD ∼ ∆ECF (By AA similarity)

4. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB. (CBSE 2024) Sol. In ∆ABE and ∆CFB, we have

∠AEB = ∠CBF (Alternate angles) F ∠A = ∠C (Opposite angles B C of a parallelogram) Therefore, ∆ABE ~ ∆CFB (By AA criterion of similarity) 5. Is the following statement true? Why? “Two quadrilaterals are similar if their corresponding angles are equal.”

Sol. False Two quadrilaterals are similar if their corresponding angles are equal and corresponding sides are also proportional. 6. Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why? Sol. True Here, the corresponding two sides and the perimeters of the two triangles are proportional, then the third side of both triangles will also be in proportion. 7. The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 9 cm long, find the length of the corresponding side of the second triangle. (CBSE 2020) 154

2   Area1   3   altitude1 3 = , given    =      Area 2   5   altitude2 5 9 6 = ≠ 25 5

So, the given statement is not correct 9. Is it true to say that, if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reason for your answer. Sol. False Because, according to SAS similarity criterion, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. Here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct. 10. In the figure, if AB || DC and AC, PQ intersect each other at the point O, prove that: OA ⋅CQ = OC ⋅ AP A

P O

Sol. Given: AC and PQ intersect each other at the point O, and AB || DC. To prove: OA ⋅CQ = OC ⋅ AP Proof: In ∆AOP and ∆COQ, ∠AOP = ∠COQ

(Vertically opposite angles)

∠APO = ∠CQO (Since AB || DC and PQ are a transversal, so alternate angles are equal) Thus, ∆AOP ∼ ∆COQ (by AAA similarity criterion) Since corresponding sides of similar triangles are proportional, OA AP = Multiplying both sides by OC × CQ, we get: OC CQ OA ⋅CQ = OC ⋅ AP Hence proved. 11. If ∆ABC ~ ∆DEF, given: = AB 4= cm, DE 6= cm, EF 9= cm, FD 12 cm. Find the perimeter of ∆ABC. Sol. Given, = AB 4= cm, DE 6= cm, EF 9= cm, FD 12 cm Since ∆ABC ~ ∆DEF, we use the similarity property: AB BC AC = = DE EF DF 4 BC AC = = 6 9 12 On taking the first two terms, we get

(i) [BPT Theorem] (ii) [BPT Theorem]

2. If D is a point on the side BC = 12 cm of a ∆ABC such that BD = 9 and ∠ADC = ∠BAC then find the length of AC. Sol. In ∆ADC and ∆ABC ∠ADC = ∠BAC (given) ∠ C = ∠ C (common) ∴∆ADC ∼ ∆BAC (By AA similarity) AC CD = BC AC ⇒ AC 2 =BC × CD ⇒ AC 2 = 12 × ( BC − BD ) = 12 (12 − 9 )

4 BC = 6 9 4×9 BC = 6 cm = 6 Taking first and last term 4 AC = 6 12 6 × 12 AC = 8 cm = 9 Perimeter of ∆ABC = AB + BC + AC

BD BE ⇒ = AD EC In ∆ABP , since DC  AP BD BC ⇒ = AD CP From (i) and (ii), BE BC = EC CP

⇒ AC 2 = 36 ⇒ AC = 6 cm 3. AD and PM are medians of triangles ∆ABC and ∆PQR respectively, where ∆ABC ∼ ∆PQR . Prove that

AB AD = PQ PM

(CBSE 2019)

Sol. Given, ∆ABC ∼ ∆PQR. AD and PM are medians of ∆ABC and ∆PQR respectively.

= 4 + 6 + 8 = 18 cm

Short Answer Questions 1. In the given figure, DE  AC and DC  AP . Prove that BE BC = EC CP A

Since, ∆ABC ∼ ∆PQR

D P

AB BC AC = = PQ QR PR (1) D is the midppont of BC. (AD is median) Q R M M is the midppont of QR. (PM is median) ∴BC = 2BD also QR = 2QM (2)

P (CBSE 2020)

Sol. In, ∆ABC , DE  AC A

Chapter 6 • Triangles

(From 1)

AB 2BD ∴ − PQ 2QM

(From 2)

AB BD ⇒ = ⇒ ∆ABD  ∆PQM PQ QM

AB BC ∴ − PQ QR

AB AD ∴ = Hence proved. PQ PM 155

4. A vertical pole of length 6 m casts a shadow 4 m long on the ground, and at the same time, a tower casts a shadow 28 m long. Find the height of the tower. Sol. Here, the pole and the tower both are perpendicular to the ground, and they cast shadows at the same time. So, the Sun makes an equal angle with the ground for both triangles. Now, use the property of similar triangles and find the height of the tower. Let AB be a pole of length 6 m, BC = 4 m be the length of its shadow, and θ be the angle which the Sun makes with the ground. In ∆ABC and ∆DAC , we have ∠BAC = ∠ADC [given ]

And ∠ACB = ∠DCA [common angle] ∴∆ABC ∼ ∆DAC [ by AA similarity criterion ] AC CB ⇒ = DC CA

⇒ CA 2 =CB × CD Hence proved. 5. ABCD is a parallelogram. Point P divides AB in the ratio 2 : 3, and point Q divides DC in the ratio 4 : 1. Prove that OC is half of OA. (CBSE SQP 2024) Sol. Let ABCD be a parallelogram where AB = DC = a. Point P divides AB in the ratio 2 : 3.

4 1 = a , CQ a 5 5

Considering ∆APO ~ ∆CQO by AA similarity:

156

Point Q divides DC in the ratio 4 : 1.

2 a AO 5 2 = = CO 1 a 1 5 1 ⇒ OC = OA 2 Hence proved.

Show that ∆BAE ~ ∆CAD.

(CBSE SQP 2023)

Sol. In ∆ABD, ∠1 = ∠2 (Given) ⇒ AB = BD (i) (Equal angles have equal opposite sides) Given, AD AC = AE BD Using equation (i), we get AD AC = (ii) AE AB In ∆BAE and ∆CAD,

Sol. In ∆ABC, D is a point on side BC such that ∠ADC = ∠BAC

⇒ CA × CA = CB × CD

AP PO AO = = CQ QO CO

7. D is a point on the side BC of a triangle ABC such that 2 ∠ADC = ∠BAC. Prove that CA= CB ⋅CD (CBSE 2023)

CA CB or = CD CA

= DQ

AD AC = AE BD and, ∠1= ∠ 2

AC AD ∠ A = ∠ A ( common ) (By equation (ii)) = AB AE ∴∆BAE ∼ ∆CAD ( By SAS similarity criterion )

[since corresponding sides of two similar triangles are proportional]

2 3 = AP = a , BP a 5 5

6. In the given figure below,

In ∆CBA and ∆CDA , ∠ C = ∠ C (Common)

∠ BAC = ∠ ADC (Given) ∆CBA ∼ ∆CAD (By AA similarity) ⇒ Their corresponding sides are proportional. ⇒ Their corresponding sides are proportional. CA CD = ⇒ CA 2 =CB ⋅CD BC CA 8. In the given figure, l || m and line segments AB, CD, and EF are concurrent at point P. AE AC CE Prove that: = = BF BD FD Sol. Given: l || m, and Line segments AB, CD, and EF are concurrent at point P. AE AC CE To Prove: = = BF BD FD Proof: In ∆APC and ∆BPD, we have: ∠APC = ∠BPD (vertically opposite angles) ∠PAC = ∠PBD (alternate angles) Since two pairs of corresponding angles are equal, we can say:

APC ∼ BPD (by AAA similarity criterion) Thus, for similarity

4= AB 2 4 AD 2 + AC 2

∠ APE = ∠ BPF (vertically opposite angles)

∠ PAE = ∠ PBF (alternate angles) ∴∆APE ∼ ∆BPF (by AAA similarity criterion)

AP AE PE Thus, = = (ii) PB BF PF In ∆PEC and ∆PFD,

∠ EPC = ∠ FPD (vertically opposite angles)

= 3 + 4 = 7 cm

ar∆PST PS 2 32 9 = = = 2 2 ar∆PRQ PR 49 7 Hence, the required ratio = 9 : 49 10. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. (CBSE 2020) A

In right-angled ∆ABD, 2 AB = AD 2 + BD 2

BC 2 AB = AD + 4 2

Chapter 6 • Triangles

12 .5

(NCERT Exemplar)

By the converse of the Basic Proportionality Theorem, AB || QR

12. The diagonals of a quadrilateral ABCD intersect AO CO each other at the point O such that = . BO DO Show that ABCD is a trapezium.

PR = PS + SR

Sol. Let ABC be an equilateral triangle, in which AD is the perpendicular bisector of BC. BC BD = 2 And AB = BC = CA .

PA 5 50 2 = = = (i) AQ 7.5 75 3 PB 4 2 and, = = (ii) BR 6 3 From (i) and (ii),

AP AC AE PE EC = = = = PB BD BF PF FD AE AC CE ∴ = = BF BD FD Hence proved.

9. In the given figure, ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of ∆PST to the area of ∆PRQ. (CBSE 2018)

P m

PE PC EC Thus, = = (iii) PF PD FD From Equations (i), (ii), and (iii),

Then, QA= QP − PA= 12.5 − 5= 7.5 cm Now,

∴∆PEC ∼ ∆PFD (by AAA similarity criterion)

Sol. PS = 3 cm, SR = 4 cm and ST || RQ.

11. A and B are respectively the points on the sides PQ and PR of a ∆PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm, and PB = 4 cm. Is AB || QR? Give a reason for your answer. Sol. True Given: = PQ 12.5= cm, PA 5= cm, BR 6= cm, PB 4 cm

∠ PCE = ∠ PDF (alternate angles)

Hence, 3 AB 2 = 4 AD 2 .

cm 5c m

AP AC PC = = (i) PB BD PD In ∆APE and ∆BPF,

4 AB 2 = 4 AD 2 + AB 2 [Since AB = AC ]

(Using Pythagoras Theorem)

AO CO Sol. Given, = BO DO AO BO ⇒ = (i) CO DO In ∆ABD, EO || AB (Construction) AE BO = (By BPT) (ii) ED DO From equations (i) and (ii), D C AE AO = O ED CO E ⇒ EO  DC (Converse of BPT) A But, EO  AB

⇒ AB  DC (Construction) ∴In quadrilateral ABCD, since AB || DC, ⇒ ABCD is a trapezium 13. In the given figure, XZ is parallel to BC. AZ = 3 cm, ZC = 2 cm, BM = 3 cm, and MC = 5 cm. Find the length of XY.

(CBSE 2023)

X B

Z C

Sol. Since XZ || BC, we have XY || BM and YZ || MC. In ∆AXZ and ∆ABC,

157

∠AXZ = ∠ABC (Since XZ || BC, corresponding angles are equal)

Sol. Since ∆ABC ~ ∆PQR, we have: ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

∠AZX = ∠ACB So, ∆AXZ ∼ ∆ABC ( By AA similarity )

ar ( ∆ABC ) AB 2 = (i) ar ( ∆PQR ) PQ 2

AX AZ XZ = = AB AC BC AX AZ 3 3 ⇒ = = = AB AZ + ZC 3 + 2 5

Since AD and PS are bisectors, ∠A = ∠P 1 1 ∠A = ∠P 2 2 Thus,

Similarly, in ABM ∆ ABM,

∆AXY ∼ ∆ABM ( as XY  BM )

∠BAD = ∠QPS

C Q

Since ∆BAD ~ ∆QPS (by AA similarity),

AX XY = AB BM 3 XY ⇒ = 5 3 9 cm ⇒ XY = 5

BA AD = (ii) QP PS By equations (i) and (ii),

14. If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 3, prove that one of the parallel sides is three times the other. (NCERT Exemplar) Sol. Given: DE : EB = 1 : 3

∠ 1 = ∠ 2 (Alternate angles)

∠ 3 = ∠ 4 (Vertically Opposite Angles) D

AB BE ⇒ = [By CPST] CD DE

4 2

Long Answer Questions B E F G

D A

Sol. Given CD is perpendicular bisector of AB.

Hence proved.

1. In the given figure, CD is the perpendicular bisector of AB. EF is perpendicular to CD. EF C intersects CD at G. Prove that CF FG = (CBSE 2023) CD DG

In ∆AEB and ∆CED

∴ ∆AEB ∼ ∆CED [By AA similarity]

ar ( ∆ABC ) AD 2 = ar ( ∆PQR ) PS 2

AB 3 [Since DE : BE 1: 3] = ⇒ = CD 1

⇒ AB = 3CD

EF ⊥ CD AE intersects CD at G. CF FG = To Prove: CD DG In ∆EFG and ∆ADG.

15. In ∆ABC, given AB = 6 3 cm, AC = 12 cm, and BC = 6 cm, find ∠B. (CBSE 2020)

∠EFG = ∠ADG = 90°

Sol. Given, AB = 6 3 cm, AC = 12 cm, and BC = 6 cm.

∴ ∆EFG ≈ ∆ADG (By AA Similarity) EF FG = (1) AD DG

It can be observed that:

= AB 2 108 = cm 2 , AC 2 144 = cm 2 , BC 2 36 cm 2 Now, AB 2 + BC 2 = 108 + 36 = 144 = AC 2 i .e . AB 2 + BC 2 = AC 2 which satisfies the Pythagoras theorem. So, ∠ B = 90. 16. If ∆ABC ~ ∆PQR and AD and PS are bisectors of corresponding angles A and P, then prove that ar ( ∆ABC )

ar ( ∆PQR ) 158

AD 2 PS 2

(NCERT Exemplar)

∠EGF = ∠AGD

(Given) (Vertically Opposite Angles)

In ∆ECF and ∆BCD. ∠EFC = ∠BDC = 90°

(Given)

∠ECF = ∠BCD

(Vertically Opposite Angles)

∴∆ECF ≈ ∆BCD

(By AA Similarity)

EF CF ⇒ (2) = BD CD But, AD = BD (CD bisect AB) EF EF CF ⇒ = = (3) AD BD CD

From (1) and (3), we get EF CF FE CF FE = ⇒ = = AD CD DG CD DG Hence proved.

From (i) and (ii), we get ∆GDB ∼ ∆FCE Since corresponding sides of two similar triangles are proportional:

2. Prove that, if a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side. Sol. Given, In ∆ABC AP AQ = BP QC

GD BG = FC EF GD × EF = BG × FC 2 FG= BG × FC

A P

To Prove: PQ  BC

Construction: Let PE  BC

Hence proved.

E Q

Proof: In, ∆ABC , since PE  BC (by construction), we apply the Basic Proportionality Theorem (BPT): AP AE = (i) PB EC Also, we are given: AP AQ = (ii) BP QC From (i) and (ii), we equate: AE AQ = EC QC

4. ∆PQR is right-angled at Q. QX ⊥ PR, XY ⊥ RQ, and XZ ⊥ PQ are drawn. Prove that: 2 XZ= PZ × ZQ

Q R

Sol. Here, RQ ⊥ PQ and XZ ⊥ PQ, or, XZ || YQ.

Similarly,

XY || ZQ.

Y 1

Since XYQZ is a rectangle (∠PQR = 90°), we have: In ∆PZX, ∠ 3 + ∠ 4 =90° (ii)

Since XQ ⊥ PR, ∠ 2 + ∠ 3 =90° (iii) By equations (i) and (iii), we get: ∠1= ∠ 3

A D

(CBSE 2020)

Sol. Given, DEFG is a square and ∠BAC = 90° 2

To prove: FG= BG × FC Since DEFG is a square, therefore we can write: DE = EF = FG = GD In ∆AGE and ∆GDB, ∠A = ∠DGB = 90° ∠ADE = ∠GBD (Corresponding angles) Thus, by AA similarity, ∆EAD ∼ ∆GDB (i) Now, in ∆AGF and ∆FCE,

By equations (ii) and (iii), we get: ∠2=∠4 Thus, by AA similarity, ∆PZX ∼ ∆XZQ So, by proportionality, PZ XZ = XZ ZQ 2 XZ= PZ × ZQ

Hence proved. 5. O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB, meeting AD in P and BC in Q. Prove that PO = QO. Sol. Given: ABCD is a trapezium. Diagonals AC and BD intersect at O.

∠A = ∠EFC = 90°

PQ || AB || DC.

(Corresponding angles)

Proof:

∠AED = ∠FCE

Thus, by AA similarity, ∆AED ∼ ∆FCE (ii) Chapter 6 • Triangles

In ∆XZQ, ∠ 1 + ∠ 2 =90° (i)

This implies that E and Q coincide. ∴ PQ  BC Hence proved. 3. In the given figure, DEFG is a square and ∠BAC = 90°. 2 Show that: FG= BG × FC

To prove: PO = QO In ∆ABD and ∆POD, PO || AB [∴ PQ || AB]

B O

159

∠D = ∠D [common angle]

In ∆TSM:

∠ABD = ∠POD [corresponding angles]

∠T + ∠S + ∠M =180

Thus, ∆ABD ∼ ∆POD (by AAA similarity criterion)

∠T + 25 + 100 = 180

So,

OP PD = (i) AB AD

In ∆ABC and ∆OQC,

(

∠C = ∠C [common angle] ∠BAC = ∠QOC [corresponding angles] Thus, ∆ABC ∼ ∆OQC (by AAA similarity criterion) OQ QC = (ii) AB BC Now, in ∆ADC, So,

OP || DC, so by Basic Proportionality Theorem AP OA = (iii) PP OC In ∆ABC, OQ || AB, so by Basic Proportionality Theorem: BQ OA = (iv) QC OC From (iii) and (iv), we get: AP BQ = PD QC Adding 1 to both sides: AP BQ + = 1 +1 PD QC AP + PD BQ + QC = PD QC AD BC = PD QC

25°

)

In ∆PQR and ∆TSM: ∠Q = ∠S

∠R = ∠M

55° 25°

25°

100°

6. In ∆PQR and ∆MST, given: ∠P = 55°, ∠Q = 25°, ∠M = 100°, ∠S = 25°. Is ∆QPR ~ ∆TSM? Why? Sol. False We know that the sum of the three angles of a triangle is 180°. In ∆PQR:

Since all corresponding angles are equal, ∆PQR ∼ ∆TSM However, the correct correspondence is: P ↔ T , Q ↔ S, R ↔ M Thus, ∆QPR is not similar to ∆TSM because their correct correspondence is different. 7. In ∆PQR, N is a point on PR, such that QN ⊥ PR. If PN. NR = QN², then prove that ∠PQR = 90°. Sol. Given, In ∆PQR, N is a point on PR, such that QN ⊥ PR and PN ⋅ NR = QN 2 To prove: ∠PQR = 90°. Proof: We have, PN ⋅ NR = QN 2 PN ⋅ NR = QN ⋅QN PN QN = QN NR PN QN = (i) QN NR In ∆QNP and ∆RNQ, ∠PNQ = ∠RNQ (each equal to 90°) ∆QNP ~ ∆RNQ (by SAS similarity criterion) Thus, ∆QNP and ∆RNQ are equiangular, i.e., ∠PQN = ∠QRN ∠RQN = ∠QPN Adding both sides: ∠PQN + ∠RQN = ∠QRN + ∠QPN ∠PQR = ∠QRN + ∠QPN (ii)

∠P + ∠Q + ∠R =180

We know that the sum of the angles of a triangle is 180°.

55 + 25 + ∠R =180

In ∆PQR: ∠ PQR + ∠ QPN + ∠ QRP = 180

)

 ∠R = 180 − 55 + 25= 180 − 80= 100

160

Dividing both sides by QN · NR:

From (i) and (ii): OP OQ = AB BC Since AB = BC, we get: OP = OQ Hence proved.

(

 ∠T= 180 − 25 + 100= 180 − 125= 55

∠P = ∠T

OQ || AB [∴ OQ || AB]

55°

From Eq. (ii): ∠QRP = ∠QRN , and ∠QPR = ∠QPN

25°

PS OS = (i) AB OB In ∆CQR and ∆CAB,

Thus, ∠PQR + ∠PQR =180

So,

2∠PQR = 180

180 = 90 2 Hence proved. ∠PQR =

QR || PS || AB ∠QCR = ∠ACB (common angle)

8. In the given figure, if ∠ACB = ∠CDA, AC = 8 cm, and AD = 3 cm, then find BD. Sol. Given, From the figure, ∠CDA = 90 Thus, ∠ACB = ∠CDA = 90

From equations (i) and (ii): OS CR = OB CB OB CB = OS CR

In the right-angled ∆ADC, C

2 (8) = (3)2 + CD 2

64= 9 + CD 2 CD 2 = 64 − 9 = 55

Subtracting 1 from both sides: A

CD = 55 cm ∠BDC = ∠ADC (each 90°)

BS BR = OS CR

∠DBC = ∠DCA (each equal to 90 − ∠ A ) Thus,

By the converse of the basic proportionality theorem, SR ｜｜OC

∆CDB ∼ ∆ADC By similarity property CD AD = BD CD CD 2 CD 2 = AD × BDBD = AD 2 ( 55 ) 55 BD = cm = 3 3

Hence proved. 10. A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, then find how far she is away from the base of the pole.

9. In the given figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR. Sol. Given: PQRS is a parallelogram, so PQ || SR and PS || QR. Also, AB || PS.

Sol. Let A be the position of the street bulb fixed on a pole, where: = AB 6= m, CD 1.5 m .

Let the woman’s shadow be ED = 3 m.

Let the distance between the pole and the woman be x m.

P O R

Proof: In ∆OPS and ∆OAB, PS || AB ∠POS = ∠AOB (common angle) ∠OSP = ∠OBA (corresponding angles) Thus, ∆OPS ∼ ∆OAB (by AAA similarity criterion) Chapter 6 • Triangles

OB CB − 1= −1 OS CR OB − OS CB − CR = OS CR

In ∆CDB and ∆ADC,

To prove: OC ｜｜SR

Thus, ∆QCR ∼ ∆CAB QR AB PS So, = = (ii) CR CB CR (Since PQRS is a parallelogram, so PS = QR)

AC = 8 cm, AD = 3 cm, and ∠ACB = ∠CDA

2 AC = AD 2 + CD 2

∠CRQ = ∠CBA (corresponding angles)

Since the woman and the pole both are standing vertically, CD || AB ∆CDE ∼ ∆ABE (by AAA similarity criterion)

Using similarity property: ED CD = EB AB 3 1.5 = 3+x 6

3 ×= 6 1.5 ( 3 + x ) 1= 8 4.5 + 1.5x 1.5x = 18 − 4.5

C 1.5 m

161

13.5 x = 9m = 1.5 Hence, she is at the distance of 9 m from the base of the pole.

2 AB = AD 2 + (3x )2 2 AB = AD 2 + 9x 2

5= AB 2 5 AD 2 + 45x 2 (i)

11. If in ∆ABC, AD is the median and AE ⊥ BC, then 1 prove that AB 2 + AC 2 = 2 AD 2 + BC 2 2 1 Sol. To prove: AB 2 + AC 2 = 2 AD 2 + (BC )2 2 Construction: Draw A AE ⊥ BC

Applying Pythagoras Theorem in ∆ADC

Applying Pythagoras Theorem in ∆ABE

5 AB 2 − 5 AC 2 = 5 AD 2 + 45x 2 − 5 AD 2 + 20x 2

AB = AE + BE

2 AC = AD 2 + CD 2 2 AC = AD 2 + (2x )2 2 AC = AD 2 + 4x 2

5= AC 2 5 AD 2 + 20x 2 (ii) Subtracting Equation (ii) from Equation (i)

(

or, AB 2 = AD 2 − DE 2 + (BD − DE )2 AB 2 = AD 2 + BD 2 − 2BD × DE (i) Applying Pythagoras Theorem in ∆AEC 2

AC = AE + EC or AC 2 =

( AD − ED ) + (ED + DC ) 2

AC 2 = AD 2 − ED 2 + ED 2 + DC 2 + 2ED × DC AC 2 = AD 2 + DC 2 + 2DC × ED

(ii)

Adding Equations (i) and (ii) AB2 + AC2 = 2(AD2 + BD2 – 2BD × DE + DC2 + 2DC × ED) Since BD = DC,

Proof: Since, 2DB = 3CD DB 3 Or, = CD 2 Let DB = 3x, CD = 2x, so BC = 5x. Applying Pythagoras Theorem in ∆ADB

5= AB 2 5 AC 2 + BC 2 Hence proved.

Sol. Given:

∠A = ∠P ∠B = ∠Q, ∠C = ∠R Proof:

C Q

Or, 2∠B = 180 − x 180 − x Or, ∠B = (i) 2 In ∆PQR, ∠P= + ∠Q + ∠R 180 (Given, ∠Q = ∠R ) Or, x  + ∠Q + ∠Q =180 Or, 2∠Q= 180 − x

AD ⊥ BC

162

5 AB 2 − 5 AC 2 = 25x 2

Or, x  = + ∠B + ∠B 180 (given, ∠B = ∠C )

Sol. Given:

2 AB = AD 2 + BD 2

)

In ∆ABC, ∠A + ∠B + ∠C =180

12. In ∆ABC, AD ⊥ BC and point D lies on BC such that 2DB = 3CD. Prove that: 5= AB 2 5 AC 2 + BC 2

To Prove: 5= AB 2 5 AC 2 + BC 2

5 AB − 5 AC =45x − 20x

Let ∠A = ∠P be x.

1  AB 2 + AC 2 = 2 AD 2 + 2  BC  (as BD = ½ BC) 2  1 = 2 AD 2 + BC 2 2 Hence proved.

2DB = 3CD

13. Vertical angles of two isosceles triangles are equal. If their areas are in the ratio 16 : 25, then find the ratio of their altitudes drawn from vertex to the opposite side.

Since BC = 5x,

AB 2 = AD 2 − DE 2 + BD 2 + DE 2 − 2BD × DE

180 − x Or, ∠Q = (ii) 2 In ∆ABC and ∆PQR, ∠A = ∠P [Given] ∠B = ∠Q [from equations (i) and (ii)] Thus, ∆ABC ~ ∆PQR (by AA similarity) ar ( ∆ABC ) AD 2 = ar ( ∆PQR ) PE 2 16 AD 2 = 25 PE 2

4 AD = 5 PE AD 4 Or, = PE 5 Hence proved. 14. Through the midpoint M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that EL = 2BL. (CBSE 2023, 2024) Sol. In ∆BMC and ∆EMD, we have ∠CMB = ∠DME (Vertically opposite angles) ∠MBC = ∠MED (Alternate angles) D

(a) ∆ABD ~ ∆CBD

Diagonal BD bisects ∠B and ∠D

To prove:

Proof:

∴ ∆BMC ≅ ∆EMD (By AAS congruency) ⇒ BC = DE (By CPCT) Also, BC = AD (∴ ABCD is a parallelogram) Now, in ∆AEL and ∆CBL, we have ∠ALE = ∠CLB (Vertically opposite angles) ∠EAL = ∠BCL (Alternate angles) ∴ ∆AEL ∼ ∆CBL (By AA similarity) EL AE = BL CB

(CBSE 2024)

Sol: (a) Given:

15. In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠B and ∠D both. Prove that: (b) AB = BC

MC = MD (∴M is the midpoint of CD)

EL 2BC = BL BC (∴ AE = AD + DE = BC + BC = 2BC) EL = 2 BL EL = 2BL Hence proved.

C D 3

∆ABD ~ ∆CBD In ∆ABD and ∆CBD,

2 B

∠ABD = ∠CBD (BD bisects ∠B)

∠ADB = ∠CDB (BD bisects ∠D)

Therefore, ∆ABD ~ ∆CBD (by AA rule) Hence proved.

(b) Since ∆ABD ~ ∆CBD Therefore, AB BC = (by CPCT) BD BD

Hence AB = BC

Hence proved.

Competency Based Questions Multiple Choice Questions 1. Leela has a triangular cabinet that fits under his staircase. There are four parallel shelves as shown below.

y cm

(a) 10 cm

3y cm

(b) 12 cm (c) 14 cm (d) 16 cm

Chapter 6 • Triangles

(d) 86.4 cm

AB 5 = . If XY = 4 AX 3 cm and BY bisects ∠ XYC then AY equals to

2y cm Book

2y cm 144 cm

The total height of the cabinet is 144 cm. What is the maximum height of a book that can stand upright on the bottom-most shelf? (CBSE QB) (a) 18 cm (b) 36 cm

1 5

4 3

163

3. Shown below N 10 cm are two triangles M ∆MNO and ∆PQR . Dimensions of their two sides are marked in the figure.

6 cm Q

15 cm

What should be the value of QR if ∆MNO is similar to ∆PQR ? (CBSE QB) (a) 9 cm

(b) 11 cm

(d) 25 cm

4. Shown below is a rectangle PQRS . ∠UTQ is a right angle.

un its

its un

Which is these is true about ∆UST and ∆TRQ ?

U Q

(a) The perimeters are in the ratio 16 : 81.

Select Response Type MCQs 1. Shown below are two triangles such that the length of two sides of each is known. M Along with the given information, which of these is 3 units sufficient to conclude whether K L P ∆KLM is similar to ∆PQR ? 4 units

(i) ∠KLM = ∠PQR

(d) The ratio of US : QR is 4 : 9.

2x + 10

(d) 4

A 2x

(b) 20

(d) 28

N θ S

164

(7x + 1)°

(3x + 29)°

Which of the following statements helps in proving that triangle AOB is similar to triangle DOC? (i) ∠ B = 70 and (ii) ∠ C = 70 (a) Statement (i) alone is sufficient, but statement (ii) alone is not sufficient.

7. Sonali is standing on one side of a 7 m wide road as shown below. She wants to estimate the distance (D) between two light poles on the other side without crossing the road. D

8 units

60°

6. Two triangles, ∆ABC and ∆PQR, are similar such that their corresponding sides are in the ratio 3 : 5. The sides of ∆ABC form an arithmetic progression (AP) with a common difference of 4. If the perimeter of ∆PQR is 120 units, what is the length of the smallest side of ∆ABC? (a) 4

2. Consider the figure below:

x+7

(d) the given information is enough to conclude that ∆KLM ∼ ∆PQR as the ratio of sides is known.

5. In the given figure, DE || AC. Find the value of x. (a) 1

6 units

3 units

(b) only (ii)

(ii) Ratio of KM : PR M (a) only (i)

(b) The ratio of ∠SUT : ∠RTQ is 4 : 9.

(b) 2

Which of the following expressions represent D in terms of p and r? (CBSE PP 2023) 7r pr m (a) m (b) p 7 r (p + 7 ) pr m m (c) (d) p +7 p

(b) Statement (ii) alone is sufficient, but statement (i) alone is not sufficient. (c) Either (i) or (ii) statement alone is sufficient. (d) Both statements together are sufficient, but neither statement alone is sufficient.

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and Reason (R). Evaluate these statements and select the appropriate option:

At the same time, the tower casts a shadow 50 m long on the ground and the house of Ajay casts 20 m shadow on the ground.

AD AE = . AB AC

Reason (R): If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. 2. Assertion (A): ABCD is a trapezium with DC || AB. E and F are points on AD and BC respectively, such AE BF that EF || AB. Then = ED FC

Vijay's House

Ajay's House

Tower

(a) What is the height of the tower? (i) 20 m (iii) 100 m

(ii) 50 m (iv) 200 m

(b) What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12 m? (i) 75 m (ii) 50 m (iii) 45 m (iv) 60 m (c) What is the height of Ajay’s house? (i) 30 m (ii) 40 m (iii) 50 m

(iv) 20 m

Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally. (CBSE 2024)

(d) When the tower casts a shadow of 40 m, at the same time, what will be the length of the shadow of Ajay’s house? (i) 16 m (ii) 32 m

Case Study Based Questions

(iii) 20 m

1. Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles. The height of Vijay’s house is 20 m when Vijay’s house casts a shadow 10 m long on the ground.

(iv) 8 m

(v) When the tower casts a shadow of 40 m, at the same time, what will be the length of the shadow of Vijay’s house? (i) 15 m (ii) 32 m (iii) 16 m (iv) 8 m

Answers Multiple Choice Questions 1. (c) 54 cm Given, total height of the cabinet is 144 cm. Let height of the book = h cm. Now, by Thales theorem, 144 − h y + 2 y + 2 y = 3y h 144 − h 5 ⇒ = ⇒ 432 = 8h h 3

2. Given that ∆ABC ∼ ∆AXY, we have: ∠4 = ∠5 Since XY || BC, AB BC = AX XY BC 5 = 4 3 20 BC = cm B 3 Since XY || BC,

1 5

4 3

∠1 = ∠2 (Alternate interior angles)

∴h = 54 cm Chapter 6 • Triangles

165

Also, BY bisects ∠XYC, ∠1= ∠ 3 ∠2=∠3 Thus, BC = CY (In a ∆, sides opposite to equal angles are equal), Let AY = x cm. Since ∆ABC ∼ ∆AXY, we use similarity properties: BC AC = XY AY 20 20 3 = 3+x 4 x 20 80 = + 4x x 3 3 20 80 = + 4x x 3 3 20x − 12x 80 = 3 3 8x = 80 x = 10 cm ∴ Option (a) is correct.

∠SUT = ∠RTQ = x° ∠VST = ∠TRQ = 90° Now, ∆TRQ ∼ ∆UST [by AA criterian]

US 4 ∴ = TR 9 [ ratio of the corresponding sides of two similar triangles is equal] 5. (a) 1 In ∆ABC, DE || AC BD BE ∴ (Basic Proportional theorem) = DA EC 2x + 10 x + 7 ∴ = 3x 2x ⇒ 2x(2x + 10) = 3x(x + 7)

⇒ 4x2 + 20x = 3x2 + 21x

⇒ x2 – x = 0 ⇒ x(x – 1) = 0

⇒x–1=0⇒x=1

6. (b) 20 Given the common difference d = 4, the sides can be expressed as: = (a – 4), a, (a + 4) Since ∆ABC ∼ ∆PQR with a side ratio 3 : 5 their perimeters also follow the same ratio:

3. (a) 9 cm Given, ∆MNO ∼ ∆PQR MN NO MO ∴ = = PQ QR PR

and it is also given, MN = 10 cm NO = 15 cm PQ = 6 cm PR = 12 cm

7. Let us represent the given figure as P is pole 1; Q is pole 2; S represents the position of Sonali. Now, as ∆MNS ∼ ∆PQS (By AA similarity)

On taking first two terms of Eq. (i), we get MN NO PQ × ON 6 × 15 ⇒ QR= = 9 cm ⇒ QR = = PQ QR 10 MN Hence, the length of QR is 9 cm 4. (c) The ratio of US : TR is 4 : 9 Given, UT = 4 units

T 4

Perimeter of ∆ABC 3 = Perimeter of ∆PQR 5 3a 3 ⇒ = ⇒ 15a = 360 ⇒ a = 24 120 5 Sides of ∆ABC = (24 – 4), 24, (24 + 1) = 20, 24, 28 The smallest side = 20

x° 9

MS MN = PS PQ p r ⇒ = p +7 D

Let ∠PQT = x

r (p + 7 ) p θ ⇒D = m p S ∴ Option (d) is correct.

⇒ ∠RTQ = ∠PQT = x  [alternate interior angles]

Select Response Type MCQs

Now, ∠STU + ∠UTQ + ∠RTQ =180 [linear pair]

 ⇒ ∠STU + 90 + x= 180°

⇒ ∠STU = 90 − x  ∠SUT = 180° – LUST – ∠STU ∠SUT = 180° – 90° – (90° – x) ∠SUT = 180° – 90° – 90° + x = x

1. In ∆KLM and ∆PQR ML 1 KL = = RQ 2 PQ

TQ = 9 units

166

U P

x°

Q 7

r N

Thus, either ∠ KLM = ∠ PQR or KM : PR is known, we can conclude about the similarity of the triangles. ∴Option (c) is correct.

2. Statement (i) ∴ ∠ B =70⇒ ∠ A =180− ( 60+70) =50 7x + 1 = 50 ⇒ 7x = 49 ⇒ x = 7 Now, ∠ D =3x + 29 =3 × 7 + 29 =21 + 29 =50 ∴∠ = A ∠ D , ∠ AOB = ∠ COD = 60

1. (a) (iii) 100 m When two corresponding angles of two triangles are similar, then ratio of sides are equal.

Now, if we take statement (ii), ∠C = 70°, then the same result will be found, i.e., ∆AOB ∼ ∆DOC ∴ Option (c) is correct.

Assertion-Reason Based Questions

1. (a) Both A and R are true, and R is the correct explanation of A A

We find that statement-2, being well known BP T, is true. Using statement-2, we obtain D E AD AE DB EC = ⇒ = DB EC AD AE DB EC B C ⇒ +1 = +1 AD AE DB + AD EC + AE AB AC = ⇒ = ⇒ AD AE AD AE AD AE ⇒ = AB AC So, assertion is true. Also, Reason is a correct explanation for assertion. Hence, option (a) is correct

2. (a) Both A and R are true, and R is the correct explanation of A

Given: Trapezium ABCD in which AB || DC and EF || AB.

 EO  AB ( As EF  AB ) ∴

AE BO (i) = ED DO ( By BPT )

E A

Using previous answer, we get:

50 x = ⇒ x = 12 × 5 = 60 10 12

(c) (ii) 40 m Height of Vijay’s House Length of Vijay’s House Shadow Height of Ajay’s house = Length of Ajay’s house shadow

20 m Height of Ajay’s house = 10 m 20 m

Height of Ajay’s house =

20 m Height of tower = 10 m 50 m 20 × 50 1000 Height of tower = = = 100 m. 10 10

(b) (iv) 60 m x Length of tower’s shadow = Length of Vijay’s house shadow 12

∴A B || DC and EF || AB ⇒ EF || CD Now in ∆ABD,

Height of Vijay’s House Length of Vijay’s House Shadow Height of tower = Length of tower’s shadow

F B

Length of tower’s shadow 40 = Length of Ajay’s house shadow y Using previous answers, we get: 50 40 40 × 20 = ⇒ y= = 16 20 50 y

Similarly, in ∆BCD

(e) (d) 8 m

BF BO = (ii) FC DO ( By BPT )

From (i) and (ii), we get

AE BF = ED FC

Chapter 6 • Triangles

20 × 20 400 = = 40 m. 10 10

(d) (i) 16 m

Construction: Join DB which intersects EF at point O.

Option (a) is correct.

Case Study Based Questions

And the third angle, ∠= B ∠= C 70 ∴∆AOB ∼ ∆DOC ( By AAA similarity criteria )

Thus, both A and R are true and R is the correct explanation of A.

Length of tower’s shadow 40 = Length of Vijay’s house shadow z Using previous answer, we get: 50 40 40 × 10 = ⇒ z= = 8m 10 50 z

167

Additional Practice Questions 1. In the given figure, ∠A = ∠C, AB = 6 cm, AP = 12 cm, CP = 4 cm. Then length of CD is (CBSE 2023) A

medians of ∆ABC and ∆PQR . If ∆ABC ∼ ∆PQR then prove that: C

12 cm

6 cm

(a) 2 cm (c) 3 cm

(b) AD × PR = AC × PS

7. In figure, ∆FEC ≅ ∆GBD and ∠1 = ∠2. Prove that ∆ADE ∼ ∆ABC. A

3.2 cm 2 cm

4.8 cm R

x cm

(a) 7 cm

(b) 6 cm

(d) 2.4 cm

3. In ∆ABC and ∆PQR we have AB = 4.5 cm, BC = 5 cm, CA = 6√2 cm, PQ = 10 cm, QR = 9 cm, PR = 12√2 cm. If ∠A = 75° and ∠B = 55°, then ∠P = (a) 75° (c) 50°

1 2

ar ( ∆ABC )

ar ( ∆DBC )

50° 3 cm

5 cm 30° P 2.5 cm

(a) 50°

(b) 30°

(d) 100°

5. If ∆PQR ∼ ∆ABC = ; PQ 6 = cm, AB 8 cm and the perimeter of ∆ABC is 36 cm, then the perimeter of ∆PQR is (a) 20.25 cm (c) 48 cm

(b) 27 cm (d) 64 cm

6 cm

AO . DO

(b) 55° (d) 130°

8. In the given figure, ∆ABC and ∆DBC are on the same base BC . If AD intersects BC at O, prove that

4. In the figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to:

168

(a) ∆ADC ∼ ∆PSR

2. In the given figure, value of x (in cm) is:

(b) 6 cm (d) 18 cm

2.4 cm

4c D m

3.6 cm

6. In the given figure, CD and RS are respectively the

9. Through the vertex D of a parallelogram ABCD , a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that DA FB FC = = AE BE CD 10. In Figure, if ∆ABC ∼ ∆DEF and their sides of length (in cm) are marked along them, then find the lengths of sides of both triangle. A 2x – 1

D 3x

2x + 1

18 C

6x 3x + 1

Challenge Yourself

Scan me

1. In Figure, PA , QB and RC are each perpendicular to 1 1 1 AC. Prove that + =. x z y P

F B

4 cm

Q y A

z C

2. In Figure, we have AB  CD  EF . If = AB 6= cm, CD x= cm, EF 10= cm, BD 4 cm and DE = y cm, calculate the values of x and y .

10 cm

y cm

x cm

R x

6 cm A

3. If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar.

Answers Additional Practice Questions 1. (a) 2 cm

Challenge Yourself = = cm; y 5.17 cm 2. x 7.75

2. (c) 3 cm 3. (c) 50° 4. (d) 100° 5. (b) 27 cm 10. AB = 9 cm, BC = 11 cm, AC = 15 cm; DE = 18 cm, EF = 16 cm, DF = 30 cm

Scan me for Exemplar Solutions Chapter 6 • Triangles

169

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions 1. In the given figure, DE  BC . If AD = 2 units, DB = AE = 3 units and EC = x units , then the value of x is (CBSE 2023) (a) 2

(b) 3

(d)

[3 × 1 = 3 Marks]

A 2

E x

9 2

2. Ankit is 5 feet tall. He places a mirror on the ground and moves until he can see the top of a building. At the instant when Ankit is 2 feet from the mirror, the building is 48 feet from the mirror. How tall is the building? (a) 96 feet

(b) 120 feet

(d) 240 feet

3. In ∆ABC , AD = DB , DE is parallel to BC and the area of ∆ABC is 40 sq. units. The area of ∆ADE is (a) 10

(b) 15

(d) 30 B

Assertion-Reason Based Questions

C [2 × 1 = 2 Marks]

(b) Both A and R are true, but R is not the correct explanation of A.

(d) A is false, but R is true.

A 3 cm D

BC , AD 3= cm. BD 4 cm and BC = 14 cm, then DE 4. Assertion (A): In the given figure, if DE = equals 6 cm. Reason (R): In ∆ABC , if DE  BC then ∆ADE ∼ ∆ABC .

4 cm B

14 cm

16 cm

5. Assertion (A): Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square 16 FDGB that can be inscribed in the triangle ABE is cm. 3 Reason (R): In two similar triangles, the ratio of their perimeters need not be equal to the ratio of their sides.

G 8 cm

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. In the given figure, AD = 2 cm, BD = 3 cm, AE = 3.5 cm, and AC = 7 cm. Is DE parallel to BC ?

(CBSE SQP 2020)

7. It is given that ∆DEF ∼ ∆RPQ . Is it true to say that ∠D = ∠R and ∠F = ∠P ? D

170

(NCERT Exemplar)

Short Answer Questions

[4 × 3 = 12 Marks]

8. ABCD is a quadrilateral; P, Q, R, and S are the points of trisection of sides AB, BC, CD, and DA respectively and are adjacent to A and C; prove that PQRS is a parallelogram. 5 cm

AO BO 1 9. In the given figure, = = and AB = 5 cm. OC OD 2

C Q

S A

Find the value of DC C

10. Ankit of height 160 cm is going away from the lamp post at a speed of 2 m /sec. If the lamp post is 3.2 m above the ground, find the length of his shadow after 5 seconds

– 3x

3x − 19, OB = x − 4, OC = x − 3, and OD = 4, find x . 11. In given figure, AB  CD . If OA =

3 x–

4 19 O

x–

Long Answer Questions

B [3 × 5 = 15 Marks]

12. In the given figure, CD is the perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD at G.

E C

Prove that: CF = FG CD DG

F G

(CBSE 2023)

A D

13. In the given figure, ∠ ABC = ∠ ACB and

BC BD = BE AC

Show that ∆ABE ∼ ∆DBC and AE  DC

(CBSE 2024)

14. In the given figure ∠1 =∠2 and ∠3 =∠4 . Show that PT . QR = PS .ST . 4

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A boy is standing on the top of a lighthouse. He observed that Boat P and Boat Q are approaching the lighthouse from opposite directions. He finds that the angle of depression of Boat P is 45° and the angle of depression of Boat Q is 30°. He also knows that the height of the lighthouse is 100 m. Based on the above information, answer the following questions. (a) What is the measure of ∠APD? (b) If ∠YAQ = 30°, then ∠AQD is also 30°. Why?

(ii) How far is Boat Q from the lighthouse?

Self-Assessment

30°

100 m

A 45°

Q (CBSE SQP 2023)

171

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

1. In given figure, two line segments AC and BD intersect each other at the point P

6 cm

(a) 50 2. If ∆ABC ∼ ∆QRP , and

(NCERT Exemplar)

(b) 30 ar ( ∆ABC )

ar ( ∆QRP )

(a) 10 cm

D 30°

P 50° 2.5 cm m C 3c

30 . such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50 and ∠CDP = Then, ∠PBA is equal to

5 cm

(d) 100

9 with AB = 18 cm and BC = 15 cm, then PR is equal to: 4

(b) 12 cm

(c)

20 cm 3

(d) 8 cm

3. If S is a point on side PQ of a ∆PQR such that PS = QS = RS, then (a) PR · QR = RS²

(b) QS² + RS² = QR²

(d) PS² + RS² = PR²

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(b) Both A and R are true, but R is not the correct explanation of A.

(d) A is false, but R is true.

4. Assertion (A): In ∆ABC , DE  BC such that AD = ( 7x − 4 ) cm, AE = ( 5x − 2 ) cm, DB = ( 3x + 4 ) cm and EC = 3x cm then x equal to 5 Reason (R): If a line is drawn parallel to one side of a triangle to intersect the other two sides in different point, then the other two sides are divided in the same ratio

A 3 cm

5. Assertion (A): In the given figure, if PQ  BC then x = 4 cm.

Reason (R): Two similar triangles are always congruent.

Very Short Answer Questions

7. In the figure, if ∠D = ∠C, then is it true that ∆ADE ∼ ∆ACB? Why?

12 cm

5 cm

P C

6 cm

10 cm

Short Answer Questions

[4 × 3 = 12 Marks]

8. Infigure, ABCD is a trapezium with AB || DC. If ∆AED is similar to ∆BEC, prove that AD = BC.

D 172

[2 × 2 = 4 Marks]

E B

1 cm

12 cm

6. In the figure, BD and CE intersect each other at the point P. Is ∆PBC ∼ ∆PDE? Why?

2x +1

9. In figure, ∆FEC ≅ ∆GDB and ∠1 = ∠2. Prove that ∆ADE ∼ ∆ABC.

1 2

D D

(CBSE 2024)

10. In Figure., P is the mid-point of EF and Q is the mid-point of DP. 1 If EQ when produced meets DF at R , prove that RD = DF . 3

AD AE 11. In figure, ∠D = ∠E and . prove that BAC is an isosceles triangle. = DB EC

E (CBSE 2020)

Long Answer Questions

C (3 × 5 = 15 Marks)

12. Prove that the area of the equilateral triangle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle. A

13. In the figure, AP = 3 cm, AR = 4.5 cm, AQ = 6 cm,

AB = 5 cm, and AC = 10 cm. Find the length of AD.

A B

14. In the figure, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ || BA and PR || BD. Prove that QR || AD.

Case Study Based Questions

(CBSE 2012)

(1 × 4 = 4 Marks)

15. In a room a bulb is fixed at a point O on the ceiling. Just below the bulb a large table is placed as shown in given figure. A cardboard is cut in the form of quadrilateral ABCD and is fixed between the bulb and the table. When bulb is switched on, shadow A ′B ′C ′D ′ of cardboard ABCD is formed on the top of the table such that quadrilateral A ′B ′C ′D ′ is an enlargement of quadrilateral ABCD with scale factor 1: 2. If AB = 1.5 cm, BC = 2.5 cm, CD = 2.4 cm, A 105 , ∠= B 100 , ∠= C 70 and ∠D = AD = 2.1 cm, ∠= 85 cm , answer the following questions: (a) The measurement of ∠A ′ is (i) 105

(ii) 100

(iii) 70

(iv) 80

(b) The sum of the angles ∠A ′ and ∠C ′ of quadrilateral A ′B ′C ′D ′ is (i) 185 (ii) 205 (iii) 175

(iv) 155

(iii) 10 cm

(d) The length of side A ′B ′ of quadrilateral A ′B ′C ′D ′ is (i) 1.5 cm (ii) 3 cm

(iii) 2.5 cm

85°

(iv) 5 cm

OR (e) The sum of the angles C ′ and D ′ of quadrilateral A ′B ′C ′D ′ is (i) 105 (ii) 100 (iii) 155 Self-Assessment

100° B

105° A

(iv) 17 cm

70°

C C'

(iv) 140 173

Coordinate Geometry

Chapter at a Glance Introduction to Coordinate Geometry ●

●

y-axis

artesian Plane: A two-dimensional plane divided into four quadrants by the C x-axis (horizontal) and y-axis (vertical). oordinates of a Point: Denoted as (a, b), where a is the abscissa (distance C along the x-axis) and b is the ordinate (distance along the y-axis).

a b

(a, b) b

x-axis

O (Origin)

Distance Formula ●

It is used to find the distance between two points (x1, y1) and (x2, y2) in a plane. d =

y-axis

( x 2 − x1 )2 + ( y 2 − y1 )2

Q (x2, y2)

Example: Find the distance between the points ( 3, 4 ) and ( 7, 1).

(x1, y1)

Here two points: ( x1 , y1 ) = (3, 4), ( x 2 , y 2 ) = (7, 1) d=

(7 − 3)2 + (1 − 4)2 =

16 + 9=

T y2 y1

25= 5

x1 O

x-axis

Distance from the Origin ●

d The distance of a point ( x , y ) from the origin ( 0, 0 ) is given by:=

x2 + y2

Example: Find the distance from the origin to the point P ( 3, 4 ). d = 32 + 42 = 9 + 16 = 5

Collinearity of Points ●

Three points A, B and C are collinear if AB + BC = AC.

Section Formula ●

●

I t is used to find the coordinates of a point dividing a line segment joining two points ( x1 , y1 ) and ( x 2 , y 2 ) in a given ratio m : n. Formula:  

174

mx 2 + nx1 my 2 + ny1  , m +n m + n 

Example: F ind P ( x , y ) which divides the line joining A (1, 2 ) and B ( 7, 8 ) in the ratio 4 : 3.

Here, two points ( x1 , y= 1)

x = = x Note:

m 4,= n 3 (1, 2 ) , ( x 2 , y= ( 7, 8 ) ;= 2)

mx 2 + nx1 my 2 + ny1 = ;y m +n m +n

( 4 × 7 ) + ( 3 × 1) 4+3

31 = ;y 7

( 4 × 8) + (3 × 2) 4+3

 31 38  38 ⇒ P (x, y ) =  ,  7  7 7 

 x + x 2 y1 + y 2  For the midpoint, the ratio is 1:1 , so the midpoint formula becomes:  1 , 2   2

NCERT Zone NCERT Exercise 7.1 1. F ind the distance between the following pairs of points: (a) (2, 3), (4, 1)

(b) (−5, 7), (−1, 3)

Sol. Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively.

Let A =(1, 5 ) , B =( 2, 3 ) , C =( −2, − 11)

Sol. (a) Distance between the two points is given by

∴ AB =

( x1 − x 2 )2 + ( y1 − y 2 )2

Therefore, distance between (2, 3) and (4, 1) is given by l =

( 2 − 4 ) + ( 3 − 1)

l = ( −2 ) + ( 2 ) l = 8l = 2 2

(b) Distance between ( −5, 7 ) and ( −1, 3 ) is given by l =

( −5 − ( −1) ) + ( 7 − 3)2

= l

(a − ( −a ) ) + (b − ( −b ) )

= l

2 b ) l 2 a2 + b2 ( 2a )2 + ( 2=

2. F ind the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B. Sol. Distance between (0, 0) and (36, 15) is given by 2 2 l = ( 36 − 0 ) + (15 − 0 )

= l 1296 + 225l = 1521l = 39 Yes, we can find the distance between the given towns A and B. Assume town A at origin point (0, 0). Therefore, town B will be at point (36, 15) with respect to town A. And hence, as calculated above, the distance between town A and B will be 39 km. 3. D etermine if the points (1, 5), (2, 3) and (−2, −11) are collinear. Chapter 7 • Coordinate Geometry

2 AB = ( −1) + ( 2AB )2=

Similarly BC = = BC

= CA

( 2 − ( −2 ) ) + ( −3 − ( −11) )

= ( 4 )2 + (14BC )2

Same way CA=

1+ AB 4= 5

16 + 196 BC = 212 2

(1 − ( −2 ) ) + ( 5 − ( −11) )

2 ( 3)2 + (16CA )=

9 + 256 CA = 265

Since AB + BC ≠ AC

l = ( −4 )2 + ( 4 )2l = 32l = 4 2 2

(1 − 2 )2 + ( 5 − 3)2

Therefore, the points (1, 5), (2, 3) and (–2, –11) are not collinear. 4. C heck whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle. Sol. L et the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given triangle respectively. ∴ AB= AB =

( 5 − 6 )2 + ( −2 − 4 )2 ( −1)2 + ( −6AB )2=

Similarly BC = BC =

( 6 − 7))2 + ( 4 − ( −2 ) )

( −1)2 + ( −6BC )2=

Same way CA = 2

1 +AB 36 = 37 2

1 +BC 36 = 37

( 5 − 7 )2 + ( −2 − ( −2 ) ) 2

CA = ( −2 ) + ( 0CA )=

4 CA +0=2

Therefore, AB = BC Thus, ABC is an isosceles triangle.

175

5. I n a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct. 10 9

2 1

4 5 6 Columns

9 10

( 3 − 6 )2 + ( 4 − 7 )2

Similarly BC =

9 CB + 9 = 18

( 3 − 6 )2 + ( 4 − 1)2 2

AD = ( −3 ) + ( 3AD )= Diagonal AC = 2

= BD

9AD + 9 = 18

( 3 − 9 )2 + ( 4 − 4 )2 2

= AC = ( −6 ) + ( 0AC ) Diagonal BD =

9 BC + 9 = 18

( 9 − 6 )2 + ( 4 − 1)2

2 ( 3)2 + ( 3CB )=

Same way AD =

9 AB + 9 = 18

( 6 − 9))2 + ( 7 − 4 )2

2 ( −3)2 + ( −3BC )=

Same way CB = = CB

36 + 0 = 6

( 6 − 6 )2 + ( 7 − 1)2

2 ( 0 )2 + ( 6BD )=

0 +BD 36 = 6

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length. Therefore, ABCD is a square and hence, Champa is correct. 6. N ame the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (a) (−1, −2), (1, 0), (−1, 2), (−3, 0) (b) (−3, 5), (3, 1), (0, 3), (−1, −4) (c) (4, 5), (7, 6), (4, 3), (1, 2) Sol. (a) Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B, C, and D of the 176

Similarly BC=

AD =

Sol. I t can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.

2 ( −3 )2 + ( −3AB )=

2 ( −2 )2 + ( −2AB )=

AC =

4 BC +4 = 8 2

( −1 − ( −3) ) + ( 2 − 0 )2 4CD +4 = 8 2

( −1 − ( −3) ) + ( 2 − 0 )2

2 ( 2 )2 + ( −2AD )=

4 AD +4 = 8 2

( −1 − ( −1) ) + ( −2 − 2 )2

2 ( 0 )2 + ( −4AC )=

Diagonal BD=

= BD

2 ( 2 )2 + ( 2CD )=

Diagonal AC =

4+ AB 4= 8

(1 − ( −1) ) + ( 0 − 2 )2

2 ( 2 )2 + ( −2BC )=

Same way AD =

BC =

= CD

AB =

( −1 − 1)2 + ( −2 − 0 )2

Same way CD =

6 Rows 5

∴ AB =

BC =

8 7

given quadrilateral respectively.

0 + 16 = 4 2

(1 − ( −3) ) + ( 0 − 0 )2

= ( 4 )2 + ( 0BD )2

16 BD +0 = 4

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square. (b) Let the points (−3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

( −3 − 3)2 + ( 5 − 1)2

∴ AB =

AB AB = ( −6 ) + ( 4= )

Similarly BC =

BC =

AD =

9 BC + 4 = 13 2

( 0 − ( −1) ) + ( 3 − ( −4 ) )

(1)2 + ( 7CD )2=

Same way AD =

36 +AB 16 = AB 52 = 2 13

( 3 − 0))2 + (1 − 3)2

2 ( 3)2 + ( −2BC )=

Same way CD =

= CD

1 +CD 49 = CD 50 = 5 2 2

( −3 − ( −1) ) + ( 5 − ( −4 ) )

2 ( −2 )2 + ( −9AD )=

4 +AD 81 = 85

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle. (c) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

( 4 − 7 )2 + ( 5 − 6 )2

∴ AB = AB =

2 ( −3)2 + ( −1AB )=

2 ( 3)2 + ( 3BC )=

Same way CD = = CD

±6 36 ⇒ ( y + 3 ) = ( y + 3 )2 =

9BC + 9 = 18

y +3= 6 or y + 3 =−6 y 3 or −9 Therefore,=

9CD + 9 = 18

9. I f Q (0, 1) is equidistant from P (5, −3) and R (x, 6),

( 4 − 4 )2 + ( 5 − 3 )2

find the values of x. Also find the distance QR and PR.

2 ( 0 )2 + ( 2AD )=

0 AD + 4 =AD 4 =2 2

( 7 − 1) + ( 6 − 2 )

Diagonal CD = = AC

2 2 64 + ( y + 3 ) = 100 ⇒ ( y + 3 ) = 100 − 64

( 4 − 1)2 + ( 3 − 2 )2

2 3 )= ( 3)2 + (CD

Same way AC = = AD

Squaring on both sides, we get

9AB + 1 = 10

( 7 − 4 )2 + ( 6 − 3 )2

Similarly BC = = BC

( −8 )2 + ( 3 + y )2 = 10

= ( 6 )2 + ( 4AC )2

Sol. PQ = QR

2 ( 5 − 0 )2 + ( −3 − 1)=

36 +AC 16 = AC 52 = 2 13

It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram. 7. F ind the point on the x-axis which is equidistant from (2, −5) and (−2, 9). Sol. W e have to find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be (x , 0) . Distance between ( x , 0 ) and ( 2, − 5 ) 2

( x − 2 ) + ( 0 − ( −5) ) = ( x − 2 ) + ( 5) Distance between ( x , 0 ) and ( −2, 9 ) 2

( x − ( −2 ) + ( 0 − ( 9) ) = ( x + 2 ) + ( 9) 2

( x + 2 )2 + ( 9 )2

Squaring on both side, we get 2

( x − 2 ) + ( 5) = ( x + 2 ) + ( 9)

x 2 + 4 − 4x + 25 = x 2 + 4 + 4x + 81 8= x 25 − 81 ⇒ 8x = −56 ⇒ x = −7 Therefore, the point is (-7, 0). 8. F ind the values of y for which the distance between the points P (2, −3) and Q (10, y ) is 10 units. Sol. I t is given that the distance between ( 2, − 3 ) and (10, y ) is 10. Therefore,

( 2 − 10 )2 + ( −3 − y )2 =10

Chapter 7 • Coordinate Geometry

25 + 16 =

( −x )2 + ( −5)2

41 x 2 + 25 ⇒ =

x 2 + 25

Squaring on both sides, we get 41 = x 2 + 25 ⇒ 16 = x 2 ⇒ x = ±4

Therefore, point R is ( 4, 6 ) or ( −4, 6 ). When point R is ( −4, 6 )

When point R is (4, 6)

(5 − ( −4 ) + ( −3 − 6 )

PR = = =

QR = =

( 5 − 4 )2 + ( −3 + 6 )2

PR =

= 12 + 32 ( 9)2 + ( −9 )2 = 1+ 9

81 + 81

= 9 2

By the given condition, these distances are equal in measure. ( x − 2 ) + ( 5) =

( 5)2 + ( −4 )2 =

( 0 − x )2 + (1 − 6 )2

= 2

(0 − ( −4 ) + (1 − 6 )

( −7 − 0 )2 + ( 6 − 1)2

QR =

= 42 + 52

( 4 )2 + ( −5)2

= 16 + 25 =

= 16 + 25 = 41

10. F ind a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4). Sol. P oint ( x , y ) is equidistant from ( 3, 6 ) and ( −3, 4 ) . 2 2 ∴ ( x − 3) + ( y − 6 ) =

( x − 3 )2 + ( y − 6 )2 =

( x − ( −3) ) + ( y − 4 )2 ( x + 3 )2 + ( y − 4 )2

Squaring on both sides, we get

( x − 3 )2 + ( y − 6 )2 = ( x + 3 )2 + ( y − 4 )2 x 2 + 9 − 6x + y 2 + 36 − 12 y = x 2 + 9 + 6x + y 2 + 16 − 8 y 36 − 16 = 6x + 6x + 12 y − 8 y = 20 12x + 4 y ⇒ 3x + y = 5 ⇒ 3x + y − 5 = 0

177

NCERT Exercise 7.2

each other along AD, as shown in figure. Niharika

1. F ind the coordinates of the point which divides the join of ( –1, 7 ) and ( 4, –3 ) in the ratio 2:3. −1, x 2 = 4, y1 = 7, y 2 = −3, m1 = 2, m2 = 3 Sol. Let x1 =

Using section formula, we get = x

m1x 2 + m2 x1 = m1 + m2

y =

m1 y 2 + m2 y1 = m1 + m2

1)} 8 − 3 {2 × 4 + 3 × ( −= = 1 2+3

{2 × ( −3) + 3 × 7=} −6 + 21 15 = = 3 2+3

Sol. I t can be observed that Niharika posted the green 1 flag at th th of the distance AD i.e., there are 100 4 flower pots. It means she stops at 25th flower pot.

Therefore, the coordinates of point is (1, 3). 2. F ind the coordinates of the points of trisection of the line segment joining ( 4, –1) and ( –2, –3 ).

The coordinates of point where she stops are (2 m, 25 m). 1 Similarly, Preet posted red flag at th of the 5 distance AD i.e., there are 100 flower pots. It means she stops at 20th flower pot.

Sol. L et C ( x1 , y1 ) and D ( x 2 , y 2 ) are the points of trisection of the line segment joining the given points i.e., AC = CD = DB Therefore, point C divides AB internally in the ratio 1: 2. Using section formula to find coordinates of point C which divides join of (4, -1) and (-2, -3) in the ratio 1:2. We get

{1× ( −2 ) + 3 × 4}= −2 + 8= 6= 2 x1= 1+ 2 3 3

= y1

= y2

−1)} −6 − 1 −7 {2 × ( −3) + 1 × (= =

1+ 2

1+ 2 3 3 5  Therefore, coordinates of point C are  2, −  and 3  7 coordinates of point D are  0, −  . 3  3. T o conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from 178

( 2 − 8 )2 + ( 25 − 20 )2

Rashmi posts blue flag exactly halfway the line

Using section formula to find coordinates of point D which divides join of ( 4, − 1) and ( −2, − 3 ) in the ratio 2 :1. We get

{2 × ( −2 ) + 1× 4}= −4 + 4= 0= 0

Using distance formula to find distance, we get

x 2=

The coordinates of point where she stops are (8, 20).

= ( −6 ) + 52 = 36 + 25 = 61 m

( −1)} −2 − 2 −5 {1× ( −3) + 2 × = = 1+ 2

1 th the distance AD on the 2nd line and posts 4 1 a green flag. Preet runs th the distance AD on the 5 eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? runs

segment joining the two flags. Using section formula, we get 2 + 8 10 = x = = 5 2 2 25 + 20 45 = y = 2 2

Therefore, coordinates of point, where Rashmi posts 45  her flag  5, .  2 

It means she posts her flag in 5th line after covering 45 = 22.5 m of distance. 2 4. F ind the ratio in which the line segment joining the points ( –3, 10 ) and ( 6, –8 ) is divided by ( –1, 6 ). Sol. L et (−1, 6) divides line segment joining points ( −3, 10 ) and ( 6, − 8 ) in 1 : k Using section formula,

2 1 1 2 3 4 5 6 7 8 9 10

( −3 ) × 1 + 6 × k  −1 = k +1

7. F ind the coordinates of a point A, where AB is the diameter of a circle whose centre is ( 2, –3 ) and B is (1, 4 ).

2 ⇒ −k − 1 = ( −3 + 6k ) ⇒ −7k = −2 ⇒ k = 7 2 Therefore, ratio is :1 which is equivalent to 2 : 7. 7

Sol. L et the coordinates of point A be ( x , y ).

Therefore, ( −1, 6 ) divides line segment joining the

Mid-point of AB is ( 2, − 3 ) , which is the centre of the circle.

points ( −3, 10 ) and ( 6, − 8 ) in ( 2 : 7 ) . 5. F ind the ratio in which the line segment joining A (1, – 5 ) and B ( – 4, 5 ) is divided by the x-axis. Also find the coordinates of the point of division? Sol. L et the ratio in which the line segment joining A (1, − 5 ) and B ( −4, 5 ) is divided by x-axis be k :1 Acc. to section formula,

1 × 1 + ( −4 ) × k  1 − 4k = x = k +1 k +1 ( −5 ) × 1 + 5k   and y = k +1

8. I f A and B are ( –2, –2 ) and ( 2, – 4 ) , respectively, find the coordinates of P such that AP = (i)

( −5 ) × 1 + 5k  1 = ⇒ 5 5k ⇒ k = k +1 Put value of k in (1), we get 0

1 × 1 + ( −4 ) × 1 1 − 4 3 x=  = =− 1+1 2 2  3  Therefore, point  − , 0  on X axis divides line  2  segment joining A (1, − 5 ) and B ( −4, 5 ) in1 :1. 6. I f (1, 2 ) , ( 4, y ) , ( x , 6 ) and ( 3, 5 ) are the vertices of a parallelogram taken in order, find x and y. Sol.= L et A

= 2 ) , B ( 4, = y ) , C ( x , 6 )= and D ( 3, 5 ) (1,

Using section formula, coordinates of midpoint AC are: 1+ x 2 + 6 1+ x , = ,4 2 2 2 Using section formula, coordinates of midpoint BD are: 4+3 5+ y 7 5+ y , = , 2 2 2 2 According to parallelogram properties, we have 1+ x 7 = = ⇒ (1 + x ) 7 = ⇒x 6 2 2 Again, according to properties, we have 5+ y 4= ⇒ 8 =5+ y ⇒ y =3 2 = x 6= and y 3 Therefore,

Chapter 7 • Coordinate Geometry

Using section formula, x +1 2= ⇒ 4 = x +1 ⇒x =3 2 Using section formula, we get 4+y −3 = ⇒ −6 = 4 + y ⇒ y = −10 2 Therefore, coordinates of point A are (3, − 10) .

on the line segment AB.

3 AB and P lies 7

Sol. A =( −2, − 2 ) and B =( 2, − 4 ) 3 It is given that AP = AB 7 3 4 PB = AB − AP = AB − AB = AB 7 7 We have AP : PB = 3 : 4 Let coordinates of P be (x, y) Using section formula to find coordinates of P, we get ( −2 ) × 4 + 2 × 3 6 − 8 2 x= = =− 3+4 7 7 ( −2 ) × 4 + ( −4 ) × 3 −8 − 12 20 = =− y= 3+4 7 7 2 20   Coordinates of P are  − , −  7   7 9. F ind the coordinates of the points which divide the line segment joining A ( –2, 2 ) and B ( 2, 8 ) into four equal parts. Sol. L et coordinates of point P = ( x1 , y1 ), Q = ( x 2 , y 2 ) and R = ( x 3 , y 3 ) We know AP = PQ = QR = RS It means P divides line segment AB in 1:3. Using section formula to find coordinates of P, x1 =

{ ( −2 ) × 3 + 2 × 1} = −6 + 2 = −4 = −1

1+ 3 4 4 {2 × 3 + 8 × 1} 6 + 8 14 7 = y1 = = = 1+ 3 4 4 2

179

10. F ind the area of a rhombus if its vertices are

Since, AP = PQ = QR = RS

( 3, 0 ) , ( 4, 5) , ( –1, 4 ) and ( –2, –1) taken in order. [Hint :

It means, Q is midpoint of AB. Using section formula to find coordinates of Q, x 2= y= 2

{( −2 ) × 1 + 2 × 1}= −2 + 2= 0= 0 1+ 1

{2 × 1 + 8 × 1}

1 (product of its diagonals)] 2 Sol. L et A =( 3, 0 ) , B =( 4, 5 ) , C =( −1, 4 ) and D =( −2, − 1) Area of a rhombus =

Using distance formulas to find length of diagonal AC , we get

2 + 8 10 = = = 5 1+ 1 2 2

AC =

= PQ = QR = RS Because AP It means R divides line segment AB in 3:1. Using section formula to find coordinates of P, x 3=

{( −2 ) × 1 + 2 × 3}= −2 + 6= 4= 1

1+ 3 4 4 {2 × 1 + 8 × 3} 2 + 24 26 13 y= = = = 3 1+ 3 4 4 2

(a) x-axis

(b) y-axis

(d) None of these

2. The distance between ( 8, 17 ) and ( 3, 5 ) is: (b) 5

(d) 25

3. A bird starts its flight from a tree at point (7,24) and heads straight to its nest at the origin (0,0). What is the distance it travels? (CBSE 2024) (a) 25

(b) 49

(d) 7

4. A point equidistant from ( 0, 4 ) and ( 0, − 4 ) lies: (NCERT Exemplar) (a) On the y-axis

(b) At the origin

(d) None of these

5. The midpoint of the line joining the points ( 2, 3 )

and ( 4, 5 ) is: (CBSE 2024) (a) (3, 4)

(b) (3, 3)

6. The coordinates of a point dividing the line joining (CBSE 2025) ( 3, 4 ) and ( 6, 8 ) in the ratio 1:1 are: (a) (4.5, 6)

(b) (4, 6)

(d) (5, 6)

7. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the midpoint of PQ, then the coordinates of P and Q are respectively. 180

= 16 + 16

= 4 2 32

Using distance formula to find length of diagonal BD, we get

Multiple Choice Questions

(a) 12

42 + ( −4 )=

BD =

7   13  Therefore, P =  −1,  , Q = ( 0, 5 ) and R =  1,  2    2

1. The point ( 7, 0 ) lies on:

3 − ( −1)  + ( 0 − 4 )

36 + 36 =

72 = 6 2

 4 − ( −2)  + 5 − ( −1) 

62 + 62 =

Area of rhombus =

1 (product of its diagonals) 2

1 1 = AC × BD = × 4 2 × 6 2 = 24 sq. units 2 2 (a) (0, -5) and (2, 0)

(b) (0, 10) and (-4, 0)

(d) (0, -10) and (4, 0) (NCERT Exemplar)

8. XOYZ is a rectangle with vertices X (-3, 0), O(0, 0), Y (0, 4) and Z (x, y). The length of its each diagonal is (a) 5 units

(b) 5 units

(d) 4 units

9. The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3)is: (a) (0, 1)

(b) (0,-1)

(d) (1, 0)

10. The area of a triangle with vertices A(3,0), B(7,0) and C (8, 4) is: (NCERT Exemplar) (a) 14

(b) 28

(d) 6

11. The point which divides the line segment joining

the points ( 7, − 6 ) and ( 3, 4 ) in ratio 1:2 internally lies in the

(NCERT Exemplar)

(a) I quadrant

(b) II quadrant

(d) IV quadrant

12. The point which lies on the perpendicular bisector of the line segment joining the points A(-2, -5) and B (2, 5)is (NCERT Exemplar) (a) (0, 0)

(b) (0, 2)

(d) ( −2, 0 )

13. If P  a , 4  is the mid-point of the line segment 3  joining the points Q ( −6, 5 ) and R ( −2, 3 ) , then the value of a is (NCERT Exemplar) (a) -4

(b) -12

(d) -6

14. If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then (NCERT Exemplar) (a) a = b

(b) a = 2b

(d) a = −b

15. The vertices of a ∆OAB are O (0, 0), A ( 4, 0 ) and B ( 0, 6 ) . The median AD is drawn on OB . The length of AD is (CBSE 2021) Y

(a) 52 units

B (0, 6)

(b) 5 units (c) 25 units

(d) 10 units X'

(4, 0) O

Answers

1. (a) x-axis

2. (c) 13

9. (b) (0,-1)

10. (c) 8

3. (a) 25

4. (b) At the origin

11. (d) IV quadrant

12. (a) (0,0)

5. (a) (3,4)

6. (a) (4.5,6)

13. (b) -12

14. (c) 2a = b

7. (d) (0, -10) and (4, 0)

8. (a) 5 units

15. (b) 5 units

Constructed Response Questions Very Short Answer Questions 1. The point ( −4, 0 ) lies on which axis?

4. What is the midpoint of the line segment joining (CBSE 2023)

Sol. The midpoint M ( x , y ) between ( x1 , y1 ) and ( x 2 , y 2 ) is x1 + x 2 y1 + y 2 Mx = = , My 2 2

Sol. Since the y-coordinate is 0 and the x-coordinate is −4, this point lies on the x-axis. 2. Find the distance of the point ( 6, 8 ) from the origin.

Plugging in ( x1 , y1 ) =

Sol. The distance d from a point ( x , y ) to the origin ( 0, 0 ) x2 + y2

is given by the formula:= d 2

d = 6 +8 =

36 + 64 =

100 = 10 units.

3. What is the distance between ( 2, − 3 ) and ( −1, 4 ) ? Sol. The distance between two points ( x1 , y1 ) and

( x 2 , y 2 ) is:

( 2, − 3) and ( x 2 , y 2 ) = ( −1, 4 ) :

( ( −1) − 2 ) + ( 4 − ( −3) )

= Mx

( −2, 4 ) and ( x 2 , y 2 ) = ( 6, 8 )

−2 + 6 4+8 = 2, = = 6 My 2 2

5. A point divides the line joining ( 2, 3 ) and ( 6, 7 ) in the ratio 3 :1. Find its coordinates. (NCERT Exemplar) Sol. The section formula in the ratio m : n gives the coordinates ( x , y ) as

x =

mx 2 + nx1 my 2 + ny1 = ,y m +n m +n

Plugging in ( x1 , y= 1)

2 2 d = ( x 2 − x1 ) + ( y 2 − y1 )

Plugging in ( x1 , y= 1)

( −2, 4 ) and ( 6, 8 ) ? (CBSE 2024)

x =

m 3,= n 1 ( 2, 3) , ( x 2 , y= 2 ) ( 6, 7 ) ,=

3 × 6 + 1 × 2 18 + 2 3 × 7 + 1 × 3 21 + 3 = == 5, y = = 6 3+1 4 4 4

The coordinates are ( 5, 6 ).

= ( −3 ) + 72 = 9 + 49 = 58

Chapter 7 • Coordinate Geometry

181

6. Find the type of triangle ABC formed whose vertices are A(1, 0), B(-5, 0) and C(-2, 5). (CBSE 2024) Sol. Given A (1, 0 ) , B ( −5, 0 ) , C ( −2, 5 ) Distance between two points ( x1 , y1 ) and ( x 2 , y 2 ) : = (x 2 − x1 )2 + ( y 2 − y1 )2 AB =

( −5 − 1)2 + (0 − 0)2

= ( −6)2 = 36 =6 BC = ( −2 + 5)2 + (5 − 0)2

Sol. To find the length of the median BE in triangle ABC where E is the midpoint of side AC,  x + x 2 y1 + y 2  To find the mid-point E of AC: E  1 , 2   2  3 −1 0 + 3   2 3  E ,= ,  (1,1.5 )  = 2  2 2  2 Finding distance between BE

= 32 + 52 = 9 + 25 = 34 AC = ( −2 − 1)2 + (5 − 0)2 = ( −3)2 + 52 = 9 + 25 = 34 Since two sides are equal: BC = AC =

8. A(3, 0), B(6, 4) and C(–1, 3) are vertices of a triangle ABC. Find length of its median BE.

(x 2 − x1 )2 + ( y 2 − y1 )2

Distance formula: d = d = (1 − 6)2 + (1.5 − 4)2 34

The triangle is an isosceles. 7. In what ratio is the line segment joining the points ( 3, − 5) and ( −1, 6 ) divided by the line y = x ? Sol. Let the point of division be P ( x , x ) , since it lies on the line y = x .

( −5)2 + ( −2.5)2 =

31.25

Hence, the length of the median BE = 31.25 9. Show that the points (−2, 3), (8, 3), and (6, 7) are the vertices of a right-angled triangle. (CBSE 2023) Sol. Let given points are A ( −2, 3 ) , B ( 8, 3 ) , and C ( 6, 7 ) A (-2, 3)

We’ll use the section formula. Let the line divide the segment AB in the ratio m : n, then the coordinates of the point P are:  m ( −1) + n ( 3 ) m ( 6 ) + n ( −5 )  P = ,  m +n m +n   Since this point lies on the line y = x , equating the coordinates: m ( −1) + n ( 3 ) m ( 6 ) + n ( −5 ) = m +n m +n ⇒ −1m + 3= n 6m − 5n m 8 ⇒ −7m = −8n ⇒ = n 7 The line y = x divides the line segment joining ( 3, − 5) and ( −1, 6 ) in the ratio 8 :7 .

C (6, 7)

AB 2 = (8 + 2)2 + (3 − 3)2 = 100 BC 2 =(6 − 8)2 + (7 − 3)2 =4 + 16 =20 AC 2 = (6 + 2)2 + (7 − 3)2 = 64 + 16 = 80 2 Clearly, AB = BC 2 + AC 2

Thus, ∆ABC is a right angle triangle.

10. If the distances of P (x, y) from A (5, 1) and B (–1, 5) are equal, then prove that 3x = 2y. Topper’s Answer

182

B (8, 3)

(CBSE 2017)

11. Point P and Q trisect the line segment joining the points A (-2, 0) and B (0, 8), such that P is nearer to A. Find the co-ordinates of points P and Q.

(CBSE 2019)

Sol. P divides AB in the ratio 1:2. 1 A (-2, 0)

2 P

B (0, 8)

Here, coordinates are A ( −2, 0 ) and B ( 0, 8 ) . P divides AB in the ratio 1:2 mx 2 + nx1 my 2 + ny1  By using section formula:  , m + n   m +n Here, = m 1,= n 2

AC =

(a + 3a )2 + (a − 3a )2

1a 2 + 3a 2 + 2 3a 2 + 3a 2 − 2 3a 2

= = 8a 2 2 2a unit = BC = AC Since, AB Therefore, ABC is an equilateral triangle. 14. If two adjacent vertices of a parallelogram are (3, 2) and (−1, 0), and the diagonals intersect at (2, −5), then find the coordinates of the other two vertices. Sol. Let other two vertices have coordinates ( x , y ) and ( x ′, y ′ ). D (x', y')

C (x, y)

Coordinates of P are  1 × 0 + 2 × ( −2 ) 1 × 8 + 2 × 0   4 8  ,   =  − , . 1+ 2 1+ 2   3 3  

O (2, -5)

Q divides AB in the ratio 2:1.  mx 2 + nx1 my 2 + ny1  By using section formula:  , m + n   m +n Here,= m 2,= n 1  2 × 0 + 1 × ( −2 ) 2 × 8 + 1 × 0   2 16  ,  =  − , . 2 +1 2 +1   3 3   12. Find the coordinates of the midpoint of the line segment joining ( −3, 5 ) and ( 5, − 3 ) . (NCERT Exemplar) Sol. To find the coordinates of the midpoint of a line segment joining two points ( x1 , y1 ) and ( x 2 , y 2 ) , we use the midpoint formula, which is given by:  x + x 2 y1 + y 2  M (x, y ) =  1 , 2   2 For the points ( −3, 5 ) and ( 5, − 3 ) :

5−3 2 −3 + 5 2 M x= = = 1; M y= = = 1 2 2 2 2 Thus, the coordinates of the midpoint are (1,1) . 13. Show that the points ( a , a ) , ( −a , − a ) and − 3a , 3a are the vertices of an equilateral triangle.

)

(

Sol. Let A ( a , a ) , B ( −a , − a ) and C − 3a , 3a

BC = = =

)

(a + a )2 + (a + a )2 = 2 2a unit

Here, AB =

( −a + 3a )2 + ( −a − 3a )2

( 2 + 3a − 2 3a + 2 3a + 3a ) 2

= 8a 2 2 2a unit

B (-1, 0)

O is mid point of AC and BD.

2+ y x +3 =2 and =−5 ⇒ x =1and y =−12 2 2 Again, 0 + y′ −1 + x ′ = 2 and = −5 ⇒ x ′ = 5 and y ′ = −10 2 2 Hence, co-ordinates are (1, − 12 ) and ( 5, − 10 ). ∴

Coordinates of Q are

(

A (3, 2)

15. Name the type of triangle formed by the points A ( −5, 6 ) , B ( −4, − 2 ) and C ( 7, 5 ) . Sol. To find the type of triangle, first we determine the length of all three sides. Now, using distance formula between two points, AB =

( −4 + 5)2 + ( −2 − 6)2 =

12 + ( −8)2

= 1 + 64 = 65 BC = =

(7 + 4)2 + (5 + 2)2 = 121 + 49=

112 + 72

170

CA =

2 (7 + 5)2 + (5 − 6)=

144 += 1

122 + ( −1)2

145

We see that, AB ≠ BC ≠ CA Hence, the required triangle is scalene because all of its sides are not equal i.e., different to each other.

Short Answer Questions 1. If ( a , b ) is the mid-point of the line segment joining the point A (10, − 6 ) and B (k , 4 ) , and a − 2b = 18. Find the value of k and the distance AB.

(NCERT Exemplar)

Chapter 7 • Coordinate Geometry

183

4. What are the coordinates of the point that divides the line joining (1, 2 ) and ( 7, 8 ) in the ratio 4 : 3?

Sol. Since, ( a , b ) is midpoint of (10, − 6 ) and (k , 4 ) Therefore, a =

−6 + 4 10 + k and b = 2 2

Sol. Given points:

= A ( x1 , y1 )

10 + k ⇒a = −1 ,b= 2

m :n = 4 : 3

10 + k − 2 × ( −1) =18 2 ⇒ 10 + k + 4 = 36 ⇒ k = 36 − 14 = 22

18 ⇒ Also given, a − 2b = 2

Distance AB =

So, m = 4 and n = 3 Using section formula, we get:

(k − 10) + (4 + 6)

= = (22 − 10) + (10)

144 += 100

= 244 2 61units

2. F ind the distance between (1, 2 ) and ( 4, 6 ) , and verify if the points form the hypotenuse of a right triangle with the origin. (CBSE SQP 2025) Sol. Using distance formula, we get, 2

( 4 − 1) + ( 6 − 2 ) =

3 +4 =

9 + 16 =

25 = 5

1+ 4 =

Distance from (0,0) to (1,2): d1 =

(1 − 0 )2 + ( 2 − 0 )2 =

12 + 22 =

= 2 ) , B ( x 2 , y 2 ) ( 7, 8 ) (1,

3 × 1) 28 + 3 31 ( 4 × 7 ) + (= =

x= y=

4+3

× 2 ) 32 + 6 38 ( 4 × 8 ) + ( 3= =

 31 38  The coordinates of the required point are  , .  7 7  5. In the given figure ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices. (CBSE 2015, 2017) Y

Distance from (0,0) to (4,6):

d2 =

( 4 − 0 )2 + ( 6 − 0 )2 =

d12 + d 2 = d 2 ⇒

42 + 62 =

( 5 ) + ( 52 ) = ( 5 ) 2

16 + 36 =

⇒ 5 + 52 = 25 Since, it does not hold the condition of Pythagoras. Therefore, the points do not form a right triangle with the origin. 3. A point is equidistant from ( 3, 4 ) and ( 5, 8 ) . Find its x-coordinate when the y-coordinate is 6.

(CBSE 2024)

Sol. Let the required point be P ( x , 6 ) . Since P is equidistant from A ( 3, 4 ) and B ( 5, 8 ) , we set the distances equal:

( x − 3 )2 + ( 6 − 4 )2 =

( x − 5 )2 + ( 6 − 8 )2

( x − 5 )2 + 4 2

( x − 3) + 4 = ( x − 5) + 4

( x − 3)2 =( x − 5)2 x 2 − 6x + 9 = x 2 − 10x + 25 −6x + 9 =−10x + 25 −6x + 10x = 25 − 9 ⇒ 4x = 16 ⇒ x = 4 The point is (4, 6).

184

Sol. Since A is at ( 2, 0 ) and AB = 3, B will be 3 units to the right of A. Thus, B = ( 2 + 3, 0 ) =( 5, 0 ) Let co-ordinates of C be (x, y) Since AC 2 = BC 2 (sides of equilateral triangle) (x − 2)2 + ( y − 0)2 = (x − 5)2 + ( y − 0)2 Or, x 2 + 4 − 4x + y 2 = x 2 + 25 − 10x + y 2 Or, 6x = 21 7 x= 2 9 And, (x − 2)2 + ( y − 0)2 = 2

Squaring both sides to remove the square roots: 2

A (2,0)

⇒ 57 ≠ 25

( x − 3)2 + 4=

7  2 9  − 2 + y = 2  2 9 3 2 9 Or,   + y 2 = 9⇒ +y = 4 2 9 27 Or, y 2= 9 − ⇒ y 2 = 4 4 3 3 Or, y = ± (+ ve sign to be taken) 2 7 3 3  Hence, C =  ,  2 2   

6. Find a point which is equidistant from the points

A ( −5, 4 ) and B ( −1, 6 ) . How many such points are

there? Sol. Let P ( x , y ) be a point

P(x, y)

equidistant from A ( −5, 4 )

⇒ PA = PB 2

⇒ ( x + 5 ) + ( y − 4 ) = ( x + 1) + ( y − 6 ) 2

B(-1, 6) 2

⇒ x + y + 10x − 8 y + 41 = x + y + 2x − 12 y + 37 ⇒ 8x + 4 y + 4 = 0 ⇒ 2x + y + 1 = 0 ... (i) The equation of a line which is the perpendicular bisector of AB. Every point on this line is equidistant from A and B. So, there are infinitely many points equidistant from A and B. Putting x = 1 in (i), we obtain: 2 + y + 1 =0 ⇒ y =−3. Therefore, (1, − 3 ) is equidistant from A and B. Clearly, P ( x , y ) lies on the perpendicular bisector of line segment AB. Therefore, there are infinitely many other points equidistant from A and B.

7. Verify if the point (4, 6) lies on the line joining (2, 3) and (8, 9). Sol. Given: A ( 2, 3 ) , B ( 8, 9 ) , P ( 4, 6 )

(CBSE SQP 2025)

Let ( 4, 6 ) divide ( 2, 3 ) and ( 8, 9 ) in ratio k :1 Using section formula, we get, For x-coordinate

4 + 4 = 2 2 units

BC =

(2 − 0)2 + (0 − 2)2 =

4 + 4 = 2 2 units

(

)

(( −2 − 2)2 + 0 − 0)2 =

16= 4 units

DE =

(0 + 4)2 + (4 − 0)2 =

16 + 16 = 4 2 units

EF =

(4 − 0)2 + (0 − 4)2 =

16 + 16 = 4 2 units

FD=

((−4 − 4) + (0 − 0) )= 2

64= 8 units

Since, the ratio of the corresponding sides of two similar triangles is equal. i.e., AB BC CA 2 2 4 1 = = = = = DE EF FD 4 2 8 2 OR 2 2 2 2 4 1 = = = 4 2 4 2 8 2 Therefore, ∆ABC ∼ ∆DEF 9. If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio 3:4, find the value of x 2 + y 2. Sol. Given that

AC 3 = CB 4

Applying section formula for x coordinate: 3x + 4 ( 2 ) or − 7 = 3x + 8or ⇒ = x = −5 −1 = 3+4 Similarly for y coordinate:

k ( 8 ) + 1( 2 ) ⇒ 4 (k + 1) = 8k + 2 4= k +1 4k + 4 = 8k + 2 ⇒ = 4 4k − 2 ⇒ 6 = 4k 6 3 ... (i) k= = 4 2 Now using the value of k from eqn (i) y-coordinate will be

3y + 4 ( 5) ⇒ or1= 4 3 y + 20 ⇒ or y = −2 2= 3+4 Hence, ( x , y ) is ( −5, − 2 ) and x 2 + y 2 = ( −5)2 + ( −2)2 = 25 + 4 = 29 10. Find the points on the x-axis which are at a distance of 2 5 from the point ( 7, − 4 ) . How many such points are there?

3 ( 9) + 1( 3) 27 + 6 33 2 = = = y 3 5 5 +1 2

Since the y coordinate calculated by using the value of k is not the same as given in the question. Therefore, the point does not lie on the line joining (2, 3) and (8, 9). 8. Show that ∆ABC with vertices A ( −2, 0 ) , B ( 0, 2 ) and C ( 2, 0 ) is similar to ∆DEF with vertices D ( −4, 0 ) , E ( 0, 4 ) , and F ( 4, 0 ) .

(CBSE 2017)

Chapter 7 • Coordinate Geometry

(0 + 2)2 + (2 − 0)2 =

and A(-5, 4)

AB =

CA=

and B ( −1, 6 ) . Then, PA = PB

Sol. Using distance formula, we get,

Sol. Let P ( x , 0 ) the point on the x-axis have 2 5 distance from the point Q ( 7, − 4 ) . By given condition, PQ = 2 5 ⇒ (PQ )2 = 20 (x − 7)2 + (0 + 4)2 = 20 x 2 − 14x + 49 + 16 = 20 ⇒ x 2 − 14x + 45 = 0 [by factorisation method ]

185

Long Answer Questions

x 2 − 9x − 5x + 45 = 0 x ( x − 9) − 5( x − 9) = 0

1. A point divides the line segment joining ( −5, 7 ) and (15, − 3) in the ratio 2 : 3. Find the coordinates of the point and prove it is equidistant from ( 0, 0 ) and (10, 2 ) . (NCERT Exemplar)

0 ⇒ x = 9, 5 ( x − 9)( x − 5) = Hence, there are two points lie on the axis, which are ( 5, 0 ) and ( 9, 0 ) .

Sol. Given points:

11. If the centre of a circle is ( 2a , a − 7 ) , then find the values of a, if the circle passes through the point (11, − 9) and has a diameter of 10 2 units.

A ( x1 , y1 ) = ( −5, 7 ) , B ( x 2 , y 2 ) = (15, − 3) m :n = 2: 3

Sol. Distance between the centre C ( 2a , a − 7 ) and the point P (11, − 9 ) , which lie on the circle = Radius of (11 − 2a )2 + ( −9 − ( a − 7 ))2

circle =

(i)

10 2 = 5 2 2 Put this value in Eq. (i), we get Length of radius =

5= 2

d1 = =

50= 121 − 44a + 4a 2 + 4 + a 2 + 4a

d= 2

0 ⇒ a = 5, 3 (a − 5)(a − 3) =

Hence, the required values of a are 5 and 3. 12. Find the coordinates of the point R on the line segment joining the points P(−1, 3) and Q(2, 5) such 3 that PR = PQ. 5 R,

PQ 5 PR + RQ 5 Thus, = = ⇒ PR 3 PR 3 RQ 5 RQ 5 PR RQ 5 ⇒ 1+ ⇒ = = −1 + = PR 3 PR 3 PR PR 3 RQ 2 = PR 3

Suppose, R(x, y) be the point which divides the line segment joining the points P(−1, 3) and Q(2, 5) in the ratio 3:2.  3 ( 2 ) + 2 ( −1) 3 ( 5 ) + 2 ( 3 )  ,  3+2 3+2  

(x, y ) = 

 6 − 2 15 + 6   4 21  = ,  = ,  5  5 5   5  4 21  Hence, the point R is  ,  . 5 5 

186

−6 + 21 15 = = 3 5 5

( 3 − 0 )2 + ( 3 − 0 )2 = 9+9 =

32 + 32

18 = 3 2

Distance of ( 3, 3 ) from (10, 2 ) is d 2

5a 2 − 40a + 75 = 0 ⇒ a 2 − 8a + 15 = 0

Thus,

Distance of ( 3, 3 ) from ( 0, 0 ) is d1

50 = (11 − 2a )2 + (2 + a )2

2+3

So, the required point is ( 3, 3 ) .

Squaring on both sides, we get

PR 3 = PQ 5

( 2 × −3) + ( 3 × 7 )

= y

(11 − 2a )2 + ( −2 − a )2

Sol. Given that,

Therefore, using section formula the x-coordinate: ( 2 × 15 ) + ( 3 × −5 ) 30 − 15 15 = x = = = 3 2+3 5 5 Therefore, the y-coordinate:

2 (10 − 3)2 + ( 2 − 3)=

49 + 1=

72 + ( −1)

50= 5 2

Since d1 = d2 , the point (3, 3) is equidistant from (0, 0) and (10, 2). 2. A point divides the line joining ( 6, 2 ) and ( −4,10 ) in the ratio 4 :1. Find the coordinates of the point and 12 . verify if it lies on the line 2x + 3 y = Sol. Given: A ( x1 , y1 ) =

( 6, 2 ) , B ( x 2 , y 2 ) = ( −4,10 )

m : n = 4 :1 Using section formula, we get, x-coordinate: ( 4 × −4 ) + (1× 6 ) −16 + 6 −10 = = = −2 x= 4 +1 5 5 y-coordinate: ( 4 × 10 ) + (1× 2 ) 40 + 2 42 = = = = 8.4 y 4 +1 5 5 Thus, the required point is ( −2, 8. 4 ) .

To check if the point ( −2, 8. 4 ) satisfies the equation of the line: 2x + 3 y = 12

Substituting x = −2 and y = 8.4 : 2 ( −2 ) + 3 ( 8.4 ) =−4 + 25.2 =21.2 ≠ 12 Since 21.2 ≠ 12, the point ( −2, 8. 4 ) does not lie on the line. 3. (a) Find the ratio in which the y axis divides the line segment joining the points (6, –4) and (–2, –7). Also, find the point of intersection. (b) S how that the points (7, 10), (–2, 5) and (3, –4) are vertices of an isosceles right triangle. (CBSE 2020) Sol. (a) We are given the points A ( 6, − 4 ) and B ( −2, − 7 ) . Since the y-axis divides the line segment, the x-coordinate of the point of intersection must be 0.

Using section formula, we get,

Setting x = 0 :

0 =

m ( −2 ) + n ( 6 ) m +n

Multiplying both sides by m + n :

m 6 0 = −2m + 6n ⇒ 2m = 6n ⇒ = = 3 n 2 Thus, the y-axis divides the line segment in the ratio 3 :1. Using the section formula for y : y=

( 3 × −7 ) + (1× −4 ) 3 +1

−21 − 4 −25 = = −6.25 4 4

Thus, the point of intersection is ( 0, −6.25 ). (b) L et the given points be:

A ( 7,10 ) , B ( −2, 5 ) , C ( 3, − 4 )

Using distance formula, we get,

AB = = BC= =

AC =

( −2 − 7 )2 + ( 5 − 10 )2 ( −9)2 + ( −5)2 =

81 + 25 =

106

Sol. Given that point P divides the line segment AB such that: AP 1 = AB 3 This means P divides AB in the ratio 1:2 (since AP:PB =1:2). Given: A ( x1 , y1= )

( 2,1) , B ( x 2 , y 2=) ( 5, − 8 )

m : n = 1: 2 x-coordinate of P : (1× 5) + ( 2 × 2 ) 5 + 4 9 x= = = = 3 1+ 2 3 3 y-coordinate of P : (1× −8 ) + ( 2 × 1) −8 + 2 −6 = = = −2 y= 1+ 2 3 3 Thus, P ( 3, − 2 ) . The equation of the line is: 2x − y + k = 0 Since point P ( 3, − 2 ) lies on this line, it must satisfy the equation: 2 ( 3 ) − ( −2 ) + k =0 ⇒ 6 + 2 + k = 0 8 +k = 0 ⇒ k = −8 5. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, –3). The origin is the mid-point of the base. Find the coordinates of the point A and B. Also find the coordinates of another point D such that BACD is a rhombus. (CBSE 2015) y

Sol.

2 2 ( 3 + 2 )2 + ( −9)2 ( 3 − ( −2 ) ) + ( −4 − 5)= 2

5 + ( −9 )= 2

25 + 81 =

( 3 − 7 ) + ( −4 − 10 ) =

106 2

( −4 ) + ( −14 )

= 16 + 196 = 212

Since AB = BC =

106 , the triangle is isosceles.

To check if the triangle is a right triangle, we use the Pythagorean theorem: AB 2 + BC 2 = 212 AC 2 ⇒ 106 + 106 =

4. Point P divides the line segment joining the points AP 1 A(2, 1) and B(5, -8) such that = . If P lies on the AB 3 0 , find the value of k . line 2x − y + k = (CBSE 2019)

212 = 212

Since the Pythagorean theorem holds, the given points form an isosceles right triangle. Chapter 7 • Coordinate Geometry

B x'

C (0, -3) y'

Since O(0,0) is the midpoint of BC, using the midpoint formula: Substituting values: −3 + y 0+x = 0 = , 0 2 2 = x 0, = y 3 So, coordinates of B = (0, 3).

187

The height of an equilateral triangle with side length s is given by: 3 Height = s 2 Here, BC = 3 - (-3) = 6

⇒ (4 + 4)2 + (3 − 3)2 = (x − 4)2 + ( y − 3)2

3 Height= × 6= 3 3 2 Since A lies on the x-axis (symmetrically), its x-coordinate is ±3 3 , and the y-coordinate remains 0.

AB 2 = CA 2

(

) (

)

⇒ 64 + 0 = x 2 − 8x + 16 + y 2 − 6 y + 9 ⇒ x 2 + y 2 − 8x − 6 y = 39 (ii) Now, taking first and third part, ⇒ (4 + 4)2 + (3 − 3)2 =− ( 4 − x )2 + (3 − y )2

⇒ 64 + 0 = 16 + x 2 + 8x + y 2 − 6 y + 9

So, coordinates of A = 3 3, 0 or −3 3, 0 .

⇒ x 2 + y 2 + 8x − 6 y = 39 ... (iii)

A rhombus has equal sides and diagonals bisect each other at right angles. Since A and D should be symmetric about the y-axis, the coordinates of D will be the reflection of A over the y-axis.

On subtracting Eq. (ii) from Eq. (iii), we get

(

)

(

So, if A = 3 3, 0 , then D = −3 3, 0

(

)

(

)

⇒ 16x = 0 ⇒x = 0 Now, put the value of x in Eq. (ii), we get 0 + y 2 − 0 − 6y = 39 ⇒ y 2 − 6 y − 39 = 0

If A = −3 3, 0 , then D = 3 3, 0 . 6. If ( −4, 3 ) and ( 4, 3 ) are two vertices of an equilateral triangle, then find the coordinates of the third vertex, given that the origin lies in the interior of the triangle. Sol. Let the third vertex of an equilateral triangle be C (x , y ). ∴ AB = BC = CA 2 2 AB = BC = CA 2

(

x 2 + y 2 + 8x − 6 y − x 2 + y 2 − 8x − 6 y = 39 − 39

(i)

6 ± (6)2 − 4 (1)( −39 ) ∴y = 2 ×1 6 ± 36 + 156 6 ± 192 6 ± 2 48 ⇒y= ⇒y= ⇒y= 2 2 2 ⇒ y = 3 ± 48 ⇒ y = 3 ± 4 3 The origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero. Then, y-coordinate of third vertex should be negative. Hence, the required coordinate of third vertex, C, is

Now, taking first two parts.

(0,3 − 4 3 ).

AB 2 = BC 2

Competency Based Questions Multiple Choice Questions 1. Two poles are to be installed on an elevated road as shown in figure. The diagram also shows the starting and ending points of the road.

(8,8)

(14,11)

Which of the following are the coordinates of the poles?

188

(a) Q (10, 9 ) and R (12, 8 )

(b) Q (10, 9 ) andR (12,10 )

(d) Q ( −10, 9 ) and R ( 0,11)

2. If a car travels from location (x, -3) to (4, 5) and the distance covered is 10 units, then what is the value of x? (a) 0 (b) -1 (c) 1

(d) None of these

3. A city planner designs a park exactly at the midpoint of landmarks located at (a, 3) and (5, b). If the park's coordinates are (3, 2), what are the values of a and b ? (a) a = 1, b = 1 (b) a = 0, b = 1 (c) a = 1, b = 0 (d) a = 0, b = 0 4. A lost hiker finds themselves 5 units away from the coordinates (3, 6). If they are currently at (a, 2), then what are the possible locations they could be at? (a) 0 and 6 (b) 1 and 5 (c) -1 and 7

(d) -2 and 8

5. A quadrilateral ABOC is drawn on a coordinate grid as shown A (-3, 4)

What are the coordinates of point A?

below. O is the origin and the coordinates of A are ( −3, 4 )

and ∠ COA = 45.

45°

What is the perimeter of ABOC ? (a) 14 + 3 2 units

(b) 17 + 3 2 units (c) 14 + 4 2 units

(d) Cannot be found using the given information. 6. The distance between the points ( acos θ + bsin θ , 0 ) and ( 0, asinθ − bcos θ ) , is 2

(a) a + b

(b) a − b

(d) a 2 − b 2

7. A rectangle’s vertices are at the points (1, 2), (5, 2), (5, 6), and (1, 6). What is the length of its diagonal? (a) 6 units

(b) 4 2 units

(d) 8 units

(NCERT Exemplar)

(b) 12

(d) 7 + 5

9. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then (NCERT Exemplar)

(a) AP =

1 AB 3

(b) AP = PB

1 1 AB (d) AP = AB 2 3 10. Shown below are 2 identical rectangles such that their breadth is half their length. (c) PB =

Y X

(b) ( −4, − 6 )

(d) ( −2, − 9 )

Select Response Type MCQs 1. ∆PQR is a triangle such that PQ:PR = 1:2. Point P lies on the X-axis and the coordinates of Q and R are known. Which of the following formula can definitely be used to find the coordinates of P? (i) Section formula

(ii) Distance formula

(a) Only (i)

(b) Only (ii)

(d) Neither (i) nor (ii)

2. If the point P(k - 1, 2) is equidistant from the points A(3, k) and B(k, 5), then determine the possible values of k. Which of the following are correct? (i) 1 (ii) 3 (iii) 5 (iv) 2 (a) (i) and (ii) (b) (ii) and (iii)

8. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (a) 5

(a) ( 4, − 5 )

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) B oth A and R are true, and R is the correct explanation of A. (b) B oth A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): The distance between ( 3, 4 ) and ( 7,1) is 5 2 . Reason (R): The distance between two points

B (4,-3)

( x1 , y1 ) and ( x 2 , y 2 ) is calculated using the formula ( x 2 − x1 )2 + ( y 2 − y1 )2 .

(CBSE SQP 2025)

2. Assertion (A): If the points A (4, 3) and B (x, 5) lie on a circle with centre O(2, 3), then the value of x is 2. Reason (R): Centre of a circle is the mid-point of each chord of the circle.

(CBSE 2023)

Chapter 7 • Coordinate Geometry

189

3. Assertion (A): C is the mid-point of PQ, if P is (4, x), C is (y, -1) and Q is (-2, 4), then x and y respectively are – 6 and 1. Reason (R): The mid-point of the line segment joining the points P(x₁, y₁) and Q(x₂, y₂) is  x1 + x 2 y1 + y 2   2 , 2   

backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground. Each team plays with 11 players on the field during the game including the goalie. Positions you might play include: Forward: As shown by players A, B, C and D; Midfielders: As shown by players E, F and G;

1. Ryan, from a very young age, was fascinated by the twinkling of stars and the vastness of space. He always dreamt of becoming an astronaut one day. So, he started to sketch his own rocket designs on the graph sheet. One such design is given below: Y 5 4

Fullbacks: As shown by players H, I and J; Goalie: As shown by player K.

E F 0 -6 -5 -4 -3 -2 -1 C

-6 -5 -4 -3 -2 -1

1 -1

-2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 -1 C -2 -3 G -4 D -5 -6 100 yards

the following questions:

Based on the above, answer the following questions: (a) Find the mid-point of the segment joining F and G. (b) (i) What is the distance between the points A and C? OR (ii) Find the coordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally.

2. A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf. It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the

190

Using the picture of a hockey field below, answer

-5 Y'

-3 -4

8 7 6 5 H 4 3 2 0 1

60 yards

Case Study Based Questions

(a) If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by 3 (i)  − , 2   2 

3  (ii)  2, −  2 

 3 (iii)  2,   2

(iv) (-2, -3)

(b) The point on x-axis equidistant from I and E is

1  (i)  , 0  2 

1  (ii)  0, −  2 

 1  (iii)  − , 0   2 

 1 (iv)  0,   2

(c) What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K, Q and E are collinear? (i) (1, 0)

(ii) (0, 1)

(iii) (-2, 1)

(iv) (-1, 0)

(d) The point on y-axis equidistant from B and C is (i) (-1, 0)

(ii) (0, -1)

(iii) (1, 0)

(iv) (0, 1)

Answers Multiple Choice Questions

9 ( 3 − a )2 = 3−a = ±3

Sol. 1. (b) For the coordinates of point Q (x₁, y₁), we have AQ : QB = 1 : 2 as shown in figure. 1 A (8,8)

1 Q (x1, y1)

1 R (x2, y2)

B (14,11)

Using section formula, we have 1 × 14 + 2 × 8 30 x= = = 10 1 1+ 2 3 y= 1

a= 6 0, 5. (a) Given, coordinates of A =

( −3, 4 ) . Let P be the point

at which AC intersects the X-axis. Here, AB = x – coordinate of A= 3= PO and AP = y – coordinate of A= 4= BO . A (-3, 4)

1 × 11 + 2 × 8 27 = = 9 1+ 2 3

Y B

Coordinate of point Q ( x1 , y1 ) be (10, 9 ). Now for point R ratio be 2:1. X'

Using section formula x= 2 y= 2

2 × 14 + 1 × 8 36 = = 12 2 +1 3

O 45°

2 × 11 + 1 × 8 30 = = 10 2 +1 3

C Y'

Hence, coordinate of point R be (12, 10 ). In ∆CPO ,

2. (d) The equation, based on the distance formula, is:

( 4 − x )2

( 5 − ( −3)) =10 2

100 ( 4 − x )2 + 8 2 = 36 ( 4 − x )2 =

4−x = ±6 x =4 ± 6 =−2,10 3. (a)

Using the midpoint formula:  a + 5 3+b  ,   = ( 3, 2 ) 2   2 a +5 3 +b and 2 = 3= 2 2 a = + 5 6 and = 3 +b 4

sin 45 = 1

PO OC

3 OC

OC = 3 2 units tan 45 =

PO 3 = 1= PC PC

PC = 3 units Now, required perimeter = AB + BO + OC + PC + AP = 3 + 4 + 3 2 + 3 + 4 = 14 + 3 2 units 6. (c) Given points are ( a cosθ + b sinθ , 0 ) and ( 0, asinθ − b cosθ )

= a 1= and b 1

Using distance formula,

4. (a) Using the distance formula:

= (a cosθ + b sinθ )2 + (b cosθ − a sinθ )2

5 ( 3 − a )2 + ( 6 − 2 )2 = 25 ( 3 − a )2 + 16 =

Chapter 7 • Coordinate Geometry

2 = (a cosθ + b sinθ − 0)2 + ( 0 − (a sinθ − b cosθ ))

a 2 cos2θ + b 2 sin 2θ + 2ab cosθ sinθ + b 2 cos2θ + a 2 sin 2θ − 2ab sinθ cosθ 191

(

)

(

cos2θ a 2 + b 2 + sin 2θ a 2 + b 2

(

= a +b

Distance between A(4,2) and B(8,4)

)

AB =

)(cos θ + sin θ )

32 = 4 2 units

8. (b) 12 We plot the vertices of a triangle i.e., (0, 4), (0, 0) and (3, 0) on the paper shown as given below Y

2 1

(0, 0) O

B (3,0)

Now, perimeter of ∆OAB = Sum of the length of all its sides = d(AO) + d(OB) + d(AB) Distance between two points formula:

(0 − 0) + (0 − 4) = 2

(3 − 0) + (0 − 0) =

10. (c)

Given figure,

(3 − 0)2 + (0 − 4)2 =

0 + 16 = 4

AE = OH = 2 units

LA = LH + HA =−3 − 4 =−7 Thus, the coordinates of the points A = ( −2, − 7 ) .

25 = 5

Hence, the required perimeter of triangle is 4+3+5= 12. 1 9. (d) AP = AB 2 Given that, the point P(2,1) lies on the line segment joining the points A(4,2) and B(8,4), which shows in the figure below: P (2,1)

In rectangle ∆AEOH ,

The coordinate of H is ( −2, − 3 ) ,

9+0 = 3 9 + 16 =

Then, according to the question, 1 Breadth of rectangle = × Length 2 1 = ×4 2 = 2 units

(in negative direction of Y -axis)

Distance between A(0, 4) and B(3, 0): =

Therefore,= AB 2= 5 2 AP AB Thus, AP = 2

AH = 4 units

Distance between O(0, 0) and B(3, 0): 2

45= 3 5

(in negative direction of X -axis)

(x 2 − x1 )2 + ( y 2 − y1 )2

Distance between A(0, 4) and O(0, 0): =

36 + 9=

Length of rectangle (OB) = 4 units

62 + 32=

In rectangle OBCD,

(0, 4) A

20= 2 5

BP = (8 − 2)2 + (4 − 1)2 =

16 + 16 =

16 + 4=

Distance between B(8, 4) and P(2, 1),

7. (b) Using the distance formula:

(4)2 + (2)2=

= a 2 + b 2 units since cos2θ + sin 2θ =1

( 5 − 1)2 + ( 6 − 2 )2 =

(8 − 4)2 + (4 − 2)2

Select Response Type MCQs 1. (b) Only (ii) In ∆PQR, Given, PQ : PR = 1:2 and point P lies on the X-axis. Y

B (8,4)

A (4,2)

Now, distance between A(4, 2) and P(2, 1), AP = (2 − 4)2 + (1 − 2)2 =

192

( −2)2 + ( −1)2 =

4 + 1=

Also, the coordinates of Q and R are known. To find the coordinates of point P distance formula can be used.

2. (d) (i) and (iii) We need to find k such that the point P (k − 1, 2 ) is equidistant from A ( 3, k ) and B (k , 5 ) . Using the distance formula: (k − 1 − 3)2 + ( 2 − k )2=

(k − 4 ) 2 + ( 2 − k ) 2 =

(k − 1 − k ) 2 + ( 2 − 5 ) 2 ( −1)2 + ( −3)2

k 2 − 8k + 16 + k 2 − 4k + 4 =

1+ 9

2k − 12k + 20 = 10

Squaring both sides: 2k 2 − 12k + 20 = 10

2k 2 − 12k + 10 = 0

x +4 =−1 ⇒ x + 4 =−2 ⇒ x =−6 2

Case Study Based Questions  −3 + 1 0 + 4  , 1. (a) Mid-point of FG is  = 2   2

( −1, 2 )

(b) (i) AC = ( −1 − 3)2 + ( −2 − 4)2

= 52 or 2 13

(ii) T he coordinates of required point are

0 (k − 5)(k − 1) = So, the possible values of k are: = k 5= or k 1.

Assertion-Reason Based Questions 1. (d) A is false, but R is true Assertion d= (7 − 3)2 + (1 − 4)2 2 = 42 + ( −3)=

Mid-point of PQ, C(y, -1) =  4−2 x +4 ,   2   2 4−2  = y ⇒ y = 1   2

16 + 9

= = 25 5 Thus, the Assertion (A) is incorrect. The reason states that the distance formula used is: (x 2 − x1 )2 + ( y 2 − y1 )2

Thus, the Reason (R) is correct. 2. (c) A is true, but R is false. Given, A(4, 3) and B(x, 5) are two points on the circle. Centre of circle O(2, 3) Thus, OA = OB [radius of circle] ( 4 − 2 )2 + ( 3 − 3)2 =( x − 2 )2 + ( 5 − 3)2

( x − 2 )2 + 4 = 4 = ( x − 2 )2 = 0 x −2=0⇒x =2 Reason is false. 3. (a) Both A and R are true, and R is the correct explanation of A. Since, C(y, -1) is the mid-point of P(4, x) and Q(-2, 4).

 1× 3 + 3 × 3 1× 2 + 3 × 4   7  ,   =  3, 2  1+ 3 1+ 3     (c) D(-2, -5) 2. (a) (iii) The coordinates of A and G are (3, 6) and (1, -3) respectively. The mid-point of AG is equidistant from A and G. The coordinates of the mid-points are

 3 +1 6 − 3   3  ,   =  2,  2   2  2

(b) (i) Let P (x, 0) be a point on x-axis equidistant from I (-1, 1) and E (2, 1). Then, PI =PE ⇒ PI 2 =PE 2 ⇒ (x + 1)2 + (0 − 1)2 = (x − 2)2 + (0 − 1)2 ⇒ x = 1  Hence, P  , 0  is the required point. 2 

1 2

(c) (ii) The coordinates of K and E are (-4, 1) and (2, 1) respectively. Clearly, Q divides KE in the ratio 2:1. So, the coordinates of Q are  2 × 2 + 1 × −4  , 1  = ( 0,1)  2 +1   (d) (iv) The coordinates of B and C are (4, 3) and (4, -1) respectively. Let P (0, y) be a point on y-axis equidistant from B and C. Then, PB =PC ⇒ PB 2 =PC 2 ⇒ (0 − 4)2 + ( y − 3)2 = (0 − 4)2 + ( y + 1)2 ⇒ y = 1 Hence, the coordinates of P are (0, 1).

Chapter 7 • Coordinate Geometry

193

Additional Practice Questions

Hence, the coordinates of P are (0, 1).

1. AD is the median of ∆ABC with vertices

A ( 5, − 6 ) , B ( 6, 4 ) and C ( 0, 0 ). Length AD is equal to

(a) 68 units

(b) 2 15 units

(d) 10 units

(CBSE 2023)

2. The coordinates of the point that divides the line segment joining (1,1) and (5,5) internally in the ratio 2:3 are: (a) (3, 3)

 13 13  (b)  ,   5 5

(d) None of these

3. What is the distance between the points (4, -1) and (-2, 2)? (a) 2 13 units

(b) 6 2 units

(d) 7 units

endpoint is (6, 8), then what is the other endpoint? (a) (0, 0) (b) (1, 2) (c) (0, -1) (d) (1, 0) 5. The points (4, 6) and (12, 14) are connected by a line segment. A point divides this segment internally in the ratio 1:3. What are the coordinates of this (c) (6, 8)

(d) (3, 5)

6. Calculate the distance between the points (1, 1) and (4, 5). 7. Which point on the y-axis is equidistant from (2, 5) and (8, 3)? 8. What is the midpoint of the segment joining (2, 2) and (8, 8)? 9. Are the points (1, 1), (2, 2), and (3, 3) collinear? Justify your answer. 10. Using the section formula, determine the point that divides the segment joining (1, 2) and (3, 8) in the ratio 2:1 internally. 11. Point P(x, y) is equidistant from points A(5, 1) and B(1, 5). Prove that x = y. (CBSE 2023)

12. If the x-coordinate of a point is thrice its y-coordinate and the point lies on the line given by the equation 2x − 2 y = 12 , find the coordinates of this point. 194

14. Find the ratio in which y-axis divides the line segment joining the points (5, -6) and (-1, -4). (CBSE 2023)

15. If the mid-point of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x + y – 10 = 0, find the value of k. (CBSE 2020)

16. If the points A ( 6,1) , B ( p , 2 ) , C ( 9, 4 ) and D ( 7, q ) are the vertices of a parallelogram ABCD , then find the values of p and q . Hence, check whether ABCD is a rectangle or not. (CBSE 2025)

4. If the midpoint of a line segment is (3, 4) and one

dividing point? (a) (0, 2) (b) (1, 3)

13. A line intersects y-axis and x-axis at point P and Q respectively. If R (2, 5) is the mid-point of line segment PQ, then find the coordinates of P and Q.

17. (1, -1), (0, 4) and (-5, 3) are vertices of a triangle. Check whether it is a scalene triangle, isosceles triangle or an equilateral triangle. Also, find the length of its median joining the vertex (1, -1) the mid-point of the opposite side. (CBSE 2025)

18. Determine the coordinates of a point P that divides the line segment joining A(3, 4) and B(12, 8) in the ratio 1:3 internally. Then, calculate the distance from P to the midpoint of AB. 19. The points ( 3, − 4 ) and ( −6, 5 ) are the end points of a diagonal of a parallelogram. If one of the end points of the second diagonal is ( −2,1) , then find its other end point. Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 20. Assertion (A): The midpoint of the segment joining the points (2, 6) and (4, 10) is (3, 8). Reason (R): The midpoint formula

 x1 + x 2 y1 + y 2  was applied correctly to calculate  2 , 2   

the midpoint.

21. Assertion (A): The distance between the points P ( 3, − 2 ) and Q ( 7, 4 ) is 10 units.

Reason (R): The formula for the distance between any two points ( x1 , y1 ) and ( x 2 , y 2 ) is (x 2 − x1 )2 + ( y 2 − y1 )2 .

Q D

22. A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of 1 m from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class 10. They made a chart for it which looks like the figure.

Based on the above, answer the following questions: (a) If A is taken as origin, what are the coordinates of the vertices of ∆PQR? (b) (i) Find the distances PQ and QR. OR (ii) F ind the coordinates of the point which divides the line segment joining points P and R in the ratio 2:1 internally. (c) Find out if ∆PQR is an isosceles triangle.

Challenge Yourself 1. Find the centroid of the triangle formed by the lines = x 0,= y 0 and x + y = 0 as sides.

Scan me

2. Find the area of the region formed by 4 x + 3 y = 12 . (Hint: |x| = x for all positive or negative values of x.)

  x1 + x 2 + x 3 y1 + y 2 + y 3   ,  Hint : Centroid of ∆ABC   3 3      where ( x1 , y1 ) , ( x 2 , y 2 )     and ( x 3 , y 3 ) are the vertex of triangle  

Q2.

Answers

Additional Practice Questions

 21  18. P  , 5 Distance  4 

97 19. D ( −1, 0 ) 4 20. (a) Both A and R are true, and R is the correct explanation of A.

68 units

 13 13  2.  ,   5 5

3. 3 5 units

4. (a) ( 0, 0 )

5. (c) (6,8)

6. 5 units

7. y = −11

8. ( 5, 5 )

9. yes

7 10.  , 6  3 

11. Proved

12. ( 9, 3 )

Challenge Yourself

13. P = ( 4, 0 ) and Q = ( 0, 10 )

14. 5 :1

15. k = 7

p 8,= q 3 16.=

10 10 1.  ,  2. 24 square units  3 3

17. Isosceles Triangle, 32.5 Chapter 7 • Coordinate Geometry

21. (d) A is false, but R is true. 22. (a) P (4, 6), Q (3, 2), and R (6, 5)

 16 16  (b) (i) 3 2 (b) (ii)  ,   3 3  (c) ∆PQR is an isosceles triangle

Scan me for Exemplar Solutions 195

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. In what ratio, does x-axis divide the line segment joining the points A(3, 6) and B(-12, -3)? (a) 1:2

(b) 1:4

(CBSE 2024)

(d) 2:1

2. Which condition must be true for the points (x, y), (x + 2, y + 2), and (x + 4, y + 4) to be collinear? (a) x = y

(b) x + y = 0

(d) x - y = 0

3. If the midpoint of the line segment joining the points (1, -1) and (k, 3) is (4, 1), what is the value of k? (a) 7

(b) 8

(d) 6

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(CBSE SQP 2025)

5. Assertion (A): The coordinates of a point dividing the line joining ( 2, 4 ) and ( 6, 8 ) in the ratio 1:1 are ( 4, 6 ) .

Reason (R): The coordinates of a point dividing the line joining ( x1 , y1 ) and ( x 2 , y 2 ) in the ratio m:n are (CBSE 2024)  mx 2 + nx1 my 2 + ny1  ,  m +n m + n  

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. Determine the coordinates of the midpoint of the segment joining the points ( −5, 7 ) and ( 3, − 1) . 7. Prove that the points ( 3, 0 ) , ( 6, 4 ) and ( −1, 3 ) are the vertices of a right isosceles triangle.

Short Answer Questions 8. A point is equidistant from ( 3, 4 ) and ( 7, 8 ) . Find its x-coordinate if the y-coordinate is given as 5.

[4 × 3 = 12 Marks] (CBSE 2024)

9. Show that (1, − 1) is the centre of the circle circumscribing the triangle whose vertices are ( 4, 3 ) , ( −2, 3 ) and ( 6, − 1). 10. A point divides the line segment joining (1, 1) and ( 5, 5 ) in the ratio 1:1 . Find its coordinates. 11. Do the points A (1, 1), B (2, 4), and C (5, 3) form a triangle? If so, name the type of triangle formed?

196

(CBSE 2023)

Long Answer Questions

[3 × 5 = 15 Marks]

5  12. The line segment AB joining the points A ( 3, − 4 ) and B (1, 2 ) is trisected at the points P ( p , − 2 ) and Q  , q  . Find the 3  values of p and q. (CBSE 2013)

13. Find the ratio in which the y-axis divides the line segment joining the points ( −1, − 4 ) and ( 5, − 6 ) . Also, find the coordinates of the point of intersection. (CBSE 2019)

14. The base PQ of an equilateral triangle PQR lies on the x-axis. The coordinates of point P are (−4,0), and the origin is the midpoint of the base. Find the coordinates of points Q and R. Also, determine the coordinates of another point S such that PQRS forms a rhombus.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Jagdish has a field which is in the shape of a right-angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O. Y

A B S

P (-200, 0)

Q (200, 0)

Based on the above information, answer the following questions: (a) Taking O as origin, coordinates of P are (−200, 0) and of Q are (200, 0). PQRS being a square, what are the coordinates of R and S ? (b) (i) What is the area of square PQRS ? OR

(ii) What is the length of diagonal PR in square PQRS ?

Self-Assessment

197

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. The perimeter of a triangle with vertices (0, 4), (0, 0), and (3, 0) is: (a) 5 units

(b) 12 units

(d) 14 units

2. If the points (1, 1), (5, 3), and (x, 5) lie on a line, what is the value of x? (a) 7

(b) 9

(CBSE 2021)

(d) 13

3. If a line divides the segment joining ( a , b ) and (c , d ) in the ratio 1:3, the coordinates of the point of division are: (a)  

3a + c 3b + d  ,  4 4 

 a + 3c b + 3d  (b)  ,  4   4

 c +d a +b  (d)  ,  4   4

a +c b +d  (c)  ,  4   4

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

( x 2 − x1 )2 + ( y 2 − y1 )2

(CBSE 2024)

5. Assertion (A): The point ( 4, 7 ) lies on the line joining (1, 3 ) and ( 7, 11) . Reason (R): A point lies on a line segment if its coordinates satisfy the equation of the line passing through the endpoints of the segment.

Very Short Answer Questions 6. Show that the points ( −2, 3 ) , (8, 3) and (6, 7) are the vertices of a right-angled triangle.

[2 × 2 = 4 Marks]

(CBSE 2024)

7. If the point P (k , 0 ) divides the line segment joining the points A ( 2, −2 ) and B ( −7, 4 ) in the ratio 1:2, then find the value of k. (CBSE 2020)

Short Answer Questions 8. If the point P (k − 1, 2 ) is equidistant from the points A ( 3, k ) and B (k , 5 ) , find the values of k.

[4 × 3 = 12 Marks]

9. Find the coordinates of the point on the y-axis which is equidistant from the points A ( 5, 3 ) and B (1, − 5 ) .

198

10. The two opposite vertices of a square are ( –1, 2 ) and ( 3, 2 ) . Find the coordinates of other two vertices. 11. Are the points (3, 2), ( 6, −1) , and ( 9, − 4 ) collinear? Justify your answer.

Long Answer Questions

[3 × 5 = 15 Marks]

12. If ABCD be a rectangle and P be any point in the plane of the rectangle, then prove that PA 2 + PC 2 = PB 2 + PD 2 13. Find the coordinates of the point that divides the line segment joining ( 2, − 3 ) and (10, 5 ) in the ratio 3 : 2. Verify if 1 the point lies on the line= (CBSE SQP 2025) y x − 2 . 2 14. Find the distance of the point ( 6, 8 ) from the origin and prove that it is the hypotenuse of a right triangle whose other two points are ( 6, 0 ) and ( 0, 8 ) .

(CBSE 2024)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Use of mobile screen for long hours makes your eye sight weak and give you headaches. Children who are addicted to play “PUBG” can get easily stressed out. To raise social awareness about ill effects of playing PUBG, a school decided to start ‘BAN PUBG’ campaign, in which students are asked to prepare campaign board in the shape of a rectangle: One such campaign board made by class X student of the school is shown in the figure. Y

D(1, 5)

BAN PUBG

C (7, 5)

& SAVE KIDS

X' A (1, 1)

B (7, 1)

(a) Find the coordinates of the point of intersection of diagonals AC and BD. (b) Find the length of the diagonal AC. (c) (i) Find the area of the campaign Board ABCD.

(ii) Find the ratio of the length of side AB to the length of the diagonal AC. (CBSE 2024)

Self-Assessment

199

Introduction to Trigonometry

Chapter at a Glance Trigonometric Ratios ●

Trigonometric ratios are ratios of the sides of a right-angled triangle with respect to one of its acute angles.

●

For a triangle with ∠θ as the reference angle: Opposite Side AB = Hypotenuse AC

◆

Sine (sin): sin θ =

◆

Cosine (cos): cosθ =

◆

Opposite Side AB Tangent (tan): tan θ = = Adjacent Side BC

◆

Cosecant (cosec): cosec θ =

◆

Secant (sec): secθ =

◆

Adjacent Side BC = Cotangent (cot): cot θ = Opposite Side AB

Adjacent Side BC = Hypotenuse AC

Hy p

Opposite

AC Hypotenuse = Opposite Side AB

AC Hypotenuse = Adjacent Side BC

Adjacent

Standard values for trigonometric ratios at specific angles 0°, 30°, 45°, 60°, 90°: Angle (θ)

sin θ

cos θ

tan θ

cosec θ

sec θ

cot θ

0°

∞

30°

1 2

3 2

60°

3 2

1 2

90°

∞

45°

2 2 2

2 2 ∞

Trigonometric Identities ●

These are equations that are true for all values of the angle: ♦ sin 2θ + cos2θ = 1

♦ 1 + tan 2θ = sec2θ (θ ≠ 90 )

♦ 1 + cot 2θ = cosec2θ (θ ≠ 0 ) 200

sin θ =

Values of Trigonometric Ratios ●

3 1 1

4 3C

NCERT Zone NCERT Exercise 8.1

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.

1. In ∆ABC right angled at B , AB = 24 cm, BC = 7 m. Determine (a) sin A, cos A

Applying Pythagoras theorem in ∆ABC, we obtain 2

( AB )2 + ( BC )2 2 2 2 4k ) ( AB ) + ( 3k ) (= 16k 2 − 9k 2 = ( AB )2 ( AB )2 = 7k 2 = ( AC )

(b) sin C , cos C

Sol. Applying Pythagoras theorem for ∆ABC , we obtain 2

= ( AC )

( AB )2 + ( BC )2

2 2 = ( 24 ) + ( 7 )

= 576 + 49 = 625

24 cm

∴ AC = 625 AC = 25 cm C 7 cm Side opposite to ∠A BC 7 (a) sin A = = = Hypotenuse AC 25

QR = 5 cm tan P =

Side adjacent to ∠A AB = Side opposite to ∠A BC

It is given that, 8 cot A = 15 AB 8 = BC 15 Let AB be 8k. Therefore, BC will be 15k, where k is a positive integer.

PR ) (=

(QR )2 = 25

3k = 7k

cot A =

2. In the given figure, find tan P − cot R Sol. Applying Pythagoras theorem P for ∆PQR, we obtain

169 = 144 + (QR )

Side opposite to ∠A BC = = Side adjacent to ∠A AB

Sol. Consider a right-angled triangle, right-angled at B.

Side adjacent to ∠C BC 7 = = cos C = Hypotenuse AC 25

7 4

4. Given 15 cot A = 8. Find sin A and sec A

cos A =

( PQ )2 + (QR )2 = (13 )2 (12 )2 + (QR )2

7k = 4k

tan A =

Side adjacent to ∠A AB 24 = = Hypotenuse AC 25 Side opposite to ∠C AB 24 = = (b) sin C = Hypotenuse AC 25

AB = 7k Side adjacent to ∠A AB = = cos A = Hypotenuse AC

Applying Pythagoras theorem in ∆ABC , we obtain

12 cm

13 cm

( AB )2 + ( BC )2 = ( AC )2 ( 8k )2 + (15k )2 = ( AC )2 64k 2 + 225k 2 ( AC )2 = 289k 2 = ( AC )

Side opposite to ∠P QR 5 = = Side adjacent to ∠P PQ 12

AC = 17k

Side adjacent to ∠R QR 5 = = cot R = Side opposite to ∠R PQ 12

sin A =

Side opposite to ∠A BC 15k 15 = = = Hypotenuse AC 17k 17

tan P − cot R =

sec A =

Hypotenuse AC 17 = = Side adjacent to ∠A AB 8

5 5 − =0 12 12

3 , calculate cos A and tan A. 4 Sol. Let ∆ABC be a right-angled triangle, right-angled at point B. Given that, 3 sin A = 4 BC 3 A = AC 4 3. If sin A =

Chapter 8 • Introduction to Trigonometry

5. Given sec θ = C

13 , calculate all other trigonometric 12

ratios. Sol. Consider a right-angle triangle ∆ABC, right-angled at point B. Hypotenuse sec θ = Side adjacent to ∠θ 13 AC = 12 AB 201

From equations (iii), (iv), and (v), we obtain ∠ACD = ∠BCD (vi)

If AC is 13k , AB will be 12k , where k is a positive integer. Applying Pythagoras theorem in ∆ABC, we obtain 2

∠CDA = ∠CDB each 90° Therefore, the remaining angles should be equal.

= ( AC )

( AB ) + ( BC )

2 = k) (13

(12k )2 + ( BC )2

∴∠CAD = ∠CBD ∠A =∠B

2 169 = k 2 144k 2 + ( BC )

7 7. If cot θ = , evalute 8

= ( BC )2 169k 2 − 144k 2 2

( BC ) = 25k 2

(a)

BC = 5k

Side adjacent to ∠θ AB 12k 12 = = = Hypotenuse AC 13k 13

tan= θ

Side opposite to ∠θ BC 5k 5 = = = Side adjacent to ∠θ AB 12k 12

cot= θ

Side adjacent to ∠θ AB 12k 12 = = = Side opposite to ∠θ BC 5k 5

cot = θ

Side adjacent to ∠θ BC 7 = = Side opposite to ∠θ AB 8

= 64k 2 + 49k 2

( AC )2 = 113k 2 AC = 113 k AB = AC BC cos = θ = AC B P

(a)

AD AC = (BC = CP by Construction) BD CP (ii) A By using the converse of B.P.T,

CD || BP ∠ACD = ∠CPB

(Corresponding angle)

(iii)

∠BCD = ∠CBP

(Alternate interior angle)

(iv)

By construction, we have BC = CP . ∴∠CBP = ∠CPB (Angle opposite side of equal side)

(v)

8k = 113 k 7k = 113 k

113 7

113

(1 + sinθ )(1 − sinθ ) 1 − sin 2 θ = (1 + cosθ )(1 − cosθ ) 1 − cos2 θ 2

 8  64 49 1−   1− 49 113   113 = = 113 = 2 64 64  7  1 − 49 1−   113 113  113 

2 (b) cot = θ

2 2 = ( 8k ) + ( 7k )

sin= θ

= ( AC ) ( AB ) + ( BC )

We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP .

202

Applying Pythagoras theorem in ∆ABC , we obtain

6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

AD AC = BD BC

If BC is 7k, then AB will be 8k, where k is a positive integer.

Hypotenuse AC 13k 13 = = = Side opposite to ∠θ BC 5k 5

From equation (i), we obtain

Sol. Let us consider a right triangle ABC, right-angled at point B.

cos = θ

Sol. Let us consider a triangle ABC in which CD ⊥ AB . It is given that, cos A = cos B AD BD (i) = AC BC

(1 + sinθ )(1 − sinθ ) (1 + cosθ )(1 − cosθ )

(b) cot 2 θ

Side opposite to ∠θ BC 5k 5 sin= θ = = = Hypotenuse AC 13k 13

cosec = θ

7 8

2   θ ) = ( cot= 

49 64

8. If 3 cot A = 4, check whether 1 − tan 2 A = cos2 A − sin 2 A or not. 2 1 + tan A Sol. It is given that 3 cot A = 4 3 cot A = , Considering a right triangle. B at right angle. 4 Side adjacent to∠A cot A = Side opposite to ∠A

AB 4 = BC 3 If AB is 4 k, then BC will be 3k, where k is a positive integer. In ∆ABC = ( AC )2

( AB )2 + ( BC )2

= ( 4k ) + ( 3k= ) 16k + 9k

25 Sol. Given that, PR + QR =

 4 3 cos2 A − sin 2 A =  −   5 5

PQ = 5 Let PR be x. = 25 − x Therefore, QR

2 PR ) (=

∴ L.H.S = R.H.S 1 − tan A = cos2 A − sin 2 A 1 + tan 2 A

50 x = 650 1

. Find the 9. In ∆ABC , right angled at B. If tan A = 3 value of (a) sin A cosC + cos A sinC (b) cos A cosC − sin A sinC 1

BC 1 = AB 3 If BC is k, then AB will be 3k, where k is positive integer.

) 3k + k ) + (k= 2

( AC )2 = 4k 2 ⇒ AC = 2k BC k 1 sin A = = = AC 2k 2

Chapter 8 • Introduction to Trigonometry

x 2 = 25 + 625 + x 2 − 50x

( AB )2 + ( BC )2

( PQ )2 + (QR )2

x 2 = ( 5 ) + ( 25 − x )

In ∆ABC

Applying Pythagoras theorem in ∆PQR , we get

7 16 9 = = − 25 25 25

 3  1   1  3  3 3 − =0     −     = 2 2 2 2 4 4      

10. In ∆PQR , right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P R and tan P.

3 9 7 1−   1− 2 1 − tan A 7 4  16 16 = = = 2 = 2 9 25 25 1 + tan A 3 1+ 1+   16 16 4

(

 1  1   3  3  1 3 4   = + = = 1    +   2  2   2   2  4 4 4

(b) cos A cos C − sin A sinC

AB 4k 4 cos A = = = AC 5k 5 BC 3k 3 sin A = = = AC 5k 5 BC 3k 3 tan A = = = AB 4 k 4

= 3k

BC k 1 cos C = = = AC 2k 2

AC = 5k

= ( AC )2

3 2 3 2

(a) sin A cos C + cos A sinC

( AC )2 = 25k 2

tan A =

AB 3k cos A == = AC 2k AB 3k sin C = = = AC 2k

x = 13 Therefore, PR = 13 cm QR = 25 − 13 = 12 cm QR 12 = PR 13 PQ 5 cos P = = PR 13 tan P = QR = 12 PR 5 sin P =

11. State whether the following are true or false. Justify your answer. (a) The value of tan A is always less than 1. 12 (b) sec A = for some value of angle A. 5 (c) cos A is the abbreviation used for the cosecant of angle A. (d) cot A is the product of cot and A 4 (e) sin θ = for some angle θ 3 203

Sol. (a) Consider a ∆ABC , right-angled at B.

BC 12 = AB 5

tan A =

But

∴ tan A > 1

12 >1 5

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sinθ is not possible. Hence, the given statement is false.

NCERT Exercise 8.2

So, tan A < 1 is not always true. A

Hence, the given statement is false.

12 (b) sec A = 5 AC 12 sec A = = AB 5

AC 12 = AB 5 Let AC be 12k, AB will be 5k, where k is a positive integer.

1. Evaluate the following: (a) sin60° cos30°+ sin60° cos60° (b) 2tan 2 45° + cos2 30° − sin 2 60° (c)

cos45° sec30° + cosec 30°

(d)

sin 30° + tan 45° – cosec 60° sec30° + cos60° + cot 45°

(e) 5cos 60 + 4sec 30° – tan 45° sin 2 30° + cos2 30° Sol. (a) sin60° cos30°+ sin60° cos60°

Applying Pythagoras theorem in ∆ABC , we obtain

 3  3   1  1     2   2  +  2   2    

= ( AC )2 ( AB )2 + ( BC )2 2 2 2 k ) ( 5k ) + ( BC ) (12=

144 = k 25k + ( BC ) 2

( BC ) = 119k 2 BC = 10.9k It can be observed that for given two sides AC = 12k and AB = 5k , BC should be such that, AC − AB < BC < AC + AB

12k − 5k < BC < 12k + 5k

7k < BC < 17k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible. Hence, the given statement is true. (c) Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false. (d) cot A is not the product of cot and A. It is the cotangent of ∠A. Hence, the given statement is false. 4 3 We know that in a right-angled triangle, (e) sin θ =

204

sin θ =

Side opposite to ∠θ Hypotenuse

3 1 + =1 4 4

(b) 2tan 2 45° + cos2 30° − sin 2 60° 2

 3  3 = 2 (1)2 +   2  −  2      3 3 = 2 + − = 2 4 4 cos45° (c) sec30° + cosec 30° 1 3 2 = = 2 2 2 2+2 3 + 3 1

(

(d)

(

)

24 − 8

3 2 6 −2 2

) (2 6 + 2 2 )

( 2 6 + 2 2 )( 2 6 − 2 2 ) 2 18 − 2 6 3 2− 6) = = ( sin 30°+ tan 45° – cosec 60° sec30°+ cos60°+ cot 45°

3 2 1 2 − +1− 2 3 3 =2 = 2 3 2 1 + + +1 3 2 3 2 3 3−4 2 3 3 3−4 = = 3 3+4 3 3+4 2 3

( 3 3 − 4 )( 3 3 − 4 ) = ( 3 3 − 4 ) = ( 3 3 + 4 )( 3 3 − 4 ) ( 3 3 ) − ( 4 ) ( 27 + 16 − 24 3 ) =

= = A is true when A (c) sin 2 A 2sin

( 27 ) − (16 )

= (e)

11 2

5cos 60° + 4sec2 30° – tan 2 45° sin 2 30° + cos2 30°

 2  2 1 5  + 4   – (1) 3 2   2 1  3    +   2   2 

 5   16    +  – 1 4 3 =    1 3  +  4 4

 15 + 64 − 12    12  67 =  = 4 12   4   2tan 30°

1 + tan 2 30° (i) sin 60° (iii) tan 60° (b)

(ii) cos 60° (iv) sin 30°

(ii) 1 (iv) 0

2tan 30°

1 − tan 2 30°

(i) cos 60° (iii) tan 60°

(ii) sin 60° (iv) sin 30°

2tan 30°

 1  2   3

 2   2       3 =  3 = = 2 1 4  1  1+ 1+  3 3  3

( 2 )( 3 ) 4 3

 3 = =  2  sin60°   2

(b)

1 3 and tan ( A − B ) = ; 3. If tan ( A + B ) = 3 0° < A + B ≤ 90° ; 0° ; A > B, find A and B. tan ( A + B=) tan60° And

1 tan ( A − B ) = 3

On adding both equations, we obtain 2 A = 90 ⇒ A =

90 ⇒ A = 45 2

From equation (1) , we obtain 45 + B = 60 ⇒ B = 15 4. State whether the following are true or false. Justify your answer. (a) sin⁡ (A + B) = sin A ⁡ + sin B (b) The value of sin ⁡θ increases as θ increases (c) The value of cos⁡ θ increases as θ increases

1 + tan 2 30°

== 3 tan 60°

tan ( A − B= 30 (2) ) tan 30° ⇒ A − B =

1 + tan 2 45°

= = A is true when A (c) sin 2 A 2sin (i) 0° (ii) 30° (iii) 45° (iv) 60°

Sol. (a)

 2   2      3  = ( 2 )( 3 ) =  3  =  2 1 2 3 1− 3 3

A +B = 60 (1)

1 − tan 2 45°

(i) tan 90° (iii) sin 45°

(d)

 1  2   3 = (d) 2 1 − tan 2 30°  1  1−   3

3 Sol. tan ( A + B ) =

2. Choose the correct option and justify your choice: (a)

and 2 sin A = 2sin 0° = 2(0) = 0 2tan 30°

( 43 − 24 3 )

Out of the given alternatives, only A = 0° is correct. As sin 2A = sin 0° = 0

1 − tan 2 45° = 1 − (1) 2 =0 1 + tan 2 45° 1 + (1)

Chapter 8 • Introduction to Trigonometry

(d) sin ⁡θ = cos⁡ θ for all values of θ (e) cot⁡ A is not defined for A = 0° Sol. (a) sin⁡ (A + B) = sin⁡ A + sin B Let A = 30° and B = 60° sin (A + B) = sin (30° + 60°) = sin 90° = 1 And sin A + sin B = sin 30° + sin 60°

1 3 1+ 3 + = 2 2 2 205

Clearly, sin (A + B) ≠ sin A + sin B. Hence, the given statement is false. (b) The value of sin⁡ θ increases as θ increases

NCERT Exercise 8.3 1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

The value of sin⁡ θ increases as θ increases in the interval of 0°< θ < 90° as

Sol. For tan A

sin0° =0

We know that,

sin 30°= 1= 0.5 2 1 sin45° = = 0.707 2

tan A =

1 cot A

for sin A

We know that,

3 sin60° = = 0.866 2

1 + cot 2 A = cosec2 A

sin90° =1

cosec A = ± 1 + cot 2 A (Taking square root both sides)

cosec2 A = 1 + cot 2 A

Hence, the given statement is true. (c) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as cos0°= 0 cos30= ° cos45° =

3 = 0.866 2 1

Here, A is acute (i.e., less than 90°) and cosec A is positive when A is acute. So, cosecA =

1 + cot 2 A

 sin A =

1 cosecA

1 ∴ sin A = 1 + cot 2 A

= 0.707

for sec A We know that

1 cos60° = = 0.5 2 cos90° =1

1 + tan 2 A = sec2 A or sec2 A = 1 + tan 2 A

It can be observed that the value of cos θ does not increased in the interval of 0° < θ < 90°.

± 1 + tan 2 A (Taking square root both sides) ∴ sec A =

Hence, the given statement is false.

Here, A is acute (i.e., less than 90°) and sec⁡ A is positive when A is acute.

(d) sin⁡θ = cos⁡ θ for all values of θ

⇒ sec A =+ 1 tan 2 A

sin 30° = 1 , cos 30° = 3 2 2

1 ⇒ sec A =+

sin45° = sin60° =

, cos 45° =

3 , cos 60° = 1 2 2

sin90° =1, cos 90° =1

It can be observed that sin⁡ θ = cos⁡ θ, Only for θ = 45° Hence, the given statement is false. (e) cot⁡ A is not defined for A = 0° Here for A = 0° sin0° 1 cos 0° = = = Undifened cos0° 0 Hence, the given statement is true. 206

cot A

(As tan A =

1 ) cotA

cot 2 A + 1 ⇒ sec A = 2 cot A cot 2 A + 1 ⇒ sec A = cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A. 1 Sol. We know that cos A = sec A We know that 1 + tan 2 A = sec2 A ∴ tan 2 A = sec2 A − 1 tan A = ± sec2 A − 1

A is acute (i.e., less than 90°) and tan⁡ A is positive when A is acute. Thus

(c)

= tan A

(d)

sec2 A − 1

(i) sec A ⁡

(

sec2 A − 1

(iv) tan2⁡A

)

(b) (1 + tan θ + secθ )(1 + cot θ − cosecθ ) Writing in terms of sin⁡ θ and cos⁡ θ sin θ 1  cosθ 1   =  1 + + −  1 +  sin θ sin θ   cosθ cosθ 

⇒ cosec2 A = 1 + cot 2 A   1 1+  ⇒ cosec2 A =    2  sec A − 1 

( cosθ + sinθ + 1) ( sinθ + cosθ − 1)

= {(

1 ⇒ cosec2 A = 1+ 2 sec A − 1

sec2 A − 1 + 1 ⇒ cosec2 A = sec2 A − 1

cosθ + sin θ ) + 1}{( sin θ + cosθ ) −1} cosθ sin θ

( cosθ + sinθ )2 − (1)2 {As(a + b)(a - b) = a2 - b2} cosθ sin θ 2

cosθ sin θ 2

{As ( a + b ) = a 2 + b 2 + 2ab }

sec A ⇒ cosec A = ± sec2 A − 1

(1 + 2 cosθ sinθ −1)

sec A ⇒ cosec A = ± sec2 A − 1

Here, A is acute (i.e., less than 90°) and cosec A is positive when A is acute. Thus, sec A cosec A = sec2 A − 1 We know that

1 cosec A

cosθ sin θ

= ( sin 2 A + cos2 A 1)

2 ( cosθ sin θ ) cosθ sin θ

= 2 Hence, option (iii) is correct. (c)

( sec A + tan A )(1 − sin A )

Writing in terms of sin and cos sin A   1 =  +  (1 – sin A )  cos A cos A  1 + sin A )(1 – sin A ) = ( cos A

1 ∴ sin A = sec A 2

sec A − 1 sec A − 1 ⇒ sin A = sec A

(1 − sin A ) = (cos A ) = cos A 2

cos A cos A Hence, (iv) is the correct answer. 2

3. Choose the correct option. Justify your choice. (a) 9sec2 A −9tan 2 A (iii) 8

(iv) 0

(b) (1 + tan θ + secθ )(1 + cot θ − cosecθ ) (ii) 1

sin θ

(cos θ + sin θ +2cosθ sinθ −1) =

(ii) 9

cosθ

sec2 A ⇒ cosec2 A = sec2 A − 1

(i) 0

(iii) cot2⁡A

= 9 sec2 A − tan 2 A = 9(1) = 9 [As sec2⁡A - tan2⁡A = 1] Hence, alternative (ii) is correct.

(i) 1

(iii) cosec A (iv) cos⁡ A

1 + tan 2 A

Sol. (a) 9sec2 A − 9tan 2 A

We know that 1 + cot 2 A = cosec2 A

sin A =

(ii) sin⁡ A

1 + cot 2 A (i) sec2⁡ A (ii) -1

We know that 1 cot A = tan A Putting value of tan⁡ A found above cot A =

( sec A + tan A )(1 − sin A )

(iii) 2

Chapter 8 • Introduction to Trigonometry

(iv) -1

(d) 1 + tan A 1 + cot 2 A  sin 2 A   1 +  cos2 A   =  cos2 A   1 +  sin 2 A  

207

sin A cos A   = and cot A  tan A =  cos A sin A  

(

)

(

)

cos2 A + 12 + sin 2 A + 2 sin A = (1 + sin A )( cos A )

 cos2 A + sin 2 A   1      cos2 A    cos2 A  = =  sin 2 A + cos2 A   1      2 sin 2 A    sin A 

cos2 A + sin 2 A ) + 1 + 2 sin A = (1 + sin A )( cos A )

(1 + 1 + 2 sin A ) = = ( sin 2 A + cos2 A 1) 2 ( sin A + cos A = 1) (1 + sin A )( cos A ) 2

( 2 + 2sin A ) = (1 + sin A )( cos A )

2 2  1  sin A sin A =  × =  1  cos2 A  cos2 A

2 (1 + sin A ) 1 2 = = 2× = = 2sec A cos A (1 + sin A )( cos A ) cos A

 sin A  2 =   = tan A cos A   Hence, option (iv) is correct.

= R.H.S

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. 1 − cosθ 2 (a) ( cosecθ − cot θ ) = 1 + cosθ

 1 − cosθ  =   sin θ  2 1 − cosθ ) (

sin 2θ

tan θ cot θ + = 1 + secθ cosecθ 1 – cot 1 – θ ( ) ( tanθ )

tan θ cot θ + (1 – cot θ ) (1 – tanθ )

We need to make it in terms of cosθ⁡ and sinθ cosθ   1 =  −   sin θ sin θ 

(c)

Sol. Taking L.H.S

Sol. Solving L.H.S.

( cosec θ − cot θ )

Hence proved.

sin θ cosθ θ cos = + sin θ sin θ   cosθ   1 −  1 − θ θ  sin cos   

sin θ cosθ θ cos sin θ = + 2 θ θ θ − − sin θ  sin cos cos    (1 − cosθ ) ( sin 2θ = 1 − cos2θ )     = sin θ cosθ     1 − cos2θ

sin 2θ cos2θ (1 − cosθ )(1 − cosθ ) {a 2 − b 2 = a + b = + = ( )(a − b ) } cosθ ( sin θ − cosθ ) sin θ ( cosθ − sin θ ) (1 − cosθ )(1 + cosθ ) sin 2θ cos2θ − = cosθ ( sin θ − cosθ ) sin θ ( sin θ − cosθ )

1 −cos θ = 1 + cosθ = R.H.S (b)

{ ( cosθ − sin θ ) = − ( sin θ − cosθ )}

(1 + sin A ) cos A + = 2 sec A A 1 + sin ( ) cos A

Sol. Taking L.H.S. (1 + sin A ) cos A + 1 + sin A ( ) cos A

 sin 2θ cos2θ  1 = −   ( sinθ − cosθ )  cosθ sinθ 

( cos A )( cos A ) + (1 + sin A )(1 + sin A ) = (1 + sin A )( cos A )

(

cos A + (1 + sin A ) = (1 + sin A )( cos A )

208

sin 2θ cos2θ − = cosθ ( sinθ − cosθ ) sin θ ( sin θ − cosθ )

)

 sin 3θ − cos3θ  1 =   ( sinθ − cosθ )  cosθ sinθ  sin 3θ − cos3θ = ( sinθ − cosθ ) cosθ sinθ

( sinθ − cosθ )(sin 2θ + cos2θ + cosθ sinθ ) = ( sinθ − cosθ ) cosθ sinθ

(

)

{ a 3 − b 3 = ( a − b ) a 2 + b 2 + ab } = 1 + cosθ sin θ ( sin 2 A + cos2 A = 1) cosθ sin θ 1 cosθ sin θ = + cosθ sin θ cosθ sin θ 1 = +1 cosθ sin θ

{( cot A ) – (1 − cosec A )} {( cot A ) – (1 − cosec A )} = {(cot A ) + (1 − cosec A )} {( cot A ) – (1 − cosec A )} 2

( cot A ) − (1 − cosec A )

{ ( a + b )( a − b ) = a 2 − b 2 }

{(cot A ) + (1 − cosec A ) − 2(cot A )(1 − cosec A )} = 2

(

(cot 2 A ) − 1 + cosec2 A − 2cosec A

= R.H.S ∴ L.H.S = R.H.S Hence proved.

)

{As ( a + b ) = a 2 + b 2 + 2ab }

{cot A +1 + cosec A − 2cosec A − 2cot A + 2cot Acosec A} = 2

cot 2 A − 1 − cosec2 A + 2cosec A

1 + sec A ( sin 2 A ) = sec A (1 − cos A )

{As ( a − b ) = a 2 + b 2 − 2ab }

Sol. Taking L.H.S.

{cosec A + cosec A + 2cot Acosec A − 2cosec A − 2cotA} = 2

1 + sec A sec A 1 1 + cos A 1+ cos A = cos A = 1 1 cos A cos A cos A (1 + cos A ) = = (1 + cos A ) cos A

cot 2 A − cosec2 A − 1 + 2cosec A

{As cot 2 A +1 =cosec2 A }

{

2cosec2 A + 2cot A cosec A − 2cosec A − 2cot A = −1 − 1 + 2cosec A

}

2cosec A ( cosec A + cot A ) −2 ( cosec A + cot A ) = 2cosec A − 2

Taking R.H.S sin 2 A 1 − cos2 A ( sin 2 A = 1 − cos2 A ) = = (1 − cos A ) (1 – cos A )

(1 − cos A )(1 + cos A ) = {a 2 − b 2 = ( a + b )( a − b ) } (1 − cos A ) = (1 + cos A )

( 2cosec A − 2 )( cosec A + cot A ) = ( 2cosec A − 2 ) = ( cosec A + cot A ) = R.H.S Hence proved. (f)

∴ L.H.S = R.H.S Hence proved.

1 + sin A = sec A + tan A 1 − sin A

Sol. Taking L.H.S

( cos A − sin A + 1)

= cosec A + cot A cos A + sin A − 1

Sol. Using the identity cosec2 A = 1 + cot 2 A Taking L.H.S

( cos A – sin A + 1) cos A + sin A –1 Divide both numerator and denominator by sin A . cos A

( cos A – sin A + 1) = sin A − 1 + sin A cos A + sin A –1

{( cot A ) – (1 − cosec A )}

= 1 + secθ cosec θ

(e)

Multiplying Numerator and denominator by

{( cot A ) – (1 − cosec A )} =

1 1 = +1 × cosθ sin θ = secθ × cosec θ + 1

(d)

( cot A –1 + cosec A ) ( cot A ) – (1 − cosec A ) = = cot A + 1 − cosec A (cot A ) + (1 − cosec A )

cos A 1 +1 − sin A sin A

Chapter 8 • Introduction to Trigonometry

1 + sin A 1 − sin A Multiplying numerator and denominator by 1 + sin A

1 + sin A 1 + sin A × 1 − sin A 1 + sin A

( 1 + sin A )

12 − sin 2 A

1 + sin A cos2 A

(cos θ = 1 − sin θ ) 2

209

Sol. Taking L.H.S ( cosec A – sin A )( sec A – cos A )

1 + sin A cos A

 1  1  = − sin A  − cos A   sin A cos A   

1 sin A + cos A cos A sec A + tan A = R.H.S Hence proved. (g)

sin θ − 2sin θ

2cos3 θ − cosθ

(

= tan θ

sin θ − 2sin 3 θ 2cos3 θ − cosθ

( (

(1 − 2sin θ ) {2(1 − sin θ ) − 1} (1 − 2sin θ ) = tan θ × ( 2 − 2sin θ − 1) (1 − 2sin θ ) = tan θ = R.H.S = tan θ × (1 − 2sin θ ) 2

cosA sin A sin A .cos A 2 = = sin A cos A ( sin 2 A + cos= A 1) 1 = L.H.S Hence proved.

2 2

 1 + tan 2 A   1 − tan A 2 2 (j)  =  tan A  1 + cot 2 A  = A 1 − cot    Sol. Solving L.H.S

Hence proved. 2

(h) ( sin A + cosec A ) + ( cos A + sec A )

= 7 + tan 2 A + cot 2 A

 1 + tan 2 A    2  1 + cot A 

Sol. Taking L.H.S ( sin A +cosec A )2 + ( cos A + sec A )2 2

(

)

1   = sin 2 A + cosec2 A + 2sin A . + sin A  1   2 2  cos A + sec A +2cos A .  cos A  

) (

= sin 2 A + cosec2 A + 2 + cos2 A + sec2 A + 2

) ) (

(

)

) )

= sin 2 A + 1 + cot 2 A + 2 + cos2 A + 1 + tan 2 A + 2 = sin 2 A + cos2 A + 1 + cot 2 A + 2 + 1 + tan 2 A + 2

(

)

= sin A + cos A + cot A + tan A + (1 + 2 + 1 + 2 ) 2

= 1 + cot A + tan A + 6 = 7 + cot 2 A + tan 2 A = R.H.S Hence proved.

(i) ( cosec A – sin A )( sec A – cos A ) = 210

) ( (

) )

1 + tan 2 A 1 + tan 2 A = = 1   tan 2 A + 1 + 1   2  tan A  tan 2 A tan 2 A 1 + tan 2 A = tan 2 A = R.H.S = 2 1 + tan A

=sin A + cosec A +2sin Acosec A + cos2 A + sec2 A + 2cos A .sec A

(

1 tan A + cot A 1 1 = = sin A ( sin A ) + cos A ( cos A ) sin A cos A + cos A sin A cos A sin A 1 sin A .cos A = = sin 2 A + cos2 A sin 2 A + cos2 A

) )

= tan θ ×

(

(cos θ = 1 − sin θ )

Taking R.H.S

sin θ 1 − 2sin 2 θ = cosθ 2cos2 θ –1

( (

)

cos2 A sin 2 A = × sin A cos A  cos2 A = 1 − sin 2 A and     sin 2 A = 1 − cos2 A    = sin A cos A

Sol. Taking L.H.S

) (

1 − sin 2 A 1 − cos2 A = × sin A cos A

1 tan A + cot A

(

)

 1 − tan A  Now,    1 − cot A  2  1 − tan A    = 1 1−  tan A  

)

 1 − tan A   tan A 1 − tan A 2 ( ) =  ( tan A − 1)  =      tan A − 1   tan A  2

 tan A (1 − tan A )  tan 2 A =  ( − tan A )2 =  = 1 1 tan − − A ( )   Hence proved.

Multiple Choice Questions 1. If sin A = (a)

4 5

3 , what is cosA ? 5 3 5 (b) (c) 5 3

1 3 and sinφ = then tan (θ + φ ) = 2 2 1 (a) 3 (b) 0 (c) 1 (d) not defined

9. If cosθ =

2 5

(d)

(CBSE SQP 2025)

2. What is the value of sin45 × cos45 ? 

1 (a) 2

(b)



3. If cot⁡ A = 1, then A = (a) 30° (b) 45

(NCERT Exemplar)

1 (d) 3

1 (c) 4

(CBSE 2021, 2023)

(d) 90

(CBSE 2023)

(a) (c) 5.

a2 +b a

a2 +b2 2

(b)



sin 2 30 + sin 2 60 (b) 64 12

a2 +b2

15 12

(d) 5 12

(CBSE 2024)

7. If ∆ABC is right angled at C , then the value of cos ( A + B ) is 1 3 (a) 0 (b) 1 (c) (d) 2 2

p2 +1 2p

(b)

p2 −1 2p

(c)

p2 −1

p2 +1

(d)

p2 +1 p2 −1

(CBSE 2022)

3  13.  tan 2 30 − sec2 45 + sin 2 60  is equal to 4  5 1 −3 (a) -1 (b) (c) (d) 6 6 2

(a)

3 4

(b)

4 5

(c)

5 3

(d)

4 3

= y tan60° − cot 60°, then 15. If = x cos30 − sin 30 and (a) x = y (b) x > y

(CBSE 2024)

(d) x > 1, y < 1

(CBSE 2025)

1 16. In a right triangle ABC , right-angled at A, if sinB = , 4 then the value of secB is

1, then sin 2 A + sin 4 A = 8. If cosA + cos A = (a) -1 (b) 0 (c) 1

(d) b − a

14. If 4secθ − 5 = 0 then the value of cotθ is

(d) 90°

(CBSE 2022)

(a)

1 6. The value of θ for which 2sin 2= ; 0° ≤ θ ≤ 90° is: θ 2 (a) 30° (b) 60° (c) 45°

(b) b 2 − a 2

p then tan θ is: 12. If secθ + tanθ =

(c)

(a) a 2 − b 2

a2 −b2

(d)

5cos 60 + 4sec 30 − tan 45 (a) 67 12

a2 +b2

(CBSE 2024)

p and b cot θ + acosecθ = q, 11. If acotθ + bcosec θ = 2 2 then p − q =

a sin θ + bcosθ a 4. If tan θ = then the value of is: a sin θ −bcosθ b a2 −b2

= sin θ cosθ , (0° < θ < 90°), then the value of 10. If secθ × sinθ is: 1 (a) (b) 2 (c) 1 (d) 0 2

(a) 4

(d) none of these

(b)

15 4

(d)

(CBSE 2025)

Answers 4 5 67 5. (a) 12 1. (a)

9. (a)

2. (a)

1 2

a2 +b2

3. (b) 45

4. (b)

6. (a) 30°

7. (a) 0

8. (c) 1

10. (c) 1

11. (b) b2 – a2

Chapter 8 • Introduction to Trigonometry

a2 −b2

12. (b)

4 p2 −1 13. (a) -1 14. (d) 3 2p

15. (c) x < y 16. (d)

211

Constructed Response Questions Very Short Answer Questions 1. Write the value of cosec 30 . 1 1  = = 2 Sol. cosec 30=  1 sin30 2 2. If sinθ + cosθ =

(CBSE 2020)

3, then find the value of sinθ ⋅ cosθ . (CBSE 2023)

CBSE Marking Scheme 2023

3 2  2  2  7. If 4cot 45 − sec 60 + sin 60 + p =, then find the 4 value of p. 3 Sol. Given: 4cot 2 45 − sec2 60 + sin 2 60 + p = 4 2  3 3 ⇒ 4 × (1)2 − (2)2 +   + p = 4  2  ⇒4−4+

Sol. sin θ + cos θ = 3 sin 2θ + cos2θ + 2sinθ cosθ = 3

⇒ 1 + 2sin θ cosθ = 3

1 1  8. If sin ( A − B ) = , cos ( A + B ) = , 0 < A + B ≤ 90 , 2 2 A > B ; find ∠A and ∠B . (CBSE 2024)

⇒ sinθ cosθ = 1

Sol. cot 60 − tan 30 =

−

CBSE Marking Scheme 2024

(CBSE 2024)

A − B = 30° (i) cos ( A + B= ) cos60°

Adding (i) and (ii) A − B + A + B= 30° + 60° 2 A= 90° ⇒ A= 45° ⇒ 45° + B= 60° ⇒ B= 15°

CBSE Marking Scheme 2023 Sol. sec2 θ =1 + tan 2 θ =1 + Again since cotθ = 7

1 8 = 7 7

1 2 1

)

(

)

(

Sol. 1 + tan 2θ (1 − sinθ )(1 += sinθ ) sec θ 1 − sin θ 2 = sec2θ .cos θ =

cos θ

)

6. If sin 2 A + cos2 A = 1 , find the value of tan A + 1 1. Sol. Given sin 2 A + cos2 A = Divide the entire equation by cos2 A sin 2 A cos2 A 1 + = 2 2 cos A cos A cos2 A 2

tan A + 1 = sec A 212

9. Prove that: 1 + Sol. L.H.S = 1 +

=1+

cot 2α cosecα = 1 + cosecα 2

(NCERT Exemplar)

(CBSE 2020)

cot α cosec α − 1 = 1+ 1 + cosecα 1 + cosecα

( cosecα + 1)( cosecα − 1) 1 + cosecα

( cosecα − 1) 1

= 1 + cosecα − 1 = cosecα

= L.H.S cosec = α R.H.S 10. Show that tan 4θ + tan 2θ = sec4θ − sec2θ .

⋅ cos2θ = 1



= A 45 = , B 15

= 1+

5. Find the value of 1 + tan θ (1 − sinθ )(1 + sinθ ) 2



1 2

1    tanθ =  7 

⇒ cosec2θ =1 + cot 2θ =1 + 7 =8 8 8− 48 3 7 ∴ L.H.S = == = R.H.S 8 64 4 8+ 7

(

1 2

A + B = 60° (ii)

1 cosec2θ − sec2θ 3 = , then show that 7 cosec2θ + sec2θ 4

4. If tan θ

1 2

Sol. sin ( A − B= ) sin 30°

3. Write the value of cot 60 − tan 30 . 1

3 3 3 3 + p = ⇒ p = − =0 ⇒ p =0 4 4 4 4

(

)

Sol. L.H.S = tan 4 θ + tan 2 θ = tan 2 θ tan 2 θ + 1

(

)(

)

= sec2θ − 1 sec2θ = sec4θ − sec2θ = R.H.S 11. Prove that ( sin α + cosα )( tan α + cot α ) =secα + cosecα . Sol. L.H.S = ( sin α + cosα )( tan α + cot α ) sin α cosα  = + ( sin α + cosα )  cos α sin α  

 sin 2 α + cos2 α  = ( sin α + cosα )    cosα sin α  = ( sin α + cosα ) ×

⇒ cos3θ = cos60 ⇒ 3θ = 60 ⇒ θ = 20

1 cosα ⋅ sin α

15. Prove this:

sinα cosα = + cosα sinα cosα sinα

Sol. L.H.S:

Sol. L.H.S =

secA − 1 secA + 1 + = 2cosecA secA + 1 secA − 1

(CBSE 2023)

secA − 1 secA + 1 + secA + 1 secA − 1

(secA − 1)2 (secA + 1)2 = + 2 sec A − 1 sec2 A − 1

1 cos A = 2 × 1 sin A sin A cos A

2×

(CBSE 2023)

1 cos A 1 1 cos A

(1 − cos A )(1 + cos A ) 1 = 1 + cos A = (1 − cos A ) 1 − cos2 A = 1 − cosA

) (

2 cos4 60 + sin 4 30 − tan 2 60 + cot 2 45

)

1 + 3sec2 30 = 4 4

1 + secA sin 2 A = secA 1 − cosA

cos A + 1 = cos A 1 cos A

= 2cosec = A R.H.S

1. Prove that:

1 + sec A Sol. = L.H.S = sec A

secA − 1 secA + 1 secA − 1 + secA + 1 = + = tanA tanA tanA

(

2 = 2sec = θ R.H.S

CBSE Marking Scheme 2023

(secA − 1)2 (secA + 1)2 = + 2 tan A tan 2 A

13. Show that:

2 2 = = 2 1 − sin θ cos2θ

Short Answer Questions

 secA − 1   secA − 1   secA + 1   secA + 1  =  × +  ×   secA + 1   secA − 1   secA − 1   secA + 1 

2sec A = = tan A

1 1 + 1 + sinθ 1 − sinθ

1 − sinθ + 1 + sinθ = (1 + sinθ )(1 − sinθ )

1 1 = + = secα + cosecα = R.H.S cosα sinα 12. Prove that :

1 1 + = 2sec2θ 1 + sinθ 1 − sinθ

 1 1  2  Sol.= 2   +    − ( 3)2 + (1)2 + 3   2 2      3  

{

}

1  4  1 = 2  +  − ( 3 + 1) + 3 × 16 16 3   1 1 = 2× − 4 + 4 = 8 4 14. Find the value of θ : 2cos3θ = 1 1 1 ⇒ cos3θ = Sol. 2cos3θ = 2

Chapter 8 • Introduction to Trigonometry

sin 2 A = = R.H.S 1 1 − cosA 2. If sin ⁡θ + cos ⁡θ = p and sec⁡ θ + cosec θ = q, show that

(

)

q p2 −1 = 2p . Sol. Given, sinθ + cosθ = p (i) ⇒ (sinθ + cosθ )2 = p2

[Squaring on both sides]

⇒ sin 2 θ + cos2 θ + 2sin θ ⋅ cosθ = p2 ⇒ 1 + 2sin θ ⋅ cosθ = p2 p2 −1 ⇒ sin θ cosθ = (ii) 2 Also, secθ + cosecθ = q

213

1 1 ⇒ + = q cosθ sin θ

sin 2 A cos2 A 2 2 = cos A + sin A 1 1 cos2 A sin 2 A

sinθ + cosθ ⇒ = q sinθ ⋅ cosθ

sin 2 A cos2 A cos2 A sin 2 A = × + × 1 1 cos2 A sin 2 A

p 2p ⇒ 2 = = q⇒ 2 q Using ( i ) and ( ii )  p −1 p −1 2

(

= sin 2 A + cos2 A =1

)

⇒ q p2 −1 = 2p 3. Simplify: Sol.

L.H.S = 1 = R.H.S. Hence proved.

1 − sinA cosA + cosA 1 − sinA

1 = = θ , prove that tanθ 1 or . 6. If 1 + sin 2θ 3sin θ cos 2

(NCERT Exemplar)

(1 − sinA )2 + cos2 A 1 − sinA cos A = + cosA (1 − sinA ) cos A 1 − sinA

(CBSE 2020)

Sol. 1 + sin θ = 3sinθ cosθ

1 − 2sinA + sin 2 A + cos2 A 1 − 2sinA + 1 = = cosA (1 − sinA ) cosA (1 − sinA ) 2 2 − 2sinA = = 2sec A = cosA (1 − sinA ) cos A

Sol. Let secθ + tanθ = λ (i) We know that, sec2 θ − tan 2 θ = 1

1 ⇒ λ ( secθ − tanθ ) = 1 ⇒ secθ − tanθ = (ii)

Adding equation (i) and (ii), we get: 1

1  1  ⇒ 2 x +  =λ + λ 4x  λ 

⇒ secθ + tanθ = 2x or

Sol. L.H.S

214

sec A

tan A 2

sec A

⇒ ( cotθ − 2 )( cotθ − 1) = 0 = cot θ 2= or cotθ 1 1 2

If cotθ = 1: tanθ = 1 tanθ = 1 or

1 2

= 7 + tan 2θ + cot 2θ

(CBSE 2019)

Sol. L.H.S: (sinθ + cosecθ )2 + (cosθ + secθ )2

On comparing, we= get λ 2= x or λ

tan 2 A

⇒ cot 2θ − 2cotθ − cotθ + 2 = 0

7. Prove that: (sinθ + cosecθ )2 + (cosθ + secθ )2

1 1 ⇒ 2x + =λ + λ 2x

5. Prove that

+1 = 3cotθ

⇒ cot 2θ − 3cotθ + 2 = 0

If cotθ = 2: tanθ =

⇒ ( secθ + tan θ )( secθ − tanθ ) = 1

2secθ = λ +

⇒ cosec2θ + 1 = 3cotθ

sin 2θ

⇒ 1 + cot 2θ + 1 = 3cotθ

1 1 , prove that secθ + tanθ = 2x or . 4x 2x

x+ 4. If secθ =

Divide both sides by sin 2θ :

= sin 2θ + cosec2θ + 2sinθ cosecθ + cos2θ + sec2θ + 2cosθ secθ

1 2x

= sin 2θ + cos2θ + cosec2θ + sec2θ +

cot 2 A 2

cosec A 2

cot A

1 2x

cosec2 A

= 1.

(CBSE 2023)

= 1 + 1 + cot 2θ + 1 + tan 2θ + 2 + 2 = 7 + tan 2θ + cot 2θ = R.H.S Hence proved.

2sin θ 2cosθ + sin θ cosθ

sin θ m and a sinθ + b= cosθ n , prove that: a 2 + b 2 = m 2 + n 2 8. If a cosθ + b=

(CBSE 2023)

Topper’s Answer

(

)

sin A + cos A sin A − cos A 2 cos2 x − sin 2 x + = cos2 x − sin 2 x sin A − cos A sin A + cos A 2sin 2 A= = −1 sin x 1 − tan x (CBSE 2024) 1− cos x sin A + cos A sin A − cos A Sol. + [ cos2 x + sin 2 x = 1] sin A − cos A sin A + cos A cos2 x − sin 2 x = (sinA + cosA )2 + (sinA − cosA )2 = cos x − sin x sin 2 A − cos2 A cos x sin 2 A + cos2 A + 2sin AcosA + sin 2 A + cos2 A − 2sin AcosA = cos x cos2 x − sin 2 x sin 2 A − 1 − sin 2 A = ( cos x − sin x ) 1+1 = cos x [cos x + sin x ][cos x − sin x ] sin 2 A − 1 + sin 2 A = ( cos x − sin x ) 2 = = R.H.S 2sin 2 A − 1 = cos x [cos x + sin x ] cot x + 1) ( cos4 x − sin 4 x 10. Prove that : =  cos x sin x  1 − tan x sec x cosecx = sin x cos x  +   sin x sin x  4 4 cos x − sin x Sol. L.H.S = [Multiplying and dividing by sinx ] 1 − tan x cotx + 1 cos2 x + sin 2 x cos2 x − sin 2 x = = R.H.S secx cosecx = (1 − tan x )

9. Prove that:

(

)(

Chapter 8 • Introduction to Trigonometry

)

(

)

215

11. Prove that:

(

) (

)(

sin 8θ − cos8θ = sin 2θ − cos2θ 1 − 2sin 2θ cos2θ

(

13. Simplify: 5sin 2 60° + 3cos2 30° − sec2 45°

)

Sol. Given 5sin2 60° + 3cos2 30° – sec2 45 [Substituting with all the appropriate values.]

) − (cos θ ) = (sin θ − cos θ )(sin θ + cos θ ) = (sin θ − cos θ )(sin θ + cos θ )(sin θ + cos θ ) Sol. L.H.S = sin 8 θ − cos8 θ = sin 4 θ 4

)(

) (

= ∴ sin 60° 2

(

)

(

){(sin θ + cos θ ) − 2sin θ cos θ }

(

)(

=sin 2θ − cos2θ

3 and sec 45° = 2

( )

5 sin2 60° + 3 cos2 30° – sec2 45° = 4

)

=sin 2θ − cos2θ 1 − 2sin 2θ cos2θ =R.H.S

14. Prove that:

p2 −1 12. = If cosecθ + cot θ p , then= prove that cosθ . p2 +1

tan θ cot θ + = 1 + secθ cosecθ 1 − cot θ 1 − tan θ

(CBSE 2023)

sin θ cosθ tan θ cot θ Sol. L.H.S: + = cosθ + sin θ 1 − cotθ 1 − tanθ 1 − cosθ 1 − sin θ sin θ cosθ

Sol. We know that cosec θ − cot θ

( cosec θ − cot θ ) ( cosec θ + cot θ ) × (1) ( cosec θ + cot θ )

sin θ cosθ tan θ cot θ cosθ sin θ + = + 1 − cotθ 1 − tanθ sinθ − cosθ cosθ − sinθ sin θ cosθ

cosec2 θ − cot 2 θ 1 = cosecθ + cot θ cosecθ + cot θ

∴ cosecθ + cot θ = p (i) sinθ ⋅ sinθ cosθ ⋅ cosθ = + cosθ ( sinθ − cosθ ) sinθ ( cosθ − sinθ ) 1 (ii) ⇒ cosecθ − cot θ = p sin 2θ cos2θ = + cosθ ( sinθ − cosθ ) sinθ ( cosθ − sinθ ) Adding (i) and (ii), we get 2cosecθ= p +

1 p

p2 +1 ⇒ cosecθ = 2p

sin 2θ cos2θ = − cosθ ( sinθ − cosθ ) sinθ ( sinθ − cosθ )

Subtracting (i) and (ii), we get

p2 −1 ⇒ cot θ = (iv) 2p

p 2 − 1 2p p2 −1 ⇒ cosθ = × 2 = 2 2p p +1 p +1

216

(

( sinθ − cosθ )(1 + sinθ cosθ ) = cosθ sinθ ( sinθ − cosθ )

1 p

= θ cot θ × sin θ Now, cos

sin 3θ − cos3θ cosθ sinθ ( sinθ − cosθ ) ( sinθ − cosθ ) sin 2θ + sinθ cosθ + cos2θ = cosθ sinθ ( sinθ − cosθ ) =

2p ⇒ sin θ = (iii) p2 +1

2cot θ= p −

2  3  3 ⇒ 5 + 3 − 2  2   2      15 9 3 3 ⇒ 5  + 3  − 2 ⇒ + −2 4 4 4 4 24 ⇒ −2=6−2= 4 4

 sin 2θ 2 + cos2θ 2 + 2sin 2θ cos2θ    = sin θ − cos θ   −2sin 2θ cos2θ  2

3 ; cos 30° = 2

1 + sinθ cosθ = cosθ sinθ 1 sinθ cosθ = + cosθ sinθ cosθ sinθ

= secθ cosec = θ + 1 R.H.S Hence proved.

)

17. Use the identity: sin 2 A + cos2 A = 1 to prove that tan 2 A + 1 = sec2 A . Hence, find the value of tan A, 5 when sec A = , where A is an acute angle. 3

 4sinθ − cosθ + 1  15. If 4tan θ = 3, evaluate    4sinθ + cosθ − 1  3 3 ⇒ tanθ = Sol. Given : 4tanθ = 4 In ∆ABC AB 3k tan = θ = BC 4k

(CBSE 2025)

Sol. Given: sin A + cos A = 1

3k B

= AB 3= k , BC 4k So,

Dividing both sides by cos2 A : C

2 = AB 2 + BC 2 By Pythagoras: AC

5 Value of tan A when sec A = 3

2 AC = 9k 2 + 16k 2

= θ Thus, sin

25k 2 = 5k

AB 3k 3 BC 4k 4 = = ,cos = θ = = AC 5k 5 AC 5k 5

3 4 4 ⋅ − +1 4sinθ − cosθ + 1 = 5 5 Now: 4sinθ + cosθ − 1 4 ⋅ 3 + 4 − 1 5 5 12 − 4 + 5 13 5 = = 5 13 = 12 + 4 − 5 11 11 5 5

then find the value of A and B.

, 0° < A + B < 90°,

Sol. Given: tan= ( A + B ) 1and tan= (A − B)

(CBSE 2019)

tan ( A + B ) = tan45 A +B = 45 (1) 

A −B = 30 (2) Add equations (1) and (2) A + B + A − B = 45 + 30 

5 So, tan 2 A + 1 =   3 25 tan 2 A =−1 + 9 16 2 tan A = 9 4 tan A = 3

Long Answer Questions

1 and tan ( A −= 16. If tan ( A + B ) = B)

tan ( A − B ) = tan 30

sin 2 A cos2 A 1 + = 2 2 cos A cos A cos2 A tan 2 A + 1 = sec2 A

2 AC = (3k )2 + (4k )2

AC 2 = 25k 2 ⇒AC =



2 A = 75 

75 = A = 37.5 2 Substitute A into equation (1) A +B = 45 37.5 + B = 45 B= 45 − 37.5 = 7.5  = A 37.5 = , B 7.5

Chapter 8 • Introduction to Trigonometry

1. In an acute-angled triangle ABC, if 1 1 sin ( A + B − C ) = and cos ( B + C − A ) = , find 2 2 ∠ A , ∠ B and ∠ C . Sol. Given: sin ( A + B − C ) = cos ( B + C − A )=

1 = sin30, 2

= cos45

Thus, A + B − C = 30 (i) B +C − A = 45 (ii) Adding (i) and (ii) ( A + B − C ) + ( B + C − A )= 30 + 45 2B = 75 B = 37.5 Subtracting (ii) from (i) ( A + B − C ) − ( B + C − A )= 30 − 45 2 A − 2C = −15 A −C = −7.5 (iii) 180 Angle Sum Property A + B + C = A + C + 37.5 = 180 A +C = 142.5 (iv)

217

Adding equations (iii) and (iv):

tan 2θ = x 2 −

−7.5 + 142.5 ( A −C ) + ( A +C ) = 



2 A = 135

tan θ=

A = 67.5

secθ + tan θ = x +

67.5 + C = 142.5 C = 75 2. If tan A n= = tan B and sin A m sin B , prove that m −1 n2 −1

1 16x 4 − 8x 2 + 1 secθ + tan θ =x + + 4x 16x 2

Sol. Now, tanA = ntanB

secθ + tan θ =x +

n 1 tan A ⇒ cot B= n tan A

⇒ tan B=

(

m 1 sin A ⇒ cosecB= m sin A

(

m 2 − n 2 cos2 A ⇒ = 1 sin 2 A

(n − 1) cos A 2

m2 −1 cos2 A ⇒ 2 = n −1 1 1 2x or 3. If secθ= x + , prove that secθ + tanθ = . 4x 2x (CBSE 2019)

1 (i) 4x

1  1  1 1 2 + sec θ = x + 2x ⋅  = x + + 4x  4x  2 16x 2

2 tan = θ sec2θ − 1 Thus,

tan 2θ = x 2 +

218

1 1 + −1 2 16x 2

4. Prove that

sinA − cosA + 1 1 = sinA + cosA − 1 secA − tanA

sinA − cosA + 1 Sol. L.H.S: sinA + cosA − 1

(CBSE 2019)

Multiply numerator and denominator by (1 + sin A ) :

Hence proved.

)

4x 2 + 1 − 4x 2 − 1 For − case: secθ + tan θ = 4x 2 1 = = 4x 2 x Hence proved.

⇒ m 2 − n 2 cos2 A = sin 2 A ⇒ m 2 − n 2 cos2 A = 1 − cos2 A

1   sec2= θ x +  4x  

)

(

m2 n2 m2 n 2 cos2 A 1 1 ⇒ − = ⇒ − = sin 2 A tan 2 A sin 2 A sin 2 A

Sol. Given: secθ= x +

)

4 x 2 + 1 + 4 x 2 − 1 8x 2 = =2 x For + case: secθ + tan θ = 4x 4x

Substituting the values of cotB and cosecB in 2 cosec B − cot 2 B = 1, we get

⇒= m 2 − 1 n 2 cos2 A − cos2 A ⇒ m 2 − 1=

1 (4x 2 − 1)2 + 4x ( 4 x )2

2 2 1 4x 2 − 1 4x + 1 ± 4x − 1 ± = secθ + tan θ =x + 4x 4x 4x

and sin A = m sin B ⇒ sin B=

1 1 1 + x2 − + 4x 2 16x 2

1 16x 4 8x 2 1 secθ + tan θ =x + + − + 2 2 4x 16x 16x 16x 2

  = ∠ A 67.5 = ,∠ B 37.5 = ,∠ C 75 Thus,

cos2 A =

1 1 (ii) + 2 16x 2

From (i) and (ii)

A +C = 142.5

x2 −

1 1 + 2 16x 2

( sinA − cosA + 1)(1 + sinA ) = ( sinA + cosA − 1)(1 + sinA ) ( sinA − cosA + 1)(1 + sinA ) = sin A + cos A − 1 + sin 2 A + sin AcosA − sin A ( sin A − cos A + 1)(1 + sin A ) = cos A − 1 + sin 2 A + sin A cos A ( sinA − cosA + 1)(1 + sinA ) = cos A − 1 + 1 − cos2 A + sin AcosA ( sinA − cosA + 1)(1 + sinA ) = cos A − cos2 A + sin AcosA ( sinA − cosA + 1)(1 + sinA ) = cos A (1 − cos A + sin A )

1 cosθ 1 + cosθ = + = R.H.S = cosecθ + cotθ = sin θ sin θ sin θ

1 + sin A 1 sin A = = + =sec A + tan A cos A cos A cos A (sec A + tan A ) = × (sec A − tan A ) (sec A − tan A )

Hence proved.

(sec2 A − tan 2 A ) = (sec A − tan A )

tan 2 A cosec2 A 1 + = 2 2 2 tan A − 1 sec A − cosec A 1 − 2cos2 A

1 = (sec A − tan A )

Sol. L.H.S:

7. Prove that:

= R.H.S Hence proved. prove that x 2 + y 2 = 1. Sol. We have, x sin 3 θ + y cos3 θ = sinθ cosθ ⇒ ( xsinθ ) sin 2 θ + ( ycosθ ) cos2 θ = sin θ cosθ

)

⇒ xsinθ sin 2 θ + ( xsinθ ) cos2 θ = sin θ cosθ ( xsinθ = ycosθ )

(

)

⇒ xsinθ sin 2 θ + cos2 θ = sin θ cosθ ⇒= x θ sinθ cos= θ ⇒ x cosθ sin Now, we have xsinθ = y cosθ ⇒ cosθ sin = θ y cosθ ⇒ = y sin θ ( = x cosθ )

6. Prove that: Sol. L.H.S:

cosθ − sinθ + 1 = cosecθ + cotθ cosθ + sinθ − 1

cosθ − sinθ + 1 cosθ + sinθ − 1

Multiply numerator and denominator by sinθ ( cosθ + sinθ − 1) :

tan 2 A − 1 sec2 A − cosec2 A

cosec2 A

sec2 A − cosec2 A

(i)

1 sin 2 A

1 1 − cos2 A sin 2 A

1 cos2 A sin 2 A = = sin 2 A − cos2 A sin 2 A − cos2 A sin 2 A cos2 A

(ii)

Now we can combine both terms (i) and (ii)

2 Hence, x 2 + y= cos2θ + sin 2θ= 1.

Hence proved.

cosec2 A

sin 2 A sin 2 A tan 2 A cos2 A cos2 A = = 2 2 2 tan A − 1 sin A sin A − cos2 A − 1 cos2 A cos2 A sin 2 A = sin 2 A − cos2 A

5. If x sin 3 = θ + y cos3 θ sinθ cosθ= and xsinθ ycosθ ,

(

tan 2 A

sin 2 A

cos2 A

sin 2 A − cos2 A sin 2 A − cos2 A sin 2 A + cos2 A 1 = = 2 2 2 sin A − cos A sin A − cos2 A

(CBSE 2018)

1 = 2 A 1 − cos − cos2 A 1 = = R.H.S 1 − 2cos2 A Hence proved. l and secθ − cosθ = 8. If cosecθ − sinθ = m , prove that

sinθ ( cosθ − sinθ + 1) = sinθ ( cosθ + sinθ − 1)

(

)

l 2m 2 l 2 + m 2 + 3 = 1.

(sinθ cosθ − sin θ + sinθ ) = 2

cosecθ − sinθ and m = secθ − cosθ Sol. We have, l =

sinθ ( cosθ + sinθ − 1)

(

∴l 2m 2 l 2 + m 2 + 3

(sinθ cosθ + sinθ − (1 − cos θ )) = 2

sinθ ( cosθ + sinθ − 1)

sin θ (1 + cosθ ) − (1 + cosθ )(1 − cosθ ) = sinθ ( cosθ + sinθ − 1)

(1 + cosθ )( cosθ + sinθ − 1) = sinθ ( cosθ + sinθ − 1) Chapter 8 • Introduction to Trigonometry

)

2 (cosecθ − sinθ ) +  = (cosecθ − sinθ )2 (secθ − cosθ )2   2 (secθ − cosθ ) + 3  2  1   − sinθ  +     1   1   sinθ sinθ   =− − cosθ     2 sin cos θ θ      1    cosθ − cosθ  + 3    2

2 

219

Sol. We have,

 1 − sin 2θ 2    +   2 2   1 − sin 2θ   1 − cos2θ   sinθ  =       2  sinθ   cosθ   1 − cos2θ     + 3   cosθ  

 cosA sinA  + 1 +  ( sinA − cosA ) sinA cosA  L.H.S =  1   1 −   3  cos A sin 3 A   cos2 A + sin 2 A   1 +  ( sinA − cosA ) sinAcosA   =  sin 3 A − cos3 A   sin 3 Acos3 A   

2 2 2 2   cos2θ   sin 2θ   cos2θ   sin 2θ   +         + 3 sin cos sin cos θ θ θ θ         

cos4θ sin 4θ  cos4θ sin 4θ  = × + + 3 2 2  2 2 sin θ cos θ  sin θ cos θ 

1   3 3 1 +  ( sinA − cosA ) sin Acos A sinAcosA   = sin 3 A − cos3 A

6 6 2 2  cos θ + sin θ + 3cos θ sin θ  = cos2θ × sin 2θ   cos2θ sin 2θ  

(

{(

= cos2θ

) + (sin θ ) } + 3cos θ sin θ (cos θ + sin θ ) 2

= (cos2 q +sin2 q)3 = 1

[(a + b)3 = a3 + b3 + 3ab(a + b)]

9. Prove that:

)

 a 3 − b 3 = ( a − b ) a 2 + b 2 + ab   

= cos6θ + sin 6θ + 3cos2θ sin 2θ 3

)

( sinAcosA + 1)( sinA − cosA ) sin 2 Acos2 A = ( sinA − cosA ) sin 2 A + cos2 A + sinAcosA

(

)

( sinA cosA + 1) sin 2 Acos2 A 2 = = sin = Acos2 A R.H.S (1 + sinAcosA )

(1 + cotA + tanA )( sinA − cosA ) = sin 2 Acos2 A sec3 A − cosec3 A

Hence proved.

Competency Based Questions Multiple Choice Questions

4. In given figure, AM = MC and ∠C is a right angle, then sin 2α − cos2α is equal to

1. If A is an acute angle in a right triangle ABC, right angled a B, then the value of sin⁡A + cos⁡A is (a) equal to 1 (b) greater than 1 (c) less than 1

(a)

(d) 2

(b)

2. In given figure, ABCD is an isosceles trapezium, its perimeter is

( ) (b) (10 + 2 2 ) units (c) (10 + 4 2 ) units (d) (11 + 4 2 ) units (a) 8 + 4 2 units

3 units

(c)

(d)

2 units 45° D

45° E

3. A pendulum of length 3 m is attached to a point 2.3 m from the ground. It swings through an angle of 30 on each side of the vertical. The height above the ground at ends of its path is

220

(a) 0.9 m

(b) 0.6 m

(d) 0.8 m

4b 2 − 3a 2 2

5a − 4b

5a 2 − 4b 2

4b 2 − 3a 2 2

4a − 5b

3b 2 − 4a 2 3b 2 − 4a 2 2

4a − 5b

α B

3 and tan ( A − B= 5. If tan ( A + B ) = ) the value of A is (a) 30° (b) 45° (c) 60°

, A > B , then

(d) 90°

6. The value of (1 + tan2 q)(1 + sin q)(1 – sin q) (1 + cos q) (1 – cos q)(1 + cot2 q) is (a) 1 (b) -1 (c) 0

(d) -2

7. P and Q are acute angles such that P > Q. Which of the following is definitely true? (a) sin⁡ P < sin ⁡Q (b) tan⁡ P > tan Q ⁡ (c) cos⁡ P > cos⁡ Q 8.

(d) cos⁡ P > sin⁡ Q

(cos A − sin A ) on simplification gives (CBSE 2023) 4

(a) 2sin A − 1

(b) 2sin A + 1

(d) 2cos2 A − 1

n then m 2 − n 2 m and tanx − sinx = 9. If tanx + sinx = is equal to (a) 4 mn (b) mn (c) 2 mn

(d) none of these

90 which of these is 10. In a right-angled ∆PQR , ∠Q = always 0? (a) cosP − secR (b) tanP − cotR (c) sinP − cosecR (d) Cannot be known without knowing the value of P k + 1 sec θ (1 + sin θ )(1 − sin θ ) , then k = 11. If = 2

(a) 1

(b) 2

(d) –1 6

0, then the value of sin θ + cos θ is 12. If sinθ − cosθ = (a)

2 3

(b)

1 3

(c)

3 4

(d)

1 4

(b) cosθ tanθ = sinθ

1 1 (c) cosec2θ − cot 2θ = (d) tan 2θ − sec2θ = 14. If sinα = (a)

3 , cosβ = 2 (b)

3 , then tan α tan β is 2 1

(b)

1 2

(d)

16. If angles A, B, C of a ∆ABC form an increasing arithmetic progression, then sin B ⁡ = (a)

1 2

(b) (d)

Chapter 8 • Introduction to Trigonometry

3 2 1

(iv) tan 2 30 + 1

(a) Both (i) and (ii)

(b) Both (i) and (iii)

(d) All (i), (ii), and (iii)

2. If A and B are acute angles such that A < B, which of the following is definitely true? (i) sinA < sinB (ii) tanA < tanB (iii) cosA > cosB

(iv) cotA > cotB

(a) Only (ii) and (iv) (b) Only (i), (ii), and (iii) (c) All (i), (ii), (iii), and (iv) (d) Only (iii) and (iv)

(

)

3. For the value of cosA and sin 90 − A , which of the following is definitely true? (i) Same for A = 45

(ii) Same for A = 30

(iii)Same for A = 0

(iv) Never same

(a) Only (ii)

(b) Only (i), (ii), and (iii)

(d) Only (iii)

(d) 0

(iii) cot 2 45 + 1

(b) Both A and R are true, but R is not the correct explanation of A.

α + β  0, then the value of cos  15. If cos (α + β ) =  is 2   equal to (a)

1. The value of which of the following is definitely 1? (i) sin 2 45 + cos2 45 (ii) sec2 60 − tan 2 60

Assertion-Reason Based Questions

13. If θ is an acute angle of a right angled triangle, then which of the following equation is not true? (a) sin θ cotθ = cosθ

Select Response Type MCQs

(d) A is false, but R is true. 1. Assertion (A): For 0 ≤ θ < 90 ,sec2θ + cos2θ ≥ 2. 1 Reason (R): For x > 0, x + ≥ 2 x sin θ − 2sin 3 θ = tan θ , where θ is acute 2. Assertion (A): 2cos3 θ − cosθ angle. sinA Reason (R): For acute angle A , tanA = and cosA sin 2 A + cos2 A = 1 4 3. Assertion (A): The value of sinθ = is not possible. 3 Reason (R): Hypotenuse is the longest side in any right angled triangle. 221

Case Study Based Questions 1. A trolley carries passengers from the ground level located at point A to up to the top of mountain chateau located at P as shown in given figure. The point A is at a distance of 2000 m from point C at the  = α 30 = , β 60. base of mountain. Here

(b) What will be height of the mountain? 2000 m (i) 1000 m (ii) 3

(iv) 2000 3 m

(iii) 2000 m

(iv)

4000 3

(d) What will be the length of BC ?

(i) 1000 m

(ii)

(iii) 1000 3 m (a) Assuming the cable is held tight what will be the length of cable? (i) 2000 m (ii) 2000 3 m (iii) 4000 3 m

(iv)

4000 3

(iv)

2000 m 3

1000 3

(e) What will be the distance of point A to the foot of the mountain located at B? (i) 4000 3 m (iii)

4000 3

(ii) 4000 6 m (iv)

4000 m 3

Answers Multiple Choice Questions 1. (b) greater than 1

In ∆ABC , the sum of any two sides is greater than the third side. Using (i) AB + BC ∴ AB + BC > AC ⇒ > 1 ⇒ sinA + cosA > 1 AC

(

)

2. (c) 10 + 4 2 units

In ∆AED , we obtain AE 2  tan45= 1 ⇒= ⇒ ED= 2 ⇒ CF= 2 DE DE Also, AE 1 2 sin45 = ⇒ = ⇒ AD = 2 2 AD 2 AD CD = CF + EF + ED = 2 + 3 + 2 = 7 ∴ Perimeter = AB + BC + CD + AD = 3+2 2 +7+2 2

= 222

(10 + 4 2 ) units

OM 3 OM ⇒ = OA 2 3 3 ⇒ OM = m = 1.5 m 2 ∴ AP = BR = MQ

cos30 = 4. (b)

⇒ AP = BR = OQ − OM = 2.3 m − 1.5 m = 0.8 m 5a 2 − 4b 2

30°

In ∆ABC , we find that BC AB = sinA = and cosA AC AC BC AB AB + BC + = ⇒ sin A + cos A = (i) AC AC AC

3. (d) 0.8 m In right triangle AMO, we get

3m 2.3 m

R Q Horizontal D

4b 2 − 3a 2 Using Pythagoras Theorem in right triangle ABC, we get

AB 2 = AC 2 + BC 2 ⇒ b 2 = a 2 + BC 2

⇒ BC = b 2 − a 2 Thus, in right triangle BCM, we obtain:

a BC = b 2 − a 2 and CM = . 2 Using Pythagoras Theorem in triangle BCM, we get a2 BM 2 = BC 2 + CM 2 ⇒ BM 2 = b 2 − a 2 + 4 2 2 4b − 3a = 4

a 4b 2 − 3a 2

BC cos = α = BM

= m2 − n2

4b 2 − 3a 2 ⇒ BM = 2 a CM 2 = sin = α = BM 4b 2 − 3a 2 2

2 b −a

and

(

)

4 b2 − a2 5a 2 − 4b 2 2 2 sin α − cos = α − = 4b 2 − 3a 2 4b 2 − 3a 2 4b 2 − 3a 2 a2

5. (b) 45

We have,

tan = (A + B)

= ⇒ tan ( A + B ) tan60 and = tan ( A − B ) tan 30

⇒= A + B 60 and= A − B 30

⇒ ( A + B ) + ( A − B ) = 60 + 30

3 and tan = (A − B)

Now,

mn = ( tanx + sinx )( tanx − sinx ) = tan 2 x − sin 2 x sin 2 x

 1  − sin 2 x = sin 2 x  − 1 2 cos x  cos x 

1 − cos2 x

=sin 2 x ⋅

cos2 x

sin 2 x

sin 4 x = cos2 x cos2 x

sin 4 x sin 2 x sin x = = sin x ⋅ 2 cos x cos x cosx = sin x tan x

From (i) and (ii), we get:

2 m= − n 2 4tan= x sinx 4 mn

mn =

10. (b) tan P − cot R

In ∆PQR , ∠ Q = 90

Then, sum of other two angles = 180 − 90 = 90

Then, cosP − secR= cos30 − sec60

tan P − cot R= tan 30 − cot 60

=0 3 tan60 3 3 sinP − cosecR =sin30 − cosec60



⇒ 2 A = 90 ⇒ A = 45

(1 + tan θ ) (1 + sinθ )(1 − sinθ )(1 + cosθ )(1 − cosθ ) (1 + cot θ ) =sec θ (1 − sin θ )(1 − cos θ ) cosec θ

cos2θ

× cos2θ × sin 2θ ×

sin 2θ

= 1

7. (b) tan P > tan Q In (0°, 90°) sine and tangent are increasing functions. So if P > Q ⇒ sin P > sin Q and tan P > tan Q 8. (d) 2cos2 A − 1

We have, cos4 A − sin 4 A

(cos2 A + sin2 A)(cos2 A – sin2 A)  a 2 − b 2 = ( a + b )( a − b )   

3 1 − = 2 cos60

−



3 −2≠0 2 1

−

1 1 1 2 =− =− ≠0  2 sin60 2 3

11. (c) 0

k + 1 sec2 θ (1 + sin θ )(1 − sin θ ) =

⇒ k= +1

⇒ k +1 =

12. (d)

1 4

cos2 θ cos2θ

cos2θ

(

× 1 − sin 2 θ

)

⇒ k +1 =1⇒ k = 0

sin θ − cosθ = 0 ⇒ sin θ = cosθ

⇒ tanθ =1 ⇒ θ =45

A 2cos2 A − 1 = cos2 A − 1 − cos2=

∴ sin 6θ + cos6θ= (sinθ )6 + (cosθ )6

9. (a) 4 mn Given: tan x += sin x m and tanx −= sin x n

sin 45 =

(

)

= 1 cos2 A − sin 2 A  sin 2 θ + cos2 θ= 1

(

)

(

) ( 6

+ cos45

)

1 1 1 + = 8 8 4

m − n = (m + n )(m − n )

m += n

x ) 2tanx ( tanx + sinx ) + ( tanx − sin=

1 13. (d) tan 2θ − sec2θ =

m −= n

x ) 2sinx ( tanx + sinx ) − ( tanx − sin=

Chapter 8 • Introduction to Trigonometry

(ii)

Suppose, P = 30 and Q = 60

6. (a) 1

=sin 2 x ⋅

4b 2 − 3a 2

x )( 2sin x ) 4tan x sin x (i) ( 2tan=

sin θ cot θ =sin θ ×



 1   1  =   +   2  2



cosθ =cosθ sin θ 223

So, option (a) is true.

sinθ cosθ tanθ =cosθ × =sinθ cosθ So, option (b) is true.

So, option ( c ) is true but option ( d ) is not true. 3 , cos β = 2

sin α = Given:

sinα =

3 ⇒ α = 60 2

cosβ =

3 ⇒ β = 30 2

 tan α tan60 = =

15. (a)

⇒ tanα tanβ = 3 ⋅

α + β = 90 ⇒

16. (d)

α +β

α + β 

A + B +C = (B − d ) + B + (B + d )

= 3B = 180 ⇒ B = 60

So A < B implies cot A > cot B — this is also true

(

)

(

)

(

)

cos0 = 1, sin 90 − 0 = sin90 =⇒ 1 Same

1  1 1   x−  ≥0⇒x + −2≥0⇒x + ≥2 x x x  Thus, reason is true. For 0 ≤ θ < 90 , we find that sec2 θ and cos2 ⁡θ are 1 . positive real numbers such that sec2 θ = cos2 θ Using reason, we obtain 1 ≥ 2 ⇒ sec2θ + cos2θ ≥ 2 cos2θ + 2 cos θ Thus, A is true, R is true, R is correct explanation for A.

3 2

Select Response Type MCQs

2. (a) Both A and R are true, and R is the correct explanation of A. sinθ − 2sin 3θ Assertion: L.H.S = 2cos3θ − cosθ

1. (a) Both (i) and (ii)

)

1. (a) Both A and R are true, and R is the correct explanation of A. For any x > 0, we have:

3 2 ∠ A + ∠ B + ∠ C =180° As they are in AP: A =− B d, B = B , C =+ B d

 = sinB sin60 =



Assertion-Reason Based Questions

= 45

 = cos  =  cos45  2 



Cotangent is decreasing in 0 , 90

Never same This is clearly false because we just saw it is always the same.

(

2 0 Given: cos (α + β ) =

3 2

tan β tan = 30 =

)

So if A < B , then cos A > cos B is true

3. (b) Only (i), (ii), and (iii) For A = 45 1 cos45 , sin 90 − 45 = 2 1  = sin 45 = ⇒ Same 2 For A = 30 3 cos30 , sin 90 − 30 = 2 3  = sin60 = ⇒ Same 2 For A = 0

2 2 ⇒ cosec2θ − cot = θ 1 and sec2θ − tan = θ 1

1 += cot 2 θ cosec2θ and 1 + = tan 2 θ sec2 θ

14. (c) 1

(

The cosine function is decreasing in 0 , 90 .

sin 2 45 + cos2 45 = 1

sec2 60 − tan 2 60 = 1

( ) cosθ ( 2cos θ − 1) tan 30 + 1 ≠ 1 sinθ 1 − 2 (1 − cos θ )  sin θ 1 − 2 + 2cos θ      2. (c) All (i), (ii), (iii) and (iv) = = θ cos cosθ ( 2cos θ − 1) The sine function is increasing in ( 0 , 90 ). ( 2cos θ − 1)



cot 45 + 1 ≠ 1

sinθ 1 − 2sin 2θ 2



So if A < B , then sin A < sin B is true.   The tangent function is also increasing in 0 , 90 .

So tan A < tan B is true 224



(

)

2 2    sin A = 1 − cos A From Reason 

2 [ ] θ 2cos θ − 1 tanθ = tan = 2 [2cos θ − 1]

From Reason tan θ =

sin θ cosθ

(2cos2 θ − 1) = tan = θ R.H.S (2cos2 θ − 1) Thus, A is true, R is true, R is correct explanation for A.

tanθ

3. (a) Both A and R are true, and R is the correct explanation of A.

P 4k = H 3k Here, perpendicular is greater than the hypotenuse which is not possible as in any right triangle hypotenuse is the longest side.

( )

⇒ CP = 2000 ⋅ tan 30 = 2000 ⋅

4000 m 3

In ∆PBC , angle at B = 60°, side PC =

2000 PC 3  = sin 60= BP BP

2000 2000 3 3 = 4000 m ⇒ BP= =  3 3 sin 60 2

(d) (ii)

2000 m 3

Case Study Based Questions 4000

1. (a) (iv)

2000 ⇒ AP= ( ) 2000 AP cos ( 30 )

cos 30 =

(b) (ii)

BP =

In ∆ACP , angle at A = 30°, side AC = 2000 m. adjacent AC = ( ) hypotenuse AP 

2000 2000 × 2 4000 = = m 3 3 3 2 2000 3

CP tan 30 = 2000

( ) 

2000 3

( )

In ∆PBC , angle at B = 60°, side PC =

cos 30 =

( )

sin= θ

So, both A and R are true and R is the correct explanation for A.

2000

4000 m. 3

( ) BC BP

cos 60 =

⇒

= BC

(e) (iv)

4000 m 3

AB = AC − BC ⇒ AB = 2000 −

4000 × cos 60 = BC 3

( )

4000 1 2000 ×= m 3 2 3

2000 4000 m = 3 3

Additional Practice Questions a 3. Given that sinθ = , then tanθ is equal to b b b (a) (b) 2 2 2 b − a2 a +b

1. In given figure, the value of cosφ is (a) (b)

5 4

5 3

A D

3 (c) 5 4 (d) 5

90° C

5 sinθ + cosθ is 2. If tanθ = , then the value of 12 sinθ − cosθ 17 17 17 7 (a) − (b) (c) (d) − 7 7 13 13 Chapter 8 • Introduction to Trigonometry

(c)

a 2

a −b

(

)(

(d)

a 2

b − a2

)

4. sec2θ − 1 cosec2θ − 1 is equal to (a) -1 5. If

(

(b) 1

(d) 2

3cot 2θ − 4cotθ + 3 = 0, then the value of

)

3 cot 2θ + tan 2 θ is (a) 15

(b) 7

(d) 4 225

 1 − tan 2 30  6.    2  is equal to  1 + tan 30  (a) sin60

(b) cos60

(d) cos30

23. If asecθ + b tanθ = m and b secθ + atanθ = n , prove that a 2 + n 2 = b 2 + m 2 (CBSE 2025)



24. Prove that:

2 2 3 7. If tan A = , then sin A + cos A is equal to 4 sec A 5 4 4 3 (a) (b) (c) (d) 4 5 3 5

(a)

(b)

1 + tan 2 θ

(d) 0

(sec θ − cosec θ ) 3

2sin30 cos30 + tan45

(

) ( ) 14. Show that: 2 ( cos 60 + sin 30 ) − ( tan 60 + cot 45 )

0 13. Prove that: 2 sin 6θ + cos6θ − 3 sin 4θ + cos4θ + 1 = 4



1 + 3sec2 30 = 4 15. Prove that:

cosθ − sinθ 1 − 3 = . cosθ + sinθ 1 + 3

tanA tanA 2cosecA − = 1 + secA 1 − secA

18. In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°. Calculate side BC. sinθ − cosθ + 1 1 = cosθ + sinθ − 1 secθ − tanθ

20. If each of α , β and γ is a positive acute angle such 1 1 , cos ( β + γ − α ) − and 2 2 tan ( γ + α − β ) − 1 find the values of α , β and γ .

β −γ ) that sin (α +=

x , prove that 21. Given that sinθ + cosθ =

(

) .

2 − x2 −1 4 4 sin θ + cos θ = 2 22. Prove that:

226

cosθ − 2cos θ sinθ − 2sin 3θ

+ cotθ = 0.

50 m

19. Prove that:

28. Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, shows kites flying together.

cos2θ sin 3θ + = 1 + sinθ cosθ 1 − tanθ sinθ − cosθ

16. Find an acute angle θ , when 17. Prove that:

= sin 2θ cos2θ

2 2 = = 2sin 2 A − 1 1 − 2cos2 A

5sin 2 30 + cos2 45 − 4tan 2 30



27. Prove that: sin A + cos A sin A − cos A 2 + = sin A − cos A sin A + cos A sin 2 A − cos2 A

1 , evaluate cosAcosecA + tanAsecA . 3

1 1 1 1 − = − secA + tanA cosA cosA secA − tanA

(1 + cotθ + tanθ )( sinθ − cosθ )

11. If 3cosθ − 4sinθ = 2cosθ + sinθ , find tanθ . 12. Evaluate

= secθ cosecθ − 2sin θ cosθ 1 + cot 2 θ

26. Prove that:

= = 9. If 2sin ( A + B ) 3 and cos ( A − B ) 1, then find the measure of angles A and B, where A , B ≥ 0 and ( A + B ) ≤ 90. 10. If sinA

cot 3 θ

25. Prove that:

3 then tanα ⋅ tanβ is 2

(CBSE 2025)

60 m

8.= If sinα

3 = , cosβ 2

tan 3 θ

30°

60° C

In given figure, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be 30° and 60° respectively. Taking AD = 50 m and BE = 60 m, find (a) the lengths of strings used (take them straight) for kites A and B as shown in the figure. (b) the distance ‘d’ between these two kites.

Challenge Yourself

Scan me

1. If a sec ⁡θ + b tan ⁡θ + c = 0 and p sec⁡ θ + q tan ⁡θ + r = 0,

 x 3  y 3 x a cos θ= , y b sin θ , prove that   +  = 4. If=  1. a  b  5. If sin⁡θ + cos⁡θ = x, prove that sin6⁡θ + cos6⁡θ 3

prove that (br − qc )2 − (pc − ar )2 = (aq − bp )2. 2. Find the value of tan15°. tan A − tan B    Hint: tan ( A − B ) =  1 + tan A tan B  

(

)

4 − 3 x2 −1

4 6. Find the value of:

3. Eliminate θ from the equations: = x p= sinθ and y qcosθ .

sin 2 5 + sin 2 10 + sin 2 15 + ... + sin 2 90 Hint: (sin(90° – θ ) = cos θ )

Answers Additional Practice Questions 4 1. (d) 5

3. (d)

a 2

b −a 5. (c) 10 7. (b) 4 5 9. A =30°, B =30° 1 5 16. θ= 60° 11.

Challenge Yourself

17 2. (a) − 7 4. (b) 1

6. (b) cos 60



8. (c) 1 16 2 + 3 8 5 12. 2− 3 6 18. 20 3

3 −1

3 +1

or 2 − 3

x  y  3.   +   = 1 p  q 

19 2

10.

(

)

1 1 , β 45= and γ 52 = = 20. α 37 2 2

Scan me for Exemplar Solutions Chapter 8 • Introduction to Trigonometry

227

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. What is the value of tan 30 + cot 60 ? 2 3 (a) (b) 3 3 x , then secθ = 2. If secθ + tanθ = (a)

x2 +1 x

(b)

x2 +1 2x

(c)

(d) 2

x2 −1 2x

(d)

(CBSE 2024)

x2 −1 x

1 + secx x true? 3. Which of the following option makes the statement sec cos2 x − 1 − tan 2 x (a) −cosec x tan x

(b) −sec x tan x

Assertion-Reason Based Questions

(d) −sec x cot x

[2 × 1 = 2 Marks]

1 5. Assertion (A): If sinθ = and θ is acute angle, then 3cosθ − 4cos3θ is equal to 0. 2 1 Reason (R): As sinθ = and θ is an acute angle, so θ must be 60. 2

(

Very Short Answer Questions 3 6. Write the value of sinA if cosA = . 5 tanθ − cotθ = tan 2 θ − cot 2 θ . 7. Prove that sinθ cosθ

Short Answer Questions 5 8. If secA = , find sinA and cosA. 4 9. If cotθ

228

1 1 − cos2θ 3 , show that . = 3 2 − sin 2θ 5

)

(NCERT Exemplar)

(CBSE 2023) [2 × 2 = 4 Marks]

(CBSE SQP 2025)

[4 × 3 = 12 Marks]

2  2  10. Evaluate: tan 60 + cot 60 . sec2 60 − cosec2 60

 sin 2 θ − 2sin 4 θ  11. Prove that sec2 θ −  = 1 4 2   2cos θ − cos θ 

Long Answer Questions

[3 × 5 = 15 Marks]

( )( ) 12. Prove that: 1 + cot A + tan A sin A − cos A = sin 2 A cos2 A . 3 3 sec A − cosec A

(CBSE 2024)

sinθ sinθ 13. Prove that: = 2+ cotθ + cosec θ cotθ − cosecθ 14. Prove that: 2sec2θ − sec4θ − 2cosec2θ + cosec4θ = cot 4θ − tan 4θ

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Rinku is feeling so hungry and so thought to eat something. He looked into the fridge and found some bread piece. He decided to make a sandwich. He cut the piece of bread diagonally and found that it forms a right angled triangle, with sides 3 3 m, 9 m, and 6 3 m.

6 3 A

Based on the above information, answer the following questions. (a) The value of ∠ C is (i) 30

(ii) 45

(b) The value of sin A + cos C is (i) 0 (ii) 1

(iii) 60

(i)

(ii)

(d) The value of sin2A + sin2C

(i) 0

Self-Assessment

(ii)

2 3

1 2

(iv)

3 2

(iii)

8 3

(iv)

10 3

(iii) 1

(iv) 90

(iii)

3 3

(iv) 2

229

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. Observe the figure shown. Which of these is the value of cosθ ? (a)

1 2

(b)

2 1

(c)

2 3 3

(d)

3 3 2

1 2. is equal to tanθ + cotθ (a) cosθ sinθ

θ 18

(b) secθ sinθ

(

) (

R (CBSE 2024)

(d) secθ cosecθ

)

3. The value of is −3 cos2 45° − tan 2 45° + 4 sin 4 30° + cos4 60° : (a) 3

(b) 2

Assertion-Reason Based Questions

(d)

1 2 [2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1 2 2 0 < A < 90 then value of cosA is . 4. Assertion (A): If sin A= 3 3 Reason (R): For every angle θ , sin 2θ + cos2θ = 1. 2tanθ 2 2 5. Assertion (A): cos θ − sin θ =2 is not an identity. 1 − tan θ Reason (R): An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of angles involved.

(

)

Very Short Answer Questions

[2 × 2 = 4 Marks]

sin θ 1 + cosθ + = 2cosecθ . 1 + cosθ sin θ 3 1   7.= If cosecθ 3= , then find the value of  x 2 − 2  . x and cotθ x x   6. Prove that:

Short Answer Questions 8. Prove that

1 1 + = cosec A + sec A . cos A + sin A −1 cos A + sin A + 1

(

9. Prove that sin 2 A + cos2 A

1. ) − 2sin Acos A = 2

2 2 10. Simplify: sin A + sin A . 1 + cosA 1 − cosA

230

[4 × 3 = 12 Marks]

(CBSE 2024) (CBSE SQP 2025)

5 sinθ − 2cosθ 11. If secθ = , find the value of . 4 tanθ − cotθ

Long Answer Questions

(

12. If cosec θ – sin θ = m and sec θ – cos θ = n, prove that m 2n

2 3 +

) (

[3 × 5 = 15 Marks]

2 mn 2 3 = 1

)

(CBSE SQP 2025)

x2 y2 z2 = θ cos φ , y b secθ sin φ and z = c tanθ , show that = + + 1. 13. If x a sec= a2 b2 c2

(

)

4 14. Prove that sec4 A − tan = A 2sec2 A tan 2 A + 1.

(CBSE 2023)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Rohan and his father go to meet his friend Ravi. When they reached to Ravi’s place, Rohan saw the roof of the house, which is triangular in shape. If he imagined the dimensions of the roof as given in the figure, then answer the following questions. (a) Find the value of ∠A . (i) 30

(ii) 45

(iii) 60

(iv) 90



B 6 2m A



6 2m D mid-point 12 m

(b) The value of sec2C − 1 is equal to

(i) 0

(ii) tan 2 45 − 1

tan 2 45 + 1

1 2

(iv) 1

is (ii) 1

 A +C  B  (d) The value of sin   . cos   is equal to  2  2 1 (i) 0 (ii) 4

Self-Assessment

(iii)

(iii) 2

(iv) -1

1 2

(iv) 1

(iii)

231

Some Applications of Trigonometry

Chapter at a Glance Introduction to Applications of Trigonometry ●

●

rigonometry is used to measure heights and distances T indirectly by observing angles.

B (Object)

The primary tools in these applications are: ◆

◆

ine of Sight: The straight path between an observer’s L eye and the object being viewed, unobstructed by any obstacles. ngle of Elevation: The angle formed between the A horizontal line and the line of sight when an observer looks upward at an object.

Line of sight

Height

Angle of elevation Horizontal

O (Observer) (Observer)

Horizontal Angle of depression

Angle of Depression: The angle formed between the horizontal line and the line of sight when an observer looks downward at an object.

Lin

Depth

f si

ght

Use of Trigonometric Ratios

(Object)

Trigonometric ratios are used to calculate: To Find Height of an object.

Use Formula sin θ =

Figure

Perpendicular Hypotenuse

⇒ Height ( A ) = H × sin θ The length of an inclined path.

cosθ =

Base Hypotenuse

⇒ Base ( B ) = H × cosθ The distance between two objects.

Perpendicular tan θ = Base Height ( A ) ⇒ Base ( B ) = tan θ

232

hypotenuse/ inclined path

height/ perpendicular 90°

base/distance between objects

NCERT Zone NCERT Exercise 9.1 1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. Sol. It can be observed from the figure that AB is vertical A pole.

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? Sol. It can be observed that AC and PR are the slides for younger and elder children respectively.

In ∆ ABC , AB 20 m = sin 30° AC AB 1 = 30° 20 2 B 20 = = 10 AB 2 Therefore, the height of the pole is 10 m.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Sol. Let AC was original tree. The broken part A ′B is making 30° with the ground.

In ∆ ABC , AB = sin 30° AC 1.5 1 = AC 2 AC = 1.5 × 2 = 3.0 m

Slide

1.5 m

30°

(for younger children)

Similarly, In ∆ PQR , PQ Slide = sin60° PR 3 3 60° = R (for elder children) PR 2 6 6 3 6 3 = PR = × = = 2 3m 3 3 3 3

3m Q

In ∆ A ′BC ,

Therefore, the lengths of slides are 3 m and 2 3 m.

BC = tan 30° A ′C BC 1 = 8 3

4. T he angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

8 BC = m 3

A ′C Similarly, = cos30° A ′B 8 3 3 m = ⇒ A ′B = 16 A ′B 2

= A'B + BC = 8 3 m Height of tree 8  24  16 =  + = ×  3  3 3

Hence, the height of tree is 8 3 m .

Chapter 9 • Some Applications of Trigonometry

30°

Sol. Let AB be the tower and the angle of elevation from point C (on ground) is 30°. In ∆ ABC ,

AB = tan 30° BC AB 1 30° = 30 3 B 30 m 30 3 30 3 AB = × = = 10 3 m 3 3 3 Therefore, the height of the tower is 10 3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. 233

Sol. Let K be the kite and the string is tied to point P on K the ground. In ∆ KLP , KL = sin60° 60 m KP 60 3 60° = KP 2 L 120 3 120 3 KP = × = = 40 3 m 3 3 3 Hence, the length of the string is 40 3 m.

Sol. Let the boy was standing at point S initially. He walked towards the building and reached at point T. It can be observed that PR = PQ – RQ

60°

B T

30°

A 1.5 m S

57 PR = 3 = tan60° ⇒ 2BR BR 57 3 57 3 19 3 m BR = × = = 2 2 3 3 2×3 As ST = AB  57 3 19 3  38 3 − = AR – BR =   = 19 3 m 2 2   2 Hence, he walked 19 3 m towards the building. 7. F rom a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Sol. Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

234

BC = tan 45° CD

20 = 1 = CD = 20 m CD In ∆ ACD ,

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

57 – 1.5 28.5 m m = 30 = = 2 In ∆ PAR , P PR = tan 30° AR 30 m 57 1 = 2 AR 3 R 57   3 m AR =  Q  2  In ∆ PRB ,

In ∆BCD

B 20 m

AC = tan60° CD AB + BC = 3 CD AB + 20 = 3 20 = AB

60°

( 20 3 − 20 ) m = 20 ( 3 − 1) m

Thus, the height of the tower is 20

45°

( 3 − 1) m.

8. A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Sol. In ∆ BCD , BC = tan 45° CD BC = 1 = BC = CD CD In ∆ ACD , AB + BC = tan60° CD

A 1.6 m B

60° C

AB + BC = 3 BC

45°

1.6 + BC = 3 ( BC ) 3 ( BC ) − BC = 1.6 = BC

1.6

= 0.8

( 3 + 1)

3 −1

3 +1

1.6

( 3 + 1) 3 −1

Therefore, the height of pedestal is 0.8

( 3 + 1) m .

9. T he angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Sol. Let AB be the building and CD be the tower. In ∆CDB ,

 3 + 1 CD   = 80  3  CD = 20 3 m

CD = tan60° BD

BO =

50 = 3 BD 50 BD = 3

50 m

AB = tan 30° BD

60°

30°

Therefore, the height of building is 16

2 m. 3

10. T wo poles of equal heights are standing opposite each other and either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of poles and the distance of the point from the poles. Sol. Let AB and CD be the poles and O is the point from where the elevation angles are measured. In ∆ABO, AB = tan60° BO AB = 3 BO AB BO = 3

In ∆ CDO ,

CD = tan 30° DO CD 1 = 80 − BO 3

60°

30° O 80 m

(CD ) 3 +

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. Sol. In ∆ABC, AB = tan60° BC AB = 3 BC AB BC = 3 In ∆ABD,

AB 3

= 80 3 Since the poles are of equal heights, CD = AB 1   80 CD  3 + = 3 

Chapter 9 • Some Applications of Trigonometry

AB = tan 30° BD AB 1 = BC + CD 3

30° 20 m

60° C

AB 1 AB 3 1 = ⇒ = AB 3 + AB 20 3 3 + 20 3 3 AB = AB + 20 3 ⇒ 2 AB = 20 3 ⇒ AB = 10 3 BC =

3 80 − BO (CD ) = 3 80 − (CD ) =

DO = BD − BO = 80 − 20 = 60 m Therefore, the height of the poles is 20 3 m and the point is 20 m and 60 m far from these poles.

In ∆ ABD ,

= AB BD tan 30° 50 1 50 2 = × = =16 3 3 3 3

AB CD 20 3 = = = 20 m 3 3 3

AB 10 3 = = 10 m 3 3

Therefore, the height of the tower is 10 3 m and the width of the canal is 10 m. 12. F rom the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. Sol. Let AB be a building and CD be a cable tower. In ∆ABD, AB = tan 45° BD

235

7 = 1 BD BD = 7m In ∆ACE , A AE = BD = 7m CE 7m = tan60° AE CE B = 3 ⇒ CE = 7 3 7 CD =CE + ED = 7 3 + 7= m 7

(

)

60° 45°

87 87 3 87 3 = 3 ⇒ CE = × = = 29 3 m CE 3 3 3 In ∆ BCG , 88.2 − 1.2 1 BG = = tan 30° ⇒ ⇒ 87 3 = CG CG CG 3 = EG = CG − CE Distance travelled by balloon

45°

(

( 3 +1) m

Therefore, the height of the cable tower is 7( 3 + 1) m 13. A s observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Sol. Let AB be the lighthouse and the two ships be at point C and D respectively. In ∆ABC, AB = tan 45° BC 75 =1 BC

30°45°

75 m

BC = 75 m In ∆ ABD , 45° AB B C = tan 30° BD 75 1 75 1 ⇒ = = 75 + CD BC + CD 3 3

30°

( 3 − 1) = CD Thus, required distance 75 ( 3 − 1) m.

Sol. Let the initial position A of balloon change to B after some time and CD be the girl.

88.2 − 1.2 = 3 CG 236

C 1.2 m

60° 30°

88.2 m E

Sol. Let AB be the tower. Initial position of the car is C, which changes to D after six seconds. In ∆ADB , AB = tan60° DB AB = 3 DB AB DB = 3

In ∆ABC , AB = tan 30° BC B 1 AB = BD + DC 3 AB = 3

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

30°

E 60°

60°

30°

3 BD + DC AB =

75 = 3 75 + CD ⇒ 75

In ∆ACE, AE = tan60° CE

)

= 87 3 − 29 3 m = 58 3 m

AB 3

+ DC

AB = DC AB 3 − 3  1  AB ( 3 − 1) 2 AB = AB  3 − = = 3  3 3 Time taken by the car to travel distance DC =

2 AB

6 seconds times taken by car to travel distance DB =

AB 3

AB 6 6 × == 3 seconds 2 AB 3 2 3

Multiple Choice Questions 1. From a point on the ground which is 30 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be 60°. The height (in metres) of the tower is (a) 10 3 m

(b) 30 3 m

(d) 30 m

(a) a − b m

2. The height of a tower is 20 m. The length of its shadow made on the level ground when the Sun’s altitude is 60°, is (CBSE 2023) 20 20 (a) m (b) m 3 3 (c)

3 3

(d) 20 m

(b) 22 m

(d) 24 m

4. A pole stands vertically on the floor. From a point, it is 120 m away from the foot of the pole and the angle of elevation is 45°. Then, find the height of pole. (a) 100 m

(b) 180 m

(d) 10 m

5. A slope is built against a wall which makes an angle 30° with the ground and the height of the wall is 4 m. The length of the slope is (a) 2 m

(b) 4 m

(d) 8 m

6. A tree trunk of height 6 m is erected with support of these metal chains. An angle of 60° is made by each metal chain with the tree trunk. The length of the metal chain is (a) 6 m

(b) 12 m

(d) 24 m

7. A ship has to cross a sea by making an angle of 30° and with a distance of 400 m to reach the shore, then, width of the sea is (a) 100 m

(b) 200 m

(d) 400 m

8. A man in a boat rowing away from a lighthouse 100 m high takes 2 min to change the angle of elevation of the top of lighthouse from 60° to 45°. If the speed of the boat is (a) 2

( 3 − 1) m/s, then m is (b) 3

Chapter 9 • Some Applications of Trigonometry

(d) 6

a +b m

a m b 10. Four points A, B, C and D are in a plane so that B is in the line joining A and C, B is due North of D and D is due West of C. BD = 2, ∠BDA = 45°, ∠BCD = 30°, then AD is equal to ab m

(d)

(a)

(b) 2 3 (d) None of these

11. A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is (a) 45° (b) 30° (c) 60° (d) 90° 12. A house subtends a right angle at the window of the opposite house and the angle of elevation of the window from the bottom of the first house is 60°. If the distance between the two houses is 6 m, then the height of the first house is (a) 4 3 m

(b) 6 3 m

(d) None of these

13. The length of shadow of a pole on the playground 1 is times the height of the pole, the angle of 3 elevation of the sun is (a) 30° (b) 45° (c) 60° (d) 90° 14. The length of a string between a kite and a point on the ground is 85 m, if the string makes an angle 15 θ with the ground level such that tanθ = , then at 8 what height is the kite from the ground? (a) 75 m (b) 79.41 m (c) 80 m (d) 72.5 m 15. It is found that on walking x metres towards a tower in a horizontal line through its base, the elevation of its top changes from 30° to 60°. The height of the tower is (a) 3 2x m

m 3

(b)

(c)

3. A vertical pole 10 m long casts a shadow of length 5 m on the ground. At the same time, a tower casts a shadow of length 12.5 m on the ground the height of the tower is (CBSE 2023) (a) 20 m

9. The angle of elevation of the top of the tower from two points at distance a m and b m from the base and in the same straight line with it are complementary, then the height of the tower is

(c)

3 x m 2

(b) 2 3x m (d)

x m

237

16. A ladder 18 m long makes an angle of 60° with a wall. The height of the point where the ladder reaches the wall is (a) 9 3 m

(b) 18 3 m

(d) 9 m

17. The angle of depression of a car parked on the road from the top of 150 m high tower is 30°. The distance of the car from the tower (in metres) is (a) 50 3 (b) 150 3 (c) 150 2

(d) 75

Answers 20

1. (b) 30 3 m

2. (a)

3. (c) 25 m

4. (c) 120 m

5. (d) 8 m

6. (b) 12 m

7. (b) 200 m

8. (d) 6

9. (c) ab m

10. (c) 2 2

11. (b) 30°

12. (b) 6 3 m

13. (c) 60°

14. (a) 75 m

3 xm 15. (b) 2

16. (a) 9 3 m

17. (b) 150 3

Constructed Response Questions Very Short Answer Questions 1. A pole casts a shadow 12 m long when the angle of elevation of the sun is 45. Find the height of the pole. (CBSE 2024)

Sol. Given: The length of the shadow = 12 metres. The angle of elevation of the sun = 45°. We need to find the height of the pole. height tan (θ ) = shadow length  height tan 45 = 12 height pole ⇒1= 12 45 45° shadow height = 12 metres

sun

238

opposite adjacent

( )

50 m 30°

50 x = × 3 ≈ 50 × 1.732 = 86.6 metres

2. From the top of a lighthouse, 30 m high, the angle of depression of a boat on the sea is 60. Find the distance of the boat from the base of the lighthouse.

( )

tan (θ ) =

 50 tan 30 = x 1 50 = 3 x

So, the height of the pole is 12 metres.

Let the distance from the base of the lighthouse to the boat be x. 30 tan 60 = x 30 3= x

3. The angle of depression of a car from the top of a building 50 m high is 30°. Find the distance of the car from the base of the building. (CBSE 2024) Sol. Given: Height = 50 m 30° Angle = 30°

( )

Sol. Given: Height = 30 m Angle = 60°

30 30 × 3 = = 10 3 ≈ 17.32 metres x = 3 3 Thus, required distance is approx 17.32 metres.

60° 30°

Thus, required distance is 86. 6 m. 4. The height of a tree is 15 m , and its shadow is 15 m long. Find the angle of elevation of the sun. Sol. Given: Height = 15 m Length of the shadow (adjacent side) = 15 m

15 opposite ⇒ tan (θ= ) = 1 ⇒ tan 45 = 1 15 adjacent Thus, the angle of elevation of the sun is 45°. tan (θ ) =

( )

5. A wire is tied from the top of a pole 15 m high to a point on the ground, making an angle of 30 with the ground. Find the length of the wire.

Sol. Given:

15 1 15 ⇒ = ( ) hypotenuse 2 hypotenuse

sin 30 =

Height of the pole (opposite side) = 15 metres. Angle with the ground (θ) = 30°. opposite sin (θ ) = hypotenuse

15 hypotenuse = = 15 × 2 = 30 metres 1 2 The length of the wire is 30 metres.

Wire Pole 15 m x

30°

6. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun? Topper’s Answer A

Tower AB is 30 m and shadow BC is 10 3 m. In ∆ABC which is right triangle, AB 30 tan = θ = = 3 BC 10 3 tan60°=

30 m

3 ⇒ θ= 60°

So, angle of elevation of sun is 60°.

7. Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the 6 sun is such that tanθ = . (CBSE 2023) 7 6 Sol. Given, tanθ = 7 Let length of shadow be x m. 18 P ∴ tanθ = = 18 m x B 6 18 ⇒ = 7 x θ 18 × 7 xm ⇒= x = 21 m 6

Short Answer Questions 1. A man observes the top of a building at an angle of elevation of 30 while standing 50 m away from its base. Find the height of the building. (CBSE 2025) Sol. Given: Angle θ= 30°. Distance from the base of the building = 50 m. A

( )

Tower

opposite base 1 height height ⇒ = tan 30 = 50 50 3 tan (θ ) =

50 50 3 50 3 height = = × = 3 3 3 3

30° 50 m

50 3 Thus the height of the building is metres. 3

Chapter 9 • Some Applications of Trigonometry

10 3m

2. From a point 100 m away from the base of a tower, the angle of elevation of its top is 45 . Find the height of the tower. (CBSE 2024) Sol. Given: Distance = 100 m, Angle θ= 45° . opposite tan (θ ) = adjacent height ⇒1 = ( ) height 100 100

tan 45 =

The height of the tower is 100 metres. 3. From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60° and 45°. Find the distance between two cars. (CBSE 2017) Sol. Given: Height 120 metres Angle of depression to Car 1: 60° Angle of depression to Car 2: 45° The angle of depression is equal to the angle of elevation from the ground to the top of the tower For Car 1

45°

h =120 m C

h tan ( 60° ) = d1 tan ( 60°= )

60°

45°

120 d1 239

6. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use 3 =1.73 ) (CBSE 2016)

For Car 2 120 h ⇒ d2 = 120 metres tan ( 45° ) = ⇒ 1 = d2 d2 D = d1 + d2 = 40 3 + 120 The distance between the two cars is 40 3 + 120 m.

Sol. Given: Angle θ = 60 Ladder = 10 m sin (θ ) = Opposite side Hypotenuse

Height ⇒ 3 Height sin 60 = = 10 2 10

( )

C ladder

60° A 3 =5 3 m 2 The ladder reaches at 5 3 metres on the wall.

Height =10 ⋅

wall B

5. The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60°. Find the angle of elevation of the sun at the time of the longer shadow. (CBSE 2021) CBSE Marking Scheme 2021 Sol.

B h

C 3x

60°

1 2

In ∆ABC, 1

⇒h = x 3 In ∆ABD,

h AB = tan θ ⇒ = tan θ 3x AD x 3 = 3x

1 = tan 30 3

∴θ = 30

240

45°

60° h − 50 tan 45 = x 45° C hE ⇒ x = h − 50 h h 50 m tan60 = ⇒ x = x 3 60° h B x D Hence, h − 50 = 3 50 3 3h − 50 3 = h ⇒h 3 −1 = 

(

50 3 h= = 3 −1

50 3

)

( 3 + 1) = 50 3 ( 3 + 1)

3 −1

h =+ 75 25 3 =+ 75 43.25 = 118.25 m 7. Two towers are 100 m apart. The angle of elevation of the top of the taller tower from the top of the shorter tower is 30. If the height of the shorter tower is 50 m, find the height of the taller tower. Sol. Given: Distance between the towers: 100 m Angle of elevation: 30 Height of the shorter tower: 50 m Let h be the height of the taller tower, and x be the extra height beyond the 50 m of the shorter tower. Therefore, h = 50 + x opposite side tan (θ ) = adjacent side

AB h = tan60 ⇒ = 3 AC x

Height of Building 50 m

(CBSE 2022)

Tower D

Angle of depression of the bottom 60°

Tower

4. A ladder is leaning against a wall, making an angle of 60 with the ground. If the length of the ladder is 10 m, find the height it reaches on the wall.

Sol. Given: Angle of depression of the top 45°

Building

120 120 3 = = 40 3 metres d1 = 3 3

30° x tan 30 = 100 50 m 1 x = 100 m 3 100 100 100 × 3 100 × 1.73 x == ≈ ≈ 57.67 m 3 3 3 h = 50 + x = 50 + 57.67 ≈ 107.67 m

( )

The height of the taller tower is approx 107.67 m. 1 2

8. A 7 m long flagstaff is fixed on the top of a tower standing on the horizontal plane. From point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the tower correct up to one place of decimal. (CBSE 2016) Sol. Given: Length of the flagstaff: 7 m Angle of elevation to the bottom: 45 Angle of elevation to the top: 60

x  i.e., tan45 1 = = y ⇒x = y

⇒y 3= x +7

45° 60°

(

)

Sol. Let the positions of the two planes at the given instant be A and B. B In right ∆ADC : (Let CD be x m) AD 1 3125 ⇒ = CD x 3

⇒x = 3125 3 m In right ∆BCD :

(i)

h sin φ = d= h θ  r r sin   2 d

Long Answer Questions

9. An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant. (CBSE 2022)

tan 30 = °

θ φ

θ  Thus, h = rsin φ cosec   2

3 −1

( 3 + 1) = 7 ( 2.73) = 9.6 m ( 3 − 1) ( 3 + 1) 2

= x

θ  h = rsin φ cosec   2

x 3= x + 7

( 3 − 1) x ⇒ x =

h sin φ = (ii) d From (i) and (ii), we get

Putting y = x , then

⇒ 7=

O r

In, ∆OAB

x +7 = In ∆ACD , tan60 = 3 y

θ  r = sin   = where OA d (i) 2 d

Tower

Sol. In, ∆OAP we have

BC In ∆BCD , tan 45 = CD

10. A spherical balloon of radius r subtends an angle θ at the eye of an observer. If the angle of elevation of its centre is φ , find the height of the centre of the balloon. (NCERT Exemplar)

60° A C

30°

BD AB + 3125 ⇒ 3= tan60 = CD x AB + 3125 (Using (i)) ⇒ 3= 3125 3 6250 m ⇒ 3125 × 3= AB + 3125 ⇒ AB = Thus, required distance is 6250 m.

Chapter 9 • Some Applications of Trigonometry

3125 m

1. A boy whose eye level is 1.35 m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is 60°. After 12 seconds, the angle of D E elevation reduces to 30°. If the speed of the wind is 3 m/s ? then find the 60° height of the B A 30° balloon from 1.35 m C G 36 the ground. F H (CBSE 2009, CBSE SQP 2020, 2025)

Sol. AF = 1.35 m Also, AF, BG, and CH are parallel. BG = CH = AF = 1.35 m BC = Distance travelled by balloon due to 3 m/s wind in 12 sec = Speed × Time =3 × 12 =36m ⇒ BC = 36m

241

And we need to find the height of the balloon from the ground i.e., we need to find DH In triangle EBA tan60 =

AB AB ⇒ 3= BE BE

Since, BE = CD CD ∴ AB = and AC = CD 3 3

30°, AD = 20 − x m, and BC = 20 m In ∆ABC ∠ACB =

CD = 36 CD 3 − 3

tan 30 = °

 3 × 3 −1  1   36 CD  3 − =   ⇒ 36 = CD  3 3   

⇒

 2   3 −1  36 = CD    ⇒ 36 = CD   3  3 

=20 − x ⇒ x = 20 −

2. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. (CBSE 2020)

Sol. In ∆ABD , H tan45 = 1= x In ∆AEC ,

60° 45°

7m [alternate angles] Also, let AB = DE = 45° A xm x m be the distance between building and tower. In right-angled ∆BAD , 7 AD tan 45 = ⇒ 1 = ⇒ x = 7m x AB

and in right-angled ∆CED , CE CB − BE [since CE = CB] = DE AB h −7 3= x

⇒ x ≈ 8.45 m

CBSE Marking Scheme 2024

⇒H = x (i )

4. The angles of depression of the top and the bottom of a 8 m tall building from the top of a multistoreyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (CBSE 2024)

= CD 18 = 3 31.14 m Height of balloon =DH = CD + CH = 31.14 + 1.35 = 32.49 m

Sol. From the point D, draw a DE  AB .

AB 1 20 − x ⇒ = BC 20 3

Thus, the height is approximately 8.45 metres.

36 × 3 = CD × 2

242

( 3 + 1) m .

= DE = 20 m (BC and DE are parallel lines) Thus, BC

Now, BC = AC − AB

 = tan60

Hence, the height of the tower is 7

30° Sol. Let CE = BD = xm 45° In ∆ADE 20 – x ∠AED = 45°, AD = 20 m 20 m B AD 20 tan 45 = ° ⇒ = 1 x DE DE ⇒ DE = 20

AC = CD 3

∠= EDB ∠= ABD 45

( 3 + 1) m

CD 1 CD ⇒ = AC 3 AC

Then,

= h x 3= +7 7

3. From the top of a 20 m tall building, the angles of depression of the top and bottom of a vertical lamp post are 30 and 45, respectively. Find the height of the lamp post. (CBSE 2023)

BE ⇒ AB = 3 In triangle, DAC tan 30 =

h −7 = x 3

 tan30 =

1 H −8 = x 3

1 1 2

⇒= x 3 ( H − 8 ) ( ii ) Solving equations (i) and (ii)

= x 4 3+ 3

(

and

H-8

30° 45 °

30°

C 8m

8m 45°

1 2 1 2

)

1 2 The height of the multi storey building is 4 3 + 3 m

(

= H 4 3+ 3

)

(

)

and the distance between them is 4 3 + 3 m.

)

5. Amit, standing on a horizontal plane, and a bird flying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, and the angle of elevation of the same bird to be 45°. Amit P and Deepak are x on opposite 0m 0 45° 2 D (Deepak) R sides of the 50 m 30° bird. Find the A Q Amit distance of the bird from Deepak. (CBSE 2019) Sol. Let P be the position of the bird, A be the position of Amit, D be the position of Deepak and FD be the building at which Deepak is standing at height 50 m. 

sin30 =

1 PQ PQ ⇒ = 2 200 PA

⇒ PQ =

200 = 100 m 2

(

)

4h 2h (i) = cotθ ⇒ cot θ = k k    2 In, ∆CMD CD = tan θ CM h 2h (ii) = tanθ ⇒ tan θ = k k    2 Multiplying (i) and (ii), 4h 2h 1 × = k k h2 =

k2 8

⇒ = h

k k 2 = m 4 2 2

7. The angle of depression of two ships from an aeroplane flying at the height of 7500 m are 30° and 45°. If both the ships are in the same line that one ship is exactly behind the other, find the distance between the ships. (CBSE 2017)

= PQ − FD ⇒ PR = PQ − RQ PR = (100 − 50 ) m =50 m In ∆PRD , ∠ R = 90

Sol. Let distance between two ships DC = x m

PR 1 50 sin 45 = ⇒ = PD 2 PD 

and BC = y m.

⇒ PD = 50 2 PD = 50 × 1.414 = 70.7 m Thus, the distance of the bird from Deepak is 70.7 m. 6. Two posts are k metre apart and the height of one is double that of the other. If from the mid-point of the line segment joining their feet, an observer finds the angles of elevation of their tops to be complementary, then find the height of the shorter post. (CBSE 2015) Sol. Let AB and CD be the two posts such that AB = 2CD .

Now, AB = tan 90 = − θ cot θ AM

In ∆ABC, AB = tan 45 BC 7500 ⇒ = 1 y ⇒y= 7500

A (Aeroplane)

30° 45 °

In ∆PQA , ∠ Q = 90

Let CD = h , then AB = 2h

7500 m 30° x D

In ∆ABD,

45° C x+y

7500 1 AB = = tan 30 ⇒ x+y BD 3 ⇒x+y = 7500 3

Let M be the midh point of CA. C

Let ∠ CMD = θ = 90 − θ . and ∠ AMB 1 = MA = k Clearly, CM 2

90°- θ

Chapter 9 • Some Applications of Trigonometry

⇒ x + 7500 = 7500 3 = ⇒ x 7500

1) ⇒ x 7500 (1.732 − 1) ( 3 −=

⇒x = 5490 m Hence, the distance between two ships is 5490 m.

243

Competency Based Questions Multiple Choice Questions 1. The angle of elevation of an object from a point on the level ground is α. Moving d metres on the ground towards the object, the angle of elevation is found to be β. Then the height of the object in metres is (a) dtan α (b) dcot β (c)

d cotα + cotβ

(d)

d cot α − cot β

2. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 m from the bottom of the tower. After 5s, the angle of depression becomes 30°. What is the approximate speed of the boat assuming that it is running in still water? (a) 31.5 km/h

(b) 36.5 km/h

(d) 40.5 km/h

3. From a light house, the angles of depression of two ships on opposite sides of the lighthouse are observed to be 30° and 45°. If the height of the lighthouse is h. What is the distance between the ships? (a) (c)

( 3 + 1)h 3h

(b)

( 3 − 1)h

 1  (d)  1 + h 3 

4. From the top of a lighthouse AB, the angles of depression of two stations C and D on opposite sides at a distance d apart are α and β. The height of the lighthouse is d (a) cot α . cot β

d tan α . tanβ (c) tanα + tanβ

d (b) cot α − cot β

d cot α . cotβ (d) cotα + cotβ

5. The base of a cliff is circular. From the extremities of a diameter of the base, angles of elevation of the top of the cliff are 30 and 60. If the height of the cliff be 500 m, then the diameter of the base of the cliff is 1000 2000 m (a) (b) m 3 3 (c)

244

2000 2

(d) 1000 3 m

6. The angle of depression of vertex of a regular hexagon lying in a horizontal plane, from the top of a tower of height 75 m located at the centre of the hexagon is 60. What is the length of each side of the hexagon? (a) 50 3 m (b) 75 m (c) 25 3 m (d) 25 m

Select Response Type MCQs

1 1. Choose the correct option, if sin A = , then: 2 (i) sec A = 2

(ii) tan A = 3

(iii) cot A = 3

(iv) cos A =

(a) both (i) and (iii)

3 2 (b) both (iii) and (iv)

(d) both (ii) and (iii)

2. Ananya, Suman and Dipti see a crane lifting a metal rod, as shown in the following figure. They notice that more precision must be required in operating the crane as the rod gets closer to the base of the crane, since the rod could hit the crane if it tilts.

Taking the length of the crane’s arm as x metres, and the length of the x hanging line as y metres, they find the approximate θ horizontal distance between the end of the rod and the base of the crane’s arm as follows: tanθ Ananya: metres ( y + 43)

y 43 cm

Suman: xcosθ metres Dipti: xcosecθ metres Who is/are correct? (a) only Ananya (c) only Suman and Dipti

(b) only Suman (d) only Ananya and Suman

Assertion-Reason Based Questions

1. Assertion (A): A man on the top of a cliff 100 m high, observes the angles of depression of two points on the opposite sides of the cliff as 30° and 60°, respectively. Then, the distance between the 400 two points is equal to m. 3 Reason (R): The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level. 2. Assesrion (A): If the angles of elevation of the top of a tower from the points at distances of 9 m and 16 m from the base of a tower in the same line are complementary, then the height of the tower is 12 m. Reason (R): If the angle of elevation of a tower from two points at distances of a and b from its foot and in the same straight line with it are complementary, then the height of the tower is ab .

Case Study Based Questions 1. Radio towers are used for transmitting a range of communication services including radio and television. The A tower will either act as an antenna itself or support one or more antennas on its B structure. On a similar concept, a radio station tower was built in two sections A and B. The 45° 30° tower is supported by O P 36 cm wires from a point O. The distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of section B is 30°, and the angle of elevation of the top of section A is 45°. (CBSE 2023)

2. One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was F A D Bird Bird Ball hit, a ball flew through the tree frightening the Height Tree 80 m 80 m 60° bird to fly away. ° 5 4 In 2 seconds, 30° Ground he observed the C G B E level bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60°. (a) At what distance from the foot of the tree was he observing the bird sitting on the tree? (b) (i) How far did the bird fly in the mentioned time? OR (ii) A fter hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? (c) What is the speed of the bird in m/min if it had flown 20( 3 + 1) m? (CBSE SQP 2023) 3. The Statue of Unity situated in Gujarat is the world’s largest Statue which stands over a 58 m high base. As part of the project, a student constructed an inclinometer and wishes to find the height of the Statue of Unity using it. He noted the following observations from two places:

Based on the above information, answer the following questions: (a) Find the length of the wire from the point O to the top of Section B. (b) (i) Find the distance AB. OR

(ii) Find the area of ∆OPB.

B A

Situation – I: The angle of elevation of the top of the Statue from Place A, which is 80 3 m away from the base of the Statue, is found to be 60°.

245

Situation – II:

(b) Represent the Situation – II with the help of a diagram.

The angle of elevation of the top of the Statue from Place B, which is 40 m above the ground, is found to be 30°, and the entire height of the Statue, including the base, is found to be 240 m.

(c) Calculate the height of the Statue excluding the base and also find the height including the base with the help of Situation – I.

Based on given information, answer the following questions:

(d) Find the horizontal distance of Point B (Situation – II) from the Statue and the value of tan a , where a is the angle of elevation of the top of the base of the Statue from Point B. (CBSE 2025)

(a) Represent the Situation – I with the help of a diagram.

Answers Multiple Choice Questions

= A

d cot α − cot β Then, in rt. ∆ABD ,

1. (d)

h = tanβ ⇒= x hcotβ x Then, in rt. ∆ACB ,

(i) C

h = tanα ⇒ x += d hcotα x +d

α dm

β xm

(ii)

d hcotα − hcotβ ∴ ( ii ) − ( i ) ⇒= d ⇒ h = cotα − cotβ 2. (a) 31.5 km/h Let the man be standing at the top of the tower AB of height h at A.

Let the original position B of the boat be C , 45 and then, ∠ACB = BC = 60 m .

30° 45°

In ∆PAQ , 45°

60 m

( 3 − 1)

∴ x = 60 × 0.732 = 43.8 m (approx.)

246

Let PQ be the light house whose height = h metre. Let A and B be the position of the ships on opposite sides of the lighthouse such that angle of depression for A and B are 30 and 45 respectively. P 30°

30°



tan 30 = In ∆PBQ ,

45°

B 30°

AQ + QB = x+y =

AB h In ∆ABC , tan 45 = = BC 60 h ⇒ == 1 h= 60 60 AB In ∆ADB , tan30 = x + 60 1 60 60 60 3 (∴ ⇒ x += = h 60 m ) ⇒ = 3 60 + x

∴ Speed of boat

( 3 + 1) h

Required distance between the ships = AB

Let the position of the boat after 5 seconds be D 30 and CD = x m (say) where ∠ADB =

⇒ = x 60

 43.8 18  = ×  km/h = 31.5 km/h.  5   5 3. (a)

Distance travelled 18 × km/h Time 5

45°

h 1 h ⇒ = ⇒x = h 3 x 3 x

h h ⇒1= ⇒y= h y y ∴x + = y h 3 += h h 3 +1 m

tan 45 =

(

)

d tan α . tanβ A 4. (c) α tanα + tanβ β Let the height of the light h house AB = h metres. Give β α ∠ACB = α , ∠ADB = β , CD = d. C B C d AB tanα = BC ⇒ tanα = h ⇒ BC = h (i) BC tanα In ∆ABD , tan β = AB ⇒ h = tan β ⇒ BD = h (ii) tan β BD BD Adding (i) and (ii)

2. (b) Only Suman

h h + tanα tanβ h ( tanβ + tanα ) d tan α . tanβ = ⇒d = ⇒h tanα + tanβ tanα + tanβ ∴CD = BC + BD =

Given the length of the crane’s arm is x metres.

= = m, ∠ CAE 60 , Given, E 500 500  In ∆AEC , tan60 = d1 500 ⇒ d1 = m 3 500 ⇒ d2 = 500 3 m In ∆BEC , tan30 = d2

= d1 + d2 = 6. (c)

+ 500 3 =

2000 3

cosθ =

PN MN

MN PM

⇒ MN =PM × cosθ

which is the Suman approximation. 60°

Select Response Type MCQs 1 1. Given: sinA = 2 1 Since sin30 = 2 A = 30 (ii) tanA =

43 + y ⇒ MN = tan θ

Hence, only Suman’s approximation is correct.

P 75 m

Angle of elevation = Angle 60° of depression F C x O ∴ In ∆FOP , 60° 75 tan60 = A B x 75 ⇒ x= = 25 3 m 3 In a regular hexagon, six equilateral ∆s are = OE = OD = OC = OB = OA = side formed, i.e., OF of hexagon ⇒ length of hexagon = 25 3 m.

(i) sec A = 2

= x cosθ metres

25 3 m

Let OP be the tower of height 75 m.

PQ + QN ⇒ tan θ = MN

 CBE 30= , AE d= ∠= 1 , BE d 2

500

43 m

In ∆PMN tan∠ PMN =

Length of hanging line = y metres



∴ Required diameter

Length of rod = 43 m,

2000

C m 3 500 m Let AB be the diameter of 60° 30° the circular base of the cliff. A d1 E d2 Let C be the top of the cliff.

5. (a)

Assertion-Reason Based Questions 1. (c) A is true, but R is false. Let AC = 100 m be the height of the cliff.

A (man)

In ∆ACB

tan60=

In ∆ACD

tan 30 =

⇒y= 100 3

x + y=

60°

100 100 ⇒ x= x 3 100 y

100 3

+ 100 3=

100 m B

400 3

30° y

Reason (R) correctly explains Assertion (A). The angles of elevation of the top of a tower from two points at distances 9 m and 16 m from its base are complementary. The Reason (R) states that if two angles of elevation from distances a and b are complementary, then the height h of the tower is: h = ab where a = 9 m and b = 16 m.

3 23 2 (b) both (iii) and (iv)

= h 9 × 16

Chapter 9 • Some Applications of Trigonometry

60° x

2. (a) Assertion (A) is true. Reason (R) is true and

(iii) cotA = 3 (iv) cosA =

30°

247

= h = 144 12 m Since the given reason directly provides the formula used to determine the height and correctly applies to the given problem, it is the correct explanation of the assertion.

Case Study Based Questions

80 CG 80 ⇒ 3 = CG 80 ⇒ CG = 3 Distance the ball travelled after hitting the tree

(ii) tan60 =

Distance the the tree = FA = GB = CB − CG A ball travelled after hitting 1. (a) Let OB be the length of 80 1   GB =80 − wire from O to the top of =80  1 − m 3 3  section B. (c) Speed of the bird B In ∆OPB , we have: Distance 20 ( 3 + 1) m OP Speed = cos30 = 2sec Time taken OB 45°

3 36 ⇒ = 2 OB

⇒ OB =

36 cm

30°

= O

72 3 = 24 3 cm 3

( 3 + 1) m × 60 sec = 600 ( 3 + 1) m 2 sec

min

3. (a)

⇒ OB =24 × 1.73 =41.52 cm (b) (i) Finding AB: In ∆APO , tan 45 = In ∆BPO ,

tan 30 =

⇒ PB=

AP AP ⇒ 1 = ⇒ AP = 36 cm 36 OP

240 m 30°

36 3 12 × 1.73 = 20.76 cm = 12 3 = 3 ⇒ AB =AP − PB =36 − 20.76 =15.24 cm

(ii) Finding the area of ∆OPB : 1 Area of ∆OPB = × OP × PB 2 1 = × 36 × 12 3= 18 × 12 3 = 18 × 12 × 1.73 2 = 373.68 cm 2 (c) Finding the height of section A from the base of the tower: Height of section A from base of tower

= 36 = cm, i.e., AP 36 cm. 80 2. (a) tan45 = CB 80 m ⇒ CB = 80 80 80 (b) (i) tan 30 = ⇒ 3 = ⇒ CG = CE CG 3 Distance the bird flew = AD = BE = CE − CB = 80 3 − 80 = 80 3 − 1 m

(

248

80 3

(b)

1 PB 36 3 PB ⇒ = ⇒ PB = × 36 3 3 3 OP

60°

)

40 m

(c) Angle of elevation from Point A = 60° Distance from base of Statue (horizontal distance) = 80 3 m Let the height of the Statue (excluding the base) be h.

Using the tan function in right-angled triangle:

= tan60

opposite side h = adjacent side 80 3

80 3

= h 80 3 × 3

h = 80 × 3 = 240 m

Thus, the height of the Statue including the base is 240 m. The height of the Statue excluding the base is 240 − 58 = 182 m

(d) Angle of elevation from Point B = 30°

Total height of Statue = 240 m

Height of Point B above ground = 40 m

Height of the Statue above Point B: H = 240 − 40 = 200 m Let the horizontal distance from Point B to the base of the Statue be d. 200 Using tan function: tan 30 = d

d = 200 3 m

200 d

18 9 = 200 3 100 3 Horizontal distance of Point B from the Statue = 200 3 m = tan α

Value of tanα =

100 3

Additional Practice Questions 1. If the angles of elevation of the top of a tower from two points at a distance of 4 m and 16 m from the base of a tower and in the same line are complementary, then the height of the tower is (a) 20 m (b) 12 m (c) 8 m (d) 16 m 2. Two friends Rohit and Mohit are standing on the opposite sides of a tower of height 60 metres. If their angles of depression seen from the top of the tower are 30° and 45° respectively, then the distance between the two friends is (a) 60 3 − 1 m (b) 60 3 + 1 m

( ) (c) 30 ( 3 − 1) m

( ) (d) 30 ( 3 + 1) m

3. From the top of a hill, it is observed that the angle of depression of the top of a tree and its foot are 45° and 60°, respectively. The height of the tree is 20 m. What is the height of the hill? (Note: The base of the hill and the tree are on the same level.) (a) 10 3 + 1 m (b) 20 + 20 3 + 1 m

(

)

( ) (d) 20 − 10 ( 3 + 1) m

( 3 + 1) m

4. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of the sun? (CBSE 2015, 2017) 5. An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer. (NCERT Exemplar)

A 22 m θ

1.5 m Q

20.5 m

(tower)

Chapter 9 • Some Applications of Trigonometry

6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is st . (NCERT Exemplar) A

h P

90° – θ θ B S t

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower. (NCERT Exemplar)

8. The ratio of the height of a tower and the length of its shadow on the ground is 3 :1. What is the angle of elevation of the Sun? (CBSE 2016, 2017) 9. The shadow of a vertical tower on level ground increases by 10 metres, when the altitude of the sun changes from angle of elevation 45° to 30°. Find the height of the tower, correct to one place of decimal. (Take 3 = 1.73) (NCERT Exemplar) 10. Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. (CBSE 2015, 2019, CBSE SQP 2020)

11. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are 30°and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use 3 = 1.732 ] (CBSE 2018)

249

14. A ladder against a vertical wall at an inclination a to the horizontal. Its foot is pulled away from the wall through a distance p, so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that p cos β − cosα = q sin α − sin β

(NCERT Exemplar)

15. At a local fair, three hot air balloons, X, Y, Z are flying along the same plane. At a particular instant, their positions and angle between them are as shown in the diagram below. (CBSE QB) Z

●

he horizontal distance between balloons X and T Y is equal to X’s altitude. Balloons X, Y, and Z are placed, such that ∠ XYZ = 90 .

Find the (a) altitude of balloon Y. (b) shortest distance between balloons Y and Z. (c) shortest distance between balloons X and Z. 16. Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, two kites flying together. B

60 m

13. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground. (NCERT Exemplar)

●

50 m

12. The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the hill is 30°. If the tower is 50 metres high, find the height of the hill. (CBSE 2015, 2018)

30°

60°

In the given figure, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be 30° and 60°, respectively. Taking AD = 50 m and BE = 60 m, find

Based on the above information, answer the following questions.

75° 45°

X 1500 m

Ground

(a) The lengths of strings used (take them straight) for kites A and B as shown in the figure. (b) The distance d between these two kites. (CBSE 2022)

Challenge Yourself

Scan me

h A

R 250

a a

X Y

Tower

Flagstaff

1. If the angles of depression of the upper and lower ends of a lamp post from the top of a hill of height h metres are a and β respectively, then show that the hsin ( β − α ) height of the lamp post (in metres) is: . cosα . sinβ

2. A tower, x metres, has a flagstaff at its top. The tower and the flagstaff subtend equal angles at a point distant y metres from the foot of the tower. Then find the length of the flagstaff (in metres). M

80 m

h Q

20 m θ

3. The angle of elevation of a cloud from a point h metres above the level of water in a lake is a and the angle of depression of its reflection in the lake is β. Show that the height of the cloud above the water hsin ( β + α ) level is sin ( β − α ) C P hm

15°30°

15° A

xm a β

45° 30°

5. The angle of elevation of the top of a tower observed from each of the three points A, B, and C on the ground forming a triangle is the same angle a. If R is the circumradius of triangle ABC, then show that the height of the tower is R tan α .

x+hm D

4. A tower PQ of height h subtends an angle of 45° at a point A on the horizontal plane. At another point B on AB inclined to the horizontal at an angle 30°, the elevation of the top of the tower is found to be 60°. = a h 3 −1 . If AB = a , then show that

(

h M

60°

)

Answers Additional Practice Questions 1. (c) 8 m 2. (b) 60( 3 + 1) m 3. (c) 20 + 10( 3 + 1) m 4. 60° 5. 45° 7. 25 3 m 8. 45° 9. 13.7 m 10. Height of poles = 20 3 m; Distances = 20 m and 60 m 11. 73.2 m 12. 150 m

13. 8 m 15. (a) Altitude of balloon Y = 3000 m (b) ZY = 500 6 m (c) Shortest distance between Y and Z is 500 6 m and X and Z is 1000 6 m. 16. (a) 100 m and 40 3 m (b) 121.65 m

Challenge Yourself 2. Length of flagstaff =

(

x y2 + x2 2

y −x

Scan me for Exemplar Solutions Chapter 9 • Some Applications of Trigonometry

251

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. A wire is tied from the top of a pole 10 m high to a point on the ground 10 m away from the pole. Find the length of the wire. (CBSE 2022) (a) 10 2 m

(b) 15 m

(d) 5 2 m

2. The angle of elevation of a cloud from a point 50 m above a lake is 30. If the height of the cloud above the lake is (CBSE 2023) 150 m , find the horizontal distance of the cloud from the point. (a) 200 m

(b) 300 m

(d) 173.2 m

3. From the top of a 30 m high tower, the angle of depression of a point on the ground is 60. Find the distance of the point from the base of the tower. (CBSE 2023) (a) 30 3 m

(b) 15 3 m

(d) 50 3 m

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): The height of a building can be calculated using the tangent trigonometric ratio if the distance from the observer to the base of the building and the angle of elevation are known. Reason (R): The tangent of an angle in a right-angled triangle is defined as the ratio of the opposite side to the adjacent side. 5. Assertion (A): If the angle of elevation of the sun increases, the length of the shadow of a vertical pole also increases. Reason (R): The length of the shadow is calculated using the sine trigonometric ratio.

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. A balloon is flying at a height of 120 m. The angle of elevation from a point on the ground is 45 . Find the distance of the balloon from the point of observation. (CBSE SQP 2025) 

7. A man observes the top of a pole at an angle of elevation of 45 while standing 10 m away from its base. Find the height of the pole. (CBSE 2023)

Short Answer Questions

[4 × 3 = 12 Marks]

8. From a point on the ground, the angle of elevation of the top of a tree is 30 . If the height of the tree is 10 m , find the distance of the point from the base of the tree. (CBSE 2023) 

9. A kite is flying at a height of 120 m , and the angle made by the string with the ground is 45. Find the length of the string, assuming it is straight. (CBSE 2023) 10. A man observes the top of a tower at an angle of 60 . If the height of the tower is 50 m , find the horizontal distance between the man and the tower. (CBSE 2024)  11. A hill is observed from a point 60 m away from its base. If the angle of elevation of its top is 45 , find the height of the hill. (NCERT Exemplar) 252

Long Answer Questions

[3 × 5 = 15 Marks]

12. A man standing at the top of a tower 50 m high observes a car on the ground at an angle of depression of 30 . He then observes another car on the ground at an angle of depression of 60. Find the distance between the two cars. 13. The angle of elevation of the top of a tree from two points on the ground, at distances of 40 m and 90 m from its base in the same straight line, are 45 and 30 , respectively. Find the height of the tree. (CBSE 2024) 14. A balloon is flying at a height of 150 m . From two observation points on the ground 80 m apart, the angles of elevation to the balloon are 30 and 60 , respectively. Find the horizontal distance of the balloon from each observation point.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Alice, a skilled technician, is tasked with performing routine maintenance on a wind turbine. The turbine’s central shaft is 5 metres tall. To access the required service point, Alice needs to position herself 1.3 metres below the top of this shaft. A

60° D

(a) How high above the ground is Alice’s working point on the wind turbine?

(i) 1.3 m

(ii) 5 m

(iii) 3.7 m

(iv) None of these

(b) What should be the length of the maintenance ladder, when inclined at an angle of 60° to the horizontal to reach Alice’s working point? 3.7 (i) 4.28 m (ii) m (iii) 3.7 m (iv) 7.4 m 3 (c) H ow far from the base of the wind turbine should Alice position the foot of the ladder to ensure stability and proper reach? 1 (i) 3.7 m (ii) 2.14 m (iii) m (iv) None of these 3 (d) What should be the length of the ladder, if Alice needs to reach the top of the central shaft while maintaining the inclination of 60°.

(i) 4.3 m

Self-Assessment

(ii) 5 m

(iii) 5.7 m

(iv) 10 m

253

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. A drone is flying at a height of 120 m . The angle of depression of a car on the ground is 45 . Find the distance of the car from the point directly below the drone. (CBSE SQP 2025) 

(a) 120 2 m

(b) 240 m

(d) 60 2 m

2. The height of a tower is 50 m . A man standing at a distance observes the top of the tower at an angle of 60 . Find the distance of the man from the base of the tower. (NCERT Exemplar) (a) 50 3 m

(b) 25 3 m

(d) 100 3 m

3. The height of a tree is 15 m, and the angle of elevation of its top from a point 20 m away from the base is: (a) 45°

(b) 30°

(d) 90°

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Height . Length of ladder

(CBSE SQP 2025)

5. Assertion (A): The angle of depression of an object is always greater than 90°. Reason (R): The angle of depression is measured downward from the horizontal level of the observer.

Very Short Answer Questions

(CBSE 2024)

[2 × 2 = 4 Marks]

6. The height of a kite flying at an angle of 60 with the ground is 100 m . Find the length of the string, assuming it is straight and the string is tied at ground level. (CBSE 2023) 7. The angle of elevation of the top of a tree from a point 20 m away from its base is 30°. Find the height of the tree.

Short Answer Questions

[4 × 3 = 12 Marks]

8. A man on a hill 50 m high observes a boat at an angle of depression of 30°. Find the distance of the boat from the base of the hill. (CBSE SQP 2025) 9. A person standing at a distance of 30 m from the base of a tree observes its top at an angle of 60°. Find the height of the tree. 10. The length of a shadow of a tower is equal to the height of the tower. Find the angle of elevation of the sun. 11. Two buildings are 40 m apart. The angle of elevation of the top of the taller building from the top of the shorter building is 45°. If the height of the shorter building is 60 m, find the height of the taller building. (CBSE 2023) 254

Long Answer Questions

[3 × 5 = 15 Marks]

12. A man observes the top of a tower at an angle of 30° from a distance of 50 m from its base. He then moves 30 m closer to the base and observes the top at an angle of 60°. Find the height of the tower. (CBSE SQP 2025) 13. Two buildings are 40 m apart. The angle of elevation of the top of the taller building from the shorter building is 45°, and the shorter building is 50 m high. Find the height of the taller building. (CBSE 2024) 14. A flagpole 15 m high is observed from a point on the ground at an angle of 30°. Another observation is made from the top of a 5 m building at an angle of 60. Find the distance of the flagpole from the point of observation on the ground. (CBSE 2023)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A satellite at height h is transmitting internet signals to an aeroplane and a helicopter. The aeroplane is flying at a height of 7,816 m, while the helicopter is at a height of 1,930 m. The angles of depression from the satellite to the top of the aeroplane and the helicopter are 30° and 60° respectively. If the distance between the two aircraft is 1937 km, and the satellite is positioned vertically above the midpoint of the distance between them, find the height h of the satellite.

S 30° A

60° G

h 7816 m H 1937 km

1930 m

(a) C alculate the distance of the satellite from the Aeroplane. (b) T he distance of the satellite from the top of the Helicopter is

(i) 1139.4 km

(ii) 577.52 km

(iii) 1937 km

(iv) 1025.36 km

(iii) 1937 km

(iv) 1025.36 km

(i) 1139.4 km

(ii) 577.52 km

(d) W hat is the angle of elevation if a man is standing at a distance of 7816 m from the Aeroplane?

Self-Assessment

255

Circles

Chapter at a Glance Chord ord

A line segment joining any two points on a circle is called a chord.

Diameter

d Ra

A chord passing through the centre of the circle is called diameter. It is the longest chord.

Diam

eter

gen t

Tan

Secant A secant to a circle is a line that has two points in common with the circle. It cuts the circle at two points, forming a chord of the circle.

Tangent When a line meets the circle at one point or two coincide. The line is known as points, a tangent. The tangent to a circle is perpendicular to the radius through the point of contact. The lengths of the two tangents from an external point to a circle are equal. ⇒ AP = PB

O A

Circle and Line in a Plane For a circle and a line on a plane, there can be three possibilities. (a) they can be non-intersecting (b) they can have a single common point: in this case, the line touches the circle. (c) they can have two common points: in this case, the line cuts the circle.

Important Theorems Theorem 1 A tangent to a circle is perpendicular to the radius through the point of contact. ⇒ OP ⊥ AB

Theorem 2 The length of two tangents drawn from an external point to a circle are equal. ⇒ AP = AQ

R QB

Theorem 3 If two tangents are drawn from an external Point, then: (a) they subtend equal angle at the centre ⇒ ∠AOP = ∠AOQ (b) they are equally inclined to the segment, joining the centre to that point ⇒ ∠OAP = ∠OAQ 256

P O

A Q

NCERT Zone NCERT Exercise 10.1

1. How many tangents can a circle have? Sol. A circle can have infinite number of tangents because a circle has infinite number of points on it and at every point a tangent can be drawn.

Now draw a perpendicular from O to line AB, which intersects AB at P.

2. Fill in the blanks. (a) A tangent to a circle intersects it in _____________ point(s).

(b) A line intersecting a circle in two points is called a _____________.

(d) The common point of a tangent to a circle and the circle is called _____________. Sol. (a) A tangent to a circle intersects it in one point. (b) A line intersecting a circle in two points is called a Secant. (c) A circle can have two parallel tangents at the most. (d) The common point of a tangent to a circle and the circle is called point of contact. 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (a) 12 cm (b) 13 cm (c) 8.5 cm

Now take two points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.

NCERT Exercise 10.2

1. From a point Q , the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (a) 7 cm (b) 12 cm (c) 15 cm (d) 24.5 cm

Sol. Let O be the centre of the circle. Given = that, OQ 25 = cm and PQ 24 cm

(d) 119 cm

Ans. (d) 119 cm

As the radius is perpendicular to the tangent at the point of contact,

Sol. In ∆OPQ, angle P is a right angle.

Therefore, OP ⊥ PQ

[Since radius is perpendicular to tangent]

Applying Pythagoras theorem in ΔOPQ, we obtain

Using Pythagoras theorem, 2

OQ = PQ + OP 2

OP 2 + PQ 2 = OQ 2

⇒ 12 = PQ + 5

⇒ 144 = PQ 2 + 25 ⇒ PQ 2 = 144 − 25 = 119

OP 2 + 242 = 252 O 5 cm P

12 cm

⇒ PQ = 119 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle. Sol. Consider a circle with centre O. Let AB be the given line.

Chapter 10 • Circles

OP 2 + 576 = 625

25 cm

OP = 625 − 576

OP 2 = 49

24 cm

Taking square root both sides: OP = 7 Therefore, (a) 7 cm is correct. 2. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110 , then ∠PTQ is equal to (a) 60

(b) 70

(d) 90

257

Sol. It is given that TP and TQ are tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

Therefore, radius drawn to these tangents will be perpendicular to the tangents.

∠OAR = 90°

Thus, OP ⊥ TP and OQ ⊥ TQ

∠OPT = 90° ∠OQT =° 90

In quadrilateral POQT ,

110°

∠OAS = 90° ∠OBP = 90°

∠OBQ =° 90 Q

Sum of all interior angles = 360 ∠OPT + ∠POQ + ∠OQT + ∠PTQ =360

It can be observed that ∠OAR = ∠OBQ (Alternate interior angles)

Hence, (b) is correct 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80 , then ∠POA is equal to (a) 50 (b) 60 (c) 70 (d) 80

∠OBP = 90° and ∠OAP = 90° In AOBP,

Sol. Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents. 80°

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Sol. It is given that PA and PB are tangents.

∠OAS = ∠OBP (Alternate interior angles) Since alternate interior angles are equal, lines PQ and RS will be parallel.

⇒ 90 + 110° + 90 + PTQ = 360 ⇒ PTQ = 70

Thus, OA ⊥ PA and OB ⊥ PB

Sum of all interior angles = 360 ∠OAP + ∠APB + ∠PBO + ∠BOA =360 90° + 80° + 90° + ∠BOA = 360 ∠BOA = 100

We have to prove that the line perpendicular to AB at P passes through centre O ’. We shall prove this by contradiction method. Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O '. Join OP and O 'P.

In ∆OPB and ∆OPA, AP = BP (Tangents from a point)

OA = OB ( Radii of the circle ) OP = OP ( Common side )

Therefore, ∆OPB ≅ ∆OPA ( SSS congruence criterion ) And thus, ∠POB = ∠POA

As perpendicular to AB at P passes through O', therefore, ∠O ′PB = 90 (1)

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

O is the centre of the circle and P is the point of contact. We know the line Joining the centre and the point of contact to the tangent of the circle are perpendicular to each other. ∴∠OPB = 90 (2)

Sol. Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B.

Comparing equations (1) and (2), we obtain

∠O ′PB = ∠OPS (3)

Respectively radius drawn to these tangents will be perpendicular to the tangents.

From the figure, it can be observed that,

1 100° ∠POA =∠AOB = =° 50 2 2 Hence, (a) is correct.

258

∠O ′PB < ∠OPS (4)

Therefore, ∠OPB = ∠OPB is not possible. It is only possible, when the line OP coincides with OP.

Therefore, the perpendicular to AB through P passes through centre O.

Q S

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

OB ⊥ AB ( Radius ⊥ tangent at the point of contact ) Applying Pythagoras theorem in ∆ABO ,

(i)

CR = CQ (Tangents on the circle from point C )

(ii)

BP = BQ (Tangents on the circle from point B)

(ii)

AP = AS (Tangents on the circle from point A) (iv) Adding all these equations, we obtain

CD + AB = AD + BC

16 + BO 2 = 25 BO = 3

DR = DS (Tangents on the circle from point D)

( DR + CR ) + ( BP + AP ) = ( DS + AS ) + (CQ + BQ )

AB 2 + BO 2 = OA 2

DR + CR + BP + AP = DS + CQ + BQ + AS

we obtain

BO 2 = 9

Sol. It can be observed that

In ∆ABO ,

Sol. Let us a circle centred at point O. AB is a tangent drawn on this circle from point A. Given that, = OA 5= cm and AB 4 cm

42 + BO 2 = 52

5 cm 4 cm

Hence, the radius of the circle is 3 cm. 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Sol. Let the two concentric circles be centred at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle. OA ⊥ PQ (As OA is the radius of the circle) Applying Pythagoras theorem in ∆OAP , we obtain OA 2 + AP 2 = OP 2

9. In the given figure, XY and X ’Y ’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X ’Y ’ at B . prove that ∠ AOB = 90 Sol. In ∆OPA and ∆OCA , OP = OC

(Radii of the same circle)

AO = AO

AP = AC (Tangents from point A)

X’

C B

Y’

(Common side) ∆OPA ≅ ∆OCA ( SSS congruence criterion ) ∠POA = ∠COA (i)

32 + AP 2 = 52

Similarly, ∆OQB ≅ ∆OCB

2 AP= 25 − 9 ⇒ AP 2 = 16 ⇒ AP = 4

In AOPQ , Since OA ⊥ PQ ,

∠QOB = ∠COB (ii)

AP = AQ (Perpendicular from the centre of the circle bisects the chord)

Since POQ is a diameter of the circle, it is straight line.

PQ = 2 AP = 2 × 4 = 8 Therefore, the length of the chord of the larger circle is 8 cm.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB =180

From equations ( i ) and ( ii ) , it can be observed that

A Q

2∠COA + 2∠COB = 180

8. A quadrilateral ABCD is drawn to circumscribe a circle (see given figure) Prove that

∠COA + ∠COB= 90°

AB + CD = AD + BC Chapter 10 • Circles

∠AOB = 90°

259

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

On putting the values of equations (1) and (2) in this equation, we obtain

Sol. Let us consider a circle centred at point O. Let be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at centre O of the circle.

Comparing equations (1), (2), and (3), we obtain

2 AB = 2BC

It can be observed that OA ( radius ) ⊥ PA ( tangent )

AB = BC (3) AB = BC = CD = DA Hence, ABCD is a rhombus. 12. A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC . A

Therefore, ∠OAP = 90° Similarly, OB ( radius ) ⊥ PB ( tangent )

∠OBP = 90°

In quadrilateral OAPB ,

O B

11. Prove that the parallelogram circumscribing a circle is a rhombus. Sol. Since ABCD is a parallelogram, AB = CD

(1)

BC = AD (2) It can be observed that

AP = AS (Tangents on the circle from point A)

6 cm

AE = AF = x Now, our three sides are

CD = CD + DB = 6 + 8 = 14 cm

( x + 6 ) cm AB= b= ( x + 8 ) cm Here, AC= a=

= BC c = 6 + 8 = 14 cm

DR + CR + BP + AP = DS + CQ + BQ + AS

a +b +c = s = 2

CD + AB = AD + BC

260

F 8 cm

O 6 cm

8 cm

Now, we find Area of ∆ABC using Herons formula Area of triangle

Adding all these equations, we obtain ( DR + CR ) + ( BP + AP ) = ( DS + AS ) + (CQ + BQ )

BF = BD = 8 cm

= s ( s − a )( s − b )( s − c ) Q

CE = CD = 6 cm

AB =AF + FC =( x + 8 ) cm

BP = BQ (Tangents on the circle from point B)

Hence,

AC =AE + EC =( x + 6 ) cm

DR = DS (Tangents on the circle from point D) A

From theorem 10.2, length of tangents drawn from external point are equal l. A

∠APB + ∠BOA = 180°

CR = CQ (Tangents on the circle from point C )

8 cm

Let ∆ABC circumscribe the circle Also, BD = 8 cm and CD = 6 cm

90° + ∠APB + 90° + ∠BOA = 360°

Hence proved.

6 cm

Sol. Given: A circle with centre O with OD = radius = 4 cm

∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

Sum of all interior angles = 360°

( x + 6 ) + ( x + 8 ) + 14 2

2 ( x + 14 ) 2x + 28 = = x + 14 2 2

x 2 + 14x − 7x − 98 = 0 ⇒ x ( x + 14 ) − 7 ( x + 14 ) = 0

s ( s − a )( s − b )( s − c )

Area ∆ ABC =

Putting values of a, b, c and s

0 ( x − 7 )( x + 14 ) =

= ( x + 14 )( x + 14 − x − 6 )( x + 14 − x − 8 )( x + 14 − 14 )

= So, x 7= and x –14 . Since x cannot be negative. So, x = 7 Now, AC = x + 6 = 7 + 6 = 13 cm

= ( x + 14 ) ( x + 14 − ( x + 6 ) ) ( x + 14 − ( x + 8 ) ) ( x + 14 − 14 )

= ( x + 14 )( 8 )( 6 )( = x)

( x + 14 )( 48x )

= x ( 48x ) + 14 (= 48x )

48x 2 + 672x

and AB = x + 8 = 7 + 8 = 15 cm 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

48x 2 + 672x

∴Area ∆ ABC= Join OE and OF

Sol. Let ABCD be a quadrilateral circumscribing a circle centred at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the centre of the circle.

Here, OD = OF = OE = radius = 4 cm

Also, we know that tangent is perpendicular to the radius. So,

Consider ∆OAP and ∆OAS ,

OD ⊥ BC , OF ⊥ AB and OE ⊥ AC Also, Area of ΔABC = Area ΔAOC + Area ΔAOB + Area ΔBOC 1 Area of triangle = × Base × height 2

AP = AS (Tangents from the same point) (i)

thus, ∠POA = ∠AOS

1 1 Area ∆AOB = ×AB × OE = × ( 8 + x ) × 4 = 16 + 2x 2 2

∠1 =∠8

1 1 Area ∆BOC = ×BC × OD = × ( 8 + 6 ) × 4 = 28 2 2 Now, Area of ΔABC = Area ΔAOC + Area ΔAOB + Area ΔBOC

48x 2 + 672x = ( 2x + 12 ) + ( 2x + 16 ) + 28 2

OA = OA (Common side) ∆OAP ≅ ∆OAS (SSS congruence criterion)

1 1 Area ∆AOC = ×AC × OF = × ( 6 + x ) × 4 = 12 + 2x 2 2

Putting values of Area ΔAOC, Area ΔAOB and Area ΔBOC in

OP = OS (Radii of the same circle)

(i)

Similarly, ∠2 =∠3 ∠4 =∠5 ∠6 =∠7

( ∠1 + ∠2 ) + ( ∠5 + ∠6=) 180°

48x + 672 = x 2

( 4x + 56 ) )= 2

( 4x ) + 56 + 2 × 4x × 56 2

48x + 672x = 16x + 3136 + 448x 48x 2 − 16x 2 + 672x − 448x − 3136 = 0

2∠1 + 2∠2 + 2∠5 + 2∠6= 360°

Squaring both sides

(

( ∠1 + ∠8 ) + ( ∠2 + ∠3) + ( ∠4 + ∠5) + ( ∠6 + ∠7=) 360° 2 ( ∠1 + ∠2 ) + 2 ( ∠5 + ∠6= ) 360°

48x 2 + 672x

1 2 3 O 4 6 5

8 7

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠= 8 360°

48x + 672x = 4x + 56

∠AOB + ∠COD = 180° Similarly, we can prove that BOC + DOA = 180° Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

32x 2 + 224x − 3136 = 0 ⇒ x 2 + 7x − 98 = 0

Chapter 10 • Circles

261

Multiple Choice Questions 1. Here is a circle with centre O. Manu wants to draw a tangent RS to the circle. What is the number of points at which the line RS will meet the circle?

7. In the given figure, AC and AB are tangents to a circle centred at O. If ∠COD = 120°, then ∠BAO is equal to (a) 30° (b) 60° (c) 45° (d) 90°

(a) 0 (b) 1

(a) 80° (b) 60° (c) 50°

80°

(a) 2 AB

(b) 3 AB

(d) 5 AB

4. In the given figure, PQ is tangent to the circle centred at O. If ∠AOB = 95°, then the measure of ∠ABQ will be (a) 47.5°

(b) 42.5°

95°

(CBSE 2023)

6. In the given figure, AB is a tangent to the circle centred at O. If OA = 6 cm and ∠OAB = 30°, then the radius of the circle is (a) 3 cm (c) 2 cm 3 cm

(CBSE PP 2023)

9. In figure, the pair of tangents AP and AQ drawn from an external point A to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. Then radius of the circle is (a) 10 cm P A

O Q

10. Maximum number of common tangents that can be drawn to two circles touches at a point is (a) 4 (b) 3 (c) 2 (d) 1

O A

B (CBSE 2023)

(d) segment

12. Two concentric circles are of radii 13 cm and 5 cm. The length of the chord of larger circle which touches the smaller circle is (a) 12 cm (b) 20 cm (c) 24 cm

30°

5 cm

(d) 2.5 cm

(CBSE SQP 2024)

(b) 3 3 cm

N 50° K M

11. A line through point of contact and passing through centre of circle is known as (a) tangent (b) chord

5. A quadrilateral PQRS is drawn to circumscribe a circle. If PQ = 12 cm, QR = 15 cm and RS = 14 cm, then the length of SP is (a) 15 cm (b) 14 cm (c) 12 cm (d) 11 cm

(b) 65°

(b) 7.5 cm

(d) 95°

(CBSE 2023)

(d) cannot be uniquely determined with the given information

3. Two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°, then OA is equal to

262

8. Shown below is a circle with 3 tangents KQ, KP 1 and LM, QL = 2 cm and KL = 6 cm. PM = KL. What 2 is the measure of ∠LMK? (a) 50°

(d) 40°

O D 120°

2. In the given figure, tangents PA and PB drawn from P to circle are inclined to each other at an angle of 80°. The measure of ∠PAB is

(d)

(d) 30 cm

13. In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to C (a) 45 (b) 60°

14. In the given figure, if quadrilateral PQRS circumscribes a circle, then PD + QB = P (a) PQ (b) QR

50°

(d) PS

A Q

(d) 55°

B R

Answers 1. (b) 1

2. (c) 50°

9. (c) 5 cm

10. (b) 3

3. (a) 2 AB

4. (a) 47.5°

11. (c) normal

12. (c) 24 cm

5. (d) 11 cm

6. (a) 3 cm

13. (c) 50

14. (a) PQ

7. (a) 30°

8. (c) 80°

Constructed Response Questions Very Short Answer Questions

Sol. ∠AOB= 180° − 130°= 50°

1. In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB = 75°. (CBSE SQP 2023) Sol. ∠PAO = ∠PBO = 90° (Angle between radius and tangent) A

From figure in quadrilateral PAOB, ∠AOB = 180° − 75°

∠AOB = 105° (By angle sum B property of a quadrilateral) 1 ⇒∠AQB = ×105 = ° 52.5° (Angle at circle is half the 2 angle subtended by the same arc at the centre) 2. In the given figure, PA is tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter. If ∠AOC = 130°, then find the measure of∠APB, where O is the centre of the circle. (CBSE 2023)

3. In the given figure, O is the centre of the circle. AB and AC are tangents drawn to the circle from point A. If ∠BAC = 65°, then find the measure of ∠BOC. Sol. ∠ABO = ∠OCA = 90° (Angle between radius and tangent) B

In quadrilateral ABOC, by ASP of quadrilateral

130°

∠BAC + ∠ABO + ∠ACO + ∠BOC = 360°

⇒ 65° + 90° + 90° + ∠BOC = 360° ⇒∠BOC = 360° − 245° = 115° 4. PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an isosceles triangle, then find ∠OQP. (CBSE SQP 2021)

Sol. In ∆OPQ,

A P

 ∴∠APB = 180 − (50  + 90 = ) 40

(CBSE 2023)

∠OAP = 90 (Angle between radius and tangent)

∠P + ∠Q + ∠O = 180° (∠O = ∠Q isosceles triangle) 2∠Q + ∠P = 180°

P Q

2∠Q + 90° = 180° 2∠Q = 90°

Chapter 10 • Circles

∠Q = 45° 263

5. In two concentric circles, a chord of length 24 cm of larger circle touches the smaller circle, whose radius is 5 cm. Find the radius of the larger circle.

∠OAB = ∠OBA = 60 (as OA and OB being radii) Using the sum of angles in ΔOAP:

Sol. Given: OX = 5 cm, AB = 24 cm

∠ OAP + ∠ AOP + ∠ OPA = 180

OX ⊥ AB (∴ radius drawn from the centre of a circle is ⊥ to the chord)

∠ OPA = 180 − 90

And, X is the mid-point of AB

∠ OPA = 30

∆OAX is a right-angled triangle

Solving for ∠OPA:

∠OXA = 90°

5 cm

AX = 12 cm

24 cm

By using Pythagoras theorem.

60°

9. Two concentric circles are of radii 10 cm and 6 cm, then the length of the chord of the larger circle which touches the smaller circle is Sol. Given: Larger circle radius = 10 cm.

OA = OX + AX = 25 + 144 = 169 cm

Smaller circle radius = 6 cm.

OA = 13 cm

Since DE is a chord, AB is perpendicular to DE (by the tangent-chord theorem).

6. In given figure, O is the centre of the circle, PQ is a chord and PT is tangent to the circle at P. Find ∠TPQ. (CBSE 2017) Sol. ∠OPQ = ∠OQP (OP and OQ are radii) T Q

∠TPO = 90°(Angle between radius and tangent)

Thus, DB = BE, By Applying Pythagoras Theorem in ΔABE 2 AE = AB 2 + BE 2

180° − 70° = = 55° 2

2 10= 62 + BE 2

100 = 36 + BE 2

70° O

BE = 64

6 cm

BE = 8 cm

∠TPQ = 90° − 55° = 35°

⇒ DE = 2 × 8 = 16 cm

7. In the given figure, if PT is a tangent to a circle with centre O and ∠TPO = 35°, then find the measure of ∠x.

10 cm

Sol. Given: PT is a tangent to the circle at T

10. From a point P, which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle is drawn. Find the area of the quadrilateral PQOR.

∠TPO = 35°.

= OR = 5 Sol. Given: Radius of the circle OQ

We know that the radius is perpendicular to the tangent at the point of contact.

Distance of external point P from centre O: OP = 13 cm

∠OTP = 90

PQ and PR are tangents

∠x = ∠OTP + ∠TPO 



x = 90 + 35

35°

x = 125 8. In the given figure, O is the centre of the circle and PA is a tangent to the circle. If ∠OAB = 60°, then find ∠OPA. (CBSE Basic 2023) Sol. Given: PA is a tangent at A. ∠OAB = 60°. We know that the radius is perpendicular to the tangent at the point of contact: ∠OAP = 90 264

Length of the Tangent PQ Using Pythagoras Theorem in ∆OPQ 2 OP = OQ 2 + PQ 2

2 13= 52 + PQ 2

169 = 25 + PQ PQ 2 = 144

PQ = 12 cm

5 cm Q

13 cm

Area of Quadrilateral PQOR PQOR is a rectangle (since tangents from an external point are equal and radii are perpendicular to tangents).

Sol. Since tangents drawn from an external point to a circle are equal in length.

Area of PQOR = 2 × Area of ∆OPQ 1 Area = 2 × × OQ × PQ 2

∴PA = PB = 10 cm [Given]

1 = 2 × × 5 × 12 2

Also, CA = CE and DB = DE Let = CE x= and DE y = CA x= and DB y . Then,

= 60 cm 2 11. In figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then find ∠POQ. (CBSE Basic 2020) P

Sol. Given: ∠QPR = 50 ∠OPR = 90° (angle between tangent and radius is 90°)

∠ OPQ = 90 − 50 = 40 In ∆OPQ, 

∠ OPQ + ∠ POQ + ∠ OQP = 180 (by angle sum property)

12. A line through the centre O of a circle of radius 7 cm cuts the tangent, at a point P on the circle, at Q such that PQ = 24 cm. Find OQ. Sol. Since a tangent at a point on a circle is perpendicular to the radius through the point, therefore, OP is perpendicular to PQ.

Thus, AB and CD are two chords of the larger circle such that they are equidistant from the centre. Hence, AB = CD

Short Answer Questions 1. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. (CBSE 2023, 2017)

In right triangle OPQ, we have

Sol. TP = TQ (tangents drawn from an external point to a circle are equal) P

2 OQ = OP 2 + PQ 2

⇒ ∠TPQ = ∠TQP

7 cm

Let ∠PTQ be q. Q

⇒ ∠TPQ = ∠TQP =

13. From an external point P, two tangents PA and PB are drawn to a circle with centre O. At one point E on the circle, a tangent is drawn which intersects PA and PB at C and D respectively. If PA = 10 cm, find the perimeter of ∆PCD. (NCERT Exemplar)

Now ⇒ ∠OPT = 90°

Chapter 10 • Circles

∴ OM = ON = Radius of the smaller circle.

∠ POQ = 100

Since AB and CD are tangents to the smaller circle.

40 + ∠ POQ + 40 = 180 ( ∠OPQ = ∠OQP , OP and OQ are radii of the triangle)

OQ 2 = 625 ⇒ OQ = 25 cm.

Sol. Let AB and CD be two chords of the circle which touch the inner circle at M and N respectively. Then, we have to prove that AB = CD.

∠ OPQ = 40

2 OQ= 49 + 576

=PC + (CE + DE ) + PD

14. In two concentric circles, prove that all chords B of the outer circle which touch the inner are of equal length.



∠ OPQ + 50 = 90 (Given, ∠QPR = 50 )

2 OQ= 72 + 242

= PC + CD + PD

= 20 cm Q



(10 − y ) cm.

Now, perimeter of ∆PCD

= (10 − x ) + ( x + y ) + (10 − y )

Then ∠OPQ + ∠QPR =90 

50°

∴ PC = (10 − x ) cm and PD =

24 cm

180°−θ θ = 90° − 2 2

O Q

θ θ  ⇒ ∠OPQ = 90° −  90° −  = 2 2  ⇒ ∠PTQ = 2∠OPQ Hence proved.

265

2. a, b and c are the sides of a right triangle, where c is the hypotenuse. A circle, of radius r, touches the a +b −c sides of the triangle. Prove that r = . 2 Sol. Let the circle touches sides AB, BC and CA at F, D and E respectively where

(CBSE 2016)

= AB c= , BS a and = AC b

Length of the tangents drawn from an external point are equal.

C Therefore, = AE AF , BD = BF , = CE CD = r

⇒ AQ = AR, BQ = BP, CP = CR

Perimeter of ∆ABC = AB + BC + CA = AB + (BP + PC) + (AR − CR)

D a

b –= r AE = AF (i) a –= r BD = BF (ii) Because, AB = AF + BF = c b – r + a – r

Sol. Given: A circle touching the side BC of ∆ABC at P and AB, AC produced at Q and R respectively. 1 To prove: AQ = (perimeter of ∆ABC). 2 Lengths of tangents drawn from an external point to a circle are equal.

(from (i) and (ii))

c = a + b − 2r 2r = a + b − c

= (AB + BQ) + (PC) + (AQ − PC) since [AQ = AR, BQ = BP, CP = CR]

= AQ + AQ = 2AQ

1 (Perimeter of ∆ABC) 2 ∴AQ is the half of the perimeter of ∆ABC ⇒ AQ =

5. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

∴OB > OC (OC being radius and B any point on tangent)

Sol. Given: Two tangents PQ and PR are drawn from an external point P to a circle with centre O. To prove: Centre of a circle R touching two intersecting O P lines lies on the angle bisector of the lines. Q In ∠RPQ, Construction: Join OR, and OQ. In ∆POR and ∆POQ ∠PRO = ∠PQO = 90° [tangent at any point of a circle is perpendicular to the radius through the point of contact]

⇒ OA < OB

OR = OQ [radii of circle]

a +b – c r= 2 Hence proved. 3. Prove that a tangent to a circle is perpendicular to the radius through the point of contact. (CBSE 2020)

Sol. OA = OC (Radii of circle) Now OB = OC + BC

B is an arbitrary point on the tangent. Thus, OA is shorter than any other line segment joining O to anypoint on tangent.

O C A

Shortest distance of a point from a given line is the perpendicular distance from that line. Hence, the tangent at any point of circle is perpendicular to the radius. 4. A circle touches the side BC of a ∆ABC at a point P and touches AB and AC when produced at Q and R respectively. Show that AQ = 1 (perimeter of ∆ABC). 2 (CBSE 2023)

266

Since, OP is common. ∴ ∆PRO ≅ PQO [R.H.S] Hence, ∠RPO = ∠QPO [by CPCT] Thus, O lies on angle bisector of PR and PQ. Hence proved. 6. In the given figure, PQ is a tangent from an S external point P to a circle with Q 2 centre O and OP cuts the circle O at T and QOR is a diameter. If T ∠POR = 130° and S is a point on R the circle, find ∠1 + ∠2. Sol. OQ is perpendicular to the tangent PQ. In ∆OPQ ∠POR = ∠1 + ∠RQP

130° =∠1 + 90° ∠ 1= 130° − 90°= 40° (i)

1 = × 130°= 65° (ii) 2

∠ROT is the angle subtended by arc RT at centre and ∠RST is the angle subtended by arc RT at point Son circumference.

∠1 + ∠2= 40° + 65°= 105°

∴ ∠2=

Adding (i) and (ii), we get

1 ×∠ROT 2

7. Prove that the lengths of two tangents drawn from an external point to a circle are equal. Topper’s Answer

8. If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

adding (v) and (vi) we get EQ + FQ = EP + FR

Sol. Let ABCDEF be a hexagon. It is circumscribing a circle.

⇒ EF = EP + FR

AM = AR (i) BM = BN (ii) CN = CO (iii) DO = DP (iv) EP = EQ

(v)

FQ = FR (vi) adding (i) and (ii) we get AM + BM = AR + BN

⇒ AB = AR + BN adding (iii) and (iv) we get CO + DO = CN + DP ⇒ CD = CN + DP

Chapter 10 • Circles

M B

E P

N D

AB + CD + EF = AR + (BN + CN) + (DP + EP) + FR = BC + DE + FA ∴ AB + CD + EF = BC + DE + FA Hence proved. 9. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO i.e., BO = 2BC. Sol. Given: Tangents BC 3and BD are drawn from an external point B such that ∠DBC = 120°

Adding all these we obtain

In ΔOBC and ΔOBD, we have OB = OB [common] OC = OD [Radii of same circle]

267

BC = BD [Tangents from an external point are equal in length]

(i)

OD2 = 9 OD = 3 cm Therefore, the radius of the inner circle is 3 cm.

∴ ΔOBC ≅ ΔOBD [By SSS criterion of congruence] ⇒ ∠OBC = ∠OBD (CPCT) 1 1 ∴ ∠OBC = ∠DBC = ×120° [∠CBD = 120°given] 2 2 ⇒∠OBC = 60° OC and BC are radius and tangent respectively at contact point C.

11. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD. Sol. Two tangents PA and PB are drawn to a circle with centre O from an external point P.

So, ∠OCB = 90° Now, in right angle ΔOBC = 60°

A C

BC ∴ COS 60° = BO ⇒

1 BC = 2 BO

P D

Perimeter of ∆PCD = PC + CD + PD

Hence proved (ii) part,

60° 120°

⇒ OB = 2BC

= PC + CE + ED + PD

⇒ OB = BC + BC

= PC + CA + DB + PD

⇒ OB = BC + BD [BC = BD from (i)]

= PA + PB = 2 PA = 2(10) = 20 cm

Hence proved.

[Since, CE = CA, DE = DB, PA = PB. Tangents from internal point to a circle are equal.]

10. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. Sol. Radius of the outer circle = 5 cm, The length of the chord, AC = 8 cm From the figure,

Let C₁ be the inner circle and C₂ be the outer circle with radius 5 cm. We know that the perpendicular drawn from the centre of a circle to the chord always bisects the chord. AC So, OD ⊥ AC. Also, AD = DC = 2 AC 8 = = 4 cm 2 2 So, AD = DC = 4 cm

O D

Considering triangle OAD, ODA is a right triangle with D at right angle. By Pythagoras theorem, AO ² = AD ² + OD ² AD = 4 cm So, (5)2 = (4)2 + OD 2 25 = 16 + OD 2 OD2 = 25 − 16 268

Given: Circle with centre O, PT is a tangent OC ⊥ AB To prove: PA × PB = PC 2 - AC2

12. In the given figure, PT is a tangent to the circle centred at O. OC is perpendicular to chord AB. Prove that PA × PB = PC 2 − AC 2.

P C

Perpendicular from centre to chord bisect it C

⇒ C is midpoint of AB. (∴∠OCA = 90°, given) AB ⇒ BC = CA = ⇒ AB = 2AC (1) 2 By Pythagoras Theorem, In ∆OCA, OA2 = OC2 + AC2 In ∆OCP, OP2 = OC2 + PC2 Taking L.H.S of to prove = PA × PB = PA × (PA + AB) = PA2 + PA × AB = PA2 + PA × 2AC

(From 1)

= PA + 2(PA)(AC) 2

= (PA + AC)2 - AC2 = PC 2 - AC 2 = R.H.S Hence proved.

[(a + b)2 - a2 + 2ab + b2]

13. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

∴ ∆QRS is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.]

Sol. Given chord AB, tangent AP and BP at A and B.

⇒ ∠ RSQ + ∠ SRQ + ∠ RQS =° 180 [Angle sum property of a triangle] ⇒ 75°+ 75°+ ∠ RQS = 180°

To prove: ∠PAM = ∠PBM Construction: Join centre O to P. Let OP meet AB at M. Proof: In ∆AMP and ∆BMP

⇒ 150°+ ∠ RQS = 180°

O M

∴ ∠ RQS = 30° 2. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that area ∆ABC = 90 cm2. (CBSE 2023)

B AP = BP (tangents from same point to a circle are equal)

MP = MP (Common Side) ∠APM = ∠BPM (tangents are equally inclined to line joining the point to circle’s centre) ∆AMP ≅ ∆BMP By SAS criteria ∠PAM = ∠PBM (By CPCT) Hence, tangents at endpoints of a chord make equal angles with it.

Long Answer Questions

30°

∴ ∠ PRQ = ∠ PQR [Angles opposite to equal sides are equal in a triangle] In ∆PQR , ∠ RQP + ∠ QRP + ∠ RPQ =° 180 [Angle sum property of a triangle] ⇒ 2∠ RQP + 30 =° 180°

[By Alternate Segment Theorem] Given, RS  PQ ∴ ∠ RQP = ∠ SRQ = 75° [Alternate angles] ⇒ ∠ RSQ = ∠ SRQ = 75°

Chapter 10 • Circles

F O

ar(∆ABC) = 90 cm2

10 cm

Tangent from same external point are equal in length.

4 cm D

8 cm

CD = CF = 8 cm Area of ∆ABC = Area of ∆OAB + Area of ∆OBC + Area of ∆OCA 1 1 1 ⇒ 90 = × OE × AB + × OD × BC + × OF × AC 2 2 2 AB = AE + BE, BC = BD + DC, AC = CF + AF P

1 1 1 × 4 × ( x + 10 ) + × 4 × (10 + 8 ) + × 4 × ( 8 + x ) 2 2 2 1 ⇒ 90 = × 4 × ( x + 10 + 10 + 8 + 8 + x ) 2 ⇒ 45 = 2x + 36 ⇒ 90 =

⇒ 2x = 9 ⇒ x = 4.5 AB = AE + BE = 4.5 + 10 = 14.5 cm AC = AF + CF = 4.5 + 8 = 12.5 cm 3. Two circles with centres O and O' of the radii 6 cm and 8 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

⇒ 2∠ RQP = 150° ⇒ ∠ RQP = 75°

so ∠ RQP= ∠ QRP= 75° ⇒ ∠ RQP = ∠ RSQ = 75°

BE = BD = 10 cm

Sol. It is given that, ∠ RPQ= 30° and PR and PQ are tangents drawn from P to the S R same circle.

In circle with radius r = 4 cm, BD = 10 cm, CD = 8 cm

AE = AF = x (Let it be)

1. In figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find the ∠RQS.

Hence PR = PQ [Since tangents drawn from an external point to a circle are equal in length]

Sol. Given ∆ABC

Sol. OO ′ =

(CBSE 2023)

6 + 8 = 10 cm {OP ⊥ O ′P } 2

= Let OA x , O ′A = 10 − x AP 2 = OP 2 − OA 2 = 36 − x 2 2

Also AP = 64 − (10 − x )

269

5. In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm. Find the length of the chord BC.

Therefore 36 − x 2 = 64 − (10 − x )2 ⇒ 36 − x 2 = 64 − 100 − x 2 + 20x ⇒x = 3.6 In ∆PAO , AP 2 = 36 − (3 ⋅ 6)2 = 23 ⋅ 04 ⇒ AP = 4.8 Length PQ =× 2 AP = 9.6 cm 4. AB and CD are two parallel chords of a circle such that AB = 16 cm and CD = 30 cm. If the chords are on the opposite sides of the centre and the distance between them is 23 cm, find the radius of the circle. Sol.

8 M 8 cm cm r x

15 cm N

15 cm

Let OM = x. Then AM = 5 − x

⇒ 62 =( 5 − x ) + BM 2 2

⇒ BM 2 = 36 − ( 5 − x ) (i) 2 In right ∆OMB, BO = BM 2 + MO 2

⇒= 52 BM 2 + x 2

Let O be the centre of the circle and radius r cm. Draw OM ⊥ AB and ON ⊥ CD. Then MON is a straight line and

2 ⇒ BM= 25 − x 2 (ii)

From (i) and (ii), we have 2

⇒ 36 − ( 5 − x ) = 25 − x 2

(

)

⇒ 36 − 25 + x 2 − 10x = 25 − x 2

1 1 = AM = AB 8 cm and= CN = CD 15 cm. 2 2

⇒ 11 + 10x = 25 ⇒ 10 = 25 − 11 = 14 ⇒ x =

Let OM = x cm. Then, ON =

∠AMB = 90° and BM = MC AB 2 AM 2 + MB 2 ( Pythagoras theorem ) In right ∆AMB, =

O 23 – x

Sol. Since, the angular bisector of the angle between two B equal chords of a circle passes through the centre 5 cm therefore, AO and so AM 6 cm O is the bisector of ∠BAC M and also is perpendicular A 6 cm bisector of chord BC.

From (ii),

( 23 − x ) cm.

14 = 1.4 cm 10

BM 2 = 25 − x 2 = 25 − (1.4 ) = 25 − 1.96 = 23.04

Join OA and OC. Then OA = OC = r cm.

⇒ BM =

In right ∆OMA,

Hence, length of the chord BC = 2BM = 2 × 4.8 = 9.6 cm.

OA 2 = AM 2 + OM 2 ⇒r 2 = 82 + x 2 (i) In right ∆ONC, 2

OC 2 = CN 2 + ON 2 ⇒r 2 = 152 + ( 23 − x ) (ii) From (i) and (ii), we have 82 + x 2 = 152 + ( 23 − x )

⇒ 64 + x 2 = 225 + 529 − 46x + x 2 ⇒ 46x = 754 − 64 ⇒ 46x= 690 ⇒ x=

690 = 15 cm. 46

From (i), r 2 =82 + 152 =64 + 225 =289 ⇒ = r

289 = 17

Hence, the radius of the circle is 17 cm.

23.04 = 4.8 cm

6. In the given figure, O is the centre of the circle. Determine ∠AQB and ∠AMB ,if PA and PB are 75 . tangents and ∠APB = Sol. In quadrilateral APBO, AO ⊥ PA and BO ⊥ PB [ radius is perpendicular to tangent at the point of contact] ∴ ∠ PAO = ∠ PBO = 90°  ∠ APB + ∠PAO + ∠PBO + ∠AOB = 360° [by angle sum property of a quadrilateral] ∴ 75° + 90° + 90° + ∠ AOB = 360°

∠ AOB = 105° 1 ∠ AQB = ∠ AOB 2

A P

75°

O B

270

[angle subtended by an arc at the centre is double of the angle on remaining part of the circle] 1 = × 105°= 52.5° 2 In cyclic quadrilateral AQBM, ∠ AQB + ∠ AMB = 180° [sum of opposite angles of a cyclic quadrilateral = 180°] 52.5° + ∠ AMB =° 180 ∠ AMB = 180° − 52.5° = 127.5° 7. As a part of a campaign, a huge balloon with message of “AWARENESS OF CANCER” was displayed from the terrace of a tall building. It was held by strings of length 8 m each, which inclined at an angle of 60° at the point, where it was tied as shown in the figure. (a) What is the length of AB? (b) If the perpendicular distance from the centre of the circle to the chord AB is 3 m, then find the radius of the circle. O

∴ AB = AP = BP = 8 m

(b) Given, OC = 3 m We know that if a perpendicular drawn from the centre of the circle to the chord, then it bisects the chord. ∴ AC =BC =AB =8 =4 m 2 2

In right angled ∆ACO ,

2 OA = AC 2 + OC 2

OA =

Which is the radius of the circle.

( 4 )2 + ( 3 )2 = 5 m

(c) Pythagoras theorem. 8. In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. Find the radius of the circle inscribed in the triangle. Sol. Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1, BP = BR (Tangent from B) CR = CQ (Tangent from C)

Sol. (a) Given, PA = PB = 8m

A BC is a right-angled triangle,

Clearly, PA and PB are tangents to the circle.

Now, draw OP which bisects ∠ APB and perpendicular to the chord AB. Thus, we have

∠ APC= ∠ BPC= 30°

A 60°

O C

8 m 30° 30° 8 m 60°

⇒ AC 2 = 25 + 144

In ∆ACP , ∠ APC + ∠ ACP + ∠ PAC = 180°

[∴sum of all angles of triangle is 180°]

30° + 90° + ∠ PAC = 180°

∠ PAC= 180° − 120°= 60°

Similarly, ∠ PBC= 60°

P B

O C

R 12 cm

2 AC = AB 2 + BC 2

⇒ AC 2 = 52 + 122

∠ ACP= ∠ BCP= 90°

5 cm

 by Pythagoras Theorem B

AP = AQ (Tangent from A)

Chapter 10 • Circles

60°

Thus, ∆ABP is an equilateral triangle.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining centre and the point of tangency. Property 3: Sum of all angles of a quadrilateral = 360°.

⇒ AC 2 = 169 ⇒ AC = 13 cm Clearly, AQ + QC = AC = 13 cm ⇒ AP + RC = 13 cm [ AQ = AP and QC = RC] Also, AB + BC = 5 cm + 12 cm = 17 cm

271

⇒ A P + PB + BR + RC = 17 cm [ AB = AP + PB and BC = BR + RC]

In ∆XQ S1 and ∆XO P1, ∠XS1Q = ∠XP1O = 90°

⇒ AP + RC + PB + BR = 17 cm

And ∠S1 XQ ~ ∠P1 XO

⇒ 13 cm + BR + BR = 17 cm [ AP + RC = 10 cm and PB = BR]

∆XQ S1 ~ ∆XO P1 XQ 5 = XQ + 5 + 80 + 15 15

⇒ 13 cm + 2BR = 17 cm ⇒ 2BR = 17 cm − 13 cm = 4 cm

3XQ = XQ + 100

⇒ BR = 2 cm

XQ = 50 cm

Now, ∠BPO = 90° [By property 2]

Length of OX = OQ + QX

∠BRO = 90° [By property 2]

= 100 + 150 = 150 cm

∠PBM = 90° [Given]

10. The figure shown below represents a circle with centre O and diameter 12 cm.

Now by property 3, ∠BPO + ∠BRO + ∠PBM + ∠ROP = 360°

In ∆DBA, ∠DBC = ∠BCD and ∠A = 50°.

⇒ ∠ROP = 360° − (∠BPO + ∠BRO + ∠PBM)

(a) What is the measure of ∠DCA?

⇒ ∠ROP = 360° − (90° + 90° + 90°)

(b) Dhruv said that, “The quadrilateral DBAC is a cyclic quadrilateral.” Is Dhruv correct? Give a reason to support your answer.

⇒ ∠ROP = 360° − 270° ⇒ ∠ROP = 90°

⇒ PO = BR = 2 cm

Hence, Radius = 2 cm

Sol. Given: ∠DBC = ∠BCD and ∠A = 50°

Now,  ∠ROP = 90° and BP = BR which are adjacent sides ∴ Quadrilateral PBRO is a square

C1 pulley

∠ BCD + ∠ DBC + 90= ° 180° (angle substended by diameter) 2∠ BCD= 90°

r2 Q

∠ BCD= 45°

∴∠ BCD= ∠ DBC= 45° (ii) Now, OB ⊥ AB

What is the length of OX when the diameter of C1 is 10 cm and length of OQ is 100 cm? 30 Sol. Radius of C1 = cm = 15 cm 2 10 Radius of C 2 = cm = 5 cm 2

∠OBA = 90° (iii) In quadrilateral ACDB, ∠ACD + ∠CDB + ∠DBA + ∠BAC = 360° ( ∠ACB + ∠BCD ) + 90° + ( ∠DBC + ∠CBA ) + 50° = 360° ∠ACB + 45° + 90° + 45° + 90° + 50° = 360 °

15 cm O

272

80 cm

5 cm C2

∠ACB = 40°

S1 Q S2

(i)

∠ BCD + ∠ DBC + ∠ CDB = 180°

S1 O

In ∆BCD,

belt

∠CDB = 90°

9. Ankit joins the centre of the two pulleys and observes line segments P1S1 and P2S2 when extended meet at a point X. P1

∴ ∠DCA = ∠BCD + ∠ACB = 45° + 40°= 85°

(b) No, Dhruv is not correct as the sum of the opposite angles of the quadrilateral DBAC is not 180°. Thus, the quadrilateral DBAC is not a cyclic quadrilateral. 12 (c) OB = = 6 cm 2

Also, D is an external point for circle with centre O.

∴CA = 2.5 × OB = 2.5 × 6 = 15 cm

In right-angled ∆ABC,

Hence proved.

2 AC = BC 2 + AB 2

= (15 )2 (12 )2 + AB 2

225 − 144 = AB 2

AB 2 = 81

AB = 9 cm

DA = DC (iv) So, from (i), (ii), (iii), and (iv), we get AB = BC = CD AB = CD

12. In given fig. the tangent at a point C of a circle and diameter when extended intersect at P. If ∠PCA = 110°, find ∠CBA. (NCERT Exemplar) C 110°

11. In the figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD. A

O C

Sol.

Sol. Join CO. Clearly, AB is a diameter of the circle. ∴∠ACB = 90° [Angle in a semi-circle]

⇒ ∠PCB + ∠ACB = 110°

But, ∠PCA = 110° [Given] ⇒ ∠PCB + 90° = 110° ⇒ ∠PCB = 20°

B O

O' D

Proof: We know that the lengths of tangents from an external point to a circle are equal. So, if A is an external point for circle having centre O', then

Since PC is tangent to the circle at C and OC is the radius through C. Therefore, OC ⊥ PC ⇒∠OCP = 90° ⇒ ∠OCB + ∠PCB = 90° ⇒ ∠OCB + 20° = 90° ⇒ ∠OCB = 70° (i) In ∆OCB, we have OC = OB [Radii of the same circle]

AB = AD (i)

⇒ ∠OBC = ∠OCB

If C is an external point for circle with centre O', then

⇒ ∠OBC = 70° [Using (i)]

CB = CD (ii)

⇒ ∠CBA = 70° [∠OBC = ∠CBA]

Also, B is an external point for circle with centre O. BA = BC (iii)

Chapter 10 • Circles

273

13. In the given figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30° (CBSE 2016) Topper’s Answer

Competency Based Questions Multiple Choice Questions

Which of the following statements is true? (a) There can be only one line passing through point M such that it is parallel to OM.

1. In given figure, the perimeter of ∆ABC is A

(a) 30 cm

(b) There can be only one line passing through point M such that it is perpendicular to OM.

4 cm

(b) 60 cm (c) 45 cm

R 6 cm

(d) 15 cm

P 5 cm

2. A circle with centre O is shown below. A

(c) There can be any number of lines passing through point M such that they are parallel to OM. (d) There can be any number of lines passing through point M such that they are perpendicular to OM. 3. In given figure, equal circles with centres O and O ′ touch each other at P . OO ′is produced to meet circle

274

C (O ′, r ) at A. AT is a tangent to the circle C (O , r ). AQ = If O ′Q is perpendicular to AT , then AT

(i) OQ and OR are at an angle of 90°.

T A

(ii) The tangents meet at an angle of 90°.

O’

(a)

2 3

(b)

1 2

(c)

1 4

(d)

1 3

(a) AD = AB + BC + CA (c) 3AD = AB + BC + CA

C A

(d) 4AD = AB + BC + CA

5. In a triangle DEF, O is the centre of the in circle ABC. ∠ DEF = 60, ∠ DFE = 75. Find ∠ AOB . D

(a) 75° A

60°

75°

6. In given figure, AB and CD are two chords of a circle intersecting at P. Choose the correct statement from the following: (a) ∆ADP ∼ ∆CBA D B O

(b) ∆ADP ∼ ∆BPC (c) ∆ADP ∼ ∆BCP

P C

(d) ∆ADP ∼ ∆CBP

7. In given fiure, if quadrilateral PQRS circumscribes a circle, then P B Q

 (a) = x 95 = , y 95

 = (b) x 100 = , y 90

 (d) = x 85 = , y 90

80° 95° O y

 = (c) x 100 = , y 85

Select Response Type MCQs 1. A circle has a centre O and radii OQ and OR. Two tangents, PQ and PR, are drawn from an external point, P. In addition to the above information, which of these must also be known to conclude that the quadrilateral PQOR is a square? Chapter 10 • Circles

(d) both (i) and (ii)

K (ii) The shortest distance of a tangent passing through point L from the centre O is equal to the radius of the circle, OL.

(a) (i) and (ii)

(b) (iii)

(d) (i), (ii) and (iii)

O E

(iii) One tangent can pass through two points K and L of a circle. Which statement(s) is/are correct?

(b) 45°

(b) only (ii)

2. A circle is shown below. (i) There is only one line passing through point K which makes an angle of 90° with OK.

4. In given figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then, (b) 2AD = AB + BC + CA

(a) only (i)

(b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): A tangent to a circle is perpendicular to the radius through the point of contact. Reason (R): The length of tangents drawn from an external point to a circle are equal. (CBSE 2023) 2. Assertion (A): If PA and PB are tangents drawn from an external point P to a circle O, then the quadrilateral AOBP is cyclic. Reason (R): The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. (CBSE 2023)

Case Study Based Questions 1. The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anti-clockwise around one and a half times through a circle, then releases the throw. When released, the discus travels along tangent to the circular spin orbit. 275

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and ∠ABO = 30°, PQ is parallel to OA.

rotating upright wheel with multiple passengercarrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity. After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.

(CBSE QB)

S R

Based on the above information answer the following questions. (i) Find the length of AB.

S O

(ii) Find the length of OB.

30°

(iii) (a) Find the length of AP.

30°

(a) In the given figure, find ∠ROQ.

OR (b) Find the length of PQ.

(i) 60°

(ii) 100°

(iii) 150°

(iv) 90°

(iii) 30°

(iv) 90°

(iii) 100°

(iv) 30°

(iii) 100°

(iv) 60°

(b) Find ∠RQP. O

75 cm A

30° P

2. A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a

(i) 75°

(ii) 60°

(i) 60°

(ii) 75°

(d) Find ∠ORP.

(i) 90°

(ii) 70°

Answers Multiple Choice Questions 1. (a) 30 cm

In ∆AO ′D and ∆AOC , ∠DAO ′ = ∠CAO (Common angle)

AR = AQ = 4 cm

∠ADO ′ = ∠ACO (both measure 90)

BP = BR = 6 cm

∴∆ADO’ ∼ ∆ACO......(By AA test of similarity)

CQ = CP = 5 cm

Now, perimeter = 2 ( 4 + 5 + 6 ) = 30 cm

2. (b) There can be only one line passing through point M such that it is perpendicular to OM. 3. (d)

1 3

AO ′ = O ′X = XO = OC (Since the two circles are equal.)

So, OA = AO ′ + O ′X + XO …. ( A − O ′ − X − O )

∴OA = 3O ′A

276

DO ′ O ′ A O ′ A 1 ∴ = = = CO ′ OA 3O ′ A 3 4. (c) 2AD = AB + BC + CA Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on the circle) are equal. By the above property, AE = AD (tangents from A) AB = AC (tangents from A)

CD = CF (tangents from C)

 7.= (c) x 100 = , y 85

BF = BE (tangents from B)

Given that the quadrilateral circumscribes a circle, the sum of opposite angles is 180°.

Now adding the above equations, AB + BC + CA = AB + BF + FC + CA

For angle x : x + 80° = 180°. Solving for x, we get x = 180° − 80° = 100°.

AB + BC + CA = AB + BE + CD + CA AB + BC + CA = AE + AD [AE = AB + BE and AD = AC + CD]

For angle y : y + 95° = 180°. Solving for y, we get y = 180° − 95° = 85°.

AB + BC + CA = AD + AD [AD = AE]

Select Response Type MCQs

AB + BC + CA = 2AD

1. (c) either (i) or (ii)

Hence, 2AD = AB + BC + CA 5. Given, AB is chord of the circle, AOC is the diameter of the circle. Also, ∠ACB = 50°, AT is the tangent of the circle. We know that the angle in a semicircle is a right angle. So, ∠CBA = 90° We know that the sum of all three interior angles of a triangle is always equal to 180° In triangle AOB,

Assertion-Reason Based Questions 1. (b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. 2. (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Case Study Based Questions 1. (a) Since OA is radius and BA is tangent to the circle at point A.

∠CAB + ∠CBA + ∠ACB = 180° ∠CAB + 90° + 50° = 180°

∴∠OAB = 90°

∠CAB + 140° = 180°

In ∆OAB,

∠CAB = 180° − 140° ∠CAB = 40°

2. (a) (i) and (ii)

(1)

OA 1 75 ⇒ = ⇒ AB = 75 3 cm OB AB 3 = 75 × 1.73 cm = 129.75 cm

tan 30° =

We know that radius of circle is perpendicular to the tangent at the point of contact.

(b) In ∆OAB, we have

From the figure,

sin 30° =

OA ⊥ OT

OA 1 75 ⇒ = OB 2 OB

So, ∠CAB + ∠BAT = 90°

⇒ OB = 75 × 2 cm = 150 cm

Substitute (1) in the above expression,

40° + ∠BAT = 90°

∠BAT = 90°− 40°

Now, OB = 150 cm

∠BAT = 50°

⇒ OQ + QB = 150 cm

⇒ 75 + QB = 150 cm

⇒ QB = 150 − 75 = 75 cm

∴ In right ∆PBQ,

cos 30° =

∴ AP = AB − PB = 75 3 −

Therefore, the measure of the angle BAT is 50°. Hence, option c is correct. 6. (d) ∆ADP ∼ ∆CBP

∠APD = ∠CPB [vertically opposite angle]

∠ADP = ∠CBP [Angle subtends on the same segment]

∴ AA similarity

∆ADP ∼ ∆CBP

Chapter 10 • Circles

PB 3 PB 75 3 ⇒ = ⇒ PB = cm QB 2 75 2

75 3 2 75 3 75 × 1.73 ⇒ AP = cm = = 64.875 cm 2 2

277

⇒ 2∠OQR = 180° − 150°

(ii) In right ∆QPB,

⇒ 2∠OQR = 30°

PQ 1 PQ 75 sin 30° = ⇒ = ⇒ PQ = cm = QB 2 75 2 37.5 cm

⇒ ∠OQR = 15°

Again,

∠OQP = 90° (∴ OQ ⊥ QP)

⇒∠OQR + ∠RQP = 90° − 15° = 75°

2. (a) (iii) In quadrilateral PQOR, we have

∠QPR +∠PRO + ∠PQO + ROQ = 360°

⇒ 30° + 90° + 90° + ∠ROQ = 360°

(c) (ii) we know that the angle subtended by an arc at centre is double the angle subtended by it at any other part of the circle.

[∴ radius is always perpendicular to the tangent at point of contact]

∴ 2∠RSQ = ∠RSQ 1 ⇒ ∠RSQ = ×150 ° = 75° 2 (d) (i) ∠ORP = 90° as radius is always perpendicular to the tangent at the point of contact.

⇒∠ROQ = 360° − 210° = 150°

(b) (i) In ∆QOR,

OQ = OR (radii)

∴ ∠ORQ = ∠OQR

Now, ∠ROQ + ∠ORQ + ∠OQR = 180°

Additional Practice Questions 1. In the adjoining figure, PA and PB are tangents to a circle with centre O such that ∠ P = 90.

(a) 3.8 O

If AB = 3 2 cm, then the diameter of the circle is:

(a) 3 2 cm

(b) 6 2 cm

(d) 6 cm

(CBSE 2025)

2. In the adjoining figure, TS is a tangent to a circle with centre O. The value of 2x° is: (a) 22.5 (b) 45 (c) 67.5 (d) 90

(d) 1.9 5. In figure, B and CD are two chords of a circle intersecting at P. Choose the correct statement from the following: (a) ∆ADP ∼ ∆CBA D

x°

6. In the figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If ∠QPR = 90°, then length of PQ is: Q P

4 cm

O R

(d) ∆ADP ∼ ∆CBP 3x°

P R

4. In figure,QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If T = 3.8 cm, then the length of QR (in cm) is 278

(b) ∆ADP ∼ ∆BPC

3. In the given figure, the quadrilateral PQRS circumscribes a circle. Here PA + CS is equal to (a) QR P (b) PR

(b) 7.6

(a) 3 cm

(b) 4 cm

(d) 2 2 cm

7. In the given figure, O is the centre of the circle. If ∠ AOB = 145, then find the value of x.

9. AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. [NCERT Exemplar]

10. The common tangent AB and CD to two circles with centres O and O' intersect at E between their centres. Prove that the points O, E and O' are collinear. (NCERT Exemplar)

O 145°

x C

8. In the given figure, tangents PQ and PR are drawn to a circle such that ∠ RPQ = 30. A chord RS is drawn parallel to the tangent PQ. Find the measure of ∠RQS. (CBSE 2023) S

1 2

D 3 4

O' B

R O 30°

Challenge Yourself

Scan me

1. In the given figure, AT and BT are the two tangents at A and B respectively. CDCDCD is also a tangent at P. A

C P

There are some more circles touching each other and the tangents AT and BT also. Which one of the following is true? (a) PC + CT = PD + DT (b) RG + GT = RH + HT (c) PC + QE = CE

(b) a + b

(d) a − b Chapter 10 • Circles

(Hint: Cyclic quadrilateral four corners touches the circles circumference. In cyclic quadrilateral sum of opposite angles are 180°) 4. AB is a diameter of circle. P is a point on the semi-circle APB. AH and BK are perpendiculars from A and B respectively to the tangent at P. Prove that AH + BK = AB. H

c S

r B

R x N

b–x B d T O’ b+d

5. In the given figure, ∆ABC is equilateral, P and S are mid-points of arcs AB and AC. Prove that PQ = QR = RS

M x

O a+c–x

P r

(d) All of these

2. In the given figure, PA is tangent to semi-circle SAR. PB is tangent to semi-circle RBT; SRT is a straight line, the lengths of the arcs are indicated in the figure. Angle ∠ APB is measured by 1 P (a) ( a − b ) A 2 1 (c) ( a + b ) 2

3. Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

A P

279

Answers Additional Practice Questions

Challenge Yourself

1. (d) 6 cm

1. (d) All of these

2. (b) 45

2. (b) a + b

3. (c) PS

3. 4 cm

4. (b) 7.6 5. (d) ∆ADP ∼ ∆CBP 6. (b) 4 cm 7. 107.5° 8. 30°

Scan me for Exemplar Solutions 280

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions 1. A line intersecting a circle in two distinct points is called a (a) secant (b) chord (c) diameter

[3 × 1 = 3 Marks]

(d) tangent A

2. If perimeter of given triangle is 38 cm, then length AP is equal to (a) 19 cm (b) 5 cm

8 cm C

6 cm R

3. In figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB. P (a) 4

(b) 7 A

B O

(d) 6

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(b) Both A and R are true, but R is not the correct explanation of A.

4. Assertion (A): If PA and PB are tangents drawn to a circle with centre O such that ∠AOB =110° then ∠APB = 90°.

Reason (R): The length of two tangents drawn from an external point are equal. 5. Assertion (A): In the given figure, O is the centre of a circle and AT is a tangent at point A, then ∠ BAT = 60

60° B

Reason (R): A straight line can meet a circle at one point only. A

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP. A 7. In the given figure, PA and PB are tangents from P to a circle O. If ∠AOB = 130°, then find ∠APB.

130°

B Self-Assessment

281

Short Answer Questions

[4 × 3 = 12 Marks]

8. In two concentric circles, the radii are OA = r cm and OQ = 6 cm, as shown in the figure. Chord CD of larger circle is a tangent to smaller circle at Q. PA is tangent to larger circle. If PA = 16 cm and OP = 20 cm, the length CD. 9. (a) In the given figure, PA and PB are tangents to a circle centred at O. Prove that (i) OP bisects ∠APB

O C

T P

11. In the figure, PQ is tangent to a circle with centre O. If ∆OAB = 30°, find ∠ABP and ∠AOB.

30°

Q B

10. In the given figure, from a point P, two tangents PT and PS are drawn to a circle with centre O such that ∠SPT = 120°, prove that OP = 2 PS.

Q A

(ii) OP is the right bisector of AB

Long Answer Questions

[3 × 5 = 15 Marks]

12. In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the circle. If ∠APB = 70°, find ∠ACB.

A P 70°

O B

13. In figure, the common tangent, AB and CD of two circles with centres O and O' intersect at E. Prove that the points O, E, O' collinear.

O E

14. In figure, O is the centre of circle of radius 5 cm. T is the point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are tangents to the circle.

C P

5 O

E 13

[1 × 4 = 4 Marks]

15. The picture given below shows a circular mirror hanging on the wall with a cord. The diagram represents the mirror as a circle with centre O. AP and AQ are tangents to the circle at P and Q respectively such that AP = 30 cm and ∠PAQ = 60°. A Based on the above information, answer the questions: (a) Find the length PQ. (b) Find m ∠POQ. (c) (i) Find the length OA. OR (ii) Find the radius of the mirror. 282

Case Study Based Questions

Q O

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. In figure, QR is a common tangent to the given circles, touching externally at the point T. The tangent at T meets QR at P, if PT = 3.8 cm, then the length of QR (in cm) is: (a) 3.8

P T

(b) 7.6 (c) 5.7 (d) 1.9 2. In the given figure, if ∠AOB = 125°, then ∠COD is equal to: (a) 62.5°

A B

125°

(b) 45°

3. If angle between two radii of a circle is 130°, the angle between the tangent at the ends of the radii is (a) 90° (b) 50° (c) 70° (d) 40°

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): In the given figure, a quadrilateral ABCD is drawn to circumscribe a given circle, as shown. Then AB + BC = AD + DC Reason (R): In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

5. Assertion (A): In figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50 with PQ, then ∠ POQ = 100

Reason (R): A tangent to a circle is perpendicular to the radius through the point of contact.

Very Short Answer Questions 6. In figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.

[2 × 2 = 4 Marks]

T R

70° Q

Self-Assessment

283

7. In given figure, if AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°, then find the length of AT (in cm).

O 30°

Short Answer Questions

[4 × 3 = 12 Marks]

8. A circle is inscribed in a ∆ABC having sides AB = 14 cm, BC = 10 cm and AC = 12 cm in figure. Find AD, BE and CF .

C E

9. In figure, PQL and PRM are tangents to the circle with centre O at the points Q and R , respectively and S is a point on the circle such that ∠ SQL = 50 and ∠ SRM = 60. Find ∠ QSR .

L S

Q O

10. In figure, PQ and RS are parallel tangents to a circle with center O and another tangent AB will point of contact C intersecting PQ at A and RS at B. prove that ∠AOB = 90°.

A Q

D O

C E

11. AB is a chord of circle with centre O. At B, a tangent PB is drawn such that its length is 24 cm. The distance of P from the centre is 26 cm. If the chord AB is 16 cm, find its distance from the centre.

O A

Long Answer Questions 12. In the given figure, AD is a diameter of a circle with centre O and AB is a tangent at A. C is a point on the circle such that DC produced intersects the tangent at B and ∠ABD = 50°, find ∠COA.

[3 × 5 = 15 Marks]

50°

C D

13. As a part of a campaign, a huge balloon with message of “AWARENESS OF CANCER” was displayed from the terrace of a tall building. It was held by strings of length 8 m each, which inclined at an angle of 60° at the point, where it was tied as shown in the figure. (a) What is the length of AB? (b) If the perpendicular distance from the centre of the circle to the chord AB is 3 m, then find the radius of the circle. (c) Which method should be applied to find the radius of the circle?

O A

B 60° P

284

14. A ∆ABC is drawn to circumscribe a circle of radius 2 cm such that the segment BD and DC into which BC is divided by the point of contact D are of lengths of 4 cm and 3 cm respectively. If area of ∆ABC is 21 cm2, find the length of AB and AC.

A O C 3 cm

Case Study Based Questions

B 4 cm

[1 × 4 = 4 Marks]

15. Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a ∆ABC, such that it touches the sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively. C C AB

F A

(a) Find the length of AD.

(i) 7

E D

(ii) 8

(iii) 5

(iv) 9

(ii) 5

(iii) 2

(iv) 9

(ii) 5

(iii) 2

(iv) 3

(iii) 24

(iv) 48

(iii) 100

(iv) 90

(b) Find the length of BE.

(i) 8

(i) 9

(d) If radius of the circle is 4 cm, find the area of ∆OAB.

(i) 20

(ii) 36 OR

(e) Find the area of ∆ABC

(i) 50

Self-Assessment

(ii) 60

285

Areas Related to Circles

Chapter at a Glance Circle: A circle is a plane figure defined as the set of all points in a plane that are equidistant from a fixed point called the centre. Radius: The constant distance from the centre of the circle to any point on the circle. It is also the line segment from the centre to the circle’s edge.

B Radius r

Area of a Circle: The area A of a circle of radius r is given by the formula: A = π r 2 Example: Find the Area of a Circle whose radius is 10 cm. (Use π = 3.14 ). A = π= r 2 3.14 × 102 = 314 cm 2

Minor and Major Sectors ●

●

Minor Sector: The smaller of the two sectors formed when a circle is divided by two radii. It is the sector where the central angle is less than 180°.

Major Sector

Major Sector: The larger sector when a circle is divided by two radii. This sector has a central angle greater than 180°.

Area of a Sector: The area A of a sector of angle θ in degrees in a circle of radius r is:

= A

360

r A

× πr 2

Area of sector

P Minor Sector

Example: Find the area of a sector of a circle of radius 28 cm and central angle 45°. θ 45 1 22 × π rA2 = A = × π × (28)2 = × × 28 × 28 = 308 cm 2 360 360 8 7

Minor and Major Segments ●

●

Minor Segment: The smaller of the two segments formed when a circle is divided by a chord (not passing through the centre). It corresponds to the minor sector. Major Segment: The larger of the two segments formed when a circle is divided by a chord. It corresponds to the major sector. rea of a Segment: The area of a segment in a circle can be calculated by subtracting the area A of the corresponding triangle from the area of the sector.

Major Segment O r A

r 2

M P

Area of segment = APB area of sector OAPB − area ( ∆OAB )

Minor Segment

Perimeter of Circular Figures Circumference of a Circle: The circumference C of a circle of radius r is given by: C = 2π r Length of an Arc: The length L of an arc of a sector with central angle θ in a circle of radius r is: = l

360

× 2π r

Example: Find the length of the arc when radius of circle 35 cm and central angle 60°. (Use π = 3.14). 1 θ 60 × 219.8 = 36.63 cm = × 2π r= l l × 2 × 3.14 × 35= 6 360 360

286

NCERT Zone NCERT Exercise 11.1 1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area of sector OACB

360°

Area of sector of angle θ =

× πr 2

60°

60° 22 2 1 22 132 = × (6 ) = × × 6 × 6 = cm 2 360° 7 6 7 7 Therefore, the area of the sector of the circle making 132 60° at the centre of the circle is cm2. 7 2. Find the area of a quadrant of a circle whose circumference is 22 cm. Sol. Let the radius of the circle be r. Circumference = 22 cm 2π r = 22 22 11 = r = 2π π Quadrant of circle will subtend 90° angle at the centre of the circle Area of such quadrant of the 90° × π ×r 2 circle = 360°

10 cm

Minor Segment

Sol. (a) Area of the minor sector OACB C 90° = × 3.11 × b 2= 78.5 cm 2 360° 1 1 Area of ∆OAB = × OA × OB = × 10 × 10 =50 cm 2 2 2 Area of the minor segment ACB = Area of minor sector OACB – Area of ∆OAB 2 = 78.5 − 50= 28.5 cm

(a) The length of the arc

3. The length of the minute hand of a 12 11 1 clock is 14 cm. Find the area swept 10 by the minute hand in 9 O 30° 5 minutes. 8

Chapter 11 • Areas Related to Circles

270°

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1  11  121 121 × 7 77 cm 2 = ×π ×  = = =  8 4 4π 4 × 22 π 

In 5 minutes, minute hand will rotate 360° = × 5 = 30° 60

(a) Minor segment (b) Major sector [Use π = 3.14] A

(b) Area of the major sector OADB 360° − 90°   270°  2 2 =   × π r=   × πr  360°   360°  3 = × 3.14 × 10 × 10 4 = 235.5 cm 2

90°

Sol. We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.

× πr 2 360° 30° 22 Area of sector of 30°= × × 14 × 14 360° 4 22 154 = × 2 × 14 = cm 2 12 3 154 Therefore, required area is cm2. 3 4. A chord of a circle of radius D 10 cm subtends a right angle at Major Sector the centre. Find the area of the corresponding: Area of sector of angle θ =

Sol. Let OACB be a sector of the circle making 60° angle at centre O of the circle. Therefore, the area of the sector of the circle making 60° at the centre 132 of the circle is cm2. 7

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.

(b) Area of the sector formed by the arc 2 3 4

60° 22 × 2 × × 21 360° 7 = 22 cm

360°

× 2π r O 21 cm A

21 cm

60°

D C

287

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the O 12 cm 12 cm corresponding segment of the 60° 60° circle. [Use π = 3.14 and 3 = 1.73] S T V

60° × πr 2 360° 1 22 = × × 21 × 21 6 7 = 231 cm 2

(a) Area of Sector OACB = (b) In ∆OAB,

∠OAB = ∠OBA (As OA = OB )

∠OAB + ∠AOB + ∠OBA = 180°

2∠OAB + 60= ° 180°

∠OAB = 60°

Let us draw a perpendicular OV on chord ST.

SV = VT (On is Perpendicular Bisector of ST) OV = cos 60° OS OV 1 = ⇒ OV = 6 cm 12 2

Therefore, Δ OAB is an equilateral triangle. 3 2 × ( side ) 4 3 441 3 2 = × ( 21) = cm 2 4 4 Area of segment ACB = Area of sector OACB − Area of ΔOAB  441 3  2  cm =  231 − 4  

SV SV 3 ⇒ OV = 6 3 cm = sin60° ⇒ = SO 12 2

Area of ΔOAB =

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. [Use π = 3.14 and 3 = 1.73] Sol. Radius (r) of circle = 15 cm Area of sector OPRQ 60° × πr 2 = 360° 1 2 =× 3.14 × (15 ) = 117.75 cm 2 6 In ΔOPQ, ∠OPQ = ∠OQP

O 15 cm P

60° 15 cm

∠OPQ + ∠OQP + ∠POQ = 180° 2 ∠OPQ = 120° ⇒ ∠OPQ = 60° ΔOPQ is an equilateral triangle. Area of ΔOPQ = 3 × ( side )2 4 3 2 225 3 cm cm22= 97.3125 cm2 = × (15 ) = 4 4 Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ = 117.75 − 97.3125 = 20.4375 cm2 Area of major segment PSQ = Area of circle − Area of segment PRQ 2 = π (15 ) − 20.4375

288

ST = 2SV = 2×6 3 = 12 3 cm 1 Area of ΔOST = × ST × OV 2 1 = × 12 3 × 6 = 36 3 2 = 36 × 1.73 = 62.28 cm 2 120° 2 Area of sector OSUT = × π (12 ) 360° 1 =× 3.14 × 144 = 150.72 cm 2 3 Area of segment SUT = Area of sector OSUT − Area of ΔOST = 150.72 − 62.28 = 88.44 cm2

(As OP = OQ )

= 3.14 = 706.5 × 225 − 20.4375 = 686.0625 cm 2

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see the given figure). Find: (a) The area of that part of the field in which the horse can graze. (b) The increase in the grazing area if the rope was 10 m long instead of 5 m. [Use π = 3.14]

(CBSE 2023)

Sol. From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius. 90° (a) Area that can be grazed by horse = × πr 2 360° 1 2 =× 3.14 × ( 5 ) = 19.625 m 2 4 (b) Area that can be grazed by the horse when length of rope is 10 m long.

90° 1 2 × π r 2 =× 3.14 × (10 ) = 78.5 m 2 360° 4 Increase in grazing area = (78.5 − 19.625) m2

= 58.875 m2

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find. (a) The total length of the silver wire required. (b) The area of each sector of the 22  brooch Use π = 7   Sol. The total length of wire required will be the length of 5 diameters and the circumference of the brooch. 35 mm Radius of circle = 2 Circumference of the brooch = 2π r 22 35 =2 × × = 110 mm 7 2 (a) Length of the wire required= 110 + 5 × 35 = 110 + 175 = 285 mm 360° Each sector angle is = 36°. 10° 36° (c) Therefore, area of each sector = × πr 2 360° 1 22 35 35 385 = × × × mm 2 = 10 7 2 2 4 10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two 22  consecutive ribs of the umbrella. Use π = 7  

Sol.

115°

25 cm

It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius. 115° 2 × π × ( 25 ) Area of such sector = 360° 23 22 158125 × × 25 × 25 = cm 2 72 7 252 158125 158125 Area swept by 2 blades= 2 × cm 2 = 252 126

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships warned. [Use π = 3.14]

80° × πr 2 A Sol. Area of sector OACB = 360° 2 C =× 3.14 × 16.5 × 16.5 9 = 189.97 km 2 B

45° 1 22 11 2 22275 × πr 2 = × = × (2025 cm 2 45 ) = 28 360° 8 28 7

11. A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each 22  sweep of the blades. Use π = 7   Chapter 11 • Areas Related to Circles

80°

Light O house

Sol. Consider segment APB. Chord AB is a side of a 360° hexagon. Each chord will substitute = 60° at 6 the centre of the circle. ∠OAB = ∠OBA

Area between two consecutive ribs of circle

16.

13. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ` 0.35 per cm2. Use 3 = 1.7 

In ∆OAB ,

Sol. There are 8 ribs in an umbrella. The area between 360° two consecutive ribs is subtending = 45 ° at the 8 centre of the assumed flat circle.

25 cm

115°

(As OA = OB)

∠AOB = 60° ∠OAB + ∠OBA + ∠AOB = 180 ° 2 ∠OAB = 120°

P 28

C D

∠OAB = 60° ΔOAB is an equilateral triangle. 3 × ( side= )2 3 × ( 28 )2 = 333.2 cm 2 4 4 60° × πr 2 Area of sector OAPB = 360° 1 22 1232 = × × 28 × 28 = cm 2 6 7 3 Area of ΔOAB =

289

Area of segment APB = Area of sector OAPB − Area of ΔOAB  1232  =  − 333.2  cm 2  3   1232  − 333.2  cm 2 Therefore area of design = 6 ×   3  = 464.8 cm 2 = 2464 − 1999.2 Cost of making 464.76 cm2 designs = 464.8 × 0.35 = `162.68

14. Tick the correct sol in the following: Area of a sector of angle p (in degrees) of a circle with radius R is p p ×πR2 (a) (b) × 2π R 180 180 p p × 2π R (c) (d) × 2π R 2 360 720

×πR2 360° p p Area of sector of angle P = × π= R2 × 2π R 2 360° 720° Hence, (d) is the correct answer. Sol. We know that area of sector of angle θ =

Multiple Choice Questions 1. Perimeter of a sector of a circle whose central angle is 90° and radius 7 cm is (a) 35 cm (b) 11 cm (c) 22 cm (d) 25 cm (CBSE 2024)

7 2. If the area of a sector of a circle is of the area of 20 the circle, then the angle at the centre is equal to (a) 110° (b) 130° (c) 100° (d) 126°

(CBSE 2024)

3. OACB is a quadrant of a circle with centre O and radius 7 cm, where ACB is the arc. Then, the perimeter of the quadrant is (a) 15 cm (b) 50 cm (c) 25 cm (d) 44 cm

(CBSE Compartment 2023)

4. If the radius of the circle with centre O is 7 cm and the measure of ∠AOB is 100°, what is the best approximation to the length of arc AB ? 22    Take π =  7   (a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm 5. If the circumference and the area of a circle are numerically equal, then diameter of the circle is (a)

(b) 2π

(d) 7 (CBSE SQP 2024)

6. If the difference between the circumference and 22 radius of a circle is 37 cm, then using π = , the 7 circumference (in cm) of the circle is (a) 154 (b) 44 (c) 14 (d) 7 (CBSE 2013)

7. If a wire is bent into the shape of a square, then the area of the square is 81 cm². When wire is bent into a semi-circular shape, then the area of the semi-circle will be (a) 22 cm² (b) 44 cm² (c) 77 cm² 290

(d) 154 cm²

8. A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is (a) 20 m (b) 21 m (c) 22 m (d) 24 m 9. The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be (a) 2800 (b) 4000 (c) 5500 (d) 7000 10. The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is (a) 55 m (b) 110 m (c) 220 m (d) 230 m 11. The circumference of a circle is 100 cm. The side of a square inscribed in the circle is 100 (a) 50 2 cm (b) cm (c) 50

2 cm

(d) 100 2 cm

12. The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of the three concentric circles drawn is: (a) 10 5 cm (b) 10 7 cm (c) 10 3 cm

(d) 10 2 cm

13. The area of a sector whose perimeter is four times its radius r units, is: (a)

r2 sq. units 4

(b) 2r 2 sq. units (d)

r2 sq. units 2

14. If a chord of a circle of radius 28 cm makes an angle of 90° at the centre, then the area of the major segment is: (a) 392 cm² (b) 1456 cm² (c) 1848 cm²

(d) 2240 cm²

15. If a circle inscribed in an equilateral triangle is 48π square units, then the perimeter of the triangle is: (a) 17 3 units (b) 36 units (c) 72 units

(d) 48 3 units

Answers 1. (d) 25 cm

2. (d) 126°

9. (d) 7000

3. (c) 25 cm

4. (d) 12 cm

11. (c)

5. (c) 4

6. (b) 44

13. (c) r ² sq. units

7. (c) 77 cm²

8. (b) 21 m

15. (c) 72 units

50 2

10. (c) 220 m cm

12. (c) 10 3 cm 14. (d) 2240 cm²

Constructed Response Questions Very Short Answer Questions 1. Find the area of the sector of a circle of radius 6 cm whose central angle is 30 (Take π = 3.14 ) (CBSE 2020)

Sol. Given, radius of a circle, = OB = 6 cm and central angle θ = OA ∠AOB = 30° By using formula, area of the sector of a circle

2. A race track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track.

6 cm

= × πr 2 360 30 = × 3.14 × 6 × 6 = 9.42 cm 2  360

30°

30   22 =  × 18 × 18 ×  ×7 360   7

= 594 cm 2 . 5. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its 22 centre. Find the radius of the circle. Use π = 7

(CBSE 2020)

Sol. AB is an arc of a circle. Let radius of a circle be r cm. = AB 22 = cm, θ 60 ,

r O R

Sol. 2π R 396 = = and 2π r 352 22 22 2× = × R 396 and 2 × = × r 352 7 7 7 1 7 1 R= 396 × × = 63 m and r= 352 × × = 56 m 22 2 22 2 Hence, the width of the track: 7 metres ( R − r ) metres = ( 63 − 56 ) metres =

3. Find the length of the arc of a circle which subtends an angle of 60 at the centre of the circle of radius 42 cm. (CBSE 2024) 22 60° Sol. Length of arc =2 × × 42 × = 44 cm 7 360°

Chapter 11 • Areas Related to Circles

4. The minute hand of a wall clock is 18 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes. (CBSE SQP 2025) Sol. Angle described by minute hand in 5 min = 30. length of minute hand = 18 = cm r . Area swept by minute hand in 35 minutes

Length of= arc

O 60°

× 2π r 360 A 22 cm 60 22 22 = × 2 × ×r 7 360 22 × r 22 = 21 22 × 21 = 22 × r r = 21 cm Hence, the radius of the circle (r) is 21 cm. 6. Find the perimeter of the shaded region if ABCD is a square of side 21 cm and APB and CPD are 22   semicircles  Use π =  7  

(CBSE SQP 2016)

21 cm

291

Sol. Since APB and DPC are two semicircles. 21 radius (r ) = cm 2 and length of arc APB + length of arc DPC = ( 2π r )

1 × 6 × 6 = 18 cm 2 2 So, the area of the shaded segment

The area of ∆MON =

= area of sector-area of triangle

22 21 × = 22 × 3 = 66 cm 7 2 Total perimeter of the shaded region =2 ×

= ( 9π − 18 ) cm 2 or 9 (π − 2 ) cm2

= AD + BC + lengths of the arcs ( APB + DPC ) = ( 21 + 21 + 66 ) cm = 108 cm. 7. The minute hand of a clock is 14 cm long. Find the area on the face of the clock described by the minute hand in 5 minutes. (CBSE 2024) CBSE Marking Scheme 2024 Sol.

Angle subtended in 5 min = 30



1 2

Area described by minute hand 30 22 = × × 14 × 14 360 7

154 cm 2 or 51.33 cm 2 approx. = 3

1 2

Sol. Diameter of the circle = diagonal of square = 2 × 8 = 16 cm

1 2

Let x be the side of square.

1 2

∴ x 2= 16 or, x= 8 2 2 Area of square = x= (8 2)2 = 128 cm 2 9. A semicircle MON is M N inscribed in another semicircle. Radius OL of the larger semicircle is K O 6 cm. Find the area of the shaded segment in terms of π. Draw a rough figure and show your steps.

Sol. Side of square = 50 cm Therefore, Area of square = 50 × 50 = 2500 cm²

21 cm

Radius of quadrant = 15 cm

8. Find the area of the square that can be inscribed in a circle of radius 8 cm. (CBSE 2015) CBSE Marking Scheme 2015

10. A child prepares a poster on “save water” on a square sheet whose each side measures 50 cm. At each corner of the sheet, she draws a quadrant of radius 15 cm in which she shows the ways to save water. At the centre, she draws a circle of diameter 21 cm and writes a slogan save water in it. Find the area of the remaining sheet. (CBSE 2015)

Area of 4 quadrants 1 = 4 × πr 2 = π r 2 = π × 15 × 15 4 22 = × 225 = 707.14 cm² 7 Area of circle = π r 2

15 cm A

22  21  22 × 441 = × = = 346.5 cm 2   7  2  28 asinθ + bcosθ Area of remainingasheet = Area of square – 4 (area of sinθ − bcos θ quadrant) – Area of circle= = 2500 − 707.14 − 346.5 = 1446.36 sq cm

Short Answer Questions

1. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 m × 16 m. Find the area of the field in which the cow can graze. (NCERT Exemplar) C

16 m

D Sol. Let ABCD be a rectangular L field of dimensions G 20 m × 16 m . Suppose, a E cow is tied at a point A. m 14 Let length of rope be (CBSE PP 2023) F A = AE 14 = m r (say).  20 m Sol. Join O with M and N such that M ∠ MON = 90 , asNshown in figure below. ∴ Area of the field in which the cow graze with M and N such that ∠ MON = 90 , as shown in figure below. θ 90 2 = Area of sector AFEG = × π= r2 × π (14 ) ∠ MON = 90 , as shown in figure below. ° 360 360 L K O 1 22 = × × 196 = 154 m 2 Now, MO = NO = OL = 6 cm 4 7 NO OL = 6 cm ( Radii of same semi circle ) 2. AB is a chord of a circle centred 90 O 2 at O such that ∠ AOB = 60. If The area of sector MON 9π cm 2 = × π × 6=  360 OA = 14 cm then find the area of the minor segment. (take 3 = 1.73) A

292

50 cm

Sol. Area of minor segment = Area of Sector OAB − Area of ∆OAB 60° 22 3 = × × 14 × 14 − × 14 × 14 360° 7 4 = 17.8 cm 2

Sol. Given, area of circular playground = 22176 cm 2 22 22176 ⇒ πr 2 = 22176 ⇒ r 2 = 3. Sides of a right triangular field are 25 m, 24 m and 7 22176 × 7 7 m. At the three corners of the field, a cow, a buffalo 2 7056 ⇒= r 7056 = 84 cm = ⇒r2 = 7056 ⇒ r= 22 and a horse are tied separately with ropes of 3.5 m 2 7056 ⇒= r 7056 = 84 cm each to graze in the field. Find the area of the field⇒ r= 22 that cannot be grazed by these animals. Now, circumference of the circular ground =2π r =2 × × 84 7 22 Sol. Required Area = Area of triangle – Area of sectors Now, circumference of3the circular ground =2π r =2 × × 84 =528 cm 7 1 528 × 50 Area of Triangle = × 24 × 7 = 84 m 2 `264 = =  Total cost of fencing the ground 264 2 100 7 Radius = 3.5 = m 6. A wiper blade has length 21 cm, sweeps 120°. 2 πr 2 Area of three sectors three angles of triangle ) Calculate the area swept by two blades. = × ((Sum Sum of three 360 of triangle) Topper’s Answer 22 × 7 × 7 × 180 = 7 × 2 × 2 × 360 77 = or 19.25 m 2 4 Required Area = 84 − 19.25 = 64.75 m 2 B

4. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

Q A

(CBSE 2019)

Sol. Since OABC is a square.

Then, In ∆AOB , ∠ OAB = 90 Radius of quadrant= OB

7. The side of a square is 10 cm. Find the area between inscribed and circumscribed A B circles of the square.

Sol. Here, diameter of inner circle A 15C2 + 152 ( By Pythagoras theorem ) = side of the square = 10 cm

10 Radius of inner circle = = 5 cm 2 10 152 + 152 ( By Pythagoras theorem ) D C Radius of inner circle = = 5 cm O 2 ⇒ OB = 15 2 cm Diagonal of square =BD = AB 2 + AD 2 = 102 + 102 θ 2 π Now, area of quadrant OQBP = × r 360 Diagonal of square =BD = AB 2 + AD 2 = 102 + 102 90 2 3.14 ( ) = × × OB 360 = = 200 10 2 cm 1 2 =× 3.14 × (15 2) Diameter = of outer circle Diagonal = of square 10 2 cm 4 = of outer circle Diagonal = of square 10 2 cm = 353.25 cm 2 Diameter 10 2 2 Radius of outer circle = = 5 2 cm Area of square = (OA = )2 (15) = 225 cm 2 2 Then, the required Area = Area of outer circle Hence, area of shaded region = area of quadrant

us of quadrant = OB

5. The area of a circular play ground is 22176 cm 2. Find the cost of fencing this ground at the rate of `50 per metre. (CBSE 2020)

152 + 152 ( By Pythagoras theorem )

− area of square = 128.25 cm 2 ( 353.25 − 225 ) cm 2 =

Chapter 11 • Areas Related to Circles

– Area of inner circle

(

= π 5 2

) − π (5 ) = 25π cm 2

293

8. Figure shows two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is O a semi-circle drawn on PQ as diameter with centre M. = PQ = 10 cm, show If OP

Sol. Radius of semi-circle DFE, r = 6 − 4 = 2 m

cm 5 cm

60°

5 cm Q

π  that area of shaded region is 25  3 −  cm 2. 6 

(CBSE 2016)

Sol. Given, OP = OQ = PQ = 10 cm and ∠POQ = 60° Area of segment PAQM

Area of minor segment =



× πr 2 −

3 2 a 4

360 60 3 2 = × π × (10) − × (10)2 4 360 100π 100 3 50π cm 2 = = − − 25 3 cm 2 6 4 3 π r 2 52 π 25π Area of semicircle = = = cm 2 2 2 2 25π  50π  Area of shaded region = − − 25 3  . 2  3  π  = 25  3 −  cm 2 . 6  9. In the adjoining figure, there are three semi-circles in which BC = 6 cm and 1 BA = 2 BC . What is 3 the area of the shaded

region?

22 (100 − 49 − 9 ) 22 × 42 = 22 × 3 = 66 cm 2 = 14 14 Required Shaded Area is 66 cm2

10. Find the area of shaded field shown in figure.

294

πr 2

11. A momento is made as shown in the A figure. Its base PBCR 7 cm 7 cm 90° is silver plated from the front side. Find P R 3 cm 3 cm the area which is silver plated. B C 22    Use π =  (CBSE 2015) 7   Sol. From the given figure 1 Area of right-angled ∆ABC = × 10 × 10 = cm22 50 cm 2 Area of quadrant APR of the circle of radius 7 cm 90 1 1 22 = × π r 2 = × π × (7)2 = × × 49 =38.5 cm 2  4 4 7 360

Long Answer Questions

Required Shaded Area = Area of semicircle AC – Area of semicircle AB – Area of semicircle BC πr 2 Area of semicircle = 2 22 × 102 22 × 72 22 × 32 − − Shaded Area = 7×2 7×2 7×2

π (2)2 = = 2π m² 2 2 ∴ Area of shaded region = ( 32 + 2π ) m 2 = and area of semi-circle DFE

Area of base PBCR = Area of ∆ABC – Area of 50 − 38.5 = 11.5 cm 2 quadrant APR =

Sol. Given: BC = 6 cm 7 14 cm , AC = AB + BC = 6 + 14 = 20 cm AB = × 6 = 3 Radius of semicircle with base BC = 3 cm Radius of semicircle with base AB = 7 cm Radius of semicircle with base AC = 10 cm

Now, area of rectangle ABCD = BC × AB = 8 × 4 = 32 m²

1. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle. (CBSE 2023)

CBSE Marking Scheme 2023 Sol. Area of minor segment 22 14 × 14 × 60 1 3 × − × 14 × 14 × 7 360 2 2

308 − 49 3 cm 2 or 17.9 cm 2 3

Area of major segment 22  308  × 14 × 14 −  − 49 3  = 7  3  = 616 −

308 + 49 3 3

1540 = + 49 3 cm 2 or 598.1 cm 2 3

1 1 1

2. The perimeter of a certain sector of a circle of radius 5.6 m is 20.0 m. Find the area of the sector. (CBSE 2024)

(NCERT Exemplar)

Sol. Given r = 5.6 m

Sol. Given, the side of the square = 28 cm.

Now, perimeter of sector = 20 m

Area of the square = 28 × 28 = 784 cm 2 28 Radius of each circle = = 14 cm 2 22 ∴ Area of two circles =2 × × 14 × 14 =1232 cm 2 7

( Length of arc + 2 × radius) =20 2πrθ ⇒ + 2 × 5.6 = 20 360 8.8 × 360 ⇒= πrθ = 1584 m 2 Now area of sector =

Area of 2 quadrants =

360

2 308 cm 2 × π × 142 ×=

Area the shaded region = Area of square

π r 2θ π rθ × r 1584 × 5.6 = = = 24.64m 2 360 360 360

+ Area of two circles – Area of two quadrants

Hence area of sector = 24.64 m².

= 784 + 1232 – 308 = 1708 cm2 A 4c

3. In the given figure, ∆ABC is a right angled triangle in which ∠ A = 90. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

90

(CBSE 2017)

7. In given figure, two concentric circles with centre O, have radii 60°, 21 cm and 42 cm. If ∠AOB = find the area of the shaded region. Area= Sol. Area of Sector

(CBSE 2019)

× π ×r 2

O 60° B

360 60° 22 60° 22 1 1 × = × 42 × 42 = 42×= 6 × 42 6 × 42 =924 22 × 42 =924 × 42××22 × 22=×22 360° 7 360° 7 6 6 Area of major sector of smaller circle Sol. In ∆ABC, ∠A = 90°, AB = 3 cm, and AC = 4 cm. 300° 22 = × × 21 × 21 By using Pythagoras theorem, 360° 7 5 BC = AB 2 + AC 2 = 32 + 42 = 5 cm. = × 22 × 3 × 21 =5 × 11 × 21 =1155 6 3 4 22 2 2 Area of semicircle with radius cm + Area of semicircle with Area radius cm of 2 2bigger circle = π r = 7 × 42 × 42 = 5544 cm 3 4 cm + Area of semicircle with radius cm Required Area = Area of bigger circle – Area of major 2 2 2 2 sector of smaller circle – Area of sector π  3 π  4 =   +   Required area = 5544 − 1155 − 924 = 3465 cm 2 2 2 2 2 5 Area of semicircle with radius cm − Area of ΔABC 6. Four equal circles are described at the four corners 2 2 of a square so that each touches two others. The π  5 1 =   − × 3 × 4 24 2 2 2 shaded area enclosed between the circles is cm 2. 7  25π  Find the radius of each circle. (CBSE SQP 2016) = − 6  cm 2 …( i )   8  2 2 π  3 π  4   25π  Sol. Let r cm be the radius of each circle. Area of shaded region= − 6 cm 2   +   −   2 2 2 2 8 2 2 Side of square = 2r π  3 π  4   25π  24 shaded region = − 6 cm 2   +   −  Area of square − Area of 4 sectors = cm 2  2 2 2 2 8 7  90  24  π × 9 π × 16 25π  × πr 2  = or,(2r )2 − 4  =  + − + 6  cm 2  360  7 8 8  8  Shaded area = Area of semicircle AB + Area of semicircle AC + Area of ΔABC – Area of semicircle BC

 25π 25π  2 =  − + 6=  6 cm  8 8

4. In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square, where O and O' are centres of the circle. Find the area of shaded region. Chapter 11 • Areas Related to Circles

r 2 22 24 = 7 7 2 2 28r − 22r 24 or, = 7 7 6 2 24 or, r = 7 7 2 or, r = 4 or, 4r 2 −

or, r = ±2 (CBSE 2017)

Radius of each circle is 2 cm (r cannot be negative). 295

7. An elastic belt is placed around the rim of a pulley of radius 5 cm. From one point C on the belt, elastic belt is pulled directly away from the centre O of the pulley until it is at P , 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find the shaded area. (Use π = 3.14 and 3 = 1.73) A

∴ Length of arc ADB =

10 cm

Reflex ∠AOB =240°

Sol. In right angled ∆OAP, 5 1 cos= θ = , or, = θ 60 D 10 2 So, ∠ AOB = 60 + 60 = 120

2 × 3.14 × 5 × 240 

B 

= AP

OP − OA

= 20.93 cm

[By using Pythagoras theorem=]

QR = area of sector OQR − area of ∆QOR = 252 m 2 ( 693 − 441) m 2 =

Similarly, area of flower bed with PS = 252 m²

360 Hence length of elastic in contact = 20.93 cm 2

radius (r) of each semi-circle = 21 2 m 90 × π × (21 2)2 Now, area of sector OQR = 360 1 22 = × × 882 m 2 =693 m 2 4 7 And area of right angled ∆QOR 1 1 = × OQ × OR = × 21 2 × 21 2 = 441 m 2 2 2 Area of flower bed (shaded part) with

(10)2 − (5)2=

Now, AP = 5 3 cm

Area ( ∆OAP + ∆OBP= ) Area of ∆OAP + Area of ∆OBP

= 504 m 2 ( 252 + 252 ) m 2 =

9. In the figure, ABC is a right angled triangle right angled at ∠A . Find the area of the shaded region, if AB 6= = cm, BC 10 cm and O is the centre of the incircle of the triangle ABC. (CBSE 2015) Sol. Let r be the radius of incircle. In ∆ABC , = AC

= 102 − 62 =

Here, = ∠ A 90 = , ∠ P 90 and = ∠ Q 90

(

∴OPAQ is a square. P

m x

42 m

8-r

296

r cm

6–

⇒x = 21 2 m

6 cm

⇒ x2 = 882

A m r c 90°

r cm

⇒ 2x 2 = 1764

[Using Pythagoras theorem]

x2 + x2 = (42)2

)

Then ∠ O = 360 − 90 + 90 + 90 = 90

In ∆QOR ,

42 m

64 = 8 cm 

42 m

∠ QOR = 90

100 − 36 = 

Sol. Let OR be x cm, then OQ = x cm Since, diagonals of a square bisect each other at S R right angles, then

( By using Pythagoras theorem )

BC 2 − AB 2

1 = 2 × × 5 × 5 3 = 25 3 = 43.25 cm 2 2 25 × 3.14 × 120 Area of= sector OACB = 26.16 cm 2 360 Shaded Area = 43.25 − 26.16 = 17.09 cm 2 8. In figure, PQRS is square lawn with side PQ = 42 metre. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts). (CBSE 2015)

Hence Total area of both flower beds with QR and PS

10 cm

6-r

Then, BC = 10 = 8 + r + 6 − r

( By using the tangent properties ) = 2r 4,= or r 2 cm 2r = 8 + 6 − 10= 22 88 2 ∴Area of circle = π r = × 2 × 2= = 12.57 cm 2 7 7 1 Now, area of ∆ABC = × 8 × 6 = 24 cm 2 2 ∴Area of shaded region = Area of ∆ABC – Area of circle = 24 − 12.57 = 11.43 cm 2

Competency Based Questions Multiple Choice Questions

Select Response Type MCQs

1. A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be (a) 3520 cm² (b) 6400 cm² (c) 7744 cm²

(d) 8800 cm²

2. The area of the circle that can be inscribed in a square of side 6 cm is (a) 36π cm² (b) 18π cm² (c) 12π cm²

(d) 9π cm² (NCERT Exemplar)

3. Savita has a lamp placed at the centre of her square yard, each side measuring 20 m. The light of lamp covers a circle of radius 10 m on yard. What area of the yard is not lit by the lamp? (a) 400π sq. m (b) 100π sq. m (c) (40–10π) sq. m

(d) ( 400 − 100π ) sq. m

4. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is (a) 56 cm (b) 42 cm (c) 28 cm

(d) 16 cm

(NCERT Exemplar)

5. The hour-hand of a clock is 6 cm long. The angle swept by it between 7:20 am and 7:55 am is 35  35  (a) (b) 2 4 (c) 35° (d) 70° (CBSE 2023) 6. A square of side length a is inscribed in a circle as shown in the figure. What is the area of the shaded region? (a)

( 3π − 2 ) a 2 4

( 2π + 7 ) a 2 (c) 6

(b) (d)

(π + 4 ) a

8 ( 3π + 2 ) a 2 8

7. A car has two wipers which do not overlap. Each wiper has a blade of length 24 cm sweeping through an angle of 120°. Find the total area cleaned at each sweep of the blades. (a) 192π cm 2 (b) 224π cm 2 (c) 384 π cm

(d) 240π cm 2

1. A circular park has a radius of 14 m. Which of the following statements about this park are correct? 22    Take π =  7   (i) The area of the park is 616 m² (ii) The circumference of the park is 88 m (iii) The diameter of the park is 28 m (iv) The radius of the park is 7 m (a) (i) and (ii)

(b) (i) and (iv)

(d) (i), (ii) and (iii)

2. A sector of a circle with radius 10 cm has a central angle of 90°. Which of the following are true? (Use π = 3.14) (i) Area of the sector is 78.5 cm² (ii) The arc length is 15.7 cm (iii) The sector is one-third of the circle (iv) The area of the full circle is 314 cm² (a) (i), (ii) and (iv)

(b) (i) and (iv)

(d) (i), (ii) and (iii)

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm. Reason (R): Circumference = 2π × radius of a circle. (CBSE 2024)

2. Assertion (A): If the radius of sector of a circle is reduced to its half and angle is doubled then the perimeter of the sector remains the same. Reason (R): The length of the arc subtending angle θ π rθ at the centre of a circle of radius r is 180

(CBSE SQP 2025)

Chapter 11 • Areas Related to Circles

297

4. Assertion (A): Two circles with different radii have the same area if the product of their radii is constant. Reason (R): The area of a circle is proportional to the square of its radius.

Case Study Based Questions 1. Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking. After survey, it was decided to build rectangular playground, with a semicircular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats. Based on the above information, sol the following questions: I Play Ground

7 Units

Parking Area

(ii) W hat is the ratio of area of playground to the area of parking area?

(c) Find the cost of fencing the playground and parking area at the rate of `2 per unit. (CBSE 2023) 2. A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze. 7m

20 m

3. Assertion (A): If the circumference of a circular park is increased by 10%, the area of the park increases by 21%. Reason (R): The area of a circle is directly proportional to the square of its circumference.

Based on the given information, sol the following questions. (a) Find the area of the square shaped grass field. (b) (i) F ind the area of the total field in which these horses can graze. OR

I 14 Units

(a) What is the total perimeter of the parking area? (b) (i) What is the total area of parking and the two quadrants?

(ii) I f the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse. (Use π = 3.14 )

Answers Multiple Choice Questions 1. The wire is initially bent into a circle with a radius of 56 cm. The circumference C of a circle is given by: C = 2π r = 2 × π × 56 = 112π cm P = 112π cm 4s = 112π 112π = s = 28π cm 4 22 s =28 × =88 cm 7 2 = A 88 = 7744 cm 2 (c) 7744 cm² is correct. 298

2. Side of the square s = 6 cm The circle inscribed in the square will have a diameter equal to the side of the square. Diameter= s= 6 cm Diameter 6 = = 3 cm r= 2 2 The area of a circle is given by: Area = π r 2 =π × 32 =9π cm 2 Area = 9π cm 2 (d) 9π cm²

3. Area of the square yard 2

Area of square = side = 20 = 400 m² area covered by the lamp (circle) Area of circle = π r 2 =× π 102 = 100π m 2 = Unlit area Area of square − Area of circle

= 400 − 100π

(d) (400 – 100π) sq. m 4. Circumference of first circle = 2π r1 = 2π × 18 = 36π

Circumference of second circle = 2π r2 = 2π × 10 = 20π

second circle = 2π r2 = 2π × 10 = 20π

Total circumference = 36π + 20π = 56π Let the radius of the new circle be RRR. Since its circumference is equal to the total circumference: 56π 2π R = 56π ⇒ R= = 28 2π (c) 28 cm 5. The hour hand moves 30° per hour. In 60 minutes, the hour hand sweeps 30°. 30 Angle swept per minute = = 0.5 60 Angle swept =35 × 0.5 =17.5 35 (b) 2 6. Diagonal of square = a 2 diagonal is also the diameter, Diameter = a 2 a 2 r= 2 Area of circle = π r 2 2 a 2  Area of circle = π    2  2 a ×2 = π× 4 πa2 = 2  πa2  1 Q 2 (π − 2) Minor arc area =  − a 2  × = 8  2  7

πa2

a 2 (π − 2) − Shaded Area = 2 8 4π a 2 − a 2π + 2a 2 = 8 2 3π + 2 ) a ( (d) 4 θ 7. Area of sector = × πr 2 360 120 Area of one wiper= × π × 242 360 1 2 = × π × 576 = 192π cm 3 Chapter 11 • Areas Related to Circles

Since there are two wipers: Total area = 2 × 192π = 384π cm 2 (c) 384π cm²

Select Response Type MCQs 1. (d) (i), (ii) and (iii) Radius r = 14 m 22 π= 7 22 (i) Area of the park = π r 2 = × 14 × 14 = 616 m 2 7 (ii) Circumference of the park 22 =2π r =2 × × 14 =88 m 2 7 (iii) Diameter =2r =2 × 14 =28 m 2 (iv) Radius is 7 m 2. (a) (i), (ii) and (iv) Radius r = 10 cm Central angle = 90° π = 3.14 (i) Area of sector =

× πr 2 =

360 1 = × 3.14 × 100 = 78.5 cm 2 4 = (ii) Arc length

360

× 2π= r

1 =× 62.8 = 15.7 cm 4

90 × 3.14 × 102 360

90 × 2 × 3.14 × 10 360

(iii) Sector is one-fourth of the circle: Since 2

90

360



1 4

(iv) Area of full circle: π r = 3.14 × 100 = 314 cm 2

Assertion-Reason Based Questions 1. Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm. Reason (R): Circumference = 2π r C = 2π r 176 =2 ×

r = 28

22 × r ⇒ 1232 = 44r 7

(a) Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). 2. Assertion (A): If the radius of a sector of a circle is reduced to half and the angle is doubled, then the perimeter of the sector remains the same. Reason (R): The length of the arc subtending angle πrθ θ at the centre of a circle of radius r is . 180

299

The perimeter of a sector = Length of the arc + 2 × Radius πrθ P = + 2r 180 Original Sector’s Perimeter πrθ = P1 + 2r 180 New Sector’s Perimeter r π × × 2θ πrθ 2 = Pr2 +r P2 = + × 2  2 180 180 the new perimeter is less than the original perimeter. (d) Assertion (A) is False, but Reason (R) is True. 3. Assertion (A): If the circumference of a circular park is increased by 10%, the area of the park increases by 21%. Reason (R): The area of a circle is directly proportional to the square of its circumference. C r= 2π 2 C2 C  A = π  A =  2π  4π Area A is proportional to C 2 New Circumference = 1.1C

(1.1C )2 1.21C 2 New AreaNew = Area = 4π 4π

Percentage Increase = in Area

This condition alone does not guarantee that their areas will be the same. For their areas to be equal, we would need: π r12 = π r22 This condition clearly requires r1 = r2 not just r1 × r2 = constant Thus, the assertion is False. The statement that area is proportional to the square of the radius is True. (d) A is false, but R is true.

Case Study Based Questions 1. (a) Total Perimeter of Parking Area 22 7 = 7 + × =18 units 7 2 (b) Total Area of Parking and Two Quadrants 2

1 22  7   1 22  = × ×   + 2  × × 22  =25.53 sq. units 2 7 2 4 7  (c) Area of Playground

44  642  = sq. units  98 −  sq. units = 7  7  2 1 22  7  77 Area of Parking = × ×   = sq. units 2 7 2 4

1.21C 2 − C 2 2

× 100

C 1.21 ( − 1) × 100

= 0.21 × 100 = 21%

(a) Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). 4. Assertion (A): Two circles with different radii have the same area if the product of their radii is constant. Reason (R): The area of a circle is proportional to the square of its radius. A = πr 2 The assertion claims that two circles will have the same area if the product of their radii is constant. Suppose two circles have radii r1 and r2. ● If their product is constant: r × r = 1 2 k ( constant )

 1 22   1 22  = 14 × 7 − 2  × × 22  = 14 × 7 − 2  × × 22  4 7  4 7 

(c) Ratio of Playground Area to Parking Area 642 77 2568 : = 7 4 539 (d) Cost of Fencing the Playground and Parking Area 22 7   =  14 + 14 + 7 + `292 ×  × 2=  92  7 2 2. (a) The area of the square-shaped grass field

Ratio =

= 20 × 20 m 2 = 400 m 2

(b) (i) Area of the field in which horses can graze  1 22  2 = 4  × × 7 × 7  m= 154 m 2 4 7   OR

(ii) Area grazed by one horse 1  = × 3.14 × 102  m 2 =78.5 m 2 4   (c) Area of the field that is left un-grazed

= 246 m 2 ( 400 − 154 ) m 2 =

Additional Practice Questions 1. If the perimeter of a semicircular protractor is 36 cm, then its diameter is (a) 14 cm (b) 16 cm (c) 18 cm 300

(d) 12 cm

2. The area of a quadrant of a circle whose circumference is 616 cm will be (a) 7546 cm² (b) 7500 cm² (c) 7456 cm²

(d) 7564 cm²

7 of the area of 20 the circle, then the angle at the centre is (CBSE 2024) (a) 110° (b) 130° (c) 100° (d) 126°

3. If the area of a sector of a circle is

4. Area of the largest triangle that can be inscribed in a semi-circle of radius r units is (NCERT Exemplar) 1 2 2 (a) r sq units (b) r sq units 2 2 (c) 2r sq units (d) r 2 2 sq units 5. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then (NCERT Exemplar) R R2 (a) R1 + R2 = (b) R12 + R22 = (d) R12 + R22 < R 2

6. The area of the square that can be inscribed in a circle of radius 8 cm is: (NCERT Exemplar) (a) 256 cm² (b) 128 cm² (c) 64 2 cm²

(d) 64 cm²

7. Find the length of the arc of a circle which subtends an angle of 60° at the centre of the circle of radius 42 cm. 8. The side of a square is 10 cm. Find the area between inscribed and circumscribed circles of the square. 9. The difference between the outer and inner radii of a hollow right circular cylinder of length 14 cm is 1 cm. If the volume of the metal used in making the cylinder is 176 cm³, find the outer and inner radii of the cylinder. 10. Find the area of the flower bed (with semicircular ends) in figure. 16 cm 44 cm 11. The inner circumference of a circular track is 132 m. The track is 2.5 m wide everywhere. Calculate the

Challenge Yourself

Chapter 11 • Areas Related to Circles

12. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make, so as to keep a speed of 66 km/h? 13. Two circles touch externally. The sum of their areas is 130π sq. cm. and the distance between their centres is 14 cm. Find the radii of the circles. 14. Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in figure.

Find area of floor that remains uncovered with tiles. (use π = 3.14) 15. Sameer and Madhur drew two circles circumscribed by squares of side 14 and 7 inches, respectively. Both the circles are divided into 5 equal sectors as shown below.

(NCERT Exemplar)

14 inches 7 inches

Based on the given information, answer the following questions. (a) Determine the area occupied by one sector of the larger circle. (b) Sameer says, ‘The area occupied by 2 sectors of the larger circle is equal to the area occupied by 4 sectors of the smaller circle, as the side of the larger square is twice of the smaller square.’ Do you agree with the statement? Justify your answer. 22 (Note: take π = if required) 7

(CBSE QB)

Scan me

1. The radius of a circle is so increased, that its circumference is increased by 5%. The area of the circle, then increases by (a) 12.5% (b) 10.25% (c) 10.5%

cost of putting up a fence along the outer circle at the rate of `3.50 per metre.

(d) 11.25%

2. The given diagram comprises of two sectors with angles 30° and 60°, their radii are 7 cm and 14 cm respectively. Both have centre O. The perimeter of the shaded region is 1 (a) 54 cm (b) 105 cm 3 1 (c) 46 cm (d) 77 cm 3 301

3. In the figure given alongside, the rectangle at the corner measures 10 cm × 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm? (a)10 cm (b) 40 cm (c) 50 cm

E D

(d) None of these

4. In the given figure, crescent is formed by two circles which touch at point A. O is the centre of the bigger circle. If CB = 9 cm and DE = 5 cm, the area of the shaded portion is

(a) 625 cm2

(b) 614.25 cm2

(d) 671.35 cm2

Answers Additional Practice Questions

13. Radii of the circles: 3 cm and 11 cm

1. (c) 18 cm

2. (d) 7564 cm²

14. Uncovered floor area: 4.3 m²

3. (d) 126°

4. (a) r2 sq units

2 5. (a) R= R12 + R22

6. (b) 128 cm²

7. 44 cm

8. 78.5 cm²

15. (a) Area occupied by one sector of the larger circle: 30.8 in²

Challenge Yourself

9. Outer radius = 2.5 cm, Inner radius = 1.5 cm

1. (b) 10.25%

10. Area = 649 m²

3. (c) 50 cm

12. 500 RPM

11. `517

1 cm 3 4. (c) 643.5 cm 2 2. (c) 46

Scan me for Exemplar Solutions 302

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then (NCERT Exemplar) R (a) R1 + R2 =

(b) R1 + R2 > R

(d) Nothing definite can be said about the relation among R1, R2, and R. S

2. In the figure below RT = 1 cm and OQ = 3 cm.

T R 1 cm

What is the area of the shaded region? 2

(b) ( 6.25π − 12) cm

(d) ( 6.25π − 15) cm 2

(a) (12.5π − 12) cm (c) (12.5π − 15) cm

3 cm

3. The area of the square that can be inscribed in a circle of radius 5 2 cm is (a) 50 cm²

(CBSE QB)

(b) 100 cm²

(CBSE 2024)

d) 200 cm²

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(b) Both A and R are true, but R is not the correct explanation of A.

(d) A is false, but R is true.

4. Assertion (A): If the radius of a circle is tripled, its area becomes nine times larger. Reason (R): Tripling the radius increases the area because the area is proportional to the radius squared. 5. Assertion (A): Doubling the diameter of a circle doubles the area of the circle. Reason (R): The diameter is twice the radius, and the area is a function of the radius squared.

Very Short Answer Questions

22   6. Find the area of a quadrant of a circle, where the circumference of circle is 44 cm.  Use π =  7  

[2 × 2 = 4 Marks] (CBSE 2011)

7. In the given figure, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. 22   OB is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.  Use π =  (CBSE 2013) 7   C

D Self-Assessment

303

Short Answer Questions

[4 × 3 = 12 Marks]

8. In the given figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region (Take π = 3.14) A

9. In the given figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. If arcs of equal radii 7 cm have been drawn, with centres A, B, C and D, then find the area of the shaded region. (CBSE 2015) A

10. In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm and centre O. If ∠POQ = 30°, then find the area of shaded region. (CBSE 2012) A O

30°

11. In given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles. A

Long Answer Questions

[3 × 5 = 15 Marks]

12. The length and breadth of a rectangular piece of paper are 28 cm and 14 cm respectively. A semi-circular portion is cut off from the breadth’s side and a semicircular portion is added on length’s side, as shown in figure. Find the area of the shaded region. (CBSE 2011) 22    Use π =  7   14 cm 28 cm 304

13. In given figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts). S

(CBSE 2019)

14. The diameter of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions. (NCERT Exemplar)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. The city council plans to revitalise an urban park to enhance its aesthetic appeal and functionality. The park’s design includes a large circular pond, four identical circular flower beds, a circular walking track surrounding the pond, and two small circular seating areas. Specifications: ● The pond has a radius of 20 metres. ●

Each of the four flower beds has a radius of 3 metres.

●

The walking track is uniformly 2 metres wide and encircles the pond.

●

Each seating area has a radius of 5 metres.

Questions: (a) Calculate the area of the pond. Use the formula for the area of a circle, A = π r 2, where r is the radius of the pond. (b) Determine the total area of the walking track. Consider the track as a ring with the inner radius equal to the pond’s radius and the outer radius as the pond’s radius plus the width of the track. (c) Calculate the total area covered by the four flower beds. Assume the flower beds are equally spaced around the pond. (d) Find the total area occupied by the two rest areas. Each rest area is designed to provide a comfortable seating space for visitors.

Self-Assessment

305

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. If the circumference of a circle and the perimeter of a square are equal, then

(NCERT Exemplar)

(a) Area of the circle = Area of the square (b) Area of the circle > Area of the square (c) Area of the circle < Area of the square (d) Nothing definite can be said about the relation between the areas of the circle and square 2. The circumference of two circles are in the ratio 4 : 5. The ratio of their radii is (a) 16 : 25 (b) 25 : 16 (c) 2 : 5 (d) 4 : 5

(CBSE 2023)

3. In the figure below, the square JKLM is inscribed within a circle and ∆JMN is a right-angled isosceles triangle. The point marked O is the centre of the circle. What is the area of the shaded part of the figure? J

(CBSE QB)

1 cm O

M 2

π 1 (a)  −  cm²  4 2

(b)  π − 1  cm² 2 

L 2

π (c)  1 −  cm² 4 

Assertion-Reason Based Questions

(d) (1 − π )2 cm² [2 × 1 = 2 Marks]

100π 4. Assertion (A): The area of a sector of angle 120° in a circle with radius 10 cm is equal to square cm. 3 1 2 Reason (R): The formula for the area of a sector of a circle is r θ , where θ is in radians. 2 5. Assertion (A): If two concentric circles have radii r and 2r, the area of the ring-shaped region between them is 3π r 2. Reason (R): The area of the ring-shaped region is found by the formula π (2r )2 − π r 2.

306

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. In the given figure, the shape of the top of a table is that of a

90. sector of a circle with centre O and ∠AOB = = OB = 42 cm, then find the perimeter of the top of the If AO 22  table.  Use π =  (CBSE 2012) 7  

B R

7. In the given figure, APB and CQD are semi-circles of diameter 7 cm each, while ARC and BSD are semi-circles

P 7 cm

B 7 cm C 7 cm Q

of diameter 14 cm each. Find the perimeter of the shaded 22  region.  Use π =  (CBSE 2011) 7  

Short Answer Questions 8. Find the area of shaded region in the given figure, where a circle of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm. (Use π = 3.14 and 3 = 1.73)

[4 × 3 = 12 Marks]

O L

B D

9. In figure, ABCD is a parallelogram. A semicircle with centre O and the diameter AB has been drawn and it passes through D. If AB = 12 cm and OD ⊥ AB , then find the area of the shaded region. (Use π = 3.14) A

B S

10. In the given figure, PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. (Use π = 3.14) (CBSE 2014)

A P

7 cm Q

3 cm T

11. In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles, each of radius 7 cm. Find the 22  area of the shaded region.  Use π =  7  

C P

Q A

Long Answer Questions

[3 × 5 = 15 Marks]

12. In the given figure, a sector OAP of a circle with centre O containing ∠θ is shown. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter

πθ   of shaded region is r  tan θ + secθ + − 1  180 

Self-Assessment

(CBSE 2016)

O θ

307

13. In the given figure, an equilateral triangle has been inscribed in a circle of radius 6 cm. Find the area of the shaded region.

(Use π = 3.14 )

14. From a thin metallic piece, in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°, a quarter circle BFEC is removed

(see figure). Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of the metal sheet. 22    Use π =  7  

(CBSE 2011)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooches are shown below. Observe them carefully.

(CBSE QB)

Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. A wire used for making 4 diameters which divide the circle into 8 equal parts. Design B: Brooch B is made of two colours gold and silver. Outer part is made with gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere. Based on the given information, sol the following questions. Refer to design A (a) Find the total length of silver wire required. (b) Find the area of each sector of the brooch. Refer to design B (c) (i) What is the circumference of the outer part (golden)?

308

(ii) Write the difference of areas of golden and silver parts.

Surface Areas and Volumes

Chapter at a Glance Surface Area is the total area covered by the surface of the solid. It is measured in square units. Example: The surface area of a cube is calculated by adding up the area of all six faces (6 times the area of one face, given each face is a square). Volume is the amount of space enclosed within a solid. It is measured in cubic units. The volume tells us the capacity of the solid. Example: How much liquid a cylindrical tank can hold or how much air can fill in a spherical balloon.

Cube

a = length of sides Curved / Lateral Surface Area of Cube = 4a 2 a

Total Surface Area of Cube = 6a 2

Diagonal of Cube = 3a

Volume of Cube = a 3 a

Cuboid l = length

b = breadth

h = height

Curved / Lateral Surface Area of Cuboid = 2 (lh + bh ) Total Surface Area of Cuboid= 2 (lb + lh + bh ) Volume of Cuboid =l × b × h Diagonal of Cube =

Height

Breadth

Length

l 2 + b2 + h2

Right Circular Cylinder r = radius of base h = height Curved / Lateral Surface Area of Cylinder = 2π rh

Total Surface Area of Cylinder = 2π rh + 2π r 2 Volume of Cylinder = π r 2h

Right Circular Hollow Cylinder R = External radius and r = internal radius and h = height. (i) Area of each end = p (R 2 - r 2) (ii) Curved surface area of hollow cylinder = 2p(R + r)h (iii) Total surface area = 2p(R + r)(R + h - r) (iv) Volume of Material = p(R 2 - r 2)h

Chapter 12 • Surface Areas and Volumes

h r

309

Right Circular Cone r = radius of base l = slant height h = altitude (height)

Curved / Lateral Surface Area of Cone = π rl Total Surface Area of Cone = π rl + π r 2

1 Volume of Cone = π r 2h 3

Sphere r = radius of sphere Curved / Lateral Surface Area of Sphere = 4π r 2 Total Surface Area of Sphere = 4π r

Centre

4 Volume of Sphere = π r 3 3

Hemisphere r = radius of hemisphere

Curved / Lateral Surface Area of Hemisphere = 2π r 2 Total Surface Area of Hemisphere = 3π r 2

2 Volume of Hemisphere = π r 3 3

NCERT Zone NCERT Exercise 12.1

vessel is 13 cm. Find the inner surface area of the

1. 2 cubes each of volume 64 cm 3 are joined end to end. Find the surface area of the resulting cuboids. Sol. Given that, Volume of cubes = 64 cm 3 Edge3 = 64

8 cm

Edge = 4 cm

4 cm

∴ surface area of the cuboid = 2 (lb + bh + lh ) 2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the

310

Sol. It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm). Height of hemispherical part = Radius = 7 cm

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm. = 2 ( 4 × 4 + 4 × 8 + 4 × 8 )= 160 cm

22   vessel.. Use π =  . 7  

6 cm 7 cm

13 cm

14 cm

Height of cylindrical part (h) = 13 − 7 = 6 cm Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part = 2 π rh + 2π r 2 Inner surface area of vessel = 2 × = 572 cm 2

22 22 ×7×6 + 2× ×7×7 7 7

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface 22   area of the toy. Use π =  . 7   Sol. It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).

12 cm

r 2 + h2

625 25 = 4 2

Total surface area of toy = CSA of conical part + CSA of hemispherical part = π rl + 2 π r 2 22 7 25 22 7 7 = × × + 2 × × × = 214.5 cm 2 7 2 2 7 2 2 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of 22   the solid. Use π =  . 7   Sol. From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm. Radius (r) of hemispherical part=

7 cm

7 = 3.5 cm 2

= Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part 2

22 7 7 =6 × 7 + × × = 332.5 cm 2 7 2 2 2

Chapter 12 • Surface Areas and Volumes

= 6 ( edge ) + 2π r 2 − π r 2

πl 2

4 1 = ( 24 + π )l 2 (unit)2 4 6. A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. 22   Find its surface area. Use π =  . 7   5 mm 14 mm 5 2

7 cm

Total surface area of solid =

= 6 ( edge ) + 2 × π × r 2 − π r 2

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part

= 6l 2 +

Slant height of the conical part = 49 + 144 = 4

l 2

3.5 cm

Height of conical part (h) = 15.5 − 3.5 = 12 cm

Sol. Diameter of hemisphere = Edge of cube = l Radius of hemisphere =

Height of hemispherical part 7 = 3.5 = cm = Radius (r) 2

2 7 =   + (12 ) = 2

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

5 cm

9 cm

Sol. It can be observed that Radius (r) of cylindrical part = Radius (r) of hemispherical part Length of cylindrical part (h) = Length of the entire capsule − 2 × r = 14 − 5 = 9 cm Surface area of capsule = 2 × CSA of hemispherical part + CSA of cylindrical part = 2 × 2π r 2 + 2π rh = 25π + 45π = 70π mm 2 22 = 70 × = 220 mm2 7 311

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of `500 per m2. (Note that the base of the tent will not 22   be covered with canvas.. Use π =  . 7  

Slant height of the conical part =

r 2 + h2 =

6.25 = 2.5 cm

Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base = 2π rh + π rl + π r 2 = 10.56 + 5.50 + 1.54 = 17.60 cm 2 The total surface area of the remaining solid to the nearest cm2 is 18 cm 2.

Sol. 2.8 m

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in given figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

2m 2.1 m

Sol. Given that, Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm

Given that,

Height of cylindrical part (h) = 10 cm

Height (h) of the cylindrical part = 2.1 m

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part

Diameter of the cylindrical part = 4 m Radius of the cylindrical part = 2 m Slant height (l) of conical part = 2.8 m

= 2π rh + 2π r 2

Area of canvas used = CSA of conical part + CSA of cylindrical part

= 2π × 3.5 × 10 + 2 × 2π × 3.5 × 3.5

= π rl + 2π rh

NCERT Exercise 12.2

= 2π × 2.8 + 2π × 2.1 = 44 m 2 Since, the cost of 1 m 2 canvas = `500 Cost of 44 m2 canvas =44 × 500 =` 22000 Therefore, it will cost `22000 for making such a tent. 8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm 2. Sol. Given that, Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

1.4 cm

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. Sol. Given that, Height (h) of conical part = Radius (r) of conical part = 1 cm Radius (r) of hemispherical part = Radius of conical part (r) = 1 cm

1 cm

1 cm 1 cm

Volume of solid 2.4 cm

Diameter of the cylindrical part = 1.4 cm Therefore, radius (r) of the cylindrical part = 0.7 cm 312

= 374 cm 2

= Volume of conical part + Volume of hemispherical part 1 2 = π r 2h + π r 3

3 3 1 2 = π 121 + π ×= 13 π cm 3 3 3

2 = π r 2h + 2 × × π r 3 3 22 4 22 = × 1.4 × 1.4 × 2.2 + × × 1.4 × 1.4 × 1.4 7 3 7 3 = 25.05 cm

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. if each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) Sol.

8 cm

Volume of sugar syrup = 30% of volume 30 = × 1127.25 100

2 cm

3 cm

2 cm

45 25.05 = 1,127.25 cm3 Volume of 45 gulab jamuns =×

= 338.17 cm 3 ≅ 338 cm 3 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the following figure).

12 cm

From the figure, it can be observed that Height (h1) of each conical part = 2 cm Height (h2) of cylindrical part = 12 × Height of conical part

= 12 - 2 × 2 = 8 cm

Radius (r) of cylindrical part = Radius of conical part 3 = cm 2 Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

Sol. Depth (h) of each conical depression = 1.4 cm Radius (r) of each conical depression = 0.5 cm

1 = π r 2h2 + 2 × × π r 2h1 3

Volume of wood = Volume of cuboid - 4 × Volume 0.5 cm of cones

= 18π + 3π = 21π = 66 cm 3 3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the given figure). Sol. It can be observed that Radius (r) of cylindrical part 1.4 cm = Radius (r) of hemispherical part =

2.8 = 1.4 cm 2

2.2 cm 2.8 cm 5 cm

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length (h) of cylindrical part = 5 - 2 × Length of hemispherical part

1 =l × b × h − 4 × × π r 2h 3 1 22 2 = 15 × 10 × 3.5 − 4 × × × 1.4 × ( 0.5 ) 3 7 = 525 − 1.47

= 523.53 cm 3 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. 5 cm

Sol. Height (h) of conical vessel = 8 cm

= 5 − 2 × 1.4 = 2.2 cm

Radius (r1) of conical vessel = 5 cm

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

Radius (r2) of lead shots = 0.5 cm

Chapter 12 • Surface Areas and Volumes

1.4 cm

8 cm

313

Let n number of lead shots were dropped in the vessel.

cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Volume of water spilled = Volume of dropped lead shots

Sol. Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm

1 = × volume of cone 4

Height (h2) of conical part of solid = 120 cm Height (h1) of cylinder = 180 cm

4 = n × π r23 3

120 cm 60 cm

Radius (r) of cylinder = 60 cm

25 × 8 n = = 100 3 1 16 ×   2 Hence, the number of lead shots dropped in the vessel is 100.

60 cm

Volume of water left = Volume of cylinder − Volume of solid

60 cm

2 1  6. A solid iron pole consists of a cylinder of height 220 cm = π r 2h1 −  π r 2h1 + π r 3  3 3   and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. 2 π ( 60 ) 180 − ( 40 + 40 )  = 3 Find the mass of the pole, given that 1 cm of iron has approximately 8 g mass. [Use p = 3.14] = 1131428.57 = cm 3 1.131 m 3 Sol. From the figure, it can be observed that 8. A spherical glass vessel has a cylindrical neck 8 cm Height (h1) of larger cylinder 8 cm long, 2 cm in diameter; the diameter o the spherical = 220 cm part is 8.5 cm. By measuring the amount of water it Radius (r1) of larger cylinder holds, a child finds its volume to be 345 cm 3. Check 60 cm 24 whether she is correct, taking the above as the = = 12 cm 2 inside measurements, and p = 3.14. Height (h2) of smaller cylinder Sol. Height (h) of cylindrical part 2 cm = 60 cm = 8 cm Radius (r2) of smaller cylinder 220 cm Radius (r2) of cylindrical part = 8 cm = 1 cm 8 cm Total volume of the pole 12 cm Radius ( r ) spherical part = volume of large cylinder + 1 = 4.25 cm volume of small cylinder.

(

24 cm

= π r12h1 + π r22h2

= π (144 × 220 + 64 × 60= ) 111532.8 cm

Mass of 1 cm iron = 8 g 3

Mass of 111532.8 cm3 = 111532.8 g × 8 g = 892262.4 g = 892.262 kg 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the

314

)

Volume of vessel = Volume of sphere + Volume of cylinder Volume, 4 3 = π r1 + π r22h 3 4 = × 3.14 × 76.765 + 8 × 3.14 3 = 346.512 cm 3

4.25 cm

Multiple Choice Questions 1. The shape of a gilli, in the gilli-danda game figure is a combination of

(NCERT Exemplar)

(a) two cylinders (c) two cones and a cylinder (d) two cylinders and a cone 2. From a solid cube of side 14 cm, a sphere of maximum diameter is carved out. The radius of sphere is 7 (a) 7 cm (b) 14 cm (c) cm (d) 14 cm 2

(CBSE 2024)

3. The curved surface area of a cone having height 24 cm and radius 7 cm is: (b) 1056 cm²

(d) 500 cm² (CBSE 2023)

4. Curved surface area of a cylinder of height 5 cm is 94.2 cm². Radius of the cylinder is: (a) 2 cm (b) 3 cm (c) 2.9 cm (d) 6 cm

(CBSE 2023)

5. The volume of a right circular cone whose area of the base is 156 cm² and the vertical height is 8 cm is: (a) 2496 cm³

(b) 1248 cm³

(d) 416 cm³

(b) 4p : 1

(CBSE 2023)

(d) 1 : 2p

7. A solid spherical ball fits exactly inside the cubical box of side 2a. The volume of the ball is (a)

16 3 πa 3

1 (b) π a 3 6

(c)

32 3 πa 3

4 (d) π a 3 3

8. The radius of a sphere (in cm) whose volume is 12π cm3, is (a) 3 cm

(b) 3 3 cm

2 (c) 3 3 cm

1 (d) 3 3 cm

(CBSE 2020)

Chapter 12 • Surface Areas and Volumes

(d) 14.08 m² (CBSE 2023)

1 (c) π d 2 2

3 (d) π d 2 4

11. If the area of the base of a cone is 51 cm² and its volume is 85 cm³, then the vertical height of the cone is given as (a)

5 cm 6

(b)

5 cm 3

(c)

5 cm 2

(d) 5 cm (CBSE 2023)

12. If the ratio of the side lengths of two cubes is 1 : 4, what is the ratio of their surface areas? (a) 1 : 4 (b) 1 : 8 (c) 1 : 16 (d) 1 : 64 13. A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together, is: (a) 1 : 1

6. The ratio of total surface area of a solid hemisphere to the square of its radius is: (a) 2p : 1

10. What is the total surface area of a solid hemisphere of diameter ‘d’? (a) 3π d 2 (b) 2π d 2

(b) a cone and a cylinder

(a) 528 cm²

9. The area of metal sheet required to make a closed hollow cylinder of height 2.4 m and base radius 0.7 m, is (a) 10.56 m² (b) 13.52 m²

(b) 1 : 4

(d) 3 : 2

14. The volume of the largest right circular cone that can be carved out from a solid cube of edge 2 cm is: (a)

4π cu. cm 3

(b)

5π cu. cm 3

(c)

8π cu. cm 3

(d)

2π cu. cm 3

15. Rahul makes a hemispherical clay pot with inner diameter 24 cm and 3 cm uniform thickness. Find the volume of clay used to make the pot. (a) 243π cm 3

(b) 1098π cm 3

(d) 2250π cm 3

315

Answers 1. (c) Two cones and a cylinder

2. (a) 7 cm

3. (a) 550 cm²

4. (b) 3 cm

5. (d) 416 cm3

6. (c) 3π : 1

2 8. (c) 3 3 cm

4 7. (d) π a ³ 3

11. (d) 5

3 10. (a) π d 2 4 12. (c) 1 : 16

13. (c) 2 : 3

14. (d)

9. (b) 13.52 m²

15. (b) 1098π cm 3

2π cm3 3

Constructed Response Questions Very Short Answer Questions

CBSE Marking Scheme 2019

1. Find the volume of the largest right circular cone that can be cut-out from a cube of edge 4.2 cm. (NCERT Exemplar)

1 Sol. V = π r 2h 3 1 V = π (2.1)2 ( 4.2 ) 3 1 V = π ( 4.41)( 4.2 ) 3 18.522 π V = 3

Sol. Height of cylinder = 20 – 7 = 13 cm and 7 radius = cm 2 2

4 22  7  Total volume = π r 2h + π r 3= × 3 7  2  3

4 22  7  13 + ×   = 22 × 49 × 13 + 4 × 22 × 343 3 7 2 7 4 3 7 8

77 × 53 = = 680.17cm 3 6

1 2

1 1 2

4. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m³?

V ≈ 19.4 cm 3

2. The volume of a right circular cylinder with its 1 height equal to the radius is 25 cm³. Find the 7 22   height of the cylinder.  Use π = 7  

(NCERT Exemplar)

(CBSE 2020)

1 Sol. The volume of a right circular cylinder = 25 cm3 7 176 = cm3 7 The volume of a cylinder is given by:V = π r 2h

Sol. According to the question, Average displacement by a person = 0.04 m³ Average displacement by 500 persons = 500 × 0.04 = 20 m³ Hence, the volume of water raised in pond = 20 m³

As, h = r V = πr 3 176 = 22r 3 176 r3 = 8 = 22 So, h = 2 cm 3. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume 22   of the solid.  Use π = 7   316

h Water

3.5 20 cm

13 cm

3.5 (CBSE 2019)

Pond

Volume of water raised in pond = 80 × 50 × h 20 m 3 = 4000 h = h 0.005 = m 0.5 cm Therefore, the raise in the height of water = 0.5 cm

5. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas. (CBSE 2018) Sol. The volume of a cube is V = s

Sol. The total canvas used is equal to the outer surface area of the tent.

Let the side lengths of the two cubes be s1and s 2 .

Since the volumes are in the ratio 1 : 27, we write:

s13 1 = 3 s 2 27 3

s1 1 1 = = 3 s2 27 3

Total canvas used = Curved surface area of cylinder + Curved surface area of cone

52.5 m

3m D

C 105 m

22    22  2 =  2 × 7 × 52.5 × 3  +  7 × 52.5 × 53  m    

Thus, the ratio of their side lengths is: 1: 3 The surface area of a cube is: S = 6s 2

= [: S 2πrh + πrl ]

S1 6s12 s12 = = S2 6s 22 s 22 Since we found that s1 : s 2 = 1: 3 we square both sides: s12 12 1 = = s 22 32 9

22 = × 52.5 × ( 6 + 53 ) m 2 = 9735 m 2 7

Short Answer Questions 1. The total surface area of a solid cylinder is 231 cm².

Thus, the ratio of their surface areas is: 1: 9 6. A sphere of diameter 12 cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the 5 water level in the cylindrical vessel rises by 3 cm. 9 Find the diameter of the cylindrical vessel. (CBSE 2016)

Topper’s Answer

2 If the curved surface area of this solid cylinder is 3 of its total surface area, find its radius and height. 22    Use π = 7   

Sol. The formula for Curved Surface Area (CSA) of a cylinder is: CSA = 2π rh Since CSA is

2 of TSA, we calculate: 3

2 154 cm 2 CSA = × 231 = 3 So, we get: 2π rh = 154 22 2 × ×r × h = 154 7 44 × r × h = 154 7 44rh = 1078 1078 = rh = 24.5 44

The Total Surface Area (TSA) of a cylinder is: = TSA 2π r (r + h ) 7. A circus tent is cylindrical up to a height of 3 m and conical above it. If the diameter of the base is 105 m and the slant height of the conical part is 53 m, find the total canvas used in making the tent. (CBSE 2004)

Chapter 12 • Surface Areas and Volumes

22 2 × × r (r + h ) = 231 7 44 231 × r (r + h ) = 7 1617 r (r + h= = 36.75 ) 44

317

Using h =

24.5 in the second equation: r

Volume of ice-cream on top (hemisphere): 2 2 = π r 3 = × π × 33 = 18π cm 3 3 3

24.5   = r r + 36.75 r  

Total volume of the ice-cream: = V total V ice −cream, cone + V ice −cream, top = 30π + 18π = 48π cm 3

r 2 + 24.5 = 36.75 2

r = 12.25 Now,

4. The difference between the outer and inner radii of a hollow right circular cylinder of length 14 cm is 1 cm. If the volume of the metal used in making the cylinder is 176 cm³, find the outer and inner radii of the cylinder. (CBSE 2024)

3.5h = 24.5 24.5 h = 7 cm = 3.5

Sol. The volume of a hollow cylinder is given by:

30 cm

(

= V πh R 2 − r 2 1.45 m

)

(

176 =π × 14 × R 2 − r 2

)

22   Using:  π = 7   176 =

22 × 14 × R 2 − r 2 7

(

1232 = 308 × R 2 − r 2

Sol. Let h be height of the cylinder, r be the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath

(

)

= 4 ) 1232 308

R2 −r 2 =

We have two equations: R −r = 1

= CSA of cylinder + CSA of hemisphere = 2π rh + 2π r 2

(

)

R2 −r 2 = 4 (Difference of squares formula)

= 2π r (h + r )

Using:

22 × 30 (145 × 30 ) cm 2 7 = 33000 cm 2

(

=2 ×

)

R 2 − r 2 =( R − r )( R + r )

= 3.3 m 2 3. An empty cone of radius 3 cm and height 12 cm. Ice-cream is filled in it so 1 that lower part of the cone which is of 6 the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. ( Take π = 3.14 ) (CBSE 2023) CBSE Marking Scheme 2023 Sol. Volume of the cone: 1 1 = π r 2h = × π × 32 × 12 = 36π cm 3 3 3

1 2

Volume of ice-cream in the cone: 5 30π cm 3 =× 36π = 6

1 2

318

= 48 × 3.14 = 150.72 cm 3

= r = 12.25 3.5 cm

2. Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see figure). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. 22    Use π = 7   

Substituting: R − r = 1 1× ( R + r ) = 4

Now, we solve: 4 (R + r ) =

R −r = 1, ( R + r ) = 4 Adding both equations: 2R = 5 5 R= = 2.5 cm 2 2.5 R= cm Substituting R = 2.5 cm in : R − r = 1 2.5 − r = 1 = r 2.5 − 1 r = 1.5 cm

4.2 cm

5 cm

5. A decorative block shown in figure is made of two solids—a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter 4.2 cm.

5 cm

Find the total surface area of the 22   block.  Use π = 7   Sol. Let S be the total surface area of the decorative block. Then, S = Total surface area of the cube – Base area of hemisphere + Curved surface area of hemisphere

(

S = 6 × 5 × 5 − π r + 2π r ⇒ S=

(150 + πr ) cm 2

) cm

22   ⇒= × (2.1)2  cm 2 S  150 + 7   ⇒ S=

(150 + 22 × 0.3 × 2.1) cm2

⇒ S=

2 163.86 cm 2 (150 + 13.86 ) cm=

6. A juice glass is cylindrical in shape with a hemi-spherical raised-up portion at the bottom. The inner diameter of the glass is 10 cm and its height is 14 cm. Find the capacity of the glass. (Use π = 3.14) Sol. Diameter of glass = 10 cm ⇒ r = 5 cm Height of glass = 14 cm Capacity of glass = Volume of Cylinder – Volume of Hemisphere 2 = π r 2h −   π r 3 Thus, 3 2r   = πr 2  h −  3   10   =3.14 × 52  14 −  3   32 = 3.14 × 25 × 3 = 837.33 cm 3 So, the capacity of glass in 837.33 cm3 7. A vessel is in the form of a hollow hemisphere surmounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Chapter 12 • Surface Areas and Volumes

Sol. Diameter of the hemisphere = 14 cm 14 ⟹ Radius (r ) = = 7 cm 2 Total height of the vessel = 13 cm Height of the cylindrical portion: The height of the cylinder is the total height minus the radius of the hemisphere: h = 13 − 7 = 6 cm The inner surface area consists of: Curved Surface Area (CSA) of the hollow cylinder CSA cylinder = 2π rh Inner Curved Surface Area of the Hollow Hemisphere CSA hemisphere = 2π r 2 CSA cylinder = 2 × 3.14 × 7 × 6 = 2 × 3.14 × 42 = 263.76 cm 2 2

CSA hemisphere = 2 × 3.14 × ( 7 ) = 2 × 3.14 × 49 = 307.72 cm 2 Inner = Surface Area CSA of Cylinder + CSA of Hemisphere = 263.76 + 307.72 = 571.48 cm 2 8. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (CBSE 2024) Sol. Diameter of the base = 4 cm 4 ⟹ Radius (r ) = = 2 cm 2 Height of the cone = 2 cm

Radius of the hemisphere = 2 cm (same as the cone’s base) The volume of a cone is given by: 1 V cone = π r 2h 3 1 3

= × 3.14 × ( 2 ) × 2 =

1 1 × 3.14 × 4 × 2 = × 25.12 = 8.37 cm 3 3 3

The volume of a hemisphere is: 2 V hemisphere = π r 3 3 2 2 2 3 16.75 cm 3 =× 3.14 × ( 2 ) =× 3.14 × 8 =× 25.12 = 3 3 3 V total =V cone + V hemisphere =8.37 + 16.75 =25.12 cm 3 9. A godown building is in the form as shown in figure. The vertical cross-section parallel to the width side of the building is a rectangle 7 m × 3 m, mounted by a semi-circle of radius 3.5 m. The inner measurements of the cuboidal portion of the building are 10 m × 7 m × 3 m. Find the volume of the godown 319

and the total interior surface area excluding the 22   floor (base).  Use π = 7  

3.5 m

Sol. The top of the building is in the form of half of the cylinder of radius 3.5 m, and length 10 m, split along the diameter. Let V be the volume of the godown. Then, 1 V = Volume of the cuboid + (Volume of the 2 cylinder of radius 3.5 m and length 10 m ⇒V =

1 22 (10 × 7 × 3) +  × 3.5 × 3.5 × 10  m3 2 7 

⇒V =

3 402.5 m 3 ( 210 + 192.5) m=

Let S be the total interior surface area excluding the base floor. Then 1 S = Area of four walls + (Curved surface area of the 2 cylinder) + 2(Area of the semi-circles) 1 22  ⇒ S = [(2 × 10 + 7) × 3] +  2 × × 3.5 × 10  2 7  1 22   +  × × (3.5)2  m 2 2 7  2 250.5 m 2 (102 + 110 + 38.5) m=

1. A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively and the total height of the tent is 13.5 m, find the area of the canvas required for making the tent, keeping a provision of 26 m² of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of `500 per m². (CBSE SQP 2023)

10.5 m

14 m

320

Since the total height of the tent includes both the cylinder and cone, we calculate the height of the cone as: hcone= 13.5 − 3 = 10.5 m The slant height (l) of the cone is found using the Pythagorean Theorem: 2

2 l = r 2 + hcone = (14 ) + (10.5 ) = 306.25l ≈ 17.5 m

Curved Surface Area (CSA) of the Cylinder CSA cylinder = 2π rh = 2 × CSA cone = π rl =

22 × 14 × 3 = 264 m 2 7

22 × 14 × 17.5 = 770 m 2 7

Total Canvas Area = CSA of Cylinder + CSA of Cone + Stitching and Wastage = 264 + 770 + 26 = 1060 m 2 Total Cost = Total Canvas Area × Cost per m2 = 1060 × 500 = `5,30,000 2. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour? CBSE Marking Scheme 2021

Long Answer Questions

Sol.

Total height of the tent = 13.5 m Cost per m² = `500

10 m

⇒ S=

Height of the cylindrical part = 3 m Canvas wastage and stitching provision = 26 m²

3.5 m 7m

Radius of both cylinder and cone = 14 m

Sol. For the pipe, radius r = 1 cm Length of water flowing in 1 sec, = h 0.7 = m 70 cm Cylindrical Tank, R = 40cm, Rise in water level = H

Volume of water flowing in 1 sec = π r 2h

1 2 1 2 1 2

V = π × 1 × 1 × 70 = 70π cm 3

1 2

Volume of water flowing in 60 sec = 70π × 60

Volume of water flowing in 30 minutes = 70π × 60 × 30 Volume of water in tank =π R 2 HV =π × 40 × 40 × H

1 2 1 2

Equating volume of water in tank to

1 2

volume of water flowing in 30 minutes: π × 40 × 40 × H = 70π × 60 × 30 70π × 60 × 30 H= π × 40 × 40

432π 768π 1200π V total = 9.6 × 3.14 cm 3 = 30.144 cm 3 + = 125 125 125

4. A building is in the form of cylinder surmounted by a hemispherical dome (see the figure). The base diameter of the 2 dome is equal to of the total 3

1 2

H = 78.75 cm P

m 4c

3c m

3. A right triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of double cone thus generated. ( Use π = 3.14 )

Hence total volume, i.e., volume of the double cone:

Sol. Given:

M x 5 cm

PQR ∆ is a right triangle

where = PQ 3= cm, PR 4 cm,= and QR 5 cm

( Pythagoras Theorem )

32 =x 2 + y 2 ⇒ y 2 =9 − x 2 (i) From right-angled triangle POR: OP 2 + OR 2 = PR 2 2

y 2 + ( 5 − x ) = 42 ⇒ y 2 = 16 − ( 5 − x ) (ii) Combining (i) and (ii): 2

(

9 − x 2 =16 − ( 5 − x ) ⇒ 9 − x 2 =16 − 25 + x 2 − 10x or 9 − x 2 =−9 + 10x − x 2 10x = 18 ⇒ x = 9 16 = 5 5 Substituting x back into (i):

)

9 5

∴ OR = 5 − x = 5 − 2

81 144 12 9 = y= 9−  = 9− y2 = 25 25 5 5 12 cm and height For the cone PQM , radius OP = 5 9 OQ = cm : 5 2

1 2 1  12  9 432π π r= π = = × cm 3 V h 3 3  5  5 125 12 cm and height For the cone PRM , radius OP = 5 16 cm : OR = 5

1  12  16 768π π = = × cm 3 V 3  5  5 125 Chapter 12 • Surface Areas and Volumes

H h

height of the building. Find the height of the building, if it 1 contains 67 m³ of air. 21 Sol. Let the total height of the building = H m 2 Base diameter of the dome = H 3 1 Radius of the dome and cylinder (r ) = H 3

Given volume of air = 67

From right-angled triangle POQ :

a r

Height of the cylindrical part = h

Let: OQ = x ,OR = 5 − x ,OP = y 2 PQ = OQ 2 + OP 2

1 1408 3 m³ = m 21 21

Since the total height of the building consists of the cylinder and hemisphere, we write: H= h + r 1 Since we know r = H , we substitute: 3 1 H= h + H 3 1 2 h= H − H =H 3 3 The volume of a cylinder is: V cylinder = π r 2h 1 2 Substituting r = H and h = H : 3 3 2

1 2 2 1  2 π  H  × H =× π V cylinder = H × H 3 3 9 3   2 2π = π × H 3 =H 3 27 27 The volume of a hemisphere is: 3

2 2 1  2 1 2π V hemisphere =π r 3 =π  H  =π × H 3 = H 3 3 3 3  3 27 81 2π 2π 3 V total = V cylinder + V hemisphere =H 3 + H 27 81  6π 3 2π 3  8π 3 = H + H = H 81  81  81

321

1408 Given that total volume = , we equate: 21 8π 1408 H 3 = 81 21 22 8× 176 1408 176 3 1408 7 H 3 1408 = = H3 = H 81 21 7 × 81 21 567 21

2 = S 16.81 + 6 × 3.14 cm = 22.81 cm 2

22.81 × 7 = cm 2 71.68 cm 2 S = 22 6. A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model. (CBSE 2023)

1408 × 567 797376 = H3 = 176 × 21 3696 H 3 = 216 = H 3= 216 6 m

Sol. Diameter of the model = 3 cm

5. A golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings assuming that the dimples are hemispherical.

⟹ Radius (r ) =

πr2

3 = 1.5 cm 2

Total length of the model = 12 cm

2πρ2

Height of each cone = 2 cm Since there are two cones, their total height is =2+2= 4 cm Height of the cylindrical portion:

= hcylinder Total length − Total height of cones hcylinder

 4.1  2 Sol. Surface area of the ball = 4π   = 16.81 cm  2  2

In case of each dimple, surface area equal to π r (where r is the radius of each dimple) is removed from the surface of the ball whereas the surface area of hemisphere i.e 2π r 2 is exposed to the surroundings. Let S be the total surface area exposed to the surroundings. Then, 2

S Surface area of the ball − 150 × π r + 150 × 2π r

 2  S= 16.81 + 150 × π   cm 2  10   4  2 S= 16.81 + 150 × π   cm  100  = S 16.81 + 6π cm 2

= 12 – 4 = 8 cm The volume of a cylinder is: 2

V cylinder= π r 2h = 3.14 × (1.5 ) × = 8 3.14 × 2.25 × 8

= 3.14 × 18= 56.52 cm 3

The volume of a cone is: 1 1 1 2 V cone =π r 2h =× 3.14 × (1.5 ) × 2 =× 3.14 × 2.25 × 2 3 3 3 1 14.13 = × 3.14 × 4.5 = =4.71 cm 3 3 3 Since there are two cones, the total volume of both cones is: V total cones = 2 × 4.71 = 9.42 cm 3 V total = V cylinder + V total cones = 56.52 + 9.42 = 65.94 cm 3

Competency Based Questions Multiple Choice Questions 1. If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of the original cylinder is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4

322

(d) 4 : 1

2. A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is (a) 3.5 (b) 7 (c)

80 7

(d) 5

(CBSE 2014)

3. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. How many marbles have been dropped in it if the water rises by 5.6 cm?

(a) 140 cm 2 (b) 142 cm 2

(b) 100

(d) 200

4. A container with a blue hemispherical lid has radius R cm. In figure 1, it contains water upto a height of R cm. It is then inverted as shown in figure 2. What is the height of water in figure 2?

Figure 1

(a)

7R cm 3

(b) 2R cm

5R cm 3

(d) R cm

Bottom view

7 cm

21 cm (CBSE QB)

(b) 1012 cm²

(d) 1386 cm²

(a) 80π cm 3

(b) 160π cm 3

(d) 240π cm 3

10. 6 spherical glitter balls with radius 0.5 cm are present in a cylindrical candle made with transparent wax as shown in the figure below. Find the volume of wax used to make the candle.

6. A solid spherical ball fits exactly inside the cubical

(a) 70π cm 3

16a 4 ⋅ a −1 box of side . The volume of the ball is 2a

(b) 71π cm 3

16 πa³ 3 32 (c) πa³ 3

1 (b) π a ³ 6 4 (d) π a ³ 3

(CBSE 2020)

(d) 142596 (NCERT Exemplar)

Chapter 12 • Surface Areas and Volumes

8 cm

7. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is 1 assumed that space of the cube remains unfilled. 8 Then the number of marbles that the cube can accommodate is (a) 142296 (b) 142396 (c) 142496

4.5 cm Side view

What is the volume of the foam used to make a set of 10 bullets?

7 cm

(a) 976 cm²

(a)

2 cm

8 cm

5. Abhinav made a model for his school project in the shape of a cylinder of radius 7 cm and height 21 cm, with a hemisphere surmounted on one end. He wants to cover the entire model with decorative paper. What is the area, in centimeter square, of paper that is required to cover the model? 22 (Use p = ) 7

5 cm

9. A toy company manufactures hollow foam bullets with plastic tops for their toy guns, the dimensions of which are shown in the figure.

Figure 2

(c)

Area = 4 cm2

(a) 50

8. A square hole is drilled across a cube with its length parallel to a side of the cube. What is the total surface area of the body so formed?

6 cm

Select Response Type MCQs 1. Surahi is made up of (i) a cylinder (ii) a cone (iii) a sphere (iv) a hemisphere (a) (i) and (iii)

(b) (ii) and (iii)

(d) (i) and (iv)

323

2. Surface area of a cylinder closed at both ends is equal to sum of which of the following: (i) 2π rh 2

(ii) 2π r 2h

(iii) 2π r

(iv) 4π rh

(a) (iii) and (iv)

(b) (i) and (iv)

(d) (i) and (iii)

Assertion-Reason Based Questions Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option:

Based on the above, answer the following questions: (a) Find slant height of the conical part. (b) Determine the floor area of the tent. (c) (i) Find area of the cloth used for making tent.

(ii) Find total volume of air inside an empty tent. (CBSE 2024)

2. On a Sunday, your parents took you to a fair. You could see lots of toys displayed and you wanted them to buy a RUBIK’S cube and strawberry ice cream for you. Observe the figures and answer the following questions:

(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): The total surface area of a combination of solids is always equal to the sum of the surface areas of individual solids. Reason (R): When two solids are combined, some surfaces are hidden or joined, so they do not contribute to the total surface area. 2. Assertion (A): The volume of a solid hemisphere is always half the volume of a sphere with the same radius. 4 Reason (R): The volume of a sphere is given by π r 3 3 2 3 while the volume of a hemisphere is given by π r . 3

Case Study Based Questions 1. The word ‘circus’ has the same root as ‘circle’. In a closed circular area, various entertainment acts including human skill and animal training are presented before the crowd. A circus tent is cylindrical up to a height of 8 m and conical above it. The diameter of the base is 28 m and the total height of the tent is 18.5 m.

324

(a) The length of the diagonal of rubik’s cube, if each edge measures 6 cm, is: (i) 3 3 cm

(ii) 3 6 cm

(iii) 2 3 cm

(iv) 6 3 cm

(b) Volume of the solid figure, if the length of the edge is 7 cm, is:

(i) 256 cm³

(ii) 196 cm³

(iii) 343 cm³

(iv) 434 cm³

(i) 309 cm²

(ii) 308 cm²

(iii) 803 cm²

(iv) 903 cm²

(d) Slant height of a cone if the radius is 7 cm and the height is 24 cm is:

(i) 26 cm

(ii) 25 cm

(iii) 52 cm

(iv) 62 cm

(e) The total surface area of cone with hemispherical ice cream is:

(i) 858 cm²

(ii) 885 cm²

(iii) 588 cm²

(iv) 855 cm²

Answers Multiple Choice Questions

≈ 1.435 cm3

1. (c) 1 : 4

N =

To solve the problem, we consider the formula for the volume of a right circular cylinder: V = π r 2h New radius when halved: rnew = 2

r 2 2

Ratio of the new volume to the original volume: V new = V

r2 h 1 4 = 2 πr h 4

Thus, the ratio of the volume of the cylinder obtained by halving the radius to the volume of the original cylinder is 1: 4. 2. (a) 3.5 cm Given, = C 22 = cm 2π r 22 =2 × 22 =

22 ×r 7

44 r 7

22 × 7 r = ⇒ 3.5 cm 44 3. (c) 150 Diameter of the beaker = 7 cm Radius rb=

7 = 3.5 cm 2

Rise in water level = 5.6 cm V water = π rb2 × rise = π (3.5)2 × 5.6 Marble diameter = 1.4 cm, thus the radius 1.4 r= = 0.7 m 2 4 3 4 V marble = = π rm π (0.7)3 3 3 N = number of marbles N=

V water V marble

V water = π (3.5)2 × 5.6 = 3.14 × 12.25 × 5.6 ≈ 214.67 cm 3 4 4 4 V marble =π (0.7)3 =× 3.14 × 0.343 =× 1.07622 3 3 3 Chapter 12 • Surface Areas and Volumes

Rounding this to the nearest whole number, the estimated number of marbles is approximately 150. 5R cm 3 The volume V of a spherical cap of height h in a hemisphere of radius R is given by the formula: 4. (c)

r r  V new π= = New Volume: 2 h π 4 h  

Ratio =

214.67 ≈ 149.57 1.435

1 2 = π h ( 3R − h ) V 3

For our case, h = R, so we substitute: 1 2 2 = π h ( 3R= −R) πR3 V 3 3 Setting the two volumes equal gives us: 2 π R 2h − π R 3 = πR3 3 2 2 ⇒ π R h =π R 3 + π R 3 3 2   ⇒ π R 2h = π R 2  R + R  3   5R ∴h = 3 5. (c) 1232 cm² We have, radius of cylinder = radius of hemisphere = 7 cm and height of cylinder h = 21 cm. Total Surface Area (TSA) of project = Curved Surface Area (CSA) of cylinder + CSA of Hemisphere

TSA of project = 2π rh + 2π r 2 = 2π (h + r ) = 2×

22 × 7 × 28 = 1232 cm 2 7

4 πa³ 3 The radius rrr of the sphere is half of the diameter: D 2a r = = a = 2 2 4 3 V = πa 3 6. (d)

7. (a) 142296 3 V cube = 22 = 10648 cm 3

1 × 10648 = 1331 cm 3 V unfilled = 8

V marbles = 10648 − 1331 = 9317 cm 3 325

0.5 1 = = cm r = 0.25 cm 2 4 3

= V marble

4 3 4 1 1 = πr π= π cm 3 3 3  4  48

V marble =

22 1 11 × ≈ cm 3 7 48 168

9317 = n = 142296 11 168 8. (d) 182 cm 2

Assertion-Reason Based Questions 1. (d) A is false, but R is true 2. (a) Both A and R are true, and R is the correct explanation of A.

Case Study Based Questions 1. (a) We have, r = radius of base = 14 cm, h = height of conical part = 10.5 m Let l be the slant height of the conical part. Then, r 2 + h2 =

142 + (10.5)2 =

TSA of cube: =6 × 25 =150 cm 2

l =

Minus the hollow part: (150 − 8 ) cm 2 = 142 cm 2

= 306.25 = 17.5 m

Surface area of hollow part = 4 × 5 × 2 = 40

(b) Floor area of tent = π r 2 =

Total surface area of body:142 + 40 = 182 cm 2 9. (b)160π cm 3

V inner = π × 0.25 × 8 = 2π V foam = 18π − 2π = 16π

= 2π rH + π rl , where H = Height of cylindrical part = π ( 2H + l )r =

(ii) V olume of air inside tent = Volume of cylindrical part + Volume of conical part 1 1   =π r 2 H + π r 2h =π r 2  H + h  3 3  

10. (b) 71π cm 3

V cylinder = π= r 2h π (3)2 × 8

Volume of 6 balls = 4 3 6 π × ( 0.5 ) = π V =× 3

22 × 14 × ( 2 × 8 + 17.5 ) m 2 7

= 44 × 33.5 m2 = 1474 m2

V foam, 10 bullets = 10 × 16π = 160π cm 3

V cylinder = π × 9 × 8 = 72π cm 3

22 × 142 =616 m 2 7

Volume of cylinder V = π r 2h V outer = π × 2.25 × 8 = 18π

196 + 110.25

22 1   × 142 ×  8 + × 10.5  = 22 × 28 × 11.5 = 7084 m 3 7 3  

2. (a) (iv) Using the formula, Diagonal of a cube = 3 × Side, we obtain Length of the diagonal

= 6 3 cm

(b) (iii) Volume = 73 = 343 cm³

22 2 308 cm² ×7 = 7

V total balls = π cm 3

= V wax V cylinder − V total balls

(d) (ii) Slant height = 72 + 242 = 25 cm

V wax= 72π − π= 71π cm 3

(e) (i) Total surface area =

Select Response Type MCQs 1. (a) (i) and (iii) 2. (d) (i) and (iii)

326

22 22 × 7 × 25 + 2 × × 7 × 7 7 7

= 858 cm2

Additional Practice Questions 1. W ater in a river which is 3 m deep and 40 m wide is flowing at the rate of 2 km/h. How much water will fall into the sea in 2 min? (a) 8000 m³ (b) 4000 m³ (c) 8000 m³

(d) 2000 m³

(CBSE 2023)

2. The volume of a right-circular cone whose area of the base is 156 cm² and the vertical height is 8 cm, is (a) 2496 cm³ (b) 1248 cm³ (c) 1664 cm³

(d) 416 cm³

(CBSE 2023)

3. If the diagonal of a cube is 17.32 cm, then its volume is (a) 10000 cm3 (b) 1732 cm3 (c) 1000 cm3

(d) 173.2 cm3

4. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm, and it is 1 assumed that th of the space of the cube remains 8 unfilled. Then, the number of marbles that the cube can accommodate is (a) 142244 (b) 142296 (c) 142496

(d) 142596

8. A hemispherical tank full of water is emptied by a 4 pipe at the rate of 3 litres per second. How much 7 time will it take to half empty the tank, if it is 3 m in 22   diameter?  Use π =  (CBSE 2019) 7   9. A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm which is surmounted by another cylinder of height 50 cm and radius 7 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (CBSE 2024) 10. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 4 mm. Find its surface area. Also, find its volume. (CBSE 2024) 4 mm 14 mm

11. In figure, from a cuboidal solid metallic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area 22   of the remaining block.  Use π = 7   7 cm

Calculate the rise in water level, assuming that no water overflows. (a) 1 cm (b) 2.5 cm (c) 1.75 cm

(d) 3 cm

6. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? 7. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 gm mass. (Use p 3.14) (CBSE 2019)

Chapter 12 • Surface Areas and Volumes

5 cm

5. A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. m

c 10

15 cm

12. Deepika takes a solid cylinder and attaches it to a solid cone to make the figure of a tree as shown in the figure below.

14 cm

7 cm

4 cm

6 cm

Green

Brown

She wants the conical part, including its base, to be painted green and the cylindrical part, including its base, to be painted brown to resemble a tree. (CBSE QB)

327

Based on the above information, answer the following questions: (a) Find the area to be painted green. (b) Find the area to be painted brown. 22    Take π as 7 .    13. Temper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals. 500 mL milk is packed in a cuboidal container of dimensions 15 cm × 8 cm × 5 cm. These milk packets are then packed in cuboidal cartons of dimensions 30 cm × 32 cm × 15 cm.

7 cm

5 cm

Based on the above information, answer the following questions. (a) Find the volume of the cuboidal carton. (b) (i) Find the total surface area of a milk packet.

(ii) How many milk packets can be filled in a carton?

Challenge Yourself

Scan me

1. A child consumed an ice-cream of inverted right– circular conical shape form the top and left only 12.5% of the cone for her mother. If the height of the ice-cream cone was 8 cm, what was the height of the remaining ice-cream cone? (a) 2.5 cm (b) 3 cm (c) 3.5 cm

(d) 4 cm

2. A solid cube has a square hole cut through horizontally and a circular hole cut through vertically. Both the holes are cut centrally in appropriate faces. The dimensions of the cube and the hole are shown in the diagram. Calculate the volume remaining after the holes have been cut. (Take p = 3.14) 8 cm

20 cm

B B

12 cm

A D 40

(d) 5994.2 cm³

3. Shown in figure, a closed box which is a prism of length 40 cm. The cross-section of the box is shown in with all right angles marked.

328

22 cm

18 cm D

The approximate volume of the prism is (a) 15200 m 3

(b) 14600 m 3

(d) 12500 m 3

4. Find the volume of the space covered by rotating a rectangular sheet of dimensions 16.1cm × 7.5 cm along its length. 5. A right circular cone is cut by two planes parallel to the base and trisecting the altitude. What is the ratio of the volumes of the three parts; top, middle and bottom respectively? A

(b) 5497.6 cm³

12 cm O

10 cm

(a) 4995.2 cm³

P h G

F Q h

Answers Additional Practice Questions 1. (a) 8000 m³ 2. (b) 1248 cm³ 3. (c) 1000 cm3 4. (b) 142296 5. (c) 1.75 cm 6. 1 hours 40 minutes 7. 892.26 kg 8. 16.5 minutes 9. 1047.2 kg 10. Surface area of the capsule = 176 mm², Volume of the capsule = 159.23 mm³ 11. 583 cm2

12. (a) Area to be painted green: 147.58 cm² (b) Area to be painted brown: 100.48 cm² 13. (a) 14400 cm3 (b) (i) 470 cm2 (b) (ii) 24 (c) 550 cm3

Challenge Yourself 1. (d) 4 cm 2. (b) 5497.6 cm³ 3. (b) 14600 m 3 4. 2684 cm3 5. 1: 7 :19

Scan me for Exemplar Solutions Chapter 12 • Surface Areas and Volumes

329

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. The radii of bases of cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3, then ratio between the volumes of cylinder and that of cone is (a) 8 : 9 (b) 9 : 8 (c) 5 : 7 (d) 7 : 5 2. If the volume of a sphere is (a) 2 cm

11 cm3, then the radius of the sphere is: 21 1 (b) 4 cm (c) cm 2

(d)

1 cm 4

3. From a solid cube of side 14 cm, a sphere of maximum diameter is carved out. The radius of sphere is: (a) 7 cm

(b) 14 cm

(c)

7 cm 2

(d) 14 cm (CBSE 2024)

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 9 . 4 Reason (R): The ratio of surface areas of two hemispheres is equal to the ratio of square of their respective radii. 4. Assertion (A): If the volume of two hemispheres are in the ratio 27 : 8 then the ratio of their surface areas is

(CBSE 2023)

5. Assertion (A): Total surface area of the toy is the sum of the curved surface area of the hemisphere and the curved surface area of the cone. Reason (R): Toy is obtained by fixing the plane surfaces of the hemisphere and cone together.

(CBSE 2023)

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. In given figure, a solid toy is in the form of a hemisphere surmounted bya right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. determine the volume of toy. ( Use π = 3.14 ) 2 cm

2 cm

1 7. The volume of a right circular cylinder with its height equal to the radius is 25 cm 3. Find the height of the 7 cylinder. ( Use π = 3.14 )

330

Short Answer Questions

[4 × 3 = 12 Marks]

8. A solid sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much height will the level of water rise in the cylindrical vessel? 9. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of the same height and same base removed. Find the volume of the remaining solid. (CBSE 2020) 54 10. A building model is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains m³ of 21 4 air. If the internal diameter of dome is equal to of the total height above the floor, find the height of the building. 5 11. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap? (CBSE 2018)

Long Answer Questions

[3 × 5 = 15 Marks]

12. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 4 mm. Find its surface area. Also, find its volume.

4 mm

14 mm

13. A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model. 14. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of circular ring of width 4 m to form an embankment. Find the height of the embankment.

Case Study Based Questions

[1 × 4 = 4 Marks]

15. In a coffee shop, coffee is served in two types of cups. One is cylindrical in shape with a diameter of 7 cm and height 14 cm, and the other is hemispherical with a diameter of 21 cm.

Based on the above, answer the following questions: (a) Find the area of the base of the cylindrical cup. (b) (i) What is the capacity of the hemispherical cup?

(ii) Find the capacity of the cylindrical cup.

Self-Assessment

(CBSE 2023)

331

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Multiple Choice Questions

Max. Marks: 40 [3 × 1 = 3 Marks]

1. A roller leveled an area of 165000 m2 in 125 revolutions, whose length is 28 m. Find the radius of the roller. (a) 7.5 m (b) 8.5 m (c) 6.5 m (d) 7 m 2. A cylinder, a cone and a hemisphere are on the same base and having same height. The ratio of their volumes is (a) 1: 2 : 3 (b) 2 :1: 3 (c) 3 :1: 2 (d) 3 : 2 :1 3. If a cone of greatest possible volume is hollowed out from a solid wooden cylinder, then the ratio of the volume of remaining wood to the volume of cone hollowed out is: (a) 1 : 1 (b) 1 : 3 (c) 2 : 1 (d) 3 : 1 (CBSE 2025)

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): Two cubes of edge length 10 cm are joined together. The total surface area of the newly formed cuboid is 1200 cm². Reason (R): Area of each surface of a cube of side 10 cm is 100 cm². 5. Assertion (A): The radii of two cylinders are in the ratio 2 : 5 and their heights are in the ratio 3 : 4. Then ratio of their curved surface area is 3 : 10. 4 Reason (R): Volume of sphere of radius r is π r 3. 3

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. A cuboid has a hemisphere carved out from one of its faces. If the cuboid’s dimensions are 10 cm × 8 cm × 6 cm and the hemisphere’s radius is 3 cm, find the volume of the remaining solid. 7. The water from a rectangular field measuring 22 m by 20 m is collected into a cylindrical tank with a 2 m diameter and a 3.5 m height. Assuming the tank is exactly filled by the water from the field, determine the depth of the water (in cm) that was on the field.

Short Answer Questions 8. A ball contains 5000 cm³ of air. More air is pumped into the ball at a rate of 160 cm³ per second. The ball is full of air when it becomes a sphere with radius 15 cm. How much time it will take to fill the ball? (Approx to nearest whole number) 332

[4 × 3 = 12 Marks]

9. Two solid cones A and B are placed in a cylindrical tube in given figure. The ratio of their capacities are 2 : 1. Find the heights and capacities of cones. Also, find the volume of the remaining portion 6 cm of the cylinder.

21 cm A

10. The base of a cone is fixed to the top of a cylinder to make a decoration.

22   The total volume of the decoration is 77,616 cm 3.  Use π = 7   Calculate the value of r.

11. A wall 24 m long. 0.4 m thick and 6 m high which is constructed with the bricks, each of 1 dimensions 23 × 16 × 10 (cm). If the mortar occupies th of the volume of the wall, then 10 find the number of bricks used in constructing the wall.

Long Answer Questions

[3 × 5 = 15 Marks]

12. On the day of her examination, Riya sharpened her pencil from both ends as shown below:

105.6 mm

2.8 mm

The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and length of entire pencil is 105.6 mm, find the total surface area of the pencil. (CBSE 2025) 13. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 gm mass. (Use π = 3.14) (CBSE 2012, 2019) 14. From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface 22   area of the remaining solid.  = Use π = , 5 2.2  7  

Case Study Based Questions

(CBSE 2025)

[1 × 4 = 4 Marks]

15. A golf ball is spherical with about 300-500 dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure, a golf ball has a diameter of 4.2 cm and the surface has 315 dimples (hemi-spherical) of radius 2 mm. Based on the above, answer the following questions:

(CBSE 2023)

(a) Find the surface area of one such dimple. (b) Find the volume of the material dug out to make one dimple. (c) Find the total surface area exposed to the surroundings.

Find the volume of the golf ball.

Self-Assessment

333

Statistics

Chapter at a Glance Mean Mean of Ungrouped Data: Mean of observations is the sum of the values of all the observations divided by the total number of observations. If x1 , x 2 , x 3 ,..., xn are observations with respective frequencies f1 , f 2 , f 3 ,..., fn then mean x is given by x=

x1 f1 + x 2 f 2 +  + xn fn f1 + f 2 +  + f n n i =1 i i n i =1 i

∑ x f ∑x f = x = ∑ f ∑f

i i i

Mean is sometimes called arithmetic mean and is denoted by x . Mean of Grouped Data with Continuous Class-intervals: Mean (Arithmetic mean) of grouped data is computed by any one of the following methods: (a) Direct Method Mean ( x ) =

∑ fi xi ∑ fi

Where xi (class mark) is the mid-point of the i th class interval and fi is the corresponding frequency. xi =

Upper class limit + Lower class limit 2

(b) Assumed Mean Method (short cut method) ∑ fi di Mean ( x )= a + ∑ fi xi − a are the deviations of d= xi − a (class mark) from a for each i . where a is the assumed mean and d= i i (c) Step Deviation Method ∑fu  Mean ( x )= a + h  i i   ∑ fi  Where a is the assumed mean, h is the class size, and ui =

Mode

xi − a . (Note: This formula holds even if class size varies) h

Mode is that value among the observations which occurs most often, i.e., the value of the observation having the maximum frequency Mode of Ungrouped Data: It is the value that occurs most frequently in a data set. It helps identify the most typical or common observation. Mode of Grouped Data with Continuous and Equal Class Intervals: To compute mode, we locate a class with the maximum frequency, called the modal class. Then we apply the following formula:

  f1 − f 0 Mode = l +  ×h  2 f1 − f 0 − f 2 

334

where, l = lower limit of the modal class, h = size of the class interval (all class sizes to be equal), f1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class.

Median For Ungrouped Data: First arrange the data in ascending or descending order, then use the formulas given below. If n is odd, th  n +1 Median =   term  2  If n is even, n    2 Median =  

n  term +  + 1  2   2 For Grouped Data:

term

n n and then the class in which lies. 2 2 This class is known as the median class. Median of the given distribution lies in this class. Let n = f1 + f 2 + f 3 +  + fn . First of all, find

Median of the grouped data can be calculated using the formula: n   2 − cf  Median ( M e ) = l +  ×h  f    Where, l = lower limit of median class, f = frequency of median class, n = number of observations, cf = cumulative frequency of the class preceding the median class, h = class-size or width of the class-interval. Cumulative Frequency: The cumulative frequency is found by continuously adding each frequency to the sum of all previous frequencies. This running total at any point represents the cumulative frequency up to that data point.

Empirical Relation Between Mean, Median and Mode Made = 3 Median − 2 Mean Median = Mean

Mode + 2 Mean 3

= 3 Median − Mode 2

NCERT Zone NCERT Exercise 13.1 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding

Chapter 13 • Statistics

the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number 0−2 2−4 4−6 6−8 8−10 10−12 12−14 of Plants Number 1 2 1 5 6 2 3 of Houses 335

Which method did you use for finding the mean, and why? Sol. To find the class mark (xi) for each interval, the following relation is used. Upper limit + Lower limit Class mark xi = 2 xi and fi xi can be calculated as follows. Number of Plants 0μ2 2μ4 4μ6 6μ8 8μ10 10μ12 12μ14 Total

Number of Houses 1 2 1 5 6 2 3 20

 12  24 = 545.2 = 550 +  −  = 20 550 − 50 5   Therefore, the mean daily wage of the workers of the factory is ` 545.20

fi xi

1 3 5 7 9 11 13

1×1=1 2×3=6 1×5=5 5 × 7 = 35 6 × 9 = 54 2 × 11 = 22 3 × 13 = 39 162

∑ fi xi = 162

∑ fi xi 162 Mean, x = = = 8.1 ∑ fi 20 Therefore, mean number of plants per house is 8.1. Here, direct method has been used as the values of class marks xi and fi xi are small. 2. Consider the following distribution of daily wages of 50 worker of a factory. Daily Wages 500−520 520−540 540−560 560−580 580−600 (in `) Number of 12 14 8 6 10 Workers Find the mean daily wages of the workers of the factory by using an appropriate method. Sol. To find the class mark for each interval, the following relation is used. Upper limit + Lower limit Class mark xi = 2 Class size (h) of this data = 20 Taking 550 as assured mean (a), di , ui , and fiui can be calculated as follows.

500−520 520−540 540−560 560−580 580−600 Total 336

12 14 8 6 10 50

510 530 550 570 590

di = xi -550 ui =

-40 -20 0 20 40

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is `18. Find the missing frequency f. Daily Pocket 11−13 13μ15 15μ17 17μ19 19μ21 21μ23 23μ25 Allowance (in `) Number 7 6 9 13 f 5 4 of Workers Sol. To find the class mark (xi) for each interval, the following relation is used.

∑ fi = 20

∑ fi = 50, ∑ fiui = −12 ∑fu  Mean, x = a +  i i  h  ∑ fi 

From the table, it can be observed that

Daily Wages Number of (in `) Workers ( fi )

From the table, it can be observed that

di 20

-2 -1 0 1 2

f i ui -24 -14 0 6 20 -12

xi = Upper limit + Lower limit 2 Given that, mean pocket allowance, Taking 18 as assured mean (a), di and fi di are calculated as follows. Daily Pocket Number of Class d = xi -18 Allowance (in `) Children fi Mark xi i

fi di

11-13

-6

-42

13-15

-4

-24

15-17

-2

-18

17-19

19-21

21-23

23-25

Total

∑ fi = 44 + f

From the table, we obtain fi 44 + f , ∑ fiu= 2 f − 40 ∑ = i x= a +

 2 f − 40  ∑ fi di ⇒ 18 = 18 +   ∑ fi  44 + f 

 2 f − 40  0 =  0 ⇒ f = 20  ⇒ 2 f − 40 =  44 + f  Hence, the missing frequency, f is 20.

2f - 40

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of Heart Beats 65–68 68–71 71–74 74–77 77–80 80–83 83–86 per Minute Number of 2 4 3 8 7 4 2 Women

Number of Mangoes Number of Boxes

50–52

135

ui =

di 3

f i ui

65–68

66.5

-9

-3

-6

68–71

69.5

-6

-2

-8

71–74

72.5

-3

-1

-3

74–77

75.5

77–80

78.5

80–83

81.5

83–86

84.5

Total

From the table, we obtain ∑ fi = 30, ∑ fiui = 4 ∑fu  Mean, x= a +  i i  h  ∑ fi   4  = 75.5 +   3  30  = 75.5 + 0.4 = 75.9

Therefore, mean hear beats per minute for these women are 75.9 beats per minute. 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Chapter 13 • Statistics

53–55

110

59–61

115

62–64

di = xi -75.5

135

115

x −a Midpoint ui = h x

56–58

110

Class (mangoes)

Upper limit + Lower limit xi = 2 Class size, h, of this data = 3

Number of Heart Number Beats per of Women Minute

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Sol. To find the class mark of each interval (xi), the following relation is used.

Taking 75.5 as assumed mean (a), di , ui , and fiui can be calculated as follows.

50–52 53−55 56−58 59−61 62−64

Total

−3

−45

−2

−220

−1

400

f i ui

−135

−375

Parameters: = a 60, = h 3, ∑ fi = 400, ∑ fiui = −375 ∑fu  a +  i i  ⋅h Mean formula: x =  ∑ fi   −375  x = 60 +   ⋅ 3 = 60 − 2.8125 = 57.19  400  Mean number of mangoes per box = 57.19 6. The table below shows the daily expenditure on food of 25 households in a locality. Daily Expenditure 100−150 150−200 200−250 250−300 300−350 (in `) Number of Households

Find the mean daily expenditure on food by a suitable method. Sol. To find the class mark (xi) for each interval, the following relation is used. Upper limit + Lower limit , Class size = 50 2 Taking 225 as assumed mean (a), di , ui , and fiui are calculated as follows: xi =

Daily Expenditure (in `)

100−150

125

-100

-2

-8

150−200

175

-50

-1

-5

200−250

225

250−300

275

300−350

325

100

Total

di = xi -225 ui =

di 50

f i ui

-7 337

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

From the table, we obtain ∑ fi = 25 ∑ fiui = −7

∑fu  7 Mean, x= a +  i i  h= 225 +  −  50 = 211  25   ∑ fi  Therefore, mean daily expenditure on food is ` 211. 7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: Concentration of SO2 (in ppm)

Frequency

0.00−0.04

0.04−0.08

0.08−0.12

0.12−0.16

0.16−0.20 0.20−0.24

Number of Days

0-6

6-10 10-14 14-20 20-28 28-38 38-40

Number of Students

Sol. To find the class mark of each interval, the following relation is used. xi = Upper limit + Lower limit 2 Taking 17 as assumed mean (a), di , ui , and fiui are calculated as follows: Number of Days

Number of Students fi

di = xi -17

fi di

0-6

-14

-154

6-10

-9

-90

Find the mean concentration of SO2 in the air.

10-14

-5

-35

Sol. To find the class marks for each interval, the following relation is used.

14-20

20-28

28-38

38-40

Total

Upper limit + Lower limit 2 Class size of this data = 0.04 Taking 0.14 as assumed mean (a), di , ui , and fiui are calculated as follows: xi =

di 50

Concentration of SO2 (in ppm)

0.00−0.04

0.02

-0.12

-3

-12

0.04−0.08

0.06

-0.08

-2

-18

0.08−0.12

0.10

-0.04

-1

-9

0.12−0.16

0.14

0.16−0.20

0.18

0.04

0.20−0.24

0.22

0.08

Total

di = xi -225 ui =

f i ui

-31

From the table, we obtain −31 ∑ fi =, 30 ∑ fiui = ∑fu  Mean, x= a +  i i  h= 0.14 +  − 31  0.04  ∑ fi   30  = 0.14 − 0.04133 = 0.09867 = 0.099 ppm Therefore, mean concentration of SO2 in the air is 0.099 ppm. 338

-181

From the table, we obtain −181 ∑ fi = 40, ∑ fiui =  ∑ fiui   181  Mean, x= a +   h = 17 +  − 40     ∑ fi 

= 17 − 4.525 = 12.48 Therefore, the mean number of days is 12.48 days for which a student was absent. 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Literacy Rate 45-55 (in %) Number of 3 Cities

55-65

65-75

75-85

85-95

Sol. To find the class marks, the following relation is used. Upper limit + Lower limit 2 Class size (h) for this data = 10 xi =

Taking 70 as assumed mean (a), di , ui , and fiui are calculated as follows:

Literacy Number of Rate (in %) Cities

di = xi -70

ui =

di 10

f i ui

45-55

-20

-2

-6

55-65

-10

-1

-10

65-75

75-85

85-95

Total

-2

From the table, we obtain

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45. = 35 − 45 Modal class

Frequency ( f 0 ) of class preceding the modal class = 21 Frequency ( f 2 ) of class succeeding the modal class = 14

15−25 25−35 35−45 45−55 55−65 11

∑ fi xi  430  30 +  35.38 = ∑ fi =  80 

Class size (h) = 10

1. The following table shows the ages of the patients admitted in a hospital during a year:

Number of Patients

Mean, x= a +

Frequency ( f1 ) of modal class = 23

NCERT Exercise 13.2

5-15

430 ∑ fi =, 80 ∑ fi di =

Lower limit (l) of modal class = 35

−2 ∑ fi = 35 , ∑ fiui = ∑fu   2  Mean, x= a +  i i  h = 70 +  −  10  35   ∑ fi  20 = 70 − = 70 − 0.57 = 69.43 35 Therefore, mean literacy rate is 69.43%.

Age (in years)

From the table, we obtain

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Sol. To find the class mark (xi) for each interval, the following relation is used. Class mark xi = Upper limit + Lower limit 2 Taking 30 as assured mean (a), di , ui , and fiui can be calculated as follows.

  f1 − f 0 Mode = l +   ×h  2 f1 − f 0 − f 2    23 − 21 = 35 +   × 10  2 ( 23 ) − 21 − 14 

=35 +

20 =35 + 1.81 =36.8 11

Mode represents that the age of maximum number of patients admitted in hospital was 36.8 years. 2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Lifetimes (in hours)

0-20 20-40 40-60 60-80 80-100 100-120

Frequency

Determine the modal lifetimes of the components. Sol. From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60–80.

Age (in years)

Number of Patients

di = xi -30

fi di

5-15

-20

-120

15-25

-10

-110

25-35

35-45

230

45-55

280

Frequency ( f 2 ) of class succeeding the modal class = 38

55-65

150

Total

Class size (h) = 20   f1 − f 0 Mode = l +   ×h f f f 2 − −  1 0 2

Chapter 13 • Statistics

430

Therefore, modal class = 60–80 Lower class limit (l) of modal class = 60 Frequency ( f1 ) of modal class = 61 Frequency ( f 0 ) of class preceding the modal class = 52

339

To find the class mark, the following relation is used.

  61 − 52 60 +  =  × 20  2 ( 61) − 52 − 38 

Upper limit + Lower limit 2 Class size (h) of the given data = 500 Class mark xi =

90 9   = 60 +  ) 60 +  ( 20= 16  122 − 90 

= 60 + 5.625 = 65.625

Therefore, modal lifetime of electrical components is 65.625 hours. 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure. Expenditure (in `)

Number of Families

1000−1500

1500−2000

2000−2500

2500−3000

3000−3500

3500−4000

4000−4500

4500−5000

It can be observed from the given data that the maximum class frequency is 40, belonging to 1500−2000 intervals. Therefore, modal class = 1500–2000 Lower limit (l) of modal class = 1500 Frequency ( f1 ) of modal class = 40 Frequency ( f 0 ) of class preceding modal class = 24 Frequency ( f 2 ) of class succeeding modal class = 33 Class size (h) = 500   f1 − f 0 Mode = l +   ×h  2 f1 − f 0 − f 2    40 − 24 1500 +  =  × 500  2 ( 40 ) − 24 − 33   16  = 1500 +   ( 500 )  80 − 57  = 1500 +

8000 23

= 1500 + 347.826 = 1847.83 Therefore, modal monthly expenditure was `1847.83. 340

Taking 2750 as assured mean (a), di , ui , and fiui can be calculated as follows. Expenditure Number (in `) of Families

di  d  ui =  i  = xi -2750  500 

f i ui

1000-1500

1250

-1500

-3

-72

1500-2000

1750

- 1000

-2

-80

2000-2500

2250

-500

-1

-33

2500-3000

2750

3000-3500

3250

500

3500-4000

3750

1000

4000-4500

4250

1500

4500-5000

4750

2000

Total

200

-35

From the table, we obtain ∑ fi = 200 ∑ fiui = −35

∑fu  Mean, x= a +  i i  h  ∑ fi   35  = 2750 +  −  500 = 2750 − 87.5 = 2662.5  200  Therefore, mean monthly expenditure was ` 2662.50. 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of Students per Teacher

Number of States/U.T

15-20

20-25

25-30

30-35

35-40

40-45

45-50

50-55

Sol. It can be observed from the given data that the maximum class frequency is 10, belonging to class interval 30–35. Therefore, modal class = 30–35 Lower limit (l) of modal class = 30

It represents that on an average, teacher−student ratio was 29.2. 5. The given distribution shows the number of runs scored by some top batsmen of the world in oneday international cricket matches

Frequency ( f1) of modal class = 10

Runs Scored

Number of Batsmen

Frequency ( f 0 ) of class preceding modal class = 9

3000-4000

Frequency ( f 2) of class succeeding modal class = 3

4000-5000

Class size (h) = 5

5000-6000

  f1 − f 0 Mode = l +   ×h  2 f1 − f 0 − f 2 

6000-7000

7000-8000

  10 − 9 30 +   × 5 =  2 (10 ) − 9 − 3 

8000-9000

9000-10000

10000-11000

 1  = 30 +  20 − 12  ( 5 )  

Find the mode of the data.

5 = 30 + = 30.625 8

Sol. It can be observed from the given data that the maximum class frequency is 18, belonging to class interval 4000-5000.

Mode = 30.625

It represents that most of the states/U.T have a teacher-student ratio as 30.6.

Therefore, modal class = 4000–5000

To find the class marks, the following relation is used. Upper limit + Lower limit Class mark xi = 2 Taking 32.5 as assured mean (a), di , ui , and fiui can be calculated as follows. Number of Number of Students States/U.T per Teacher

d  di = ui =  i  f i ui  5  xi - 32.5

15-20

17.5

-15

-3

-9

20-25

22.5

-10

-2

-16

25-30

27.5

-5

-1

-9

30-35

32.5

35-40

37.5

40-45

42.5

45-50

47.5

50-55

52.5

Total

∑fu  Mean, x= a +  i i  h  ∑ fi  = 32.5 +  − 23  × 5 = 29.22    35  Chapter 13 • Statistics

-23

Lower limit (l) of modal class = 4000 Frequency ( f1 ) of modal class = 18 Frequency ( f 0 ) of class preceding modal class = 4 Frequency ( f 2) of class succeeding modal class = 9 Class size (h) = 1000   f1 − f 0 Mode = l +   ×h  2 f1 − f 0 − f 2   18 − 4  4000 +  =  × 1000  2 (18 ) − 4 − 9   14000  = 4000 +    23  = 4000 + 608.695

= 4608.695

Therefore, mode of the given data is 4608.7 runs 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data: Number 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 of Cars Frequency

341

Sol. It can be observed from the given data that the maximum class frequency is 20, belonging to class interval 40–50. Therefore, modal class = 40-50 Lower limit (l) of modal class = 40

Monthly Number of Consumption Consumers (in units)

di = xi - 30 ui =

di f i ui 20

65-85

-60

-3

-12

85-105

-40

-2

-10

Frequency ( f 0 ) of class preceding modal class = 12

105-125

115

-20

-1

-13

Frequency ( f 2) of class succeeding modal class = 11

125-145

135

Class size (h) = 10

145-165

155

165-185

175

185-205

195

Total

Frequency ( f1 ) of modal class = 20

  f1 − f 0 Mode = l +  ×h  2 f1 − f 0 − f 2 

  20 − 12 40 +  =  × 10  2 ( 20 ) − 12 − 11 

 80  = 40 +    40 − 23 

From the table, we obtain ∑ fi = 68

80 = 40 + =40 + 4.7 =44.7 17

∑ fiui = 7

Therefore, mode of this data is 44.7 cars.

NCERT Exercise 13.3 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

= 135 +

140 = 137.058 68

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125–145. Modal class = 125–145

Number of Consumers

65−85

Frequency ( f1) of modal class = 20

85−105

Class size (h) = 20

105−125

125−145

Frequency ( f 0) of class preceding the modal class = 13

145−165

165−185

185−205

Sol. To find the class marks, the following relation is used. Upper limit + Lower limit 2 Taking 135 as assured mean (a), di ,ui , and fiui can be calculated as follows.

342

∑fu   7  Mean, x = a + i i ×h = 135 +   × 20  68   ∑ fi 

Monthly Consumption (in units)

Class mark xi =

Lower limit (l) of modal class = 125

Frequency ( f 2) of class succeeding the modal class = 14   f1 − f 0 Mode = l +  ×h f f f 2 − −  1 0 2   20 − 13 125 +  =  × 20  2 ( 20 ) − 13 − 14  7 × 20 = 125 + 10.76 =135.76 13 To find the median of the given data, cumulative frequency is calculated as follows. = 125 +

Monthly Consumption Number of Cumulative (in units) Consumers Frequency 65-85

85-105

105-125

125-145

145-165

165-185

185-205

class intervals with their respective cumulative frequency can be defined as below. Age (in years) 18−20

From the table, we obtain n = 68 Cumulative frequency (cf) just greater than 68 n   i.e., = 34  is 42, belonging to interval 125-145. 2 2  Therefore, median class = 125–145

Number of Policy Cumulative Frequency (cf ) Holders (fi ) 2

20−25

6−2=4

25−30

24 − 6 = 18

30−35

45 − 24 = 21

35−40

78 − 45 = 33

40−45

89 − 78 = 11

45−50

92 − 89 = 3

50−55

98 − 92 = 6

55−60

100 − 98 = 2

100

Total (n) From the table, it can be observed that n = 100 . Cumulative frequency (cf) just greater than

Lower limit (l) of median class = 125

100 n  = 50  is 78, belonging to interval 35–40.  i.e., 2 2 

Class size (h) = 20 Frequency (f) of median class = 20 Cumulative frequency (cf) of class preceding median class = 22 2. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Therefore, median class = 35–40 Lower limit (l) of median class = 35 Class size (h) = 5 Frequency (f) of median class = 33 Cumulative frequency (cf) of class preceding median class = 45  n     2  − cf    ×h Median = l +  f      

Age (in years)

Number of Policy Holders

Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Length (in mm)

Number or Leaves fi

Below 55

118−126

Below 60

100

127−135

136−144

145−153

154−162

163−171

172−180

Sol. Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, Chapter 13 • Statistics

 50 − 45  = 35 + 25 = 35.76 35 +  = ×5 33  33  Therefore, median age is 35.76 years.

3. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

343

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5…

4. Find the following table gives the distribution of the life time of 400 neon lamps: Life Time (in hours)

Number of Lamps

1500−2000

2000−2500

2500−3000

3000−3500

3500−4000

4000−4500

4500−5000

171.5 − 180.5) Sol. The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. 1 Therefore, = 0.5 has to be added and subtracted to 2 upper class limits and lower class limits respectively. Continuous class intervals with respective

Find the median life time of a lamp.

cumulative frequencies can be represented as follows.

Sol. The cumulative frequencies with their respective class intervals are as follows.

Length (in mm)

Number or Leaves fi

Cumulative Frequency

Life Time (in hours)

Number of Lamps

Cumulative Frequency

117.5−126.5

1500−2000

126.5−135.5

2000−2500

14 + 56 = 70

135.5−144.5

2500−3000

70 + 60 = 130

144.5−153.5

3000−3500

130 + 86 = 216

153.5−162.5

3500−4000

216 + 74 = 290

162.5−171.5

4000−4500

290 + 62 = 352

171.5−180.5

4500−5000

352 + 48 = 400

From the table, it can be observed that the cumulative frequency just greater than 40 n    i.e., = 20  is 29, belonging to class interval 2  2  144.5–153.5. Median class = 144.5–153.5 Lower limit (l) of median class = 144.5 Class size (h) = 9 Frequency (f) of median class =12 Cumulative frequency (cf) of class preceding median class =17  n     2  − cf  ×h Median = l +    f       9  20 − 17  144.5 +  9 144.5 += 146.75 =  ×= 4  12  Therefore, median length of leaves is 146.75 mm. 344

It can be observed that the cumulative frequency 400 n   just greater than  i.e., = 200  is 216, belonging 2  2  to class interval 3000 − 3500. Median class = 3000 − 3500 Lower limit (l ) of median class = 3000 Frequency (f ) of median class = 86 Cumulative frequency (cf ) of class preceding median class = 130 Class size (h) = 500  n     2  − cf    ×h Median = l +  f        200 − 130  3000 +  =  × 500 86   70 × 500 86 = 3406.976

= 3000 +

Therefore, median life time of lamps is 3406.98 hours.

5. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of 1−4 Letters

4−7 7−10 10−13 13−16 16−19

Number of Surnames

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. Sol. The cumulative frequencies with their respective class intervals are as follows. Number of Letters

Frequency (fi)

Cumulative Frequency

1−4

4−7

30 + 6 = 36

7−10

36 + 40 = 76

10−13

76 + 16 = 92

13−16

92 + 4 = 96

16−19

96 + 4 = 100

Total (n)

100

Taking 11.5 as assured mean (a), di , ui , and fiui can be calculated according to step deviation method as follows. Number of Frequency Letters (fi )

-9

-3

-18

4−7

5.5

-6

-2

-60

7−10

8.5

-3

-1

-40

10−13

11.5

13−16

14.5

16−19

17.5

Total (n)

100

 n     2  − cf   50 − 36  ×h = Median = l +    7+ ×3 f    40      14 × 3 = 7+ = 8.05 40 To find the class marks of the given class intervals, the following relation is used. Upper limit + Lower limit Class mark xi = 2

Chapter 13 • Statistics

-106

From the table, we obtain −106 ∑ fi = 100, ∑ fiui = ∑fu   −106  Mean, x = 11.5 +  a + i i ×h =  ×3  100   ∑ fi 

=11.5 − 3.18 = 8.32

The data in the given table can be written as Number of Letters

Frequency (fi)

1−4

4−7

7−10

10−13

13−16

16−19

Total (n)

100

Median class = 7−10

Class size (h) = 3

f i ui

2.5

100 n   = 50  is 76, belonging to  i.e., 2  2  Class interval 7–10.

Frequency (f ) of median class = 40

di 3

just greater than

Cumulative frequency (cf ) of class preceding median class = 36

di = xi -11.5 ui =

1−4

It can be observed that the cumulative frequency

Lower limit (l ) of median class = 7

From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7–10. Modal class = 7–10 Lower limit (l ) of modal class = 7 Class size (h) = 3 Frequency (f1) of modal class = 40 Frequency (f0) of class preceding the modal class = 30 Frequency (f2) of class succeeding the modal class = 16   f1 − f 0 Mode = l +  ×h  2 f1 − f 0 − f 2    40 − 30 10 7 +  =  × 3 =7 + ×3 34  2 ( 40 ) − 30 − 16 

= 7 + 0.88 = 7.88 345

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88. 6. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg) Number of Students

30 n   = 15  =15 is 19, belonging to class interval  i.e., 2  2  55 - 60. Median class = 55–60 Lower limit (l ) of median class = 55

40−45 45−50 50−55 55−60 60−65 65−70 70−75 2

Cumulative frequency just greater than

Sol. The cumulative frequencies with their respective class intervals are as follows. Weight (in kg)

Frequency (fi)

Cumulative Frequency

40−45

45−50

2+3=5

50−55

5 + 8 = 13

55−60

13 + 6 = 19

60−65

19 + 6 = 25

65−70

25 + 3 = 28

70−75

28 + 2 = 30

Total (n)

Frequency (f ) of median class = 6 Cumulative frequency (cf ) of median class = 13 Class size (h) = 5  n     2  − cf  Median = l +    ×h f      

 15 − 13  = 55 +  ×5  6 

= 55 +

10 = 56.67 6

Therefore, median weight is 56.67 kg.

Multiple Choice Questions 1. Which of the following cannot be determined graphically? (a) Mean (b) Median (c) Mode

3. Find the mode of the data 5, 3, 4, 3, 5, 3, 6, 4, 5. (a) 5 (b) 4 (c) 3

(d) none of these

2. The heights of plants in Dipti’s Garden are recorded in the table given below. The median plant height is 55 cm. Heights of Plants (in cm) Number of Plants

0–20 2x

20–40 40–60 60–80 80–100 4

Which of the following is the value of x? (a) 1 (b) 2 (c) 8 (d) The value of x cannot be found without knowing the total number of plants.

346

(d) Both (a) and (c)

4. If the mean and the mode of a distribution are 15 and 18 respectively, then the median of the distribution is (a) 17

(b) 15

(d) 18 (CBSE 2023)

5. The median of a data set is always:

(i) The most frequently occurring value

(ii) Affected by extreme values

(iii) The middle value when arranged in order

(iv) The average of all data points (a) (i) and (iii) (b) only (iii) (c) (ii) and (iv)

(d) only (iv)

6. If xi are the mid-points of the class intervals of grouped data, fi are the corresponding frequencies and x is the mean, then ∑ ( fi ( xi − x ) ) s equal to (a) 0 (b) –1 (c) 1 (d) 2 (NCERT Exemplar)

7. The mean of n observations is X . If the first item is increased by 1, second by 2 and so on, then the new mean is n (a) X + n (b) X + 2 n +1 (c) X + (d) none of these 2

12. If mode of a series exceeds its mean by 12, then mode exceeds the median by (a) 4 (b) 8 (c) 6 (d) 10 13. If the mean and the median of a data are 12 and 15 respectively, then its mode is (a) 13.5 (b) 21 (c) 6 (d) 14 14. The upper limit of the modal class of the given distribution is Height Below Below Below Below Below Below (in cm) 140 145 150 155 160 165 Number of Girls

8. If the mean of the following distribution is 2.6, then the value of y is Variable (x)

Frequency

(a) 3

(b) 8

(d) 24

(a) 165

(b) 4

Marks

(d) 10

10. If the difference of mode and median of a data is 24, then the difference of median and mean is (a) 12 (b) 24 (c) 8 (d) 36 11. If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by (a) 2 (b) 1.5 (c) 1 (d) 0.5

11 (b) 160

(d) 150

15. The distribution below gives the marks obtained by 80 students in a test:

9. If the mean of a frequency distribution is 9.1 and ∑ fi xi = 132 + 5k , ∑ fi = 20, then k = (a) 3

No. of Students

Less than 10

Less than 20

Less than 30

Less than 40

Less than 50

Less than 60

The modal class of this distribution is: (a) 10–20 (b) 20–30 (c) 30–40

(d) 50–60 (CBSE 2023)

Answers 1. (a) Mean

2. (b) 2

9. (d) 10

10. (a) 12

3. (d) Both (a) and (c)

4. (c) 16

11. (d) 0.5

12. (b) 8

5. (b) only (iii) n +1 7. (c) X + 2

6. (a) 0

13. (b) 21

14. (d) 150

8. (b) 8

15. (c) 30–40

Constructed Response Questions Very Short Answer Questions 1. Find the class-marks of the classes 10–25 and 35–55. (CBSE 2020)

Chapter 13 • Statistics

Sol. Class mark of 10–25 is calculated as: 10 + 25 35 = = 17.5 2 2

347

And, class mark of 35–55 is calculated as:

n (n + 1) n +1 30 = 15 ⇒ n + 1 = = 15 ⇒ 2 2n

35 + 55 90 = = 45 2 2

2. Find the mean of the distribution

5. Find the mode of the following distribution:

Class

1–3

3–5

5–7

7–10

Frequency

(NCERT Exemplar)

Sol. We first find the class mark xi of each class and then proceed as follows. Class

Class Marks Frequency xi fi

3–5

5–7

162

7–10

8.5

144.5

Total

i i

∑ f = 75 ∑ f x = 412.5

Heights 120–130 130–140 140–150 150–160 160–170 Total (in cm) 8

Change the above distribution to “More than type” distribution. Sol.

More More More More More Heights than than than than than Total (in cm) 120 130 140 150 160

Number of Girls

4. If the mean of the first n natural numbers is 15, then find n. (CBSE 2020)

Sol. The mean of the first n natural numbers is given by: 1 + 2 + 3 + ... + n = 15 n Sum of first n natural numbers:

348

Sol. We have, 0–10 10–20 20–30 30–40 40–50 50–60 4

7(f0) 12(f1) 5(f2)

   12 − 7  f1 − f 0 30 +  Mode = l +  × 10 ×h = × − − 2 12 7 5 2 f − f − f    1 0 2  5  30 +  =  × 10  24 − 12 

412.5 75

3. The data regarding the heights of 50 girls of class X of a school is given below:

Here, modal class is 30–40, l = 30, f1 = 12, f 0 = 7, f 2 = 5 , h = 10

∑ Means (x ) = = = 5.5

Number of Girls

0–10 10–20 20–30 30–40 40–50 50–60

Number of Students

∑

Marks

(CBSE 2020)

fi xi

1–3

fi xi

n = 29

n (n + 1) 2

= 30 +

50 = 30 + 4.16 = 34.16 12

Thus, Mode = 34.16. 6. Find the unknown values in the following table: Class Interval

Frequency

Cumulative Frequency

0–10

10–20

20–30

30–40

40–50

Sol. x1 = 5 + 7 = 12 x 2 = 18 − x1 = 18 − 12 = 6 x 3 = 18 + 5 = 23 x 4 = 30 − x 3 = 30 − 23 = 7

(CBSE 2016)

7. Find the value of p, if the mean of the following distribution is 7.5. (CBSE2020)

8. Find the mode of the following frequency distribution:

Classes

2–4

4–6

6–8

8–10 10–12 12–14

Classes 10–14 14–18 18–22 22–26 26–30 30–34 34–38

Frequency (fi )

Frequency

Sol. To find the mode of a frequency distribution, identify the class with the highest frequency, which is known as the modal class.

∑ ( fi xi ) ∑ fi

From the given data, the highest frequency is 25 for the class 26–30.

Classes

Frequency (fi )

Mid-point xi

fi xi

, f 2 22, = h 4 = l 26, = f 0 25, = f1 20=

2–4

4–6

 ( f1 − f 0 )  Mode = ×h l +  ( 2 f − f − f )  1 0 2  

6–8

105

8–10

10–12

12–14

Total

∑ fi = 41 + p

 ( 25 − 20 )  5 26 +  =  × 4 =26 +   × 4 2 × 25 − 20 − 22 ) 8 ( = 26 + 2.5 Mode = 28.5

9. Find the median of the data, using an empirical relation when it is given that Mode = 12.4 and Mean = 10.5.

∑ ( fi xi ) = 303 + 9p

Given mean x = 7.5

Sol. Median =

use the formula: x=

∑ ( fi xi ) 303 + 9p 41 + p

1 2 Mode + Mean 3 3

(CBSE 2015)

12.4 21 1 2 = + (12.4 ) + (10.5 ) 3 3 3 3

12.4 + 21 33.4 = 11.13 (approx.) = 3 3

∑ fi

7.5 =

(CBSE 2019)

Sol. To find the value of p, we need to calculate the mean using the formula for the mean of grouped data: x=

7.5 ( 41 + p ) = 303 + 9p ⇒ 307.5 + 7.5p =303 + 9p

4.5 = 1.5p ⇒= p

4.5 = 3 1.5

Therefore, the value of p is 3.

Short Answer Questions 1. In a test, the marks obtained by 100 students (out of 50) are given below: Marks Obtained

0–10

10–20

20–30

30–40

40–50

Number of Students

Find the mean marks of the students. (CBSE 2024)

Chapter 13 • Statistics

349

Sol. Topper’s Answer

2. The following table gives the number of pages written by Sarika for completing her own book for 30 days. Number of pages Written 16–18 19–21 22–24 25–27 28–30 per Day Number of Days

Find the mean number of pages written per day. (NCERT Exemplar)

Sol.

Class Mark Mid-value xi Number of Days fi fi xi

Number of Seats

100–104 104–108 108–112 112–116 116–120

Frequency

Determine the mean number of seats occupied over the flights. (NCERT Exemplar)

Sol. We first find the class mark xi of each class and then proceed as follows.

15.5–18.5

18.5–21.5

Number of Seats

Class Marks xi

21.5–24.5

100–104

102

-8

-120

24.5–27.5

234

104–108

106

-4

-80

27.5–30.5

377

108–112

a = 110

Total

780

112–116

114

116–120

118

120

Total

N= ∑ fi = 100

∑ fi di = −8

∑ fi xi 780 26 ∴ Mean ( x ) = == 30 ∑ fi

350

3. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table.

Frequency Deviation fi di = xi -a

fi di

The mode formula is:

Assumed mean, a = 110

( f1 − f 0 ) l+ Mode = ×h ( 2 f1 − f 0 − f 2 )

Class width h = 4 Total observations, N = 100 Using the assumed mean method,

∑ fi di  −8  = 110 +   ∑ fi  100  =110 − 0.08 =109.92 The mean number of seats occupied over the flights is 109.92. Mean ( x )= a +

4. If the mean of the following frequency distribution is 62.8, then find the missing frequency x : (CBSE SQP 2023)

Class

0–20 20–40 40–60 60–80 80–100 100–120

No. of Girls

(17 − 9 ) 8 12 + ×4 Mode = 12 + ×4 = 34 − 21 ( 2 (17 ) − 9 − 12 ) 8 × 4= 12 + 2.46 = 14.46 13 Hence, the mode is 14.46. = 12 +

6. Find the mode of the following frequency distribution. Class Frequency

0–10 10–20 20–30 30–40 40–50 50–60 60–70 8

(CBSE 2019) (CBSE Board 2016)

Sol. The modal class is 30−40 with a frequency of 16, which is the highest frequency. Parameters: = l 30, = f1 16, = f 0 10, = f 2 12, = h 10

Sol.

Using the mode formula:

Class 0–20 20–40 40–60 60–80 80–100 100–120 Total Class 10 30 50 70 90 110 Mark (x) No. of Girl 5 (f ) fx

∑ f = 40 + x

50 240 50x 840 630 880 ∑ fx= 2640 + 50x

Now, using the mean formula: ∑ fx Mean = ∑f 62.8 =

  f1 − f 0 Mode = l + ×h  2 f1 − f 0 − f 2  16 − 10   Mode = 30 +   × 10  2 × 16 − 10 − 12   6  Mode = 30 +   × 10  32 − 22   6  Mode = 30 +   × 10  10  Mode = 30 + 6 Mode = 36

2640 + 50x 40 + x

7. Find the mode of the following frequency distribution.

⇒ 62.8 ( 40 + x = ) 2640 + 50x

Class

15–20 20–15 25–30 30–35 35–40 40–45

⇒ 2512 + 62.8x = 2640 + 50x

Frequency

⇒ 12.8x = 128

Sol. From the given data, the highest frequency is 10 for the class 30–35 Here, modal class is 30–35 = l 30,= f 0 9,= f1 10,= f 2 3,= h 5

⇒x = 10 Hence, the missing frequency is 10. 5. Compute the mode for the following frequency distribution. (CBSE 2019, 2020)

Size of Items (in cm)

0–4 4–8 8–12 12–16 16–20 20–24 24–28

Frequency

Sol. The modal class is the class with the highest frequency, which is 12–16 (with frequency 17). l = 12 (lower boundary of modal class) f1 = 17 , f 0 = 9 , f 2 = 12 , h = 4 Chapter 13 • Statistics

Using the mode formula:   f1 − f 0  10 − 9  Mode = l + 30 +  ×h = ×5 2 f − f − f 2 × 10 − 9 − 3    1 0 2 1 =30 +   × 5= 30 + 0.625 = 30.625 8 Hence, the mode is 30.625 8. Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median. (CBSE SQP 2022) 351

Distance (in m) 0–1 Number of Students

1–2

2–3

3–4

4–5

9. The mode of the following data is 67. Find the missing frequency x. Class

Sol.

40–50 50–60 60–70 70–80 80–90

Frequency Distance (in m)

0–1

1–2

2–3

3–4

4–5

Number of Students

Cumulative Frequency (cf )

120

182

220

250

Sol. From the given table, the modal class is the class having the maximum frequency, i.e., modal class = 60–70 = l 60,= f1 15, = f 0 x= , f 2 12, = h 10 Given: Using the mode formula:

Here,

  n 250 f1 − f 0 = = 125 Mode = l + ×h 2 2  2 f1 − f 2 − f 0  ⇒ Median class = 2–3, l = 2, h = 1, cf = 120, f = 62  15 − x  67 = 60 +   × 10 Using the median formula:  30 − 12 − x  n   15 − x   2 − cf  7  =  125 − 120  ) 10 (15 − x )  × 10 ⇒ 7 × (18 − x= × 2+ 1 Median = l + ×h =   18 − x  62 f     3x 150 − 126 126 − 7x = 150 − 10x ⇒ =   =

5 129 =2 m or 2.08 m 62 62

Hence, 50% of students jumped below 2.08 m and 50% above it.

3x = 24 ⇒ x = 8

Hence, the missing frequency is 8.

10. Find the mean of the following frequency distribution. (CBSE 2023)

Class

25–30

30–35

35–40

40–45

45–50

50–55

55–60

Frequency

Sol. Topper’s Answer

352

11. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows: Speed (in km/h)

Sol. First, we construct a cumulative frequency table. Weekly Income (in `)

Number of Families fi

Cumulative Frequency cf

0–1000

250

1000–2000 = mid class

190

250 + 190 = 440

2000–3000

100

440 + 100 = 540

3000–4000

540 + 40 = 580

4000–5000

580 + 15 = 595

5000–6000

595 + 5 = 600

85–100 100–115 115–130 130–145

Number of Players

Calculate the median bowling speed. (NCERT Exemplar)

Sol. First, we construct the cumulative frequency table:

11 + 9 = 20

115–130

20 + 8 = 28

130–145

28 + 5 = 33

Given: Total number of players n = 33 n 33 = = 16.5 2 2 Since the cumulative frequency 20 includes 16.5, the median class is 100–115. From the table: l = 100, f = 9 , cf = 11, h = 15 Using the median formula: n   2 − cf   16.5 − 11  Median = l + 100 +  ×h =  × 15 f 9       5.5 × 15 82.5 = 109.17 = 100 + = 100 + 9 9 Hence, the median bowling speed is 109.17 km/h. 12. Weekly income of 600 families is tabulated below: Weekly Income (in `)

Number of Families

0–1000

250

1000–2000

190

2000–3000

100

3000–4000

4000–5000

5000–6000

Total

600

Compute the median income. (NCERT Exemplar)

Chapter 13 • Statistics

Since, cumulative frequency 440 lies in the interval 1000–2000, this is the median class. l = 1000, f = 190, cf = 250, h = 1000 Using the median formula: n   2 − cf   300 − 250  1000 +  Median = l +  ×h =  × 1000  190   f    50 50000 = 1000 + × 1000 = 1000 + 190 190 = 1000 + 263.15 = 1263.15 Hence, the median income is `1263.15. 13. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70. 35–40

30–35

100–115

n 600 = = 300 2 2

25–30

20–25

15–20

85–100

It is given that n = 600

10–15

Cumulative Frequency cf

5–10

Number of Players fi

0–5

Speed (in km/h)

Frequency 12

Class

Sol. Cumulative frequency table:

(CBSE SQP 2021)

Class

Frequency (f )

Cumulative Frequency (cf)

0–5

10–15

20–25

5–10

15–20

25–30

35–40

30–35 Total

6 70

12 + a 24 + a 39 + a

39 + a + b 45 + a + b

51 + a + b 55 + a + b

353

From the table: 55 + a + b = 70

Given mean = 53

a +b = 15 (i)

Mean =

N 70 Total frequency N = 70, so = = 35 2 2 The cumulative frequency just before 35 is in the class 10–15, and the next class (15–20) contains the median. Thus, the median class is 15–20.

⇒ 17k = 476 ⇒ k = 28 Hence, the value of k is 28. 15. The marks obtained by 100 students in an examination are given below:

Marks

Number of Students (fi)

Mid Values (xi)

Class Interval

30–35

35–40

80–100 13

32.5

-3

-42

37.5

-2

-32

40–45

42.5

-1

-28

45–50

47.5

50–55

52.5

fi xi

55–60

57.5

62.5

Class Marks (xi) 10

120

60–65

20–40

450

Total

N =Σfi =110

40–60

1600

60–80

70k

80–100

1170

Total

72+k

fi µ i

Frequency (fi)

0–20

xi − A h

(CBSE 2019)

Sol.

14. The mean of the following distribution is 53. Find the missing frequency k?

(CBSE 2019)

8 +b = 15 ⇒ b = 7

Find the mean marks of the students.

 35 − ( 24 + a )   11 − a  16 = 15 +   × 5 ⇒1=   15  3    a =8 Substituting the value of a in equation (i), we get

30–35 35–40 40–45 45–50 50–55 55–60 60–65

Number of Students

Where,= N 70, = l 15, = h 5, Median = 16

Frequency

3816 + 53k = 3340 + 70k ⇒ 70k − 53k = 3816 − 3340

Marks

0–20 20–40 40–60 60–80

i i

53 ( 72 + k = ) 3340 + 70k ⇒ 3816 + 53k = 3340 + 70k

Median formula: N   2 − cf  Median = l +  ×h  f   

Class

∑ f x ⇒ 53 = 3340 + 70k 72 + k ∑f

µi =

Σfi µi = −59

Assumed mean, A = 47.5

∑ f µ × h =47.5 +  −59  × 5

Mean = A+

3340 + 70k

i i

 110 

= 47.5 − 2.6862 = 44.818

Long Answer Questions 1. The mean of the following distribution is 18. Find the frequency f of the class 19–21. Class

11–13

13–15

15–17

17–19

19–21

21–23

23–25

Frequency

4 (CBSE 2018, 2020)

354

Frequency

(CBSE Outside Delhi 2020) (CBSE Delhi 2018)

Sol. Sol. Topper’s Answer

3. The monthly expenditure on milk in 200 families of a Housing Society is given on below: 2. The monthly expenditure milk in 200 families of 2500–3000

Number of Families

3500–4000

4000–4500

4500–5000

1500–2000 24 40 33 2000–2500

40 30 22 33

Monthly Monthly Expenditure expenditure (in `) (in `) 1000–1500 Number of families

3000–3500

2000–2500

1500–2000

1000–1500

a Housing Society is given below:

2500–3000 x median and Find the value of x and also, find the mean expenditure on milk. 3000–3500 30 Sol.

3500–4000

(CBSE 2023)

4000–4500 16 Total families = 200 4500–5000 7 ⇒ 172 + x = 200 Find the value of x and also, find the median and ⇒ x = expenditure 28 mean on milk. (CBSE 2023) Sol. Total families = 200 Chapter 13 • Statistics

⇒ 172 + x = 200 ⇒x = 28 Chapter 13 • Statistics

Number

Monthly ClassCumulative of x −A fi µi mark exp. frequency µi= i Number Monthly Class- families Cumulative h x − A (in `) (cf) x ( i) of Mark f i µi fi Exp. Frequency µi = i h Families (in `) (cf ) (xi ) fi 1000–1500 1250 24 24 -3 -72

1500–2000 1250 1750 1000–1500 2000–2500 2250 1500–2000 1750 2500–3000 2750 = A

3000–3500 2250 3250 2000–2500 3500–4000 3750 2500–3000 2750 = A 4000–4500 4250 3000–3500 4500–5000 3250 4750

40 24 33 40 f = 28

30 33 22 f = 28 16 30 7

64 24 97 64 125

-2 -3 -1 -2 0

-80 -72 -33 -80 0

177

-35 44

155 97 177 125 193 155 200

1 -1 2 0 3 14

Total 3500–4000

3750

4000–4500

4250

193

4500–5000

4750

200

Total

200

200 22

30 -33 44 0 48 30 28 48 28

-35

355

4. The table below shows the salaries of 280 persons:

N 200 = = 100 2 2 = l lower limit = 2500, = cf 97, = f 28, = h 500

Salary (in Thousand `)

No. of Persons

5–10

10–15

133

15–20

20–25

Using the median formula: N   2 − cf  Median = l + ×h  f   

 100 − 97   3  =+ 2500   × 500 = 2500 +   × 500  28   28 

25–30

= 2553.6

30–35

35–40

40–45

45–50

Thus, median expenditure = `2553.6. Using the mean formula: Mean = A+

∑ fiui −35 × 500 = × h 2750 + 200 ∑ fi

Calculate the median salary of the data. (CBSE 2018)

= 2750 − 87.5 = 2662.5

Sol.

Thus, mean expenditure = `2662.5.

Salary (in thousand `)

No. of Persons

Cumulative Frequency (cf )

5–10

10–15

133

182

15–20

245

20–25

260

25–30

266

3. The monthly income of 100 families is given below: Income (in `)

Number of Families

0–5000

5000–10000

10000–15000

30–35

273

15000–20000

35–40

277

20000–25000

40–45

279

25000–30000

45–50

280

30000–35000

Here,

35000–40000

n 280 = = 140 2 2

Calculate the modal income.

(NCERT Exemplar)

Sol. In the given data, the highest frequency is 41, which lies in the interval 10000–15000. Given: l =10000, f1 = 41, f 0 = 26, f 2 = 16, h = 5000 Using the mode formula:   f1 − f 0 Mode = l + ×h  2 f1 − f 0 − f 2    41 − 26 10000 +  =  × 5000 2 41 26 16 × − −    15   15  = 10000 +   × 5000 = 10000 +   × 5000 − 82 42    40  = 10000 + 15 × 125 = 11875 Hence, the modal income is ` 11875. 356

⇒ Median class = 10–15 Where, = l 10, N = 280, = h 5,= cf 49, = f 133 Using the median formula: n   2 − cf  Median = l + ×h  f   

 91   140 − 49  10 +  = 10 +  ×5 ×5= 133  133    = 10 +

455 = 13.42 133

Thus, median salary is `13.42 thousand or `13,420 (approx.)

5. The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class

Frequency

0–100

200–300

100–200

500–600

400–500

9+ y = 24 ⇒ y = 24 − 9 = 15 Hence, x = 9, y = 15

6. Daily wages of 110 workers, obtained in a survey, are tabulated below:

600–700

700–800

900–1000

800–900

Daily Wages (in `)

Number of Workers

100–120

120–140

140–160

160–180

180–200

200–220

220–240

7 (CBSE 2020)

Sol.

Class Interval

Frequency (f )

Cumulative Frequency (cf )

0–100

200–300

7+x

300–400

19 + x

400–500

36 + x

500–600

56 + x

600–700

56 + x + y

700–800

65 + x + y

800–900

900–1000

100–200

( 50 − 36 − x ) × 5

24: Substituting x = 9 into x + y =

300–400

 50 − 36 − x  25  =  × 100 ⇒ 25 = 20   25 = 14 − x ⇒ x = 14 − 5 = 9 Solve for y using Equation (i)

Compute the mean daily wages and modal daily wages of these workers. (CBSE 2021) Sol. Daily Wages Number of (in `) Workers (fi)

ui =

xi − a h

f i ui

100–120

110

-3

-30

120–140

130

-2

-30

72 + x + y

140–160

150

-1

-20

76 + x + y

160–180

170 = A

180–200

190

Total N = 100

200–220

210

76 + x + y = 100 ⇒ x + y= 100 − 76= 24 (i)

220–240

230

Total

110

Given, median = 525, which lies in the class 500–600.

Using the mean formula:

Median class = 500–600

A = 170, ∑fiui = 1, ∑fi = 110, h = 20

Using the median formula:

Mean daily wages = A+

∑ fiui ×h ∑ fi

= 170 +

1 × 20 110

Substituting values:  100  100 525 = 500 +  − (36 + x )  × 20  2 

= `170.19 (approx.)

 50 − ( 36 + x )  525 = 500 +   × 100 20  

Given:

N   2 − cf  Median = l + ×h  f   

Chapter 13 • Statistics

In the given data, the highest frequency is 22, which lies in the interval 160-180. l = 160

f1 = 22

f0 = 20

f2 = 18

h = 20 357

Using the mode formula:   f1 − f 0 Mode = l + ×h 2 f − f − f  1 0 2

Substituting f1 = 28 into Equation (i) f= 52 − 28 ⇒ f 2 = 24 2

Hence, the missing frequencies are: f1 = 28, f 2 = 24

22 − 20   160 +  =  × 20  2 × 22 − 20 − 18 

8. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mean and median of the following data

2 =160 +   × 20 6 1 =160 +   × 20 3 = `166.67 (approx.)

Number of Cars

Frequency (periods)

0–10

10–20

20–30

30–40

40–50

50–60

60–70

70–80

Hence, the modal wages of these worker is `166.67 (approx). 7. The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20–40 and 60–80, respectively, are not known. Find these frequencies if the sum of all the frequencies is 120. (NCERT Exemplar)

Class

0–20

Frequency

20–40 40–60 60–80 80–100 f1

Sol. First we calculate the class mark of given data Class

Frequency (fi)

0–20

20–40

40–60

80–100

60–80

Class x −a Marks ui = i h (xi) 10

fi ui

(CBSE 2023)

Sol.

Number of Cars

fi xi

0–10

10–20

210

420

20–30

-2

-34

-1

-f1

30–40

50–60

Total Σfi= 68 + f1+ f 2

40–50

60–70

Σfiui = 4 + f 2 − f1

Given that the sum of all frequencies is 120:

25 45 55 65

70–80

Total

13 20 11 15 8

100

⇒ f1 + f 2 = 52 (i)

Here, N 100 = = 50 2 2 ⇒ Median class = 40–50,

4 + l2 . l1 × 20 120 ⇒ 4 + f 2 − f1 = 0

⇒ − f 2 + f1 =4 (ii)

⇒ 50 = 50 +

Adding Equation (i) and (ii): ( f1 + f 2 ) + ( − f 2 + f1 )= 52 + 4 2 f1 = 56 ⇒ f1 = 28 358

605 975 600

4070

i i

Mean =

Using the step deviation method: ∑ fiui Mean = a+ ×h ∑ fi

900

fx 4070 ∑= = 40.7 ∑ f 100

∑ fi = 68 + f1 + f 2 = 120 Here, (a) = 50, (h) = 20

325

Where, = l 40, = N 100, = h 10, = cf 46, = f 20 Using the median formula: N   2 − cf   50 − 46  40 +  Median = l + ×h =  × 10 f  20      4   40 = 40 +   × 10 = 42 = 40 + 20  20  Thus, median is 42.

34 66 77 92

100

9. On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table: Age (in years)

Number of Students

Size of agricultural Holdings (in hec)

Number of Families

05–07

0–5

07–09

120

5–10

09–11

10–15

11–13

100

13–15

15–20

15–17

20–25

17–19

25–30

30–35

Find mean and median of the above data. Sol. Class Interval

05–07

07–09 09–11 11–13 13–15 15–17 17–19 Total

10. Size of agricultural holdings in a survey of 200 families is given in the following table

120

190

100 28

fi ui

-3

-210

-2

-240

Sol.

Size of agricultural Holdings (in hec)

Number of Families (fi)

Cumulative Frequency

0–5

-1

-32

5–10

10–15

367

15–20

135

20–25

175

25–30

195

30–35

200

400

12 16 18

Σf = 400

56 Σfiui = −366

Let a = Assumed mean = 12

∑ f u ×h ∑f

Mean x = a+

(NCERT Exemplar)

395

xi − a h

Compute median and mode size of the holdings.

222 322

ui =

(CBSE 2015)

i i

Median here, N = 200

N = Now, 100, which lies in the interval 15 − 20. 2 183  −366  Lower limit, = l 15,= h 5,= f 80 and= cf 55 Mean = 12 +  = 12 − 1.83 = 10.17  × 2 = 12 − 100  400  N  Thus, Mean of data is 10.17  2 − cf   100 − 55  Median =l +   × h =15 +  ×5 Here, f  80      N 400 = = 200 45   2 2 = 15 +   × 5 = 15 + 2.81 = 17.81 hec  80  ⇒ Median class = 09–11, Mode: In a given table 80 is the highest frequency. Where, So, the modal class is 15 − 20. = l 9,= N 400, = h 2,= cf 190, = f 32 Here, = l 15, = f1 80, = f 0 30, = f 2 40 and = h 5 Using the median formula:  fm − f 0  N  Mode = l + ×h  2 − cf  200 − 190    2 fm − f 0 − f 2  Median = l + 9+ ×2 ×h =  32    f  80 − 30    50    = 15 +  15 +  ×5= ×5  2 × 80 − 30 − 40   160 − 70  10 = 9+ × 2= 9 + 0.625 = 9.625 25  50  32 = 15 + 2.77 = 17.77 hec = 15 +   × 5 = 15 + 90 9   Thus, median is 9.625. i

Chapter 13 • Statistics

359

11. If the median of the following frequency distribution is 32.5, find the values of f1 and f2 . (CBSE 2019)

Since the median is 32.5, the median class is 30–40 (as its cf just exceeds 20).

Class

Frequency

Using the median formula:

0–10

10–20

20–30

N   2 − cf  Median = l + ×h  f   

30–40

40–50

50–60

60–70

Total

Where: l = 30,

 20 − ( f1 + 14) 32.5 = 30 +   × 10  12

Sol. The median class is the class where cumulative N 40 frequency (CF) is just greater than = = 20. 2 2 Cumulative Frequency Calculation: Class

Frequency fi

Cumulative Frequency (cf )

0–10

20–30

f1 + 14

f1 + 26 + f2

10–20 30–40 40–50

N = 20 , cf= f1 + 14 , f = 12, h = 10 2

60 − 10 f1  6 − f1  32.5 −= 30  × 10 ⇒ 2.5 =   12  12

30 = 60 − 10 f1 ⇒ 10 f1 = 30 ⇒ f1 = 3

Now, using

∑ f = 40 i

3 + 5 + 9 + 12 + f2 + 3 + 2 = 40

f1 + 5

34 + f2 = 40

f1 + 26

f1 = 3, f2 = 6

f2 = 6

Competency Based Questions Multiple Choice Questions 1. If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then the value of x is (a) 4 (b) 7 (c) 11 (d) 6 (CBSE 2024)

2. The marks obtained by a set of students in an exam are recorded in a grouped frequency table. The maximum number of students are found to be in the range of (70–80) marks. If the number of students in the ranges before and after the (70–80) range are equal, which of the following is the mode of the data? (a) 70 marks (b) 75 marks

(b) Only mean and median (c) All – mean, median, mode (d) Cannot be said without knowing the outlier 4. If the value of each observation in a data is increased by 2, then the median of the new data (a) Increased by 2 (b) Increased by 2n (c) Remains same

3. In Statistics, an outlier is a data point that differs significantly from other observations if a data set. If an outlier is included in the following data set, which measure(s) of central tendency would change? 13, 17, 22, 43, 43, 48, 52, 51 (a) Only mean

(d) Decreased by 2 (CBSE 2024)

5. The mean of five numbers is 15. If we include one more number, the mean of six numbers becomes 17. The included number is (a) 27 (b) 37 (c) 17 (d) 25 (CBSE 2024)

6. In the given number line, p, q and r represent the measures of central tendency of the data:3, 4, 6, 5, 6, 10 pq

Smallest data point

Largest data point

Which of these is true for p, q and r ? (a) p – Mean, q – Median, r – Mode (b) p – Median, q – Mean, r – Mode (c) p – Mode, q – Median, r – Mean (d) Cannot be determined unless specific numbers on the scale is provided 7. The table given below shows the literacy rate of 70 cities in a country. Literacy Rate (in %)

No. of Cities

30–40

40–50

50–60

60–70

70–80

80–90

90–100

Class

Frequency

65–85

105–125

125–145 145–165 165–185 185–205 Chapter 13 • Statistics

20 14 7 4

Select Response Type MCQs 1. A measure of central tendency that is not affected by extreme values in a data set is: (i) Mean (ii) Median (iii) Mode

(iv) Range

(a) (i) and (ii) (c) only mean

(b) (ii) and (iii) (d) only mode

(ii) It is always the middlemost value

(CBSE 2024)

for finding the mean of grouped data, di are the deviations from a of: (a) Lower limits of the classes (b) Upper limits of the classes (c) Mid-points of the classes (d) Frequencies of the class marks

Which of the following is true about the mode of the given data? (i) It is the most frequently occurring value

8. After an examination, a teacher wants to know the marks obtained by the maximum number of students in her class. She requires to calculate __________ of marks. (a) Median (b) Mode (c) Mean (d) Range

85–105

(NCERT Exemplar)

10. In the formula ∑ xi di x= a + ∑ fi

2. A class of students scored the following marks in a test: 10, 15, 20, 25, 25, 30, 35, 40, 40, 40, 45, 50, 55

What is the literacy rate for maximum number of cities? (a) 70.025 (b) 72.5 (c) 75 (d) 112.5

9. Consider the data:

The difference of the upper limit of the median class and the lower limit of the modal class is: (a) 0 (b) 19 (c) 20 (d) 38

(iii)It is 40 (iv)It is always the average of all values (a) (i) and (iii) (c) only (iii)

(b) (ii) and (iv) (d) only (iv)

Assertion-Reason Based Questions The following questions consist of two statements— Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below: (a) Both A and R are true and R is the correct explanation for A. (b) Both A and R are true but R is not the correct explanation for A. (c) A is true but R is false. (d) A is false but R is true. 1. Assertion (A): If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, then median is 30. 361

n +1 Reason (R): Median =   value, if n is odd.  2  2. Assertion (A): If the value of mode and mean are 60 and 66 respectively, then the value of median is 64. 1 Reason (R): Median = ( mode + 2 × mean ) 3

Case Study Based Questions 1. India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country. (CBSE 2023) It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018: Rainfall (mm)

Number of Sub-divisions

200–400

600–800

400–600

(d) If a sub-division having at least 1000 mm rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall? 2. BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when calledout a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and win the game. The table given below, shows the data of one such game where 48 balls were used before Tara said ‘BINGO’. Numbers Announced Number of Times

800–1000

1000–1200

1200–1400

1400–1600

1600–1800

Based on the above information, answer the following questions:

0–15

15–30

30–45

45–60

60–75

Based on the above information, answer the following: (i) Write the median class.

(a) Write the modal class.

(ii) When first ball was picked up, what was the probability of calling out an even number?

(b) Find the median of the given data.

(iii)(a) Find median of the given data. OR

OR (c) Find the mean rainfall in this season.

(b) Find mode of the given data.

Answers Multiple Choice Questions 1. (b) 7 Mean is given by: x + x + 2) + (x + 4) + (x + 6) + (x + 8) ( = 11 5 5x + 20 = 11 5 5x + 20 = 55 5x = 35 x =7

2. (b) 75 marks We are given that the maximum number of students 362

are in the range (70–80) marks. This means the modal class is 70–80. We are also given that the number of students before and after the 70–80 range are equal. This means: f 0 (frequency of the class before 70–80) = f 2 (frequency of the class after 70–80) The formula for mode is:   f1 − f 0 Mode = l + ×h  2 f1 − f 0 − f 2  Since f 0 = f 2  f −f  Mode = l + 1 0 ×h  2 f1 − 2 f 0 

 f −f  1 =l +  1 0  × h =l +   × h  2( f − f )  2 1 0   10 = 70 + 5 = 75 2 3. (a) Only mean

= 70 +

Given set of numbers is 13, 17, 22, 43, 43, 48, 52, 51 On rearranging the given set of numbers, we get 13, 17, 22, 43, 43, 48, 51, 52 Median of given data 43 + 43 = = 43 2 In the given data, there are 8 observations. If one outlier is added, then it becomes 9 observations. Median = Middle number = 43 Mode of the given data = 43 When an outlier is added, mode will not change as its frequency would be 1. Now, mean of the given data 13 + 17 + 22 + 43 + 43 + 48 + 51 + 52 = 8 289 = = 36.125 8 Let the outlier be x. Then, new mean 289 + x ≠ 36.125 9

which will not be the same as the mean of the already given data. Hence, the required answer is “only mean”. 4. (a) Increased by 2 If each observation in a dataset is increased by 2, the new median will also increase by 2. 5. (a) 27 Given: Mean of 5 numbers = 15 Mean of 6 numbers (after including one more number) = 17 Sum of 5 numbers = 15 × 5 = 75 Sum of 6 numbers = 17 × 6 = 102 Included number = 102 − 75 = 27

Chapter 13 • Statistics

6. (b) p – Median, q – Mean, r – Mode Given, data = 3, 4, 6, 5, 6, 10 On rearranging the given data, we get 3, 4, 5, 6, 6, 10 Mean of the given data

3 + 4 + 5 + 6 + 6 + 10 34 = = = 5.66 6 6 Mode of the given data = A number which has the highest frequency = 6 Median of the given data 5+6 = = 5.5 2 From the above central tendency, 5.5 < 5.66 < 6 Median < Mean < Mode On comparing with p, q, r, we get p= q = Mean r = Mode Median Hence, p, q, and r denote the central tendency of median, mean, and mode, respectively. 7. (b) 72.5 The given table is Literacy Rate (in %)

No. of Cities (fi)

30–40

50–60

70–80

40–50 60–70 80–90

90–100

16 12 4

Maximum number of cities = 18 Modal class = 70–80 = l 70, = f1 18, = f 0 16, = f 2 12, = h 10 Mode of the given data   f1 − f 0 l + = ×h  2 f1 − f 0 − f 2 

18 − 16   = 70 +   × 10  2 × 18 − 16 − 12  2   = 70 +   × 10  36 − 16 − 12  2 = 70 + × 10 = 70 + 2.5 = 72.5 8 Hence, the literacy rate for the maximum number of cities is 72.5. 8. (b) Mode Mode is the measure of central tendency that represents the most frequently occurring value in a dataset. Since the teacher is looking for the most common marks, she needs to calculate the mode. 363

9. (c) 20

2. (a) (i) and (iii)

Find the Total Frequency (N)

●

N = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67 Class

Frequency

Cumulative Frequencies

65–85

●

105–125

●

145–165

85–105

125–145 165–185 185–205

20 7 4

●

42 63 67

The median class is the class where the cumulative frequency first exceeds: N 67 = = 33.5 2 2 The first cumulative frequency ≥ 33.5 is 42, which corresponds to the class 125 – 145. Upper limit of the median class = 145 The modal class is the class with the highest frequency. The highest frequency = 20, which corresponds to the class 125–145. Lower limit of the modal class = 125

10. (c) Mid-points of the classes The given formula for the mean of grouped data is: ∑ fi di x= a + ∑ fi d= xi − a i

he mode is not necessarily the middlemost T value (that’s the median). he most frequently appearing value is 40 T (appears 3 times). The mode is not the average of all values.

Assertion-Reason Based Questions 1. (d) A is false but R is true. Sorting the runs scored in ascending order 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52 Since there are 11 values (odd number), the median is at the

(11 + 1) 2

= 6th position

The 6th value in the sorted list is 27, not 30. Thus, Assertion (A) is incorrect because the median is 27, not 30. The given Reason (R) is correct since it correctly states the formula for median when n is odd. 2. (a) Both A and R are true and R is the correct explanation for A.

Difference = Upper limit of median class – Lower limit of modal class = 145 − 125 = 20

he mode is the most frequently occurring T value.

Given: Median =

Mode + 2 × Mean 3

Using the given formula to calculate the median Median =

60 + ( 2 × 66 ) 3

192 = 60 + 132 = = 64 3 3

where xi is the midpoint (class mark) of each class.

Both (A) and (R) are true, and (R) correctly explains (A)

Select Response Type MCQs

Case Study Based Questions

1. (b) (ii) and (iii) Mean is affected by extreme values as it considers all data points. Median is the middle value and is not affected by extreme values. Mode is the most frequent value, also not affected by extreme values. Range is affected since it depends on the highest and lowest values. 364

1. (a) The modal class is the class with the highest frequency. Given frequencies: [2, 4, 7, 4, 2, 3, 1, 1]

he maximum frequency is 7, which T corresponds to the class 600–800 mm. Modal class: 600–800 mm

(b) Finding the Median

The total number of sub-divisions (N) is:

N = 2 + 4 + 7 + 4 + 2 + 3 + 1 + 1 = 24

2. (a) The cumulative frequency distribution is as given below:

N 24 The median position is: = = 12 2 2

xi × f i

Number Cumulative of Sub- Mid Point Frequencies divisions (x ) i (fi)

Rainfall (mm)

600

300

200–400

2000

500

400–600

4900

700

600–800

3600

900

800–1000

2200

1100

1000–1200

3900

1300

1200–1400

1500

1400–1600

1700

1600–1800

20400

Total

Since the 12th observation falls in the 600–800 mm class, this is the median class.

N   2 − cf  Now, Median = l + ×h  f1     12 − 6  × 200  Median = 600 +   7 

6  = 600 +  × 200  = 600 + 171.43 = 771.43 mm 7 

Median rainfall: 771.43 mm (c) Mean =

∑ ( fi × xi ) 20400 = = 850 mm ∑ fi 24

Mean rainfall: 850 mm

(d) A sub-division is considered to have “good rainfall” if it receives at least 1000 mm of rain.

The sub-divisions with rainfall ≥ 1000 mm are:

Cumulative Frequency

48 = N

∑ fi = 48 and We find that N =

We find that 30–45 is the median class with:

●

l = 30

●

f = 10

●

h = 15

●

F = 17

●

N = 48

N   2 −F  Median = l +  ×h  f    7×3  24 − 17  = 30 +   × 15 = 30 + 2  10  = 30 + 10.5 = 40.5

(b) Finding the Mode

We observe that 45–60 is the modal class with:

●

l = 45

●

f = 12

●

f1 = 10

●

f2 = 9

●

h = 15

●

1200–1400 mm → 3 sub-divisions

●

1400–1600 mm → 1 sub-division

●

1600–1800 mm → 1 sub-division

Chapter 13 • Statistics

Number of Times (f )

(b) There are 75 balls, out of which 37 balls are even numbered. 37 ∴ Probability of calling out an even number = 75 (c) (a) F inding the Median

1000–1200 mm → 2 sub-divisions

Number of sub-divisions with good rainfall: 7

0–15 15–30 30–45 45–60 60–75

N = 24. 2 N The cumulative frequency just greater than 2 i.e., 24 is 27, and the corresponding class is 30–45. So, 30–45 is the median class.

●

Total = 2 + 3 + 1 + 1 = 7

Numbers Announced

 f − f1  l + Mode = ×h  2 f − f1 − f 2   12 − 10  2 = 45 +   × 15 = 45 +   × 15  24 − 10 − 9  5 = 45 + 6 = 51

365

Additional Practice Questions 1. The algebraic sum of the deviations of a frequency distribution from its mean is (a) always positive (b) always negative (c) 0 (d) a non-zero number (CBSE 2024)

2. The mode of a frequency distribution can be determined graphically from (a) Histogram (b) Frequency polygon (c) Ogive (d) Frequency curve

9. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss. No. of Heads per Toss (x) No. of Tosses (f )

38 144 342 287 164 25

3. The median of first 10 prime numbers is (a) 11 (b) 12 (c) 13 (d) 14

10. Mode for the following distribution is 22. If x < y < 10 , then the value of y is

4. If the mode of the data: 64, 60, 48, x , 43, 48, 43, 34 is 43, then x + 5 = (a) 44 (b) 45 (c) 46 (d) 48

Class Interval

5. The value of x for which the class mark of the class interval 30 − x is 36 is (a) 38 (b) 40 (c) 36 (d) 42 (CBSE 2024)

uestion 6 and 7 consist of two statements—Assertion (A) Q and Reason (R). Answer these questions by selecting the appropriate option given below: (a) Both A and R are true and R is the correct explanation for A. (b) Both A and R are true but R is not the correct explanation for A. (c) A is true but R is false. (d) A is false but R is true. 6. Assertion (A): The algebraic sum of the deviations of a frequency distribution from its mean is zero. Reason (R): Mode of a frequency distribution cannot be determined graphically. 7. Assertion (A): The mean of 1, 4, 7,10, …, 301 is 151 Reason (R): The mean of the series a , a + d , a + 2d ,…, a + 2nd , is a +nd. 8. The number of telephone calls received at an exchange per interval for 250 successive oneminuteintervals are given in the following frequency table: No. of Calls No. of Intervals (f )

366

0−10 10−20 20−30 30−40 40−50 Total 5

11. The distribution below gives the marks obtained by 80 students on a test. Marks Number of Students

Less Less Less Less Less Less than than than than than than 10 20 30 40 50 60 3

The modal class of this distribution is (CBSE 2023)

12. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median: 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719 13. Find the mean marks of the students from the following cumulative frequency distribution: Marks Below

10 20 30 40 50 60 70 80 90 100

Number of 5 Students

17 29 45 60 70 78 83 85

14. The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital. Variable 10–2020–3030–4040–5050–6060–7070–80 Total

15 24 29 46 54 43 39

Compute the mean number of calls per interval.

Frequency (f )

Frequency 12

Find the average age for which maximum cases occurred.

229

15. Daily wages of 120 workers, obtained in a survey, are tabulated below: Daily Wages 100–120 120–140 140–160 160–180 180–200 200–220 220–240 (in `) Number of Workers

Compute the mean daily wages and modal daily wages of these workers. (CBSE 2021)

16. A student noted the number of cars passing through a spot on a road for 105 periods each of 3 minutes and summarised it in the table given below. Find the mean and median of the following data Number of Cars

0–10

Frequency (periods)

17. If the median of the following frequency distribution is 32.5, find the values of f1 and f 2. Frequency

0–10 10–2020–3030–4040–5050–6060–70 Total 3

(iii) 130–140

(iv) 140–150

(c) The difference of the upper limit of the median class and the lower limit of the modal class is (i) 10 (ii) 20 (iii) 30 (iv) 40 (d) The modal class is (i) 110–120

(ii) 120–130 (iv) 140–150

(e) The median value is (i) 120 (ii) 115

(CBSE 2023)

Class

(ii) 120–130

(iii) 130–140

10–20 20–30 30–40 40–50 50–60 60–70 70–80 16

(b) The median class is (i) 110–120

(CBSE 2019)

(iii) 125

(iv) 121

19. A 100 m race was organised in a school sports meet. The time was recorded with the help of a stopwatch. A table shown below describes the time in which the race was finished by the number of students Time (in sec.)

0−20

20−40

40−60

No. of Students

60−80 80−100 3

Based on the above data answer the following questions: (a) The lower limit of the modal class, is (i) 30 (ii) 40 (iii) 60 (iv) 20

18. In a wrestling championship 50 wrestlers participated. Their weights were recorded as given in the following table:

(b) The average time taken by the student to finish the race, is (i) 40 (ii) 47 (iii) 50 (iv) 30

Weight (in kg) 100−110 110−120 120−130 130−140 140−150

No. of Wrestlers

Based on the above information answer the following questions: (a) The upper limit of the modal class is (i) 110 (ii) 120 (iii) 130 (iv) 140

(iii) Median

(iv) All of these

(d) How many students finished the race within 1 minute (i) 30 (ii) 40 (iii) 19 (iv) 29

Challenge Yourself

Scan me

1. If the arithmetic mean of 7, 8, x, 11, 14 is x, then x is ___________. (a) 9

Chapter 13 • Statistics

(b) 9.5

(d) 10.5

2. If a mode exceeds a mean by 12, then the mode exceeds the median by ___________. (a) 4

(b) 8

(d) 10

367

3. If the median of the data, x1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x 8 is p, then find the median of the data x 3 , x 4 , x 5 , x 6 (where x1 < x 2 < x 3 < x 4 < x 5 < x 6 < x 7 < x 8 ):

Height (in cm)

Frequency

Cumulative Frequency

150–155

(a) p

155–160

p 2 p (c) 4

160–165

165–170

170–175

(d) Cannot be determined

175–180

Total

(b)

4. Find the unknown entries a, b, c, d, e, and f in the following distribution of heights of students in a class:

Answers Additional Practice Questions

16. Mean = 40.33, Median = 41.75 17. f 1 = 5, f 2 = 12

1. (c) 0

(a) Histogram

3. (b) 12

(d) 48

5. (d) 42

(c)

7. (a)

3.54

9. 2.47

10. 5

11. 30–40

12. 716

Challenge Yourself

13. 48.41

14. 34, 45

1. (c) 10

15. Mean = 170.17, Mode = 165

18. (i) (b) 120

(ii) (b) 120 – 130

(iv) (a) 110–120

(v) (d) 121

19. (i) (d) 20

(ii) (b) 47

(iii) (c) Median

(iv) (d) 29

2. (b) 8

(iii) (b) 20

3. p

a 12,= b 13,= c 35,= d 8,= e 5,= f 50 4.=

Scan me for Exemplar Solutions 368

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions 1. If the mean of the first n natural numbers is: (a) 5

(b) 4

[3 × 1 = 3 Marks]

5n . Find the value of n. 9 (c) 9

(CBSE 2024)

(d) 10

2. If the mean of 6, 7, p, 8, q, 14 is 9, then find the relation between p and q. 19 (b) p + q =

19 (a) p − q =

(CBSE 2024)

21 (c) p − q =

21 (d) p + q =

(d) 22

3. If the mean of the numbers 10, 12, x , 18, 20 is 16, find x . (a) 20

(b) 24

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

∑ xi n

5. Assertion (A): If the mode of a dataset is 35 and the median is 30, then the mean is 32. Reason (R): The mode, median, and mean follow the empirical formula: Mode = 3 × Median − 2 × Mean

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. Find the mode of the following frequency distribution: Class Interval

25–30

30–35

35–40

40–45

45–50

50–55

Frequency

7. Find the median class of the following distribution: Class

0–10

10–20

20–30

30–40

40–50

50–60

60–70

Frequency

Short Answer Questions

[4 × 3 = 12 Marks]

8. Find the missing frequency f, if the median of the given frequency distribution is 42. Class Interval

10–20

20–30

30–40

40–50

50–60

Frequency

Self-Assessment

369

9. The weights (in kg) of 40 students in a class are given below. Find the median weight. Weight (kg)

30–35

35–40

40–45

45–50

50–55

Number of Students

10. Find the mode of the following frequency distribution: Class Interval

50–60

60–70

70–80

80–90

90–100

Frequency

11. The fuel efficiency (km per litre) of 60 cars of the same model was tested by a manufacturer and details are given below: Mileage (km/L)

8–10

10–12

12–14

14–16

16–18

Number of Cars

Find the mean fuel efficiency of the cars.

Long Answer Questions

[3 × 5 = 15 Marks]

12. Find the mean and mode for the following data: Classes

10–20

20–30

30–40

40–50

50–60

60–70

70–80

Frequency

13. The mean of the following frequency distribution is 62.8, and the sum of all the frequencies is 50. Compute the missing frequencies f1 and f 2 . Classes

0–20

20–40

40–60

60–80

80–100

100–120

Frequency

14. Monthly expenditures on milk in 100 families of a housing society are given in the following frequency distribution: Monthly Expenditure (in `)

0–175

175–350

350–525

525–700

700–875

Number of Families

875–1050 1050–1225 7

Find the mode and median for this distribution

5 (CBSE 2016)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. Yoga is an ancient practice which is a form of meditation and exercise. By practicing yoga, we not only make our body healthy but also achieve inner peace and calmness. The International Yoga Day is celebrated on 21st June every year since 2015. To promote Yoga, Green Park Society in Pune organised a 7-day Yoga camp in their society. The number of people of different age groups who enrolled for this camp is given as follows: Monthly Expenditure (in `)

15–25

25–35

35–45

45–55

55–65

65–75

75–85

Number of Families

Based on the above information, answer the following questions: (a) Write the number of people of age above and equal to 55 years who enrolled for the camp. (b) (i) Find the median age of people enrolled for the camp. OR

(ii) If x more people of the age group 65–75 had enrolled for the camp, the mean age would have been 58. Find the value of x.

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. The time in seconds, taken by 150 athletes to run a 100m hurdle race are tabulated below:

Time (sec) Number of Athletes

The number of athletes who completed the race in less than 17 sec, is (a) 14

(b) 76

(d) 82

Class

0–5

5–10

10–15

15–20

20–25

The lower limit of the modal class is

Frequency

(a) 15

2. For the following distribution: (b) 25

3. If for a distribution:

13–14 14–15 15–16 16–17 17–18 18–19

∑ 1

(a) 3

fi xi =+ 132 5p , (b) 6

(d) 35

(CBSE 2024)

20 and mean of the distribution is 8.1, then the value of p is ∑f = i

(d) 5 (CBSE 2024)

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(b) Both A and R are true, but R is not the correct explanation of A.

(d) A is false, but R is true.

4. Assertion (A): A dataset consists of 17 observations arranged in ascending order. Then, the 9th observation represents the median of the data. th  n +1 Reason (R): When the total number of observations n is odd, then the   term is the median of the  2  arranged data. 5. Assertion (A): The modal class for the following distribution is 50–60. ∑ fi xi Reason (R): Mean = ∑ fi

Class Interval 10–20 20–30 30–40 40–50 50–60 60–70 Frequency

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. If the mean of the following distribution is 6, then find the value of a. 7. The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.

a +5

Runs Scored

0–40

Number of Batsmen

40–80 80–120 120–160 160–200 20

Short Answer Questions

[4 × 3 = 12 Marks]

8. The mode of the following frequency distribution is 38. Find the value of x. Class Interval

0–10

10–20

20–30

30–40

40–50

50–60

60–70

Frequency

Self-Assessment

(CBSE SQP 2022) 371

9. The mode of a distribution is 55 and the modal class is 45–60. The frequency preceding the modal class is 5. The frequency after the modal class is 10. Find the frequency of the modal class. 10. Find the mode of the following data. Class

0–20

20–40

40–60

60–80

80–100

Frequency

100–120 120–140 5

(CBSE Standard 2020)

11. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table below. Find the mean and median of the folllowing data: Number of cars

0–10

10–20

20–30

30–40

40–50

50–60

60–70

70–80

Frequency (Periods)

Long Answer Questions

[3 × 5 = 15 Marks]

12. The table below represents the percentage distribution of mental health disorders in Asian countries (2019). Can the median of the above data be greater than 12.5%? Give a valid reason. Percentage of Citizens with Mental Health Disorders

7.5 –10

10 –12.5

12.5 –15

15 –17.5

17.5 – 20

Number of Asian Countries

13. The mean of the following frequency distribution is 66, and the sum of all the frequencies is 60. Compute the missing frequencies f1 and f 2. Class Interval

0–20

20–40

40–60

60–80

80–100

100–120

Frequency

14. The median of the following data is 50. Find the values of p and q, if the sum of all frequencies is 90. Also, find the mode. Marks obtained

20–30

30–40

40–50

50–60

60–70

70–80

80–90

Number of Students

(CBSE SQP Standard 2023)

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A survey is conducted for 56 families of a colony. The following table gives the weekly consumption of electricity of these families. Weekly Consumption (in unit)

0–10

10–20

20–30

30–40

40–50

50–60

Number of Families

Based on the above information, answer the following questions: (a) What is the modal class? (b) Write the number of families whose weekly consumption (in units) is at least 30 units. (c) (i) What is the median weekly consumption? OR (ii) What is the mean weekly consumption? 372

Probability

Chapter at a Glance ●

Probability - A measure of the likelihood of an event occurring.

If E is an event of a random experiment. The probability of the event E is written as P(E) and Number of favourable outcomes Formula: P ( E ) = Total number of outcomes ● Random Experiment - An experiment in which all possible outcomes are known, and the exact output cannot be predicted in advance. ◆

●

Sample Space (S) - The set of all possible outcomes of a random experiment. Example: When tossing a fair coin, the sample space is S = {Heads, Tails}. Event - A subset of the sample space. It represents one or more outcomes of the experiment.

Example: If a fair dice is thrown, then all possible outcomes are 1, 2, 3, 4, 5, 6. Getting odd numbers 1, 3, 5, getting = = 3, 5} , E 2 {2, 3, 5} prime numbers 2, 3, 5 are events. Events can be represented as E1 {1,

Types of Events ●

Impossible Event: An event that cannot possibly occur; its probability is 0. Example: Rolling a 7 on a standard six-sided die.

●

Sure Event: An event that is certain to occur; its probability is 1. Example: Rolling a number between 1 and 6 when rolling a standard die.

●

Simple Event: An event consisting of only one outcome. Example: Rolling a 3 on a standard die.

●

Compound Event: An event consisting of more than one outcome. Example: Rolling an even number on a standard die (2, 4, or 6).

Complementary Events ● ●

The complement of an event A concerning sample space S is the set of all elements of S that are not in A. Formula: P ( not A )= 1 − P ( A ) or P ( A )= 1 − P ( A )

3 = 0.5, then the probability of the complement 6 1 − P( A ) = 1 − 0.5 = 0.5 (rolling not an even number) would be: P ( A ′ ) =

Example: If the probability of A (rolling an even number) is P ( A = )

Understanding Dice

Understanding Deck of Cards Red colour (26 cards)

Diamond (13 cards)

A deck of 52 playing cards Black colour (26 cards)

Chapter 14 • Probability

Heart (13 cards)

Spade (13 cards) Clubs (13 cards)

King (one) Queen (one) Each suits have

Jack (one) Ace (one) 9 number cards 2 to 10

373

NCERT Zone NCERT Exercise 14.1

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

1. Complete the following statements: (a) Probability of an event E + Probability of the event ‘not E’ = . (b) The probability of an event that cannot happen is . Such as event is called .

Sol. When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable.

Sol. (a) 1

4. Which of the following cannot be the probability of an event? 2 (a) (b) -1.5 (c) 15% (d) 0.7 3 Sol. Probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, −1.5 cannot be a probability of an event.

(b) 0, impossible event

Hence, (b)

(c) The probability of an event that is certain to happen is . Such as event is called . (d) The sum of the probabilities of all the elementary events of an experiment is

(e) The probability of an event is greater than or equal to and less than or equal to .

5. If P ( E ) = 0.05, what is the probability of ‘not E’?

(d) 1

Sol. We know that,

(e) 0, 1

P E = 1 − P(E ) = 1 – 0.05 = 0.95

2. Which of the following experiments have equally likely outcomes? Explain. (a) A driver attempts to start a car. The car starts or does not start.

( )

Therefore, the probability of ‘not E’ is 0.95. 6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(b) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(a) an orange flavoured candy?

Sol. (a) The bag contains lemon flavoured candies only. This implies that every time, she will take out only lemon flavoured candies. Therefore, event that Malini will take out an orange flavoured candy is an impossible event.

(d) A baby is born. It is a boy or a girl. Sol. (a) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. (b) It is not an equally likely event, as it depends on the player’s ability and there is no information given about that. Also, no other parameters are given in question. (c) It is an equally likely event. The answer could be right or wrong. (d) It is an equally likely event. The born baby could be a boy or a girl.

374

(b) a lemon flavoured candy?

Hence, P ( an orange flavoured candy ) = 0

(b) As the bag has lemon flavoured candies, Malini will take out only lemon flavoured candies. Therefore, event that Malini will take out a lemon flavoured candy is a sure event.

P ( a lemon flavoured candy ) = 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Sol. Probability that two students are not having same

( )

birthday P E = 0.992

(b) Will not be a `5 coin?

Probability that two students are having same birthday P ( E )= 1 − P E

( )

= 1 − 0.992

= 0.008

What is the probability that the ball drawn is (a) red? (b) not red? Sol. (a) Total number of balls in the bag = 8 Number of favourable outcomes ( Red balls ) = 3 Probability of getting a red ball =

Number of favourable outcomes 3 = Number of possible outcomes 8

(b) Probability of not getting red ball

3 5 = 8 8 9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

=1 − Probability of getting a red ball =1 −

(a) red?

(b) white?

Sol. Total number of coins in a piggy bank

= 100 + 50 + 20 + 10 = 180

(a) Number of 50 p coins =100

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag.

(a) Will be a 50 p coin?

Sol. Total number of marbles = 5 + 8 + 4 = 17

Probability of getting a 50 p coin Number of favourable outcomes 100 5 = = = 180 9 Number of possible outcomes

(b) Number of `5 coins =10 Probability of getting a Rs 5 coin Number of favourable outcomes 10 1 = = = Number of possible outcomes 180 18 Probability of not getting a `5 coin =− 1

1 17 = 18 18

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish? Sol. Total number of fish in a tank = Number of male fishes + Number of female fishes = 5 + 8 = 13 Number of favourable outcomes ( male fish ) = 5

Probability of getting a male fish Number of favourable outcomes 5 (a) Number of red marbles == 5 = Number of possible outcomes 13 Probability of getting a red marble 12. A game of chance consists of spinning an arrow 5 = Number of favourable outcomes = which comes to rest pointing at one of the numbers Number of possible outcomes 17 1, 2, 3, 4, 5, 6, 7, 8 (see the given figure), and these (b) Number of white marbles = 8 are equally likely outcomes. What is the probability Probability of getting a red marble that it will point at (a) 8? (b) an odd number? Number of favourable outcomes 8 = = Number of possible outcomes 17 (c) a number greater than 2? (d) a number less than 9? (c) Number of green marbles = 4

Probability of getting a red marble Number of favourable outcomes 4 = = Number of possible outcomes 17

Sol. Total number of possible outcomes = 8 (a) Number of favourable outcomes ( 8 ) = 1

Probability of getting 8 Number of favourable outcomes 1 Probability of not getting a green marble = = Number of possible outcomes 8 4 13 1 =− = (b) Total number of odd numbers on spinner = 4 17 17

10. A piggy bank contains hundred 50 p coins, fifty `1 coins, twenty `2 coins and ten `5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin Chapter 14 • Probability

Probability of getting an odd Number of favourable outcomes 4 1 = = = Number of possible outcomes 8 2

375

(c) The numbers greater than 2 are 3, 4, 5, 6, 7, and 8. Therefore, total numbers greater than 2 = 6 Probability of getting greater than 2 = Number of favourable outcomes= 6= 3 Number of possible outcomes 8 4

P(getting a red face card) Number of favourable outcomes 6 3 = = Number of possible outcomes 52 26

(d) Total number of Jack of hearts = 1 (d) The numbers less than 9 are 1, 2, 3, 4, 6, 7, and 8. P(getting a Jack of hearts) Therefore, total numbers less than 9 = 8 Number of favourable outcomes 1 = = Probability of getting a number less than 9 Number of possible outcomes 52 Number of favourable outcomes 8 (e) Total number of spade cards = 13 = = 1 = Number of possible outcomes 8 P(getting a spade card) 13. A die is thrown once. Find the probability of getting (a) a prime number; (b) a number lying between 2 and 6; (c) an odd number.

Number of favourable outcomes 13 1 = = Number of possible outcomes 52 4

(f) Total number of queen of diamonds = 1 P(getting a queen of diamond)

Number of favourable outcomes 1 = = Sol. The possible outcomes when a dice is thrown Number of possible outcomes 52 = {1, 2, 3, 4, 5, 6} 15. Five cards—the ten, jack, queen, king and ace Number of possible outcomes of a dice = 6 of diamonds, are well-shuffled with their face (a) Prime numbers on a dice are 2, 3, and 5. downwards. One card is then picked up at random. Total prime numbers on a dice = 3 (a) What is the probability that the card is the Probability of getting a prime number queen? Number of favourable outcomes 3 1 = = = (b) If the queen is drawn and put aside, what is the Number of possible outcomes 6 2 probability that the second card picked up is (b) Numbers lying between 2 and 6 = 3, 4, 5 (i) an ace? (ii) a queen? Total numbers lying between 2 and 6 = 3 Sol. (a) Total number of cards = 5 Probability of getting a number between 2 and 6 Total number of queens = 1 3 1 is= = P(getting a queen) 6 2 (c) Odd numbers on a dice = 1, 3, and 5 Number of favourable outcomes 1 = = Number of possible outcomes 5 Total odd numbers on a dice = 3 3 1 (b) When the queen is drawn and put aside, the Probability of getting an odd number= = 6 2 total number of remaining cards will be 4. 14. One card is drawn from a well-shuffled deck of 52 (i) Total number of aces = 1 cards. Find the probability of getting: P(getting an ace) (a) a king of red colour (b) a face card Number of favourable outcomes 1 = = (c) a red face card (d) the jack of hearts Number of possible outcomes 4 (e) a spade

(f) the queen of diamonds

Sol. Total number of cards in a well-shuffled deck = 52

(a) Total number of kings of red colour = 2 =

P(getting a king of red colour) Number of favourable outcomes 2 1 = = Number of possible outcomes 52 26

(b) Total number of face cards = 12 P(getting a face card) = 376

Number of favourable outcomes 12 3 = = Number of possible outcomes 52 13

(ii) A s queen is already drawn, therefore, the number of queens will be 0. P(getting a queen) =

Number of favourable outcomes 0 = = 0 Number of possible outcomes 4

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Sol. Total number of pens = 12 + 132 = 144 Total number of good pens = 132 P(getting a good pen) Number of favourable outcomes 132 11 = = = Number of possible outcomes 144 12 17. (a) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (b) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? Sol. (a) Total number of bulbs = 20 Total number of defective bulbs = 4 P(getting a defective bulb)

Therefore, total numbers divisible by 5 = 18

Probability of getting a number divisible by 5

Number of favourable outcomes 18 1 = = Number of possible outcomes 90 5

19. A child has a die whose six faces shows the letters as A, B, C, D, E and A: The die is thrown once. What is the probability of getting (a) A? (b) D? Sol. Total number of possible outcomes on the dice = 6 (a) Total number of faces having A on it = 2 P(getting A) =

Number of favourable outcomes 2 1 = = Number of possible outcomes 6 3

(b) Total number of faces having D on it = 1 P(getting D)

Number of favourable outcomes 4 1 Number of favourable outcomes 1 = = = = Number of possible outcomes 20 5 Number of possible outcomes 6 (b) Remaining total number of bulbs = 19 3m 20. Suppose you drop a die at Remaining total number of non-defective bulbs random on the rectangular = 16 − 1 = 15 region shown in the given figure. P(getting a not defective bulb) What is the probability that it Number of favourable outcomes 15 will land inside the circle with diameter 1m? = Number of possible outcomes 19 Sol. Area of rectangle =l × b =3 × 2 =6 m 2 . 18. A box contains 90 discs which are numbered from 1 π = π= r2 Area of circle (of diameter 1 m) to 90. If one disc is drawn at random from the box, 4 find the probability that it bears P(die will land inside the circle) =

(a) a two-digit number

(b) a perfect square number (c) a number divisible by 5 Sol. Total number of discs = 90 (a) Total number of two-digit numbers between 1 and 90 = 81 P(getting a two-digit number) =

Number of favourable outcomes 81 9 = = Number of possible outcomes 90 10

(b) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9. P(getting a perfect square) Number of favourable outcomes 9 1 = = = Number of possible outcomes 90 10 (c) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90. Chapter 14 • Probability

Favourable area 4 π = = = Total area 6 24 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (a) She will buy it? (b) She will not buy it? Sol. Total number of pens = 144 Total number of defective pens = 20 Total number of good pens = 144 − 20 = 124 (a) Probability of getting a good pen P(Nuri buys a pen) =

Number of favourable outcomes 124 31 = = Number of possible outcomes 144 36

(b) P(Nuri will not buy a pen)

=− 1 P ( Nuri will buy ) =− 1

31 5 = 36 36 377

22. (a) Complete the following table: Event: ‘Sum of 2 Dice’

1 Probability 36

10 11 12

5 36

1 36

(b) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

1 Therefore, each of them has a probability of . 11 Do you agree with this argument? Justify your Solution. Sol. If 2 dice are thrown, the possible events are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) So, the total number of events: 6 × 6 = 36 (a) It can be observed that, To get the sum as 2, possible outcomes = (1, 1) To get the sum as 3, possible outcomes = (2, 1) and (1, 2) To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2) To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2) To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2), (3, 3) To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3) To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3), (4, 4) To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4) To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5) To get the sum as 11, possible outcomes = (5, 6), (6,5) To get the sum as 12, possible outcomes = (6, 6) So, the table will be as: Event: ‘Sum of 2 Dice’ Probability

378

10 11 12

1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36

(b) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. Sol. The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH} 2 1 P(Hanif will win the game)= = 8 4 1 3 P(Hanif will lose the game) =1 − = 4 4 24. A die is thrown twice. What is the probability that (a) 5 will not come up either time? (b) 5 will come up at least once? Sol. Total number of outcomes = 6 × 6 = 36 (a) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Hence, total number of favourable cases = 11

P(5 will come up either time) =

11 36

11 25 = 36 36 (b) Total number of cases, when 5 can come at least once = 11 11 P(5 will come at least once) = 36 P(5 will not come up either time) =− 1

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. (a) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these 1 outcomes, the probability is . 3 (b) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd 1 number is . 2 Sol. (a) Incorrect, when two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T).

It can be observed that there can be one of each in two possible ways − (H, T), (T, H). Therefore, the probability of getting two heads is 1 , 4 1 the probability of getting two tails is , and the 4 probability of getting one of each is

1 . 2

It can be observed that for each outcome, the 1 probability is not . 3 (b) Correct, when a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2, 4, 6 are even numbers. Therefore, the probability of getting an odd 1 number is . 2

Multiple Choice Questions 1. Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is: 1 1 1 (a) (b) (c) 7 (d) 9 6 12 12

2. A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is: 7 1 7 5 (a) (b) (c) (d) 55 10 50 49

(NCERT Exemplar)

(a) b = 4a

(b) b = 3a

(d) b = a

(CBSE 2024)

9. A bag contains 3 red balls and 2 green balls. A ball is drawn randomly. What is the probability that it is not green?

(CBSE 2024)

3. A card is drawn at random from a deck of cards. What is the probability of drawing a queen? 1 1 1 1 (a) (b) (c) (d) 13 12 52 4 4. If two coins are tossed, what is the probability of getting at least one head? 1 1 3 (a) (b) (c) (d) 1 4 2 4 3

8. The probability of guessing the correct answer to a a certain question is . If the probability of not b 2 guessing the correct answer to this question is , 3 then

(a)

3 5

(b)

2 5

(c)

1 2

(d)

1 5

(NCERT Exemplar)

10. A school has 10 basketball players, out of which 5 are boys and rest are girls. What is the probability that a particular player selected is a girl? (a)

1 2

(b)

1 5

(c)

2 5

(d)

3 10

5. When a die is rolled, what is the probability of getting a number greater than 4? 1 2 1 (a) (b) 1 (c) (d) 3 3 6 2

11. If three coins are tossed simultaneously, what is the probability of getting at least two heads?

6. The probability of getting a bad apple in a box of 400 apples is 0.035. The total number of bad apples is

12. A letter is chosen at random from the English alphabet. What is the probability that it is a vowel?

(CBSE 2019)

(a) 7

(b) 14

(d) 28

7. A spinner is divided into 8 equal sections, each coloured differently respectively red, blue, green, white, red, blue, green. What is the probability that the spinner will stop on the red section? (a)

1 8

(b)

Chapter 14 • Probability

1 4

(c)

3 8

(d)

1 2

(a)

1 4

1 5

(b)

3 8

(c)

5 26

(c)

1 2

1 26

(d)

1 8

21 26

(NCERT Exemplar)

( )

x , then the value of 13. For an event E, P ( E ) + P E = 3 x − 3 is (a) -2

(b) 2

(d) -1 379

14. Jyoti and Dara are playing a game of tic-tac-toe. The probability of Jyoti winning the game is 0.7. What is the probability that Jyoti not winning the game? (a) 0.7 (b) 0.5 (c) 0.3

15. If P ( A ) denotes the probability of an event A, then (a) P ( A ) < 0

(b) P ( A ) > 1

(d) −1 ≤ P ( A ) ≤ 1 (NCERT Exemplar)

(d) Cannot be determined

Answers 1 13

1. (d)

1 12

2. (c)

1 10

3. (a)

4. (c)

3 4

5. (a)

6. (b) 14

7. (b)

1 4

1 3

8. (b) b = 3a

9. (a)

3 5

10. (a)

1 2

13. (a) –2

11. (c)

1 2

12. (b)

5 26

14. (c) 0.3

15. (c) 0 ≤ P ( A ) ≤ 1, since, probability of an event always lies between 0 and 1.

Constructed Response Questions Very Short Answer Questions 1. A bag contains 3 red, 2 green, and 4 blue balls. What is the probability of drawing a green ball? (CBSE SQP 2025) Sol. Total number of balls (3 + 2 + 4) = 9 Number of green balls = 2 2 So, P ( green ball ) = 9 2. When a die is rolled, what is the probability of getting a number greater than 4? (CBSE 2018) Sol. Numbers greater than 4 are 5 and 6. 2 1 Thus, P ( getting a number greater than 4 = ) = 6 3 3. Find the probability of drawing an ace from a standard deck of 52 cards. (NCERT Exemplar) Sol. There are 4 aces in a deck of 52 cards. 4 1 Therefore, P ( ace card= ) = 52 13 4. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7? Sol. Total number of possible outcomes n ( S ) = 34 Favourable number of outcomes is (7, 14, 21, 28 and 35) = 5 5 P ( multiple of 7 ) = 34 5. If a coin is tossed twice, what is the probability of getting two heads? (CBSE 2024) Sol. When a coin is tossed twice, the sample space is {HH, HT, TH, TT}, only one of these outcomes is two heads {HH}. 380

1 4 6. If a person throws a dart at a circular target and hits the bullseye with a probability of 0.2, what is the probability of missing it? (CBSE 2024) Sol. P ( miss ) =− 1 P ( hit ) =− 1 0.2 = 0.8

Thus, P ( two heads ) =

7. If three coins are tossed, then what is the probability of getting exactly one head? (CBSE 2023) Sol. When three coins are tossed, the sample space is {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. The favourable outcomes are HTT, THT, TTH. 3 Thus, P ( Exactly1 head ) = 8 8. If a number is selected from numbers 1 to 50, then what is the probability that it is divisible by 5? (NCERT Exemplar)

Sol. The total number of numbers from 1 to 50 is: n ( S ) = 50. The numbers that are divisible by 5 within this range are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 The total count of these numbers is: n ( A ) = 10 P ( number that is divisible by 5= )

n ( A ) 10 1 = = n ( S ) 50 5

9. A bag contains 6 white and 3 black balls. What is the probability of drawing a black ball? Sol. The total numbers of balls is (6 + 3) = 9. With 9 balls in total and 3 being black, 3 1 P ( black )= = 9 3 10. If a single card is drawn from a standard deck of cards, then what is the probability it is a face card? (NCERT Exemplar)

Sol. There are 12 face cards in a deck of 52 cards. 12 3 Hence, P ( face card= ) = 52 13 11. When two dice are rolled, what is the probability that the sum is 6? (CBSE 2022) Sol. When two dice are rolled, total number of outcomes 36 are ( 6 × 6 ) =

The favourable outcomes are (1, 5), (2, 4), (3, 3), (4, 2), (5, 1). 5 Thus, P ( sum is 6 ) = 36 12. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb? Sol. Out of 600 electric bulbs, one bulb can be chosen in 600 ways. Total number of elementary events = 600 =( 600 − 12 ) non-defective bulbs, out There are 588 of which one bulb can be chosen in 588 ways. Favourable number of elementary events = 588 P(Getting a non defective bu= lb)

588 49 = = 0.98 600 50

13. What is the probability of drawing a red ball from a bag containing 5 red, 6 blue, and 4 green balls? (CBSE SQP 2024)

Sol. The total number of balls is 15, with 5 being red. 5 1 Hence, P(red bal= l) = . 15 3

14. In a lottery, there are 10 prizes and 25 blanks. What is the probability of drawing a prize? (CBSE 2023)

Sol. There are a total of 35 items, with 10 being prizes. 10 2 Thus, P ( prize= ) = 35 7 15. If a letter is taken at random from the word ‘APPLE’, then what is the probability that it is a vowel? Sol. Total Letters = 5 Number of vowels = 2 Thus, P ( getting a vowel ) =

2 5

16. What is the probability that a random day selected from a week is a Sunday? Sol. Total days in a week = 7 1 P ( Sunday ) = 7

Chapter 14 • Probability

17. A student is selected at random from a class of 20 students. What is the probability that the selected student has a birthday in May? Sol. Assuming an equal probability for birthdays in any given month, Total months in a year =12 Favourable month (May) = 1 1 P ( May ) = 12 18. A box has 7 tennis balls and 14 season balls. A ball is drawn at random. Find the probability of getting a tennis ball? Sol. Total number of balls = 7 + 14 = 21 Tennis balls = 7 7 1 P ( tennis ball= ) = 21 3 19. What is the probability that a non-leap year has exactly 52 Sundays? (CBSE 2024) Sol. There are 365 days in a non-leap year. A non-leap year always has 52 full weeks, meaning it has exactly 52 Sundays, plus 1 extra day. The extra day can be one of the following: {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}. If the extra day is not Sunday, then the year will have exactly 52 Sundays. The number of favourable cases where the extra day is not Sunday is: 6(Monday to Saturday) 6 P ( 52 Sundays ) = 7

Short Answer Questions 1. A box contains 6 red, 4 green, and 3 blue balls. If a ball is picked at random, what is the probability that it is not blue? (CBSE 2022) Sol. The total number of balls in the box is (6 + 4 + 3) n ( S ) = 13 The number of non-blue balls (red+green) is: n ( A ) = 6 + 4 = 10 The probability of picking a non-blue ball is: P= (A)

n ( A ) 10 = n ( S ) 13

2. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the process is a success and find the probability of losing the game. (CBSE 2019)

381

Sol. Total number of outcomes = 8 Success (HHH, TTT), favourable number of outcomes = 2 2 1 P(success) = = 8 4 1 3 So, P(loss) = 1 − = 4 4 3. A school has 300 students. A random sample of 30 students found that 6 wear glasses. Estimate the probability that a student selected at random from the whole school wears glasses. (CBSE 2024) Sol. Probability from Sample Data Student in sample = 30 Student war glasses in sample = 6 P ( wearing glasses ) Number of students wearing glasses in sample = Total students in sample 6 1 = = 30 5 Since the selection is random, the probability remains the same for the whole school. Therefore, the probability that a randomly selected student wears glasses is 0.2. 4. A bag contains tickets numbered from 1 to 49. A ticket is drawn at random. What is the probability that it is (a) a multiple of 7? (b) not a multiple of 7? Sol. The tickets are numbered from 1 to 49, so the total number of possible outcomes is: n ( S ) = 49 (a) The numbers that are multiples of 7 within 1 to 49 are: 7, 14, 21, 28, 35, 42, 49 The total count of these numbers is: n ( A ) = 7 P(Drawing a Multiple of 7) n(A ) 7 1 = = n ( S ) 49 7 (b) P(Not Drawing a Multiple of 7)

= P ( A= )

= 1 − P ( Drawing a Multiple of 7 )

=1 −

1 6 = 7 7

5. One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn (a) is the queen of hearts. (b) is not a jack. (CBSE Standard 2024) Sol. A standard deck of cards contains 52 cards, Total number of possible outcomes is: n ( S ) = 52

382

(a) There is only one Queen of Hearts in the deck: n ( A ) = 1

P(Drawing the Queen of Hearts): n(A ) 1 P= (A) = n ( S ) 52

(b) A standard deck has 4 Jacks (one from each suit: Hearts, Diamonds, Clubs, Spades). The number of cards that are not Jacks is: n ( B ) = 52 − 4 = 48

P(Drawing a Card That Is Not a Jack): P ( B= )

n ( B ) 48 12 = = n ( S ) 52 13

6. A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects, and 3 have major defects. Varnika will buy a phone if it is good, but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is (a) acceptable to Varnika? (b) acceptable to the trader?

(NCERT Exemplar)

Sol. The total number of mobile phones in the lot is: n ( S ) = 48

The lot consists of: Good phones = 42 Phones with minor defects = 3 Phones with major defects = 3 (a) Varnika will buy a phone only if it is good. So the number of favourable outcomes is: n ( A ) = 42 P(Phone Is Acceptable to Varnika) = P ( A= )

n ( A ) 42 7 = = n ( S ) 48 8

(b) The trader will buy a phone if it has no major defect. n ( B ) = 42 + 3 = 45

P(Phone Is Acceptable to the Trader) n ( B ) 45 15 = P ( B= = = ) n ( S ) 48 16 7. Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why? (NCERT Exemplar)

Sol. Apoorv throws two dice once.

(b) Total number of balls after adding 2 red balls: 18 + 2 = 20 Total number of outcomes = 36 Total red balls: x + 2 Will get 36 if (6, 6) appears x +2 P ( red ball ) = ∴ Number of outcomes for getting product 36 = 1 20 1 Now, according to the given condition: ∴ Probability for Apoorv = 36 x + 2= 9 × x Also, Peehu throws one die. 20 8 18 Total number of outcomes = 6 x +2 1 x = = × 20 8 2 Peehu will get 36 only if (6) appears x +2 x = = ∴ Number of favourable outcomes =1 5 4 1 ⇒ 4 ( x + 2 ) = 5x ∴ Probability for Peehu = 6 1 1 ⇒ 8 = 5x − 4x Since: > 6 36 ⇒ x = 8 Hence, Peehu has a better chance of getting the number 36. Hence, the value of x is 8. 8. A coin is tossed 3 times. List the possible outcomes. Find the probability of getting (a) all heads (b) at least 2 heads

(NCERT Exemplar)

Sol. The possible outcomes if a coin is tossed 3 times is:

10. An integer is chosen between 70 and 100. Find the probability that it is: (a) a prime number

(b) divisible by 7 (CBSE 2021)

Sol. Total outcomes (numbers between 70 and 100 are 29) = n ( S ) = 29

S = {(HHH),(HTT),(HTH),(THT),(TTH),(THH), (HTH), (HHT)}

(a) Prime numbers in this range: 71, 73, 79, 83, 89, 97

n (S ) = 8

Number of favourable outcomes: n ( E ) = 6

(a) Let E1 = Event of getting all heads = {(HHH)}

n ( E1 ) = 1 n ( E1 ) 1 P= ( E1 ) = n (S ) 8

n (E 2 ) = 4

n (E 2 ) 4 1 P(E 2 = = = ) n (S ) 8 2

9. A bag contains 18 balls out of which x balls are red. (a) If one ball is drawn at random from the bag, what is the probability that it is not red? (b) If 2 more red balls are put in the bag, the 9 probability of drawing a red ball will be times 8 the probability of drawing a red ball in the first case. Find the value of x. x Sol. (a) P ( red ball ) = 18 P ( no red ball ) =− 1 P ( red ball ) =− 1

Chapter 14 • Probability

n (E ) 6 = n ( S ) 29

(b) Numbers divisible by 7 in this range: 77, 84, 91, 98

(b) Let E2 = Event of getting at least 2 heads = {(HHT), (HTH), (THH)}

P ( prime number = )

(CBSE 2015)

x 18 − x = 18 18

Number of favourable outcomes: n ( E1 ) = 4

P ( number divisible = by 7 )

n ( E1 ) 4 = n ( S ) 29

11. A bag contains 15 white and some black balls. The probability of drawing a black ball from the bag is thrice that of drawing a white ball, then find the number of black balls in the bag. (CBSE 2017) Sol. Let number of black balls = x = 15 + x So, total balls

x x + 15 15 P(Drawing White Ball) = x + 15 P(Drawing Black Ball) =

Now, according to Question

x 15 = 3 × x + 15 15 + x x = 45 Hence, the number of black balls in the bag is 45.

383

Long Answer Questions

Sol. Given that, Total number of cars = 150

1. A golfer hits a shot, and the golf ball is reported to have landed somewhere in the rectangular golf course shown in the figure. What is the probability that the ball landed inside the pond shown in the figure? 10 km

2 km 4 km

Pond

4 km

Number of defective cars = 6 To find: Number of non-defective cars = 150 - 6 = 144 (a) Let E1 = Event of selecting a car which is not defective

n ( E1 ) Number = = of non defective cars 144 n ( E1 ) 144 24 = = = P(Not Defective) n ( S ) 150 25 5.5 km

2 km

24 25 (b) Let E2 = Event of selecting a car which is defective ∴ Probabilty of not defective is

= = of defective cars 6 n ( E 2 ) Number Sol. The dimensions of the rectangular golf course are: n (E 2 ) 6 1 Required probability = = = = Length 10 = km,Width 5.5 km n ( S ) 150 25 Area of the golf course = Length × Width 1 ∴ Probabilty of defective is 2 25 Area of the golf course =10 × 5.5 =55 km 4. The King, Queen and Jack of clubs are removed The dimensions of the pond are: from a pack of 52 cards and then the remaining = Length 2= km,Width 1.5 km cards are well shuffled. A card is selected from the Area of the= pond Length × Width remaining cards. Find the probability of getting a 2 (a) of spade (b) of black king Area of pond =× 2 1.5 = 3 km The probability of the golf ball landing in the pond is given by: Area of the pond 3 P ( landing in pond ) = = Area of the golf course 55 2. A pair of dice is rolled. What is the probability that the product of the numbers on the two dice is (a) less than or equal to 12? (b) more than 12? Sol. Total number of possible outcomes = 36 (a) Favourable outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1), (6, 2) i.e., 23 in number. 23 So, P ( Less than or equal to 12 ) = 36 (b) P(More than 12) 23 13 1 P ( Less than or equal to 12 ) =− 1 =− = 36 36

3. A car factory inspects 150 cars and finds 6 defective ones. If a car is chosen at random for a detailed check, what is the probability that it is (a) not defective? (b) defective? 384

(d) of Jacks

Sol. Total number of cards is equal to 52 - 3 = 49 (a) Number of spade cards =13 13 P(spade card) = 49 (b) Number of black king (Only Spade) = 1 1 P(black king) = 49 (c) Number of club cards = 10 10 P(club) = 49 (d) Number of Jacks = 3 3 P(jack) = 49 5. A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is (NCERT Exemplar) (a) not red?

(b) white?

Sol. Given that, A bag contains total number of balls = 24 A bag contains number of red balls = x A bag contains number of white balls = 2x

and a bag contains number of blue balls = 3x By condition, x + 2x + 3x = 24

{(

6x = 24 ⇒ x = 4

Thus, number of red, white and blue balls are 4, 8, and 12 respectively.

(a) Let E1 = Event of selecting a ball which is not red i.e., can be white or blue. = n ( E1 ) Number of white balls + Number of blue balls n ( E1 ) =8 + 12 =20 n ( E1 ) 20 5 = = = Required probability n ( S ) 24 6 (b) Let E2 = Event of selecting a ball which is white. = = of white balls 8 n ( E 2 ) Number

n (E 2 ) 8 1 Required probability = = = n ( S ) 24 3

6. In a game, the entry fee is of ` 5. The game consists of tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. For tossing a coin three times, find the probability that she: (NCERT Exemplar) (a) loses the entry fee (b) gets double entry fee (c) just gets her entry fee. Sol. Total possible outcomes of tossing a coin 3 times,

( HHH ) , ( HTT ) , ( HTH ) , ( THT ) , S=  ( TTH ) , ( THH ) , ( HTH ) , ( HHT )  n (S ) = 8

(a) Let E1 = Event that Sweta loses the entry fee

She tosses tail on three times = {( TTT )}

n ( E1 ) = 1

n ( E1 ) 1 P= ( E1 ) = n (S ) 8

(b) Let E2 = Event that Sweta gets double entry fee

She tosses heads on three times = {( HHH )}

n (E 2 ) = 1

P= (E 2 )

n (E 2 ) 1 = n (S ) 8

Chapter 14 • Probability

) (

)}

n (E 3 ) = 6 n (E 3 ) 6 3 P(E 3 = = = ) n (S ) 8 4

7. From a pack of 52 playing cards, Jacks, Queens, and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the drawn card is: (CBSE 2015, 2016) (a) a black king,

(b) a card of red colour,

(c) a card of black colour. Sol. Total Removed Cards: 2 + 2 + 2 = 6 cards Remaining Cards: 52 − 6 = 46 cards (a) Probability of Drawing a Black King Favourable Outcomes: The Black Kings (King of Spades, King of Clubs) were not removed. Number of Black Kings = 2 P ( Black King ) =

Number of Black Kings 2 1 = = Total Remaining Cards 46 23

(b) Probability of Drawing a Card of Red Colour Favourable Outcomes: All red Jacks, Queens, and Kings were removed. Remaining are red numbered cards and Aces. Number of Red cards normally = 26 (13 Hearts, 13 Diamonds) Red cards removed = 6 (2 Jacks, 2 Queens, 2 Kings) Remaining Red cards = 26 - 6 = 20 P ( Red Card ) =

Remaining Red Cards 20 10 = = Total Remaining Cards 46 23

(c) Probability of Drawing a Card of Black Colour Favourable Outcomes: No black cards were removed. Number of Black cards normally = 26 (13 Spades,13 Clubs) P ( Black Card ) = Black Cards 26 13 = = Total Remaining Cards 46 23

385

8. A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is: (a) a prime number

(CBSE 2015)

(b) a composite number (c) a number divisible by 3 Sol. (a) Prime numbers in this range: 2, 3, 5, 7, 11, 13, 17, 19 Number of prime numbers: n ( E ) = 8

(b) Composite numbers in this range: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 Number of composite numbers: n ( E1 ) = 11

Composite numbers from 1 to 20: 11 P ( composite number ) = 20 (c) Numbers divisible by 3: 3, 6, 9, 12, 15, 18 Number of such numbers: n ( E 2 ) = 6

Numbers divisible by 3 from 1 to 20: 6 3 = P ( number divisible by 3 ) = 20 10

Prime numbers from 1 to 20: 8 2 P ( prime number= ) = 20 5

Competency Based Questions Multiple Choice Questions 1. Two numbers ‘a’ and ‘b’ are selected successively without replacement in that order from the integers a 1 to 10. The probability that is an integer, is b (a)

17 45

(b)

1 5

(c)

17 90

(d)

8 45

2. A number is chosen from the numbers 1, 2, 3 and denoted as x , and a number is chosen from the numbers 1, 4, 9 and denoted as y . Then, P(xy < 9) is (a)

1 9

(b)

3 9

(c)

5 9

(d)

7 9

3. The probability of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is (a) 18

(b) 180

(d) 220 (CBSE 2024)

4. At a party, there is one last pizza slice and two people who want it. To decide who gets the last slice, two fair six-sided dice are rolled. If the largest number in the roll is 1, 3, or 6, Ananya would get the last slice, and 2, 4, or 5, Pranit would get it. In a random roll of dice, who has a higher chance of getting the last pizza slice? (a) Ananya (b) Pranit (c) Both have an equal chance (d) Cannot be answered without knowing the exact numbers in a roll.

386

5. There are 50 students in a class and their results are below: Result

Pass

Fail

Number of students

If a student is chosen at random out of the class (i.e., without any bias), then find the probability that the student is not failing. (a) 0.66

(b) 0.7

(d) 1

6. In basketball, different shots have varying point values – a two-point shot is taken from inside the three-point line, while a three-point shot is taken from outside the three-point line. In a basketball match, a player shot 5 three-point shots and 9 two-point shots out of the 35 shots he made. A particular shot he took was chosen at random. What is the probability that the shot that was chosen was neither a three-point shot nor a two-point shot? 1 2 3 6 (a) (b) (c) (d) 7 5 5 7 (CBSE QB)

7. Matilda made the following pattern during art class.

If she colours a shape at random, what is the probability that she will colour a circle? (a)

1 3

(b)

3 10

(c)

3 13

(d)

1 13

8. A girl calculates that the probability of her winning the first prize in a lottery is 0.06. If 8000 tickets are sold, then how many tickets has she bought? (a) 420 (b) 480 (c) 840 (d) 48

Select Response Type MCQs

1 (i) 2

4 (ii) 3

(iii) 0

(b) Both (i) and (iii)

(d) Only (iv)

2. A bowl contains 3 red and 2 blue marbles. Ruhi wants to pick a red marble. Which of the following changes could she make so that the probability of picking a red marble is greater than it was before? (i) Adding a red marble (ii) Removing a blue marble (iii) Adding 1 red and 1 blue marble (a) Only (i) (b) Only (i) and (ii) (c) Only (i) and (iii)

Ship

7 (iv) 5

(a) Only (i)

(d) All of these

Train

Ae ro p

1. Which of the following cannot be the probability of an event?

33°

36° Car

Bus

177°

Observe the pie chart and answer the following questions: (a) If one person is selected at random, find the probability that he/she travelled by bus or ship. (b) Which is the most favourite mode of transport and how many people used it? (c) The probability that a randomly selected person 7 used an aeroplane is . Find the revenue 60 collected by the air company at the rate of ` 5,000 per person. 2. Eight Ball: This is a game played on a pool table with 15 balls numbered 1 through 15 and a cue ball that is solid white. Of the 15 numbered balls, 8 are a solid (non-white) colour and numbered 1 through 8, and seven are striped balls numbered 9 through 15.

(c) A is true, but R is false. (d) A is false, but R is true. 1. Assertion (A): In a cricket match, a batsman hits a boundary 9 times out of 45 balls he plays. The probability that in a given ball, he does not hit the 4 boundary is . 5 Reason (R): P(E) + P(not E) = 1

(CBSE 2024)

2. Assertion (A): The probability that a leap year has 53 2 Tuesdays is . 7 Reason (R): The probability that a non-leap year has 5 53 Tuesdays is . 7

Case Study Based Questions

1. In a survey on holidays, 120 people were asked to state which type of transport they used on their last holiday. The following pie chart shows the results of the survey. Chapter 14 • Probability

The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one is drawn out. Based on the above information, answer the following questions. (a) What is the probability of drawing ball bears number 8? (b) What is the probability that drawn ball is a solid coloured and bears an even number? (c) What is the probability that drawn ball bears an even number?

387

Answers Multiple Choice Questions

Number of bad eggs = 18

17 90 We have a set of natural numbers from 1 to 10 where a and b are two variables which can take values from 1 to 10. So, total number of possible a  combination of a and b so that   is a fraction 2 1  1 1 1 without replacement are:  , , ,…  2 3 4 10   Similarly we have 9 such sets of 10 elements each. So total number of possible combination, 9 × 10 = 90 a   Now the possible combination which makes   b  an integer without replacement are:

Number of good eggs = Total number of eggs Number of bad eggs

 2 3 4 5 6 7 8 9 10 4 6 6 8 8 9 10 10  , , , , , , , , , , , , , , , ,  17 = 1 1 1 1 1 1 1 1 1 2 2 3 2 4 3 2 5 

Let B = event of getting the largest number in the roll of Pranit’s number is 2, 4, or 5.

1. (c)

P( A )

Favourable outcomes 17 = Total number of outcomes 90

5 9 Given x is chosen from the numbers 1, 2, 3 and y is chosen from the numbers 1, 4, 9. To find the probability of P(xy < 9), we will make a multiplication table for x and y such that xy < 9. 2. (c)

y x=1 x=2 x=3

1 1 2 3

4 4 8 12

9 9 18 27

Total number of possible outcomes = 9 Number of favourable outcomes where xy < 9 = 5 P(xy < 9)

Number of favourable outcomes 5 = Total number of outcomes 9

3. (c) 382 Given total number of eggs = 400 The probability of getting a bad egg = 0.045 Let the number of bad eggs = x

Probability of getting a bad egg = Number of bad eggs Total number of eggs

x ⇒ 0.045 = 400 45 ⇒ × 400 = x ⇒ x = 18 1000

388

400 − 18 = 382 Therefore, the number of good eggs in the lot is 382. 4. (b) Pranit has a higher chance of getting the last pizza slice. Here, two fair six-sided dice are rolled. Let A = event of getting the largest number in the roll of Ananya’s number is 1, 3, or 6. (1,1) , (1, 3 ) , ( 2, 3 ) , ( 3, 3 ) , (1, 6 ) ,  A =  ( 2, 6 ) , ( 3, 6 ) , ( 4, 6 ) , ( 5, 6 ) , ( 6, 6 )  n ( A ) = 10

(1, 2 ) , ( 2, 1) , (1, 4 ) , ( 2, 4 ) , ( 3, 4 ) , ( 4, 4 ) , B =  (1, 5 ) , ( 2, 5 ) , ( 3, 5 ) , ( 4, 5 ) , ( 5, 5 )  n ( B ) = 11

Since n ( B ) > n ( A ), Pranit has a higher chance of getting the last pizza slice. 5. (b) 0.7 Given total number of students = 50 Number of students who are not failing (pass) = 35 35 P ( Not failing= ) = 0.7 50 3 5 Given 6. (c)

Number of three-point shots taken = 5 Number of two-point shots taken = 9 Total number of shots taken by the player = 35 Total number of shots taken by the player (twopoint and three-point shots) = 9 + 5 = 14

14 2 P ( Three-point or two-point shot= ) = 35 5 Now, the probability that the shot chosen was neither a three-point shot nor a two-point shot is: 1 − P ( Three-point or two-point shot ) =1 − Thus, the required probability is

3 . 5

2 3 = 5 5

7. (c)

Both the assertion and the reason are correct, and the reason is a correct explanation of the assertion.

3 13

The given figure is It consists of the following number of shapes: ● Number of squares = 4 ●

Number of arrows = 4

●

Number of circles = 3

●

Number of hexagons = 2

Total number of shapes in the given figure: 4+ 4+3+2 = 13 Probability that Matilda will colour a circle P ( Circle )

Number of circles 3 = Total number of shapes 13

Hence, the probability that Matilda will colour a 3 circle is . 13 8. (b) 480 Suppose she bought n tickets.

2. (c) A is true, but R is false. Analysis of Assertion: A leap year has 366 days, equivalent to 52 full weeks and 2 extra days. Out of 7 days in a week, 2 extra days. 2 P ( 53 Tuesdays in a leap year ) = 7 This is correct. Understanding Reason: A non-leap year has 365 days, equivalent to 52 full weeks and 1 extra day. Out of 7 days in a week, 1 extra days.

1 P ( 53 Tuesdays in a non-leap year ) = 7 This is incorrect. Assertion is true, but Reason is false.

Case Study Based Questions 1. (a) Since no. of people who travel by bus or ship =

Total number of tickets is 8000.

69 ×120 = 23 360

P(E) = Probability he/she travelled by bus or ship n ( No. of outcomes favourable to E ) 23 Probability of winning the first prize = = = 8000 ( Total possible outcomes of E ) 120 n ⇒ 0.06 = ⇒= n 480 8000 (b) ‘Car’ is the favourite mode of transport

Select Response Type MCQs

No. of people who travel by it = 177 ×120 = 59 people 360

1. (c) Both (ii) and (iv) 2. (b) Only (i) and (ii)

Assertion-Reason Based Questions 1. (a) Both A and R are true, and R is the correct explanation of A. Understanding the Assertion: Number of boundaries 9 1 P ( E= = = ) Total balls 45 5 1 4 P ( not E ) =1 − P ( E ) =1 − = 5 5 4 P ( not E ) = 5 Understanding Reason: The reason provided is that the sum of the probability of an event and its complement is 1. This is a fundamental principle in probability, and it holds true based on the calculations.

Chapter 14 • Probability

Out of the 120 surveyed people, 59 use cars.

= (c) No. of people who use planes 14 people

7 ×120 = 60

Thus, total revenue = ` (5000 × 14) = ` 70,000

2. (a) There is only one ball with the number 8, and there are 15 balls in total.

P ( ball number 8 ) = =

1 15

Number of balls numbered 8 Total number of balls

(b) Solid coloured balls are numbered 1 through 8. The even-numbered solid balls within this range are 2, 4, 6, and 8. Therefore, there are 4 even-numbered solid balls.

389

P ( solid and even ) = Number of even − numbered solid balls 4 = Total number of balls 15

●

Even-numbered solids: 2, 4, 6, 8 (4 balls).

●

Even-numbered stripes: 10, 12, 14 (3 balls).

Total even-numbered balls = 4 (solids) + 3 (stripes) = 7 balls.

P ( even number ) =

Total even − numbered balls 7 = Total number of balls 15

Additional Practice Questions

1. In a survey, it is found that every fifth person has a vehicle. The probability of a person NOT having a vehicle is 4 1 (a) (b) 5% (c) (d) 95% 5 5

(CBSE 2023)

2. A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the card drawn is not an ace is 9 4 1 12 (a) (b) (c) (d) 13 13 13 13

(CBSE 2023)

3. Two dice are thrown together. The probability that they show different numbers is (a)

1 6

(b)

5 6

(c)

1 3

(d)

2 3

(CBSE 2024)

4. If the probability of a player winning a game is 0.79, then the probability of his losing the same game is (a) 1.79 (b) 0.31 (c) 0.21% (d) 0.21 5. Two dice are rolled together. The probability of getting sum of numbers on the two dice as 2, 3 or 5, is 5 4 7 11 (a) (b) (c) (d) 36 9 36 36 6. A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is 1 8 1 7 (a) (b) (c) (d) 5 15 3 15

7. Find the probability that a non-leap year selected at random will contain 53 Mondays. (CBSE 2016)

8. If a number x is chosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2 <2? 9. There are 80 cards numbered from 1 to 80. One card is drawn at random from them. Find the probability that the number on the selected card is not divisible by 8. 390

10. A die is thrown once. What is the probability of getting a prime number? (CBSE 2020) 11. One card is drawn at random from a well shuffled deck of 52 cards. Find the probability that the card drawn (CBSE 2024) (a) is queen of hearts (b) is not a jack. 12. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game. 13. The probability of selecting a blue marble at random from a jar that contains only blue, black and green 1 marbles is . The probability of selecting a black 5 1 marble at random from the same jar is . If the jar 4 contains 11 green marbles, find the total number of marbles in the jar. 14. A die is thrown once. What is the probability of getting a number less than 3? (CBSE 2020) 15. A card is drawn at random from a well shuffled deck of 52 cards. Find the probability of getting neither a red card nor a queen. (CBSE SQP 2018) 16. A number is chosen at random from the numbers -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. Then find the probability that square of this number is less than or equal to 1. 17. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. 18. Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25?

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 19. Assertion (A): The probability of an impossible event is 0. Reason (R): An impossible event does not contain any favourable outcomes, hence its probability is 0. 20. Blood group describes the type of blood a person has. It is a classification of blood based on the presence or absence of inherited antigenic substances on the surface of red blood cells. Blood

types predict whether a serious reaction will occur in a blood transfusion. In a sample of 50 people, 21 had type O blood, 22 had type A, 5 had type B and rest had type AB blood group. Based on the above, answer the following questions: (a) What is the probability that a person chosen at random had type O blood? (b) What is the probability that a person chosen at random had type AB blood group? (c) What is the probability that a person chosen at random had neither type A nor type B blood group? OR What is the probability that person chosen at random had either type A or type B or type O blood group? (CBSE 2023)

Challenge Yourself

Scan me

1. A purse contains four fifty-paise coins, three tworupee coins and three five-rupee coins. If three coins are selected at random, then what is the probability of getting the minimum amount? 1 1 1 1 (a) (b) (c) (d) 10 30 15 5 (Hint: Find separate probability to get minimum amount on each selection of coin) 2. A number is selected at random from the set {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. What is the probability that it will be a root of the equation x 2 − 6x + 9 = 0?

3. An urn holds 6 pink balls and a white balls. If the probability of drawing a white ball is twice that of drawing a pink ball, determine the value of a. 4. One number is selected from the first 50 natural numbers. What is the probability that it is a root of the inequality? 256 x+ > 40? x (Hint: Solve the quadratic equation first then use the inequalities for all values)

Answers Additional Practice Questions

18. Rina

1. (c)

2. (d)

3. (b)

4. (d)

5. (a)

6. (a)

1 7. 7 10.

1 2

3 8. 5 11. (a)

1 52

3 12. 4

13. 20

15.

16.

6 13

Chapter 14 • Probability

3 11

9. (b)

7 8

12 13

1 14. 3 17. 1 2

20. (a)

21 50

19. (a) (b)

1 25

(c)

23 40

24 25

Challenge Yourself 1. (c) 4.

1 30

1 10

3. 12

1 2

Scan me for Exemplar Solutions 391

SELF-ASSESSMENT 1

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. In a school of 800 students, it is observed that 64 students are left-handed. If one student is chosen at random, what is the probability that the student is left-handed? 1 1 1 1 (a) (b) (c) (d) 25 15 10 12.5 2. A dice is rolled twice. The probability that 5 will not come up either time is 11 1 13 (a) (b) (c) 36 3 36

(d)

3. Two coins are tossed together. The probability of getting at least one tail is 1 1 3 (a) (b) (c) 2 4 4

(d) 1

25 36 (CBSE 2022)

(CBSE 2023)

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

(CBSE SQP 2025)

5. Assertion (A): The probability of getting at least one head when flipping two coins is 0.75.

Reason (R): There are three outcomes that include at least one head when two coins are flipped: one head and one tail, two heads. (CBSE 2024)

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. If two different numbers are taken from {1, 2, 3, 4}, what is the probability that their sum is odd? (CBSE 2024) 1 7. A jar contains red, blue, and green marbles. If the probability of selecting a red marble is and selecting a blue 4 1 marble is , what is the probability of not selecting a blue or red marble. (CBSE SQP 2024) 3

Short Answer Questions

[4 × 3 = 12 Marks]

8. Three coins are tossed simultaneously. What is the probability of getting (a) at least one head?

(b) exactly two tails?

a is formed, where a is the b number of the sector where the arrow stops after the first spin and b is the number of the sector where the arrow stops after the second spin. On every spin, each of the numbered sector has an equal probability of being the sector on which the arrow stops. What is the probability that the a fraction is greater than 1? (Hint: Find the number of outcomes for a > b ) b

(CBSE 2024)

9. On the disc shown below, a player spins the arrow twice. The fraction

392

1 4 5

10. What is the probability that a number selected from numbers 1 to 15 is not prime?

(NCERT Exemplar)

11. A fair die is rolled twice. What is the probability that the sum of the numbers on the two dice will be 5? (NCERT Exemplar)

Long Answer Questions

[3 × 5 = 15 Marks]

12. There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 250 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked out, what is the probability that it contains no cash prize? (NCERT Exemplar)

13. In a game, the entry fee is Rs 5 . The game consists of a tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that (a) she loses the entry fee

(b) she gets double entry fee

(c) she just gets her entry fee 14. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded. (a) How many different scores are possible?

(b) What is the probability of getting a total of 7?

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A school library categorises its 250 books into three genres: Fiction, Non-fiction, and Reference. The number of fiction, non-fiction are 80, 120 and rest are reference books.

(a) What is the probability that a randomly selected book is Fiction? (b) What is the probability that a randomly selected book is Non-fiction? (c) How many reference books are there? (d) What is the probability that a randomly selected book is reference?

Self-Assessment

393

SELF-ASSESSMENT 2

Scan me for Solutions

Time: 60 mins

Max. Marks: 40

Multiple Choice Questions

[3 × 1 = 3 Marks]

1. From the letters of the word “MOBILE”, a letter is selected at random. The probability that the selected letter is a vowel, is 1 3 1 1 (a) (b) (c) (d) 6 7 2 3 2. The probability of happening of an event is denoted by p and the probability of non-happening of the event is denoted by q. Relation between p and q is (a) p + q = 1

(b) p = 1, q = 1

(d) p + q + 1 = 0

3. A box contains cards numbered 1 to 100. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect cube, is 1 1 3 7 (a) (b) (c) (d) (CBSE 2024) 20 25 50 100

Assertion-Reason Based Questions

[2 × 1 = 2 Marks]

Directions: Each Question consists of two statements: an Assertion (A) and a Reason (R). Evaluate these statements and select the appropriate option: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true. 4. Assertion (A): If two dice are thrown simultaneously, the probability of getting a total of 7 is higher than getting a total of 6. Reason (R): There are more combinations to achieve a sum of 7 than a sum of 6 when two dice are rolled. (CBSE 2024) 5. Assertion (A): In a game, the entry fee is ` 10. The game consists of tossing a coin 3 times. If one or two heads 3 show, Anita gets her entry fee back. The probability that she gets the entry fee back is . 4 Reason (R): When three coins are tossed together, all the outcomes are (HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT). (CBSE SQP 2025)

Very Short Answer Questions

[2 × 2 = 4 Marks]

6. A class contains 15 boys and 10 girls. A student is selected at random. What is the probability it is a boy? (CBSE 2019)

7. A die is rolled twice. What is the probability that the first roll is higher than the second? (CBSE SQP 2023)

Short Answer Questions

[4 × 3 = 12 Marks]

8. A carton of 24 bulbs contains 6 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

394

9. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together, and the total score is recorded. (a) How many different scores are possible? (b) What is the probability of getting a total of 7? (NCERT Exemplar)

10. A letter is chosen at random from the word “STATISTICS”. What is the probability that it is a consonant? (CBSE 2023)

11. Box a contains 25 slips of which 19 are marked ` 1 and the others are marked ` 5 each. Box b contains 50 slips of which 45 are marked ` 1 each and the others are marked ` 13 each. Slips of both the boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than ` 1?

Long Answer Questions

[3 × 5 = 15 Marks]

12. A fair six-sided die is rolled twice. Find the probability that the sum of the numbers obtained is (a) at least 10. (b) at most 10. 13. From a deck of 52 playing cards all the face cards (King, Queen, and Jack) are removed. The remaining cards are shuffled well, and one card is drawn at random. Find the probability of getting: (a) a card of diamond

(b) a red Ace card

(d) a number card (2–10)

14. A letter is chosen at random from the letters of the word "SCIENCE". What is the probability that it is (a) a vowel? (b) a consonant?

Case Study Based Questions

[1 × 4 = 4 Marks]

15. A middle school decided to run the following spinner game as a fund-raiser on Christmas Carnival.

(R)

(G)

(R)

Red

Green

Red

(B) Blue

(Y)

(G)

Yellow

Green

Making Purple: Spin each spinner once. Blue and red make purple. So, if one spinner shows Red (R) and another Blue (B), then you win. One such outcome is written as ’RB. Based on the above, answer the following questions: (a) List all possible outcomes of the game. (b) Find the probability of ‘Making Purple’. (c) (i) For each win, a participant gets 10 rupee, but if he/she loses, he/she has to pay 5 rupee to the school. If 99 participants played, calculate how much fund could the school have collected. OR (ii) If the same amount of 5 rupee has been decided for winning or losing the game, then how much fund had been collected by school? (Number of participants = 99) (CBSE 2023)

Self-Assessment

395

Answers Self-Assessment 1 Chapter 3

Chapter 1 2. (b) 4

1. (b) No solution 3 2. (a) k = 2 3. (c)= x a= ,y b

5. (d) A is false, but R is true.

8. 1:40 pm.

4. (c) A is true, but R is false.

9. 99936

10. 35 cm

5. (d) A is false, but R is true.

11. 50

12. 17

6. Salary `1300, annual increment `50

1. (d) pq 3. (a) always irrational. 4. (c) A is true, but R is false.

13. (a) HCF of 84 and 231 = 21

Number of columns for two fields = 21

84 =4 21 231 = 11 Number of columns in smaller field = 21 15. (a) 288 (b) 4 (b) Number of columns in smaller field =

–5

(d) ab

–5

Chapter 2 1. (a) 4

2. (c) 3

–10

3. (a) -5 and 6 4. (a) Both A and R are true, and R is the correct explanation of A.

Thus, the given system is consistent.

5. (a) Both A and R are true, and R is the correct explanation of A.

9. The son’s present age is 10 years, and the father’s present age is 33 years.

6. k = 0

10. = x 3,= y 2

3 7. Other zero is . 2 8. Quadratic polynomial is x 2 + x − 20. −1, q = −6 9. p =

10.

11. x 2 − 10x + 24

12.

13. 15 15. (a) ( 20 − 2x )(16 − 2x ) x (b) 4x ( x − 10)( x − 8) (c) x = 8 cm

396

8. k = 4

11. The two numbers are 75 and 30. 12. x = −1, y = −1 15, − 15 2

3 14. 90

,3 3

13. Speed in still water = 8 km/h, Speed of the stream = 4 km/h 14. One Man 140 days, One Boy 280 days. 15. (a) For batch I, 20x + 5 y = 9000 For batch II, 5x + 25 y = 26000 (b) x = 200 (c) y − x = `800 (d) `22000

Chapter 4

Chapter 6

1. (b) No

2. (b) k ≤ 4

0 3. (b) x + x − 5 = 4. (b) Both A and R are true, but R is not the correct explanation of A. 5. (c) A is true, but R is false.

6. Not real

7. 16 8. x = 7 and x = 0

9. 12 m, 8 m

−121 ± 5049 22

11. 17 and 18

10. x =

12. k = 16 14. 21 cm 15.

3 13. 7

(a) (i) 2(x + 5) km 2

0 (b) (iii) x + 5x − 500 = (c) (i) 20 km/h (d) (iv) 16 hours

Chapter 5 1. (c) 20th 2. (b) 40

2. (b) 120 feet

3. (a) 10 4. (a) Both A and R are true, and R is the correct explanation of A. 5. (c) A is true, but R is false. 6. DE is not parallel to BC. 7. No

9. DC = 10 cm

10. 10 cm

11. 11 or 8

15. (a) 45 = ∠ APD (b) ∠ = YAQ ∠= AQD 30[Alternate angles] (c) (i) Hence, the Boat P is at a distance of 100 m from the lighthouse. (c) (ii) The Boat Q is at a distance of 100 3 m from the lighthouse.

Chapter 7 1. (d) 2:1 2. (c) No, additional condition is needed. 3. (a) 7

3. (b) 15 4. (b) Both A and R are true, but R is not the correct explanation of A 5. (a) Both A and R are true, and R is the correct explanation of A 7. 1605 8. First term (a) = 9, common difference (d) = 3 9. AP: 8, 14, 20,... 10. 25 terms and 36 terms 11. 2, 6, 10, 14 12. 1: 3, 5 : 49 13. 5 minutes 14. Angles: 75°, 85°, 95°, 105° 15. (a) (i)

9 2

1. (d)

1 m 4

(b) (iii) 12.5 m 3 (c) (ii) 50 m3 (d) (iii) 375 m2 (e) (iv) 750 m3

Answer • Self-Assessment 1

4. (a) Both A and R are true, and R is the correct explanation of A. 5. (a) Both A and R are true, and R is the correct explanation of A. 6. (–1, 3) 8. 6 10. (3, 3) 11. Isosceles Triangle 7 p = ,q 0 12.= 3 13   13. 1:5;  0, −  3  14. P = (-4, 0) Q = (4, 0) R = (0, 4 3 ) S = (8, 4 3 ) 15. (a) (200, 400); (–200, 400) (b) (i) 160000 sq. units

(ii) 400 2 units

Chapter 8 2 3 3 3. (c) −cosecx cotx 1. (a)

2. (b)

x +1 2x

4. (a) Both A and R are true, and R is the correct explanation of A. 5. (c) A is true, but R is false. 6. sinA = ±

10.

5 4

15. (a) (iii) 60

12. 40° 14. 6.6 cm 15. (a) 30 cm (b) 120°

Chapter 11 R 1. (a) R1 + R2 = 2. (b) ( 6.25π − 12 ) cm 2

3. (b) 100 cm²

4. (a) Both A and R are true, and R is the correct explanation of A.

(b) (ii) 1 (c) (iv)

11. ∠AOB = 120° and ∠ABP = 60°

4 5

4 3 , sinA = ± . 5 5

8. cosA =

8. 12 3 cm

10 3

5. (d) A is false, but R is true.

(d) (iii) 1

6. 38.5 cm2

Chapter 9 1. (a) 10 2 m

2. (d) 173.2 m

3. (b) 15 3 m

7. 66.5 cm2

8. 36.33 cm2

9. 196 cm2

10. 9.625 cm2

11. 42 cm2

12. 623 cm2

13. 504 m2

14. 560

4. (a) Both A and R are true, and R is the correct explanation of A.

15. (a) 400π m2 1256.64 m2

(b) 84π m2/263.76 m2

5. (d) A is true, but R is false.

6. 120 m

(d) 50π m2/157.08 m2

7. 10 m

8. 10 3

Chapter 12

9. 120 2 10.

1. (c) 5 : 7 1 2. (c) cm 2

50 3 m or 28.87 m 3

11. 60 m

12. 57.74 m

4. (a) Both A and R are true, and R is the correct explanation of A.

13. 40 m 14. 259.81 m and 86.60 m 15. (a) (iii) 3.7 m

(b) (i) 4.28 m

(d) (iii) 5.7 m

Chapter 10 1. (a) secant 2. (b) 5 cm 3. (c) 5 4. (d) A is false, but R is true. 5. (c) A is true, but R is false. 6. 2a 7. 50° 398

3. (a) 7 cm

5. (a) Both A and R are true, and R is the correct explanation of A. 6. 25.12 cm3 7. h = 2 cm 8. 1 cm 9. 1055.04 cm³ 10. 1.82 m 11. 528 m³, 125.6 m² 12. SA 176 mm2 Volume 159.23 mm3 13. 65.94 cm³ 14. 1.125 m

Chapter 14

15. (a) 38.5 cm2 (b) (i) 2425.5 cm or (ii) 539 cm 3

1. (d)

1 25

2. (d)

3 4 4. (d) A is false, but R is true. 3. (c)

Chapter 13 1. (c) 9 2. (d) p + q = 19

3. (a) 20

4. (a) Both A and R are true, and R explains A 5. (d) A is false, but R is true 6. 38.33

7. 30–40

8. f = 0

9. 45 kg

10. 76.15

11. 13.67 km/L

12. Mean = 42.2 Mode = 45

5. (a) Both A and R are true, and R is the correct explanation of A. 6.

2 3

8. (a) 10.

7 3 1 , (b) , (c) 8 8 2

3 5

12. 0.64 or 64%

13. f1 = 8

5 12

11.

1 9

13. (a)

1 1 3 (b) (c) 8 8 4

14. (a) 6 scores are possible, 0, 1, 2, 6, 7 and 12 (b)

f2 = 7

15. (a) P(Fiction book) =

14. Mode = 743.75 Median = 616.67 15. (a) 82

(b) 58 year (ii)

(b) x = 24

80 8 = 250 25

1 3

120 12 = 250 25 (c) Number of reference books = 250 - 80 - 120 = 50 (b) P(Non-fiction book) =

(d) P(Reference book) =

Answer • Self-Assessment 1

25 36

50 1 = 250 5

399

Answers Self-Assessment 2 Chapter 1 1. (c) q

9. 2. (a) 13.

11. 20

3. (d) 2

12. x 2 + 52x + 627

4. (a) Both A and R are true, and R is the correct explanation of A.

14. (a)

5. (a) Both A and R are true, and R is the correct explanation of A. 7. 2520

8. 80 min

13. a = 3, b = −6

3 5

(b) -5

0 15. (a) 2x 2 + 5x − 300 = (c)

(b)

25 is invalid. The width of the garden 2 represents a physical dimension and must be positive. Negative values for x are not realistic in this context.

(d) x = −

9. 75 10. HCF = 4 and LCM = 9696 12. Number of students in each group = 12

10. 29

Number of groups in music = 5

Chapter 3

Number of groups in dance = 7

1. (d) 2

Number of groups in handicrafts = 9

2. (b) 25500

Total number of rooms required = 21

3. (d) 2

13. 16 times 15. (a) (i) 45

4. (a) Both A and R are true, and R is the correct explanation of A

(b) (iii) 360

5. (d) A is false, but R is true

6. p ≠ 10

(d) (iii) 8

7. k = 7

14. 91

= and y 0 8. x 1=

(e) (iv) 6

9. = x 2,= y 3

Chapter 2 1. (b) 5 3. (b)

b 2 − 2ac

2. (b) −

3 2

c2 4. (a) Both A and R are true, and R is the correct explanation of A. 5. (a) Both A and R are true, and R is the correct explanation of A. 13 6. 7. 8 6 8. 2x 2 − 3x + 1

400

a 4,= b 8 10.=

11. 69, 96 12. (a) a unique solution 13. 60 14. 20 questions 15. (a) The fixed charges are `5. (b) The charges per km are `10. (c) Total Cost = F + 10c = 20 + 10 × 10 = 20 + 100 = `120 ` 360

Chapter 4

Chapter 6

1. (c) − 5 3

1. (d) 100°

2. (d) 3x 2 + 4 3x + 4 = 0

(d)

20 3

4. (d) A is false, but R is true. 5. (a) Both A and R are true, and R is the correct explanation of A. 6. k = 2 7. 0, real and equal roots

8. Proved

9. k = −2

10. 36

11. a = 3 and b = −7

12. 42 km/h

13. 15, 12

14. a= b= c

15. (a) (iii) (20-x) km/h Distance Time 0 (c) (iii) x ² + 30x − 400 = (b) (i) Speed =

(d) (ii) 10 km/h

Chapter 5 1. (d) 5 2. (b) 7

3. (c) PR² + QR² = PQ² 4. (d) A is false, but R is true. 5. (c) A is true, but R is false. 6. Using SAS similarity criterion 7. True

13. AD = 7.5 cm

15. (a) (i) 105° (b) (iii) 175° (c) (iv) 17 cm (d) (ii) 3 cm (e) (iii) 155°

Chapter 7 1. (b) 12 units 2. (b) 9  3a + c 3b + d  , 3. (a)   4   4 4. (a) Both A and R are true, and R is the correct explanation of A. 5. (a) Both A and R are true, and R is the correct explanation of A.

3. (c) ` 25, 25, 750

6. Proved

4. (b) Both A and R are true, but R is not the correct explanation of A

8. k = 1 or 5

5. (c) A is true, but R is false 6. Common difference (d) = 2 7. 3, 5, 7, 9… 8. First term ( a ) = 2, Common difference (d ) = −2 9. S n = n2 10. x = 50

10.

(1, 0 ) , (1, 4 )

12. Proved

14. 67, 69, 71 15. (a) (iii) 10 m (b) (ii) 2 ( 3n + 2 ) m (c) (iv) 76 m (d) (iii) n ( 3n + 7 ) m (e) (ii) 6

Answer • Self-Assessment 2

7. k = −1  1 9.  0,   2 11. Yes 13.

( 6.8, 1.8 ) ; No

14. 10 units 15. (a) (4, 3) (b)

52 units

11. 16th term, −100 is not any term of AP. 12. 2

2. (a) 10 cm.

Chapter 8 1 2 3. (b) 2 1. (a)

2. (a) cosθ sinθ

4. (a) Both A and R are true, and R is the correct explanation of A. 5. (b) Both A and R are true, but R is not the correct explanation of A. 1 7. 10. 2 9 401

12 7 15. (a) (ii) 45

(b) (iv) 1

(d) (iii) 1/2

11.

Chapter 9 1. (c) 120 m

2. (b) 25 3 m

3. (c)  1 − π  cm² 4  4. (c) A is true, but R is false. 5. (a) Both A and R are true, and R is the correct explanation of A. 6. 282 cm

7. 115.5 cm2

3. (a) 45°

8. 137.64 cm2

9. 43.74 cm2

4. (a) Both A and R are true, and R correctly explains A.

10. 31.4 cm

11. 28 cm2

5. (c) A is false, but R is true.

6. 115.47 m

13. 66.43 cm2

14. 6.125 cm2

7. 11.55 m

8. 86.60 m

15. (a) 200 mm

(b) 77 mm2

9. 51.96 m

10. 45°

11. 100 m

12. 31.75 m

13. 90 m

14. 25.98

15. (a) 1118.36

(b) (iii) 1937 km

(d) 45°

Chapter 10 1. (b) 7.6 cm

2. (d) 55°

3. (b) 50° 4. (d) Assertion is incorrect but Reason is correct.

Chapter 12 1. (a) 7.5 m

2. (c) 3 :1: 2

3. (c) 2 : 1 4. (d) A is false, but R is true 5. (b) Both A and R are true, but R is not the correct explanation of A 6. 423.43 cm³

7. 2.5 cm

8. 57 seconds 9. 14 cm, 7 cm, 132 cm3, 66 cm3, 396 cm3

5. (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

10. 21 cm

6. 55°

12. 1366.2 mm

7. 2 3 cm

(ii) 51π mm 2

11. 14087 2

8. AD 5= = cm, BE 7= cm, CF 3 cm

13. Mass of the pole = 892 kg (approx.)

9. 70°

14. Volume of the remaining solid: 2025.3 cm3

11. 6 cm

12. 80°

13. (a) 8 m (b) 5 m

14. AB = 7.5 cm and AC = 6.5 cm 15. (a) (i) 7 (b) (ii) 5 (c) (iii) 3 (d) (iv) 24 (e) (v) 60

Chapter 11 1. (b) Area of the circle > Area of the square 2. (d) 4 : 5 402

Surface area of the remaining solid: 1360.8 cm2 (approx) 15 (a) Surface area of one dimple = 0.025 cm² (b) Volume of material dug out for one dimple = 0.017 cm³ (c) Total surface area exposed to the surroundings = 95 cm² OR Volume of the golf ball = 35.34 cm³

Chapter 13 1. (c) 90 3. (b) 6

2. (a) 15

4 (a) Both A and R are true, and R is the correct explanation of A.

4. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

5. (b) Both A and R are true, but R is not the correct explanation of A.

5. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

6. a = 7

7. 110.66

8. 15

9. 15

10. 65 11. Mean = 40.7, Median = 42 12. No, the median of the given data cannot be greater than 12.5% because the 10–12.5. 13. f1 = 5 , f 2 = 25 14. p = 5, q = 7, Mode = 46.7 15. (a) 20–30

(b) 10

Answer • Self-Assessment 2

19 50

9. (a) 6 (b)

7 10 1 12. (a) (b) 1 6 10.

11.

13. (a)

1 4

14. (a)

4 3 (b) 7 7

(b)

1 1 (c) 20 4

(d)

1 3

11 75

9 10

15. (a) {RR, RB, RG, GR, GB, GG, YR, YB, YG}

Chapter 14 1 1. (b) 2 1 3. (c) 10

5 12 3 5 8. , 4 23

2. (a) p + q = 1

1 9 (c) (i) ` 330 (ii) ` 385 (b)

403

Examination Time

Series: EH5GF

SET-1

Q.P. Code 30/5/1

Roll No.

x x x x x x x x

Candidates must write the Q.P. Code on the title page of the answer-book.

CBSE Question Paper (Standard) 2024-25 Time Allowed: 3 hours �� Max Marks: 80

General Instructions: I. This question paper contains 38 questions. All questions are compulsory. II. This Question Paper is divided into FIVE Sections - Section A, B, C, D, and E. III. Section A - Question number 1 to 18 are Multiple Choice Questions (MCQs) and question number 19 & 20 are Assertion Reason based question of 1 mark each. IV. Section B - Questions number 21 to 25 are Very Short-Answer-I (SA-I) type questions of 2 marks each. V. Section C - Questions number 26 to 31 are Short Answer-II (SA-II) type questions carrying 3 marks each. VI. Section D - Questions number 32 to 35 are Long Answer (LA) type questions carrying 5 marks each. VII. Section E - Questions number 36 to 38 are Case Study/Passage based integrated units and assessment questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study. VIII. There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D, and 3 questions in Section E. 22 IX. Draw neat figures wherever required. Take π = wherever required if not stated. 7 X. Use of a calculator is NOT allowed.

SECTION A This section consists of 20 questions of 1 mark each.

( 3 + 2 ) + ( 3 − 2 ) is a/an 2

(a) positive rational number

(1) (b) negative rational number

(d) negative irrational number

2. Let x = a 2b 3c n and y = a 3b mc 2, where a, b, c are prime numbers. If LCM of x and y is a 3b 4c 3, then the value of m + n is (1) (a) 10

(b) 7

(d) 5

3. For any prime number p, if p divides a2, where a is any real number then p also divides 1 3 1 (a) a (b) a (c) a (d) a 2 2 8 4. Which of the following equations is a quadratic equation? (a) x2 + 1 = (x - 1)2

(b)

2x x (x + x ) = 2

(1)

(d) (x + 1)(x - 1) = (x + 1)2

5. If x2 + bx + b = 0 has two real and distinct roots, then the value of b can be (a) 0

(1)

(b) 4

Examination Time • CBSE Question Paper (Standard) 2024–25

(1) (d) -3 407

6. In the figure given below, points P, Q, R divides the line segment AB in four equal parts. The point Q divides PB in the ratio. (a) 1 : 3

(b) 2 : 3

(1)

(d) 1 : 1

7. A bag contains red balls and black balls in the ratio 3 : 7. A ball is drawn at random. The probability that ball so drawn is black in colour, is (1) 3 1 (a) (b) 0.3 (c) 0.7 (d) 7 7 8. Which of the following statements is false? (1) (a) Two right triangles are always similar. (b) Two squares are always similar.

9. In the adjoining figure, ABCD is a trapezium in which XY || AB || CD. If AX = (a) 2 : 3

(b) 3 : 2

(d) 1 : 2

2 AD, then CY : YB = 3

(1)

(a) Infinite number of tangents can be drawn to a circle.

(b) Infinite number of tangents can be drawn to a circle from a point outside the circle. (c) Infinite number of secants can be drawn to a circle from a point outside the circle. (d) Angle between tangent and diameter at point of contact is 90°.

11. In the adjoining figure, PA and PB are tangents to a circle with centre O. The measure of angle APB is (a) 210°

(b) 150°

210°

(d) 30°

12. 1 − tan 30° is equal to 1 + tan 2 30° 2

(a) sin⁡ 60°

(1)

(1) (b) cos⁡ 60°

(d) sec⁡ 60°

13. An observer 1.8 m tall stands away from a chimney at a distance of 38.2 m along the ground. The angle of elevation of top of chimney from the eyes of observer is 45°. The height of chimney above the ground is

(1)

14. In the adjoining figure, the sum of radii of two concentric circles is 16 cm. The length of chord AB which touches the inner circle at P is 16 cm. The difference of the radii of the given circles is

(1)

(a) 38.2 m

(b) 36.4 m

(d)

( 38.2 ) 2 m

(a) 8 cm (b) 4 cm

(d) 3 cm

15. A cone of height 12 cm and slant height 13 cm is surmounted on a hemisphere having radius equal to that of cone. The entire height of the solid is (1) (a) 17 cm

(b) 18 cm

(d) 23 cm

16. If x × median + y × mean = z × mode; is the empirical relationship between mean, median and mode, then the value of x + y + z is (1) (a) 6

408

(b) 3

(d) 1

17. Following data shows the marks obtained by 100 students in a class test: Marks Obtained Number of Students

20 6

29 28

28 24

33 15

42 2

38 4

43 1

The median will be the average of which two observations?

(a) 29 and 33

(b) 25 and 28

(1)

25 20

(d) 33 and 38

18. The probability of getting a composite number greater than 3 on throwing a die is 1 1 1 (a) (b) (c) (d) 6 3 2

2 3

(1)

Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false.

(d) Assertion is false, but Reason is true.

3 4 19. Assertion: For an acute angle θ , value of sin θ = ⇒ cosθ = − 5 5 2 2 Reason: For any value of θ, (0° ≤ 0 ≤ 90°) sin θ + cos θ = 1 20. Assertion: For an AP, 3, 6, 9, ..., 198, 10th form from the end is 168.

(1) (1)

Reason: If ‘a’ and ‘l’ are the first term and last term of an AP with common difference ‘d’, then nth term from the end of the given AP is l - (n - 1)d.

SECTION B This section has 5 very short answer type questions of 2 marks each. 21. (a) The cost of 2 kg apples and 1 kg of grapes on a day was found to be `320. The cost of 4 kg apples and 2 kg grapes was found to be `600. If cost of 1 kg of apples and 1 kg of grapes is `x and `y respectively, represent the given situation algebraically as a system of equations and check whether the system so obtained is consistent or not. (2) OR (b) Solve for x and y: 2x + 3x = 5 and 3x − 8 y = − 6 22. The coordinates of the end points of the line segment AB are A(-2, -2) and B(2, -4). P is the point on AB such that 4 (2) BP = AB . Find the coordinates of point P. 7 23. (a) It is given that sin(A - B) = sin A cos B - cos A sin B. Use it to find the value of sin15°. (2) OR (b) If sin A = y, then express cos A and tan A in terms of y. (2) 24. AD and PS are medians of triangles ABC and PQR respectively such that ΔABD ~ ΔPQS. Prove that ΔABC ~ ΔPQR. (2) 25. While shuffling a pack of 52 cards, one card was accidently dropped. Find the probability that the dropped card (2) (a) is not a face card.

(b) is a black king.

SECTION C This section has 6 short answer type questions of 3 marks each. 26. (a) Prove that 3 is an irrational number.

(3) OR

(b) State true or false for each of the following statements and justify in each case: (i) 2 × 3 × 5 × 7 + 7 is a composite number.

Examination Time • CBSE Question Paper (Standard) 2024–25

(ii) 2 × 3 × 5 × 7 + 1 is a composite number. 409

27. Obtain the zeros of the polynomial 7x2 + 18x - 9. Hence, write a polynomial each of whose zeros is twice the zeros of given polynomial. (3) 28. Solve the following system of equations graphically: 2x - y - 2 = 0

(3)

-4x + y + 4 = 0 Also, find the absolute difference between the ordinates of the points where given lines cut y axis. 29. Find a relation between x and y such that P(x, y) is equidistant from the points A(3, 5) and B(7, 1). Hence, write the coordinates of the points on x-axis and y-axis which are equidistant from points A and B. (3) 30. (a) Prove the following trigonometric identity: 1 + cosec A cos2 A = cosec A 1 − sin A

(3)

3 (b) Let 2A + B and A + 2B be acute angles such that sin ( 2 A + B ) = and tan⁡(A + 2B) = 1. Find the value of 2 cot⁡(4A - 7B). 31. In the adjoining figure, XY and X'Y' are parallel tangents to a circle with centre O.

Another tangent AB touches the circle at C intersecting XY at A and X'Y' at B.

SECTION D

(3)

Prove that AB subtends right angle at the centre of the circle; or ∠AOB = 90°.

Y' B

This section has 4 long answer questions of 5 marks each. 32. (a) A 2-digit number is seven times the sum of its digits and two (2) more than 5 times the product of its digits. Find the number. (5) OR (b) Find the value(s) of p for which the quadratic equation given as (p + 4)x2 - (p + 1)x + 1 = 0 has real and equal roots. Also, find the roots of the equation(s) so obtained. 33. If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points then it divides the two sides in the same ratio. Prove it. Also, state the converse of the above statement.

(5)

34. (a) From one of the faces of a solid wooden cube of side 14 cm, maximum number of hemispheres of diameter 1.4 cm are scooped out. Find the total number of hemispheres that can be scooped out. Also, find the total surface area of the remaining solid. (5) OR (b) From a solid cylinder of height 24 cm and radius 5 cm, two cones of height 12 cm and radius 5 cm are hollowed out. Find the volume and surface area of the remaining solid. 35. Medical check-up was carried out for 35 students of a class and their weights were recorded as follows: Weight (in kg) Number of Students

38-40 3

40-42 2

42-44 4

44-46 5

Find the difference between the mean weight and the median weight.

46-48 14

48-50 4

(5)

50-52 3

SECTION E This section has 3 case-study based questions of 4 marks each. 36. A farmer has a circular piece of land. He wishes to construct his house in the form of largest possible square within the land as shown in figure. The radius of circular piece of land is 35 m. Based on given information, answer the following questions: (i) Find the length of wire needed to fence the entire land. 410

(4)

(ii) Find the length of each side of the square land on which house will be constructed. (iii) (a) The farmer wishes to grow grass on the shaded region around the house. Find the cost of growing the grass at the rate of `50 per square metre. OR (iii) (b)

Find the ratio of area of land on which house is built to remaining area of circular piece of land.

37. In an equilateral triangle of side 10 cm, equilateral triangles of side 1 cm are formed as shown in the figure, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on. Based on given information, answer the following questions using Arithmetic Progression.

1st Row 2nd Row 3rd Row

(i) How many triangles will be there in bottom most row? (ii) How many triangles will be there in fourth row from the bottom? (iii) (a)

(4)

10th Row

Find the total number of triangles of side 1 cm each till 8th row. OR

(b)

How many more number of triangles are there from 5th row to 10th row than in first 4 rows? Show working.

38. Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding.

(4)

An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane. Based on given information, answer the questions given in part (i) and (ii). (i) Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground. (ii) Find the length of stairs used to reach the door sill. Further, answer any one of the following questions: (iii) (a) If the 20 feet long stairs is inclined at an angle of 60° to reach the door sill, then find the height of the

(

door sill above the ground. use 3 = 1.732

)

(b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed 30°? Also, find the horizontal distance of base of stair car from the plane.

SET-2 | Q.P. Code: 30/5/2 SECTION A This section consists of 20 questions of 1 mark each. 4.

Let a = p2 q3 rn and b = p3 qm r2, where p, q, r are prime numbers. If LCM of a and b is p3 q4 r3, then the value of 3n - 2m is (1) (a) -1

(b) 1

(d) -3 (1)

The point P divides the line segment AB in the ratio 3 : 1 as shown below: The value of

(a) 3

AB is PB

(b)

1 4

Examination Time • CBSE Question Paper (Standard) 2024–25

(d)

1 3 411

8. Which of the following equations is a quadratic equation?

(1) 1 (a) x = (x - 1) + 3x (b) x = (x + 1) (c) x = x (d) x + 1 = x 9. A piggy bank contains `1 coins and `2 coins in the ratio 9:11 respectively. The piggy bank is accidently dropped and a coin pops out of it. The probability that it is a `2 coin is (1) 9 1 (a) (b) 0.45 (c) 0.55 (d) 11 11 2tan60° 12. The value of is same as the value of (1) 1 − tan 2 60° 3

(a) -tan ⁡30° (b) -tan⁡ 60° (c) 2sin ⁡60° Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option:

(d) 2cos ⁡60°

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false. (d) Assertion is false, but Reason is true.

20. Assertion: For an acute angle θ, sec⁡θ = 3 ⇒ tan⁡θ = 2 2

(1)

Reason: sec ⁡θ = 1 - tan ⁡θ for all values of θ. 2

SECTION B This section has 5 very short answer type questions of 2 marks each. 21. AD and PS are angle bisectors of ∠A and ∠P of triangles ABC and PQR. If ∆ABC ∼ ∆PQR, prove that ∆ACD ∼ ∆PRS. (2) 22. All the face cards are removed from the pack of 52 cards and a card is drawn at random from the remaining cards. Find the probability that the card so drawn is (a) a spade.

(2)

(b) not an ace.

SECTION C This section has 6 short answer type questions of 3 marks each. 29. Obtain the zeros of the polynomial p(x) = 2x2 - 5x - 3. Hence, obtain a polynomial each of whose zeros is one less than each of the zero of p(x).

(3)

30. Solve the following system of equations graphically: 2x + y = 5 and 4x - y = 7. Hence, write the coordinates of the points where given lines meet y-axis. (3)

SECTION D This section has 4 long answer questions of 5 marks each. 32. During a medical checkup, height of 35 students of a class were recorded as follows: Height (in cm) Number of Students

90-100 3

100-110 2

110-120 4

120-130 5

130-140 14

(5)

140-150 7

Find the difference between the mean height and median height. 34. State the converse of basic proportionality theorem. BF AE 2 Also find in the following figure, given that AB || DC || EF and = . FC ED 3 Also, find the length of EF if AB =10 cm and DC = 15 cm.

412

A E

B G

(5) F

SET-3 | Q.P. Code: 30/5/3 SECTION A This section consists of 20 questions of 1 mark each. 3. The value of (1 - 2sin2 60°) is same as that of (a) sin ⁡30°

(1)

(b) -sin⁡ 30°

(d) -cos⁡ 30°

11. Let p = x y z and q = x y z , where x, y, z are prime numbers. If LCM(p, q) = x y z , then the value of (2m + 3n) is

(1)

13. Which of the following equation is a quadratic equation?

(1)

(a) 18

(b) 17

1 (a)  x +  = 2 x  

(b)

(d) 14

0 (c) (x + 1) = (1 - x) ( x − x ) + 2x x = 2

(d)

x ( x + 1) = 2

15. In the following figure, P and Q are points of trisection of line segment AB: AB the value of = PB (a) 1

(1) A

2 (d) 2 3 16. A bag contains red coloured, blue coloured and green coloured balls in the ratio 2 : 3 : 4. A ball is drawn at random from the given bag. The probability that the ball so drawn being not of blue colour is 1 2 8 1 (a) (b) (c) (d) 3 3 9 9 (b) 1.5

(c)

(1)

Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false. (d) Assertion is false, but Reason is true. 19. Assertion: For an acute angle θ, value of cosecθ cannot be

Reason: cosecθ ≥ 1 for 0° ≤ 0 ≤ 90°.

(1)

SECTION B This section has 5 very short answer type questions of 2 marks each. 23. In ∆ABC and ∆PQR, AD and PS are altitudes such that ∆ABD ∼ ∆PQS and ∆ACD ∼ ∆PRS. Prove that ∆ABC ∼ ∆PQR. (2) 24. From a pack of 52 cards, all aces and all kings are removed. A card is drawn at random from the remaining cards. Find the probability that the card so drawn is (2) (a) a face card.

(b) a card of red colour.

SECTION C This section has 6 short answer type questions of 3 marks each. 26. Find the zeros of the polynomial p(x) = 6x2 - 5x - 1. Hence, obtain a polynomial each of whose zeros is three times the zeros of p(x). (3) 31. Solve the following system of equations graphically: 2x + 3y = 6

(3)

x + y - 1 = 0 Also, find the sum of ordinates of the points where given lines meet y axis. Examination Time • CBSE Question Paper (Standard) 2024–25

413

SECTION D This section has 4 long answer questions of 5 marks each.

32. State the basic proportionality theorem. Use the theorem to do the following: In ∆ABC, AD is the angle bisector of angle A. BA is produced to E such that CE || AD. BD BA Prove that = DC AC 34. The following table gives the daily income of 50 cab drivers of a particular city: Income (`) Number of Drivers

500-600 12

600-700 14

Find the mean income and the modal income.

414

700-800 8

800-900 6

900-1000 10

(5)

12 B

3 D

(5)

SET-1 | Q.P. Code: 30/5/1

11. (d) 30°

SOLUTION

∠AOB = 360° - 210° = 150°

1. (a) Positive rational number

( 3 + 2) + ( 3 − 2) 2

∠APB = 180° - ∠AOB

∠APB = 180° - 150° = 30°

(a + b) + (a - b) = 2(a + b ) 2

(( 3 ) + 2 ) = 2(3 + 4) = 2 × 7 = 14 2

2. (b) 7 Given: x = a2 b3 cn, y = a3 bm c2, LCM(x, y) = a3 b4 c3 Then: max ⁡(b3, bm) = b4

⇒m=4

Max ⁡(c , c ) = c

⇒n=3

Now: m + n = 4 + 3 = 7 3. (a) a Statement: For any prime number p, if p | a2 ⇒p|a

4. (b)

10. (b) Infinite number of tangents can be drawn to a circle from a point outside the circle.

2x x (x + x ) = Given: ( x + x ) = 2x x 2

⇒ x 2 + 2x x + x = 2x x 2 ⇒ x + x = 0 this is quadratic equation. 5. (d) -3 Given: x2 + bx + b = 0 For real and distinct roots: D = b2 - 4b > 0 b2 - 4 > 0 ⇒ b(b - 4) > 0 ⇒ b < 0 or b > 4 Since, b < 0, thus -3 is correct answer. 6. (c) 1 : 2 PQ = 1, QB = QR + RB = 2 PQ 1 ⇒ Hence, Q divides PB in the ratio 1:2 = QB 2 7. (c) 0.7 Total balls = 10, Black balls = 7 7 P(black) = = 0.7 10 8. (a) Two right triangles are always similar. 9. (d) 1 : 2

2 AD 3 AX 2 DX 1 DX 1 Thus, =⇒ =⇒ = AD 3 AD 3 AX 2 Given: AX =

DX CY 1 1: 2 ⇒ = =⇒ CY : YB = AX YB 2 Examination Time • CBSE Question Paper (Standard) 2024–25

12. (d) cos 60°

 1  1 1−   1− 1 − tan 2 30° 3  =  3 = = 2 1 + tan 2 30°  1  1+ 1 1+   3  3 2 1 = = = cos 60° 4 2 13. (c) 40 m

2 3 4 3

Given: Distance = 38.2 m, Angle of elevation = 45°, Observer’s eye height = 1.8 m

height above eyes base height above eyes ⇒ 1= 38.2 ⇒ Height above eyes = 38.2 m tan 45° =

⇒ Total height = 38.2 + 1.8 = 40 m 14. (b) 4 cm Let R be radius of larger circle and r radius of smaller circle. Thus, R + r = 16 Chord AB of length 16 cm touches inner circle at point P In triangle OPA:OP = r, OA = R, AP = 8 (half of chord) Use Pythagoras: OA2 = OP 2 + AP 2 ⇒ R 2 = r 2 + 64 Using R + r = 16 (16 - r)2 = r 2 + 64 ⇒ 256 - 32r + r 2 = r 2 + 64 ⇒ -32r = -192 ⇒r=6 ⇒ R = 6 = 16 ⇒ R = 10 R - r = 10 - 6 = 4 cm 15. (a) 17 cm Given: Cone height = 12 cm, Slant height = 13 cm Use Pythagoras to find radius: 415

= r

132 − 122

169 − 144

But sinceθ is acute. 4 Thus, cosθ = 5 Assertion is false.

= 25 = 5 cm So hemisphere of radius 5 cm is placed below the cone.

Now, the identity sin2θ + cos2θ = 1 is universally true for all angles, including θ in the range 0° ≤ θ ≤ 90°.

Total height of solid:

Reason is true.

= height of cone + radius of hemisphere = 12 + 5

20. (d) Assertion is false, Reason is true. Given: AP = 3, 6, 9, ..., 198

= 17 cm

a = 3, d = 6 - 3 = 3, l = 198

16. (c) 2

nth from end = l - (n - 1)d

Given: x ⋅ median + y ⋅ mean = z ⋅ mode

⇒ 10th from end = 198 - (10 - 1) ⋅ 3 = 198 - 27

Mode = 3 ⋅ Median - 2 ⋅ Mean

⇒ 3 ⋅ Median = Mode + 2 ⋅ Mean

But Assertion says it’s 168 → Incorrect

⇒ x = 3,

Now, nth from end = l - (n - 1)d

y = -2,

Correctly states the formula.

z=1

So Reason is true, but Assertion is false.

x + y + z = 3 + (-2) + 1 = 2

21. (a) Let cost of 1 kg apples = `x,

17. (c) 28 and 29

Cost of 1 kg grapes = `y

Rearrange the table.

Marks Obtained

Number of Students

Cumulative Frequency

20 25 28 29 33 38 42 43

6 20 24 28 15 4 2 1

6 26 50 78 93 97 99 100

Total students = 100 n 100 = = = 50 2 2 Find the 50th and 51st terms for median

Composite numbers greater than 3: 4, 6 P(Composite numbers greater than 3)= 19. (d) Assertion is false, Reason is true. 3 Given: If sin⁡θ = , then 5 cos2θ = 1 - sin2θ

⇒ 416

9 16 3 = 1 −   =− 1 = 25 25 5 4 cosθ = ± 5

As per question: 2x + y = 320 and 4x + 2y = 600 a1 2 1 b1 1 c1 320 8 , = = = ,= = a 2 4 2 b2 2 c 2 600 15 1 1 8 = ≠ 2 2 15 Thus, these lines are parallel and no solution exist. So, system is inconsistent. (b)

2x + 3 y = 5 (1) 3x − 8 y = − 6 (2)

Multiply (1) by

3, and ( 2 ) by 2 :

5 3 (3) (1) × 3 : 6x + 3y =

50th and 51st terms fall between 28 and 29 1 18. (b) 3 Die outcomes: {1, 2, 3, 4, 5, 6}

= 171

2 1 = 6 3

− 12 ( 2 ) × 2 : 6x − 16 y =

6x − 4 y = −2 3 (4)

Now subtract (4) from (3):

( 6x + 3y ) − ( 6x − 4 y=) 5 3 − ( −2 3 )

7y = 7 3 y = 3

Substituting in equation 1 we get: 2x + 3

5 ( 3) =

2x + 3 = 5 2x = 2

x = 2 and y= 3

22. Given: A(-2, -2), B(2, -4)

BC = QR

4 Point P lies on segment AB such that BP = AB 7 4 3 If BP = AB , then AP = AB 7 7 So, P divides AB in the ratio: AP : PB = 3 : 4

∠ABC = ∠PQR (corresponding angles are equal) AB = PQ (corresponding sides are equal) By SAS criterion ∆ABC ∼ ∆PQR

 mx 2 + nx1 my 2 + ny1  P (x, y ) =  , m + n   m +n A(x1, y1) = (-2, -2),

25. (a) Probability that the dropped card is not a face card Non-face cards = 52 − 12 = 40 40 10 P(not a face card) = = 52 13 (b) Probability that the dropped card is a black king

B(x2, y2) = (2, -4), m = 3,

n=4 3 × 2 + 4 × ( −2 ) 6 − 8 −2 = x = = 3+ 4 7 7

Black kings = King of Spades, King of Clubs → 2 cards 2 1 P(black king) = = 52 26 26. (a) Let us assume, to the contrary, that is rational. That is, we can find integers a and b (≠ 0) such a that 3 = . b Suppose a and b do not have a common factor other than 1, then we can divide by the common factor and assume that a and b are coprime.

3 × ( −4 ) + 4 × ( −2 ) −12 − 8 −20 = = 3+4 7 7  −2 20  = P  ,−  7   7 23. (a) Given: sin⁡(A - B) = sin⁡A cosB - cos⁡A sin⁡B = y

Lets use: A = 45°, B = 30° sin⁡(45° - 30°) = sin⁡15°

= sin⁡45° cos⁡30° - cos⁡45° sin⁡30° 1

sin15= °

⋅

3 1 1 − ⋅ 2 2 2

So, b 3 = a Squaring on both sides and rearranging, we get 3b2 = a2.

1  3 −1  3 −1 =⋅  = 2  2  2 2

Therefore, a2 is divisible by 3, and by the theorem, it follows that a is also divisible by 3.

(b) Given: sin⁡A = y sin2A + cos2A = 1 ⇒ cos⁡A =

So, we can write a = 3c for some integer c.

(1 − y )

sin A tan = A = cos A

cos A =

tan A =

Substituting for a, we get

y 1− y

1− y y

1− y

3b2 = 9c2, that is, b2 = 3c2. This means that b2 is divisible by 3, and so b is also divisible by 3 (using the theorem with p = 3). Therefore, a and b have at least 3 as a common factor. However, this contradicts the principle that a and b are coprime.

24. Given: ∆ABD ∼ ∆PQS, AD and PS are medians of triangle ∆ABC and ∆PQR.

This contradicts the concept of coprimity between a and b.

To prove: ∆ABC ∼ ∆PQR

This contradiction has arisen because of our incorrect assumption that 3 is rational.

As ∆ABD ∼ ∆PQS BD = QS (corresponding sides are equal)

(1)

BD = DC (D is the midpoint of BC) (2) QS = SR (S is the midpoint of QR) (3) From 1 we get 2BD = 2QS ⇒ BD + BD = QS + QS BD + DC = QS + SR (Using 2 and 3) Examination Time • CBSE Question Paper (Standard) 2024–25

So, we conclude that 3 is irrational. (b) (i) 2 × 3 × 5 × 7 + 7

2 × 3 × 5 × 7 = 210 ⇒ 210 + 7 = 217 217 = 7 × 31

⇒ Yes, it has more than two factors

True — 217 is a composite number 417

(ii) 2 × 3 × 5 × 7 + 1

(-6x + 14x) + (-10y + 2y) + 34 - 50 = 0

2 × 3 × 5 × 7 = 210

⇒ 8x - 8y - 16 = 0

⇒ 210 + 1 = 211

211 has no divisors other than 1 and itself → it’s a prime number

False — 211 is NOT a composite number

27. Given 7x2 + 18x - 9

Coordinates on x-axis and y-axis x - 0 = 2 ⇒ x = 2 ⇒ (2, 0) 30. (a) To prove:

⇒ 7x(x + 3) - 3(x + 3) ⇒ (x + 3)(7x - 3)

3 7 3 6 Twice of the zeros: 2 × (-3) = -6, 2 × = 7 7 6 New polynomial: ( x + 6 )  x −  7  6 6 36  2 = ( x + 6 )  x −  = x − x + 6x − 7 7 7  36 36 2 2 =x + x − =7x + 36x − 36 7 7 7x2 + 36x - 36 new polynomial. So the zeros are x = −3,

-4x + y + 4 = 0

x y

Thus: 2x - y - 2 = 0

x y

⇒ y = 4x - 4

Y 3 2 1 X' -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 -7 -8 Y'

1 + cosecA 1 cosecA = + = sin A + 1 cosecA cosecA cosecA

= R.H.S

cos2 A 1 − sin 2 A (since cos2A = 1 - sin2A) = 1 − sin A 1 − sin A

(1 − sin A )(1 + sin A ) 1 − sin A

= 1 + sin A

Hence, the identity is proved.

0 -2 0 -4

1 0 1 0

(b) Given: sin⁡(2A + B) = 3 , 2 tan⁡(A + 2B) = 1 sin⁡(60°) = 3 2 ⇒ 2A + B = 60°

(1)

tan⁡(45°) = 1 ⇒ A + 2B = 45°

(2)

Multiply (2) by 2 : 2A + 4B = 90° Now subtract equation (1):

(1, 0) 1 2 3 4

(2A + 4B) - (2A + B) = 90° - 60°

5 X

⇒ 3B = 30° ⇒ B = 10° Substitute in (2): A + 2(10°) = 45°

⇒ A = 25°

cot⁡(4A - 7B) = cot⁡(4 ⋅ 25° - 7 ⋅ 10°)

Absolute difference between ordinates on y-axis: 2 29. Given A(3, 5), B(7, 1) 2

= cot⁡(30°) =

cot⁡(4A - 7B) =

In ΔOPA and ΔOCA

| -2 - (-4) | = | -2 + 4 | = 2

( x − 3) + ( y − 5) =

= cot⁡(100° - 70°)

31. Draw a line between points O and C.

Absolute difference between ordinates

P(x, y) is equidistant from A and B

Solution/Point of intersection (1, 0)

L.H.S =

This, L.H.S = R.H.S = 1 + sinA

⇒ y = 2x - 2 -4x + y + 4 = 0

1 + cosecA cos2 A = cosecA 1 − sin A

28. Given: 2x - y - 2 = 0,

( x − 7 ) + ( y − 1) 2

OP = OC (Radii of the circle) AP = A C (tangents drawn from an external point to a circle are equal.) AO = AO (Common)

O 4

C Y'

(x - 3)2 + (y - 5)2 = (x - 7)2 + (y - 1)2

By SSS congruency, ΔOPA ≅ ΔOCA

x2 - 6x + 9 + y2 - 10y + 25 = x2 - 14x + 49 + y2 - 2y + 1

⇒ ∠POA = ∠ AOC (congruent angle of congruent triangle) (1)

-6x - 10y + 34 = -14x - 2y + 50 418

Relation between x and y : x - y = 2

0 - y = 2 ⇒ y = -2 ⇒ (0, -2)

7x2 + 21x - 3x - 9

PA = PB ⇒

⇒x-y=2

Similarly, ΔOCB ≅ ΔOQB

p = 5 or p = -3

⇒ ∠COB = ∠ BOQ (congruent angle of congruent triangle) (2)

When: p = 5

PQ is a diameter, hence a straight line and

= 180°

2∠AOC + 2∠COB = 180° [From equation (1) and (2)]

33. Given, a line is drawn parallel to one side of a triangle to intersect the other two sides. To prove that the two sides are divided in the same ratio.

∴ ∠AOC + ∠COB = 90° We have, ∠AOC + ∠COB = ∠AOB ∠AOB = 90°

Consider a triangle ABC,

Hence proved. 32. (a) Let the 2-digit number be 10a + b, As per question: 10a + b = 7(a + b) (i) 10a + b = 5ab + 2

(ii)

From (i): 10a + b = 7a + 7b

⇒ a = 2b

(iii)

Substitute (iii) into (ii): ⇒ 10(2b) + b = 5(2b)(b) + 2 ⇒ 20b + b = 10b2 + 2 ⇒ 21b = 10b2 + 2 ⇒ 10b2 - 21b + 2 = 0

( −21) − 4 ⋅10 ⋅ 2 2

2 ⋅ 10 21 ± 441 − 80 20

21 ± 361 21 ± 19 = 20 20 40 2 = b = 2 or= b = 0.1 ( not valid ) 20 20 =

So, b = 2, from (iii) a = 2b = 4

Hence, the number = 10a + b = 10 × 4 + 2 = 42 The number is 42. (b) Given: (p + 4)x2 - (p + 1)x + 1 = 0 For real and equal roots: Discriminant = 0 D = [-(p + 1)]2 - 4(p + 4)(1) D = p2 + 2p + 1 - 4p - 16 = p2 - 2p - 15 p2 - 2p - 15 = 0 p=

2±

( −2 ) + 4 ⋅15 2

DE is parallel to BC. Now join BE and CD.

We know that,

⇒ 3a - 6b = 0

21 ±

D and E are the points that intersect AB and AC.

Also, draw EF and DG perpendicular to AB and AC.

⇒ 10a - 7a + b - 7b = 0

6 1 = 2⋅9 3

(1)x2 - (-2)x + 1 = x2 + 2x + 1 = 0 −2 ⇒x= = −1 2 ⋅1 1 Roots: For p = 5 : x = 3 For p = -3 : x = -1

But ∠POQ = ∠POA + ∠AOC + ∠COB + ∠BOQ

∴

⇒x=

When: p = -3

∠POQ = 180°

(9)x2 - (6)x + 1 = 0

2 ± 64 2 ± 8 = 2 2

Examination Time • CBSE Question Paper (Standard) 2024–25

1 × base × height 2 1 Now, area of ∆ADE = × AD × EF 2 1 Area of ∆BDE = × BD × EF 2 1 × AD × EF AD area of ∆ADE 2 So, (1) = = area of ∆BDE 1 × BD × EF BD 2 1 Area of ∆ADE = × AE × DG 2 1 Area of ∆DEC = × EC × DG 2 1 × AE × DG ∆ ADE AE area of So, (2) = 2= 1 EC area of ∆DEC × EC × DG 2 From the figure, ∆BDE and ∆DEC lie on the same base DE. Area of triangle =

Also, ∆BDE and ∆DEC lie between the same parallel DE and BC. So, area of ∆BDE = area of ∆DEC

Since, area of the triangles are same, then AD = AE BD EC Therefore, the two sides are divided in the same ratio. Converse Statement: If in a triangle, a line divides any two sides in the same ratio, then that line is parallel to the third side. This is the Converse of Basic Proportionality Theorem. 419

34. (a) Side of cube = 14 cm, 1.4 Radius = r = = 0.7 cm 2 The total number of hemispheres that can be 14 × 14 scooped out = = 100 1.4 × 1.4 Total surface area of cube = 6a2 = 6 × 142 = 6 × 196 = 1176 cm2

= 154 cm2

Curved surface area of one hemisphere

Net surface area of remaining solid: 1176 - 154 + 308 = 1330 cm

f = 14,

Number of hemispheres scooped out: 100 Total surface area of remaining solid: 1330 cm2 (b) Given: Cylinder : height = 24 cm, radius = 5 cm

Volume of Cylinder: pr2h = p × 52 × 24 = 600p 1 2 Volume of two Cones: 2 × pr2h = p × 52 × 12 3 3 = 200p Volume of Remaining Solid: 600p - 200p ( 400 × 22 ) ≈ 1257.14 cm3 = 7 CSA of Cylinder: 2prh = 2p × 5 × 24 = 240p 52 + 122 =

= 0.7 kg

Difference between mean and median weight is 0.7 kg. 36. Given: Radius of circular land = 35 m 22 (i) Length of wire = 2 × × 35 = 220 m 7 (ii) Let the side of the square be s. Diagonal of square = 70 m

169 = 13

CSA of One Cone: prl = p × 5 × 13 = 65p So, CSA of two cones = 2 × 65p = 130p Total Surface Area: CSA of cylinder + CSA of 2 cones = 240p + 130p = 370p

( 270 × 22 ) ≈ 1162.86 cm2

7 Volume of remaining solid: 1257.14 cm3 Surface area of remaining solid: 1162.86 cm

420

= 46.5 kg

Diagonal of the square = s 2

Slant height of cone:

 17.5 − 14   3.5  Median = 46 +   × 2 = 46 +  ⋅2  14   14  = 46 + 0.5 Difference = Median - Mean = 46.5 - 45.8

radius = 5 cm

r 2 + h2 =

h=2

Two Cones: height = 12 cm,

f ⋅ x 1603 ∑= = 45.8 kg 35 ∑f

CF = 14,

117 82 172 225 658 196 153 1603

n   2 − CF  Median = l +   ⋅h  f    Where: l = 46,

So total added = 100 × 3.08 = 308 cm2

f×x

Now, use the Median Formula:

= 2pr2 22 =2× × 0.49 7 = 3.08 cm2

Number of Cumulative Class Students (f) Frequency Mark (x) 3 3 39 2 5 41 4 9 43 5 14 45 14 28 47 4 32 49 3 35 51 35

n 35 Now, = n 35, so = = 17.5 2 2 The median class is 46-48, since 17.5 lies in its cumulative frequency range.

Area removed = 100 × pr2 = 100 ×

Weight (in kg) 38-40 40-42 42-44 44-46 46-48 48-50 50-52 Total

Mean =

Remove 100 circular bases

22 × (0.7)2 7 = 100 × 1.54

35.

Thus, 70 = s 2 Length of each side: s =

or 35 2 m 2 (iii) (a) Area on which grass is grown

= Area of two segments

1  90 22  =2 ×  × × 35 × 35 − × 35 × 35 2  360 7  2 = 700 m Cost of growing grass = 700 × 50

= `35000

(iii) (a) sin60° = opposite hypotenuse

(b) Required ratio area of square = area of circle – area of square 35 2 × 35 2 22 × 35 × 35 − 35 2 × 35 2 7 2450 7 = = 140 4

Required ratio = 7 : 4

OR 20 1 (b) = sin= 30° l 2 ⇒ l = 40 ft 20 1 = tan= 30° x 3 1 tan⁡ 30° = 3

37. Given AP is 1, 3, 5, … a = 1, d = 2 Formulas: an = a + (n - 1)d n Sn = ⋅ [2a + (n - 1)d] 2 (i) a10 = 1 + (10 - 1) × 2 = 1 + 18 = 19 19 triangles

(ii) 4th row from bottom = 7th row a7 = 1 + (7 - 1) ⋅ 2 = 1 + 12 = 13

13 triangles 8 (iii) (a) S8 = ⋅ [2 × 1 + (8 - 1) × 2] 2 = 4 × [2 + 14]

60°

⇒ x = 20 3 ft

Shortest length of stair = 40 ft

Horizontal distance = 34.64 ft

30°

SET-2 | Q.P. Code: 30/5/2 SOLUTION

= 4 × 16 = 64

4. (b) 1

64 triangles

Given: a = p2 q3 rn, b = p3 qn r2,

OR (b) Number of triangles from 5th row to 10th row = S10 - S4 4 10 = × [2 × 1 + 9 × 2] - × [2 × 1 + 3 × 2] 2 2 = 100 - 16 = 84

Number of triangles in first 4 rows, 4 S4 = × [2 × 1 + 3 × 2] = 16 2 Required number of triangles = 84 - 16 = 68

LCM (a, b) = p3 q4 r3 Then: max ⁡(q3, qm) = q4

⇒m=4

Max ⁡(r , r ) = r

⇒n=3

Now: 3n - 2m = 3(3) - 2(4) = 9 - 8 = 1 6. (c) 4 Give: AP : PB = 3 : 1 AB 1 = = 4 PB 4

⇒ AB = 3 + 1 = 4

8. (b) x3 = (x + 1)3

68 more triangles from 5th to 10th row than in first 4 rows. opposite 15 = = 1 adjacent 15 ⇒ θ = 45°

38. (i) tan= θ

x3 = x3 + 3x2 + 3x + 1 0 = 3x2 + 3x + 1 (This is a quadratic equation) 9. (c) 0.55 As per question: `1 : `2 = 9 : 11 Total = 20 coins

15 ft.

(ii) θ = 45° 15 = sin 45° l 15 1 ⇒ = l 2

20 ft.

h , and h 20 3 sin60° = 2 h 3 = sin⁡ 60° = h = 20 × 20 2 = 10 3 ≈ 10 × 1.732 = 17.32 ft =

Probability (`2 coin) =

15 ft.

15 2 ⇒l =

Examination Time • CBSE Question Paper (Standard) 2024–25

12. (b) -tan 60°

11 = 0.55 20

2tan60° 2 3 2 3 = = = − 3= − tan60° 1 − tan 2 60° 1 − 3 −2

( tan60° = 3 )

421

20. (c) Assertion is true, but Reason is false.

( x − 2 )  x +

Given: sec⁡θ = 3



3  ⇒ ( 2x − 4 )  x +  = 0 2  3  3 3   ⇒ 2x  x +  − 4  x +  = ( 2x − 4 )  x +  2 2 2       = 2x2 + 3x - 4x - 6

sec⁡ θ = 1 + tan⁡ θ 2

⇒ 9 = 1 + tan⁡ θ 2

⇒ tan⁡2θ = 8 ⇒ tan= θ

3 0 = 2

= 8 2 2

Assertion is true.

= 2x2 - x - 6

sec2θ = 1 + tan2θ

New polynomial: 2x2 - x - 6.

⇒ Incorrect given as

30. Given: 2x + y = 5,

sec2θ =1 - tan2θ

4x - y = 7

Reason is false. 21. Given: ∆ABC ∼ ∆PQR, AD and PS are angle bisectors of ∠A and ∠P To prove: ∆ACD ∼ ∆PRS

Thus: 2x + y = 5 ⇒ y = -2x + 5 4x - y = 7 ⇒ y = 4x - 7

As ∆ABC ∼ ∆PQR

Y 3 2 1

∠A = ∠P, ∠C = ∠R, ∠B = ∠Q (Corresponding angles are equal) ∠CAD = ∠RPS (AD is angle bisectors)

X' -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 -7 -8 Y'

∠ACD = ∠ PRS (PS is angle bisectors) Thus, ∆ACD ∼ ∆PRS

(By AA Criterion)

22. Total cards = 52, Face cards = 3 per suit × 4 = 12, Remaining cards = 52 - 12 = 40 cards

(a) Each suit has 13 cards → remove 3 face cards

→ 10 spades remain

10 1 = 40 4 (b) There are 4 aces total → none are face cards P(spade) =

90-100 100-110 110-120 120-130 130-140 140-150 Total

40 − 4 36 9 = = 40 40 10

29. Given: p(x) = 2x2 - 5x - 3 x=

− ( −5 ) ±

( −5) − 4 ( 2 )( −3 ) 2(2) 2

5 ± 25 + 24 = 4 5 ± 49 5 ± 7 = = 4 4 −2 12 1 = 3, x = = − x= 4 4 2

New polynomial with zeros one less than each of 1 3 these: 3 − 1 =2, and − − 1 =− 2 2

x y

0 1.75 -7 0

= x

2.5 0

(2, 1) 1 2 3 4 5 X 2x + y = 5

4x - y = 7

3 2 4 5 14 7 35

95 105 115 125 135 145

f×x 285 210 460 625 1890 1015 4485

fx 4485 ∑= = 128.14 cm ∑ f 35

l = 130, N = 35, F = 14, f = 14, h = 10 N   2 −F  Median = l +   ×h  f    3.5  17.5 − 14  Median = 130 +  × 10  × 10 = 130 + 14  14  =130 + 2.5 = 132.5 cm Median = 132.5, Mean ≈ 128.14 ⇒ 132.5 - 128.14 = 4.36 cm

422

0 5

Solution/Point of intersection (2, 1) 32. Class Interval Frequency (f) Midpoint (x)

→ all 4 remain

P(not an ace) =

x y

34. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. A Given: AB || DC || EF, E AE 2 = , AB = 10 cm, ED 3 DC = 15 cm D AE BF 2 = = (By Converse of BPT) ED FC 3

13. (b)

EF is parallel to AB and DC and lies between them in the trapezium. So, EF divides AB and DC in same ratio. Let EF = x Since, AB = 10 cm and DC = 15 cm, And EF is between them (in same proportion as AE : ED = 2 : 3). Then EF is at the same proportional distance between AB and DC. Use section formula:

3 ⋅ AB + 2 ⋅ DC 3 (10 ) + 2 (15 ) = 3+2 5 30 + 30 60 = = = 12 5 5 EF = 12 cm

= EF

SET-3 | Q.P. Code: 30/5/3

x 2 − 2x x + x + 2x x = 0 x2 + x = 0 (This is a quadratic equation)

B G

0 ; ( x − x ) + 2x x = 0 ( x − x ) + 2x x =

15. (b) 1.5 Points P and Q divide AB into three equal parts. 2 PB = AB 3 AB AB 1 3 Thus: = = = = 1.5 PB 2 AB 2 2 3 3 2 16. (c) 3 Given: Red : Blue : Green = 2 : 3 : 4, Total = 2 + 3 + 4 = 9 Red + Green 2 + 4 6 2 P(Not Blue) = = = = Total 9 9 3 19. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion. 1 We know: cosecθ = sin θ 1 ⇒ cosecθ = ⇒ sin θ = 2 2 But

2 > 1 and sin⁡θ ≤ 1, so not possible.

Assertion is true. Now, cosec⁡θ ≥ 1 for 0° < θ ≤ 90° 1 As, sin⁡θ ≤ 1 ⇒ cos= ecθ ≥1 sin θ Reason is true.

22.

SOLUTION 3. (b) -sin 30°

 3 1 - 2sin2 60° = 1 - 2    2  3 =1-2· 4 3 1 =1=2 2 = -sin⁡ 30°

11. (b) 17 Given: p = x2 y3 zn,

q = x3 ym z2

Then: max ⁡(y3, ym) = y4 ⇒ m=4 n max⁡ (z , z2) = z3 ⇒ n=3 2m + 3n = 2(4) + 3(3) =8+9 = 17

Examination Time • CBSE Question Paper (Standard) 2024–25

Given: ∆ABD ∼ ∆PQS, ∆ACD ∼ ∆PRS To prove: ∆ABC ∼ ∆PQR As ∆ABD ∼ ∆PQS

∠B = ∠Q (Congruent angle of congruent triangle) As ∆ACD ∼ ∆PRS

∠C = ∠R (Congruent angle of congruent triangle) By AA similarity criterion ∆ABC ∼ ∆PQR Hence proved.

23. Total cards = 52 Removed: All aces (4) and kings (4) Remaining = 52 - 8 = 44 cards (a) Probability of a face card Face cards = Jack, Queen, King Kings removed → remaining face cards = 4 Jacks + 4 Queens = 8 8 2 = = P(face card) 44 11 423

(b) Probability of red card Red cards = Hearts + Diamonds = 26 Removed red aces (2) and red kings (2) → 4 red cards removed Remaining red cards = 26 - 4 = 22 P(red card) = 26. p(x) = 6x - 5x - 1

22 1 = 44 2

32. Basic Proportionality Theorem: In a triangle, if a line is drawn parallel to one side of the triangle and intersects the other two sides, then it divides those two sides in the same ratio. BD BA To prove: = DC AC Given: In ∆ABC, AD is the angle bisector of ∠A BA is produced to E and CE || AD

− ( −5 ) ±

∠1 = ∠2 (AD is the angle bisector) ∠3 = ∠2 (Alternate angles)

( −5) − 4 ( 6 )( −1) 2(6 ) 2

∠BAC = ∠1 + ∠2 = 2∠2 From 1 we get

5 ± 25 + 24 12 5 ± 49 = 12 5±7 = 12 12 1 −2 = 1 or x = = − x= 12 12 6 =

∠BAC = 2∠3 ∠EAC = 180° - 2∠3 ∠AEC = ∠4 (Given) ∠ACE = ∠3 (Given) In ∆AEC. ∠AEC + ∠ACE + ∠EAC = 180° ⇒∠4 + ∠3 + 180° - 2∠3 = 180°

New polynomial with zeros 3 × each original:

⇒∠4 - ∠3 = 0

1  1 3 × 1 = 3, 3 ×  −  =− 2  6 1  Now, ( x − 3 )  x +  = 0 2 

AE = AC

34.

31. Given: 2x + 3y = 6, x+y=1

Thus: 2x + 3y = 6

( 6 − 2x )

3 x+y=1 ⇒y=1-x

(-3, 4)

x y

0 2

3 0

x y

0 1

1 0

Y 5 4 3 2 1

X' -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6 Y'

1 2 3 4

5 X

Solution/Point of intersection (-3, 4) Sum of Ordinates where lines meet y-axis = 2 + 1 = 3 424

(2)

Now, from BPT: In triangle EABC, since CE || AD,

= 2x3 - 5x - 3 required polynomial

⇒ y=

⇒ ∠4 = ∠3

Thus, ∆AEC is isosceles triangle

1  ⇒ ( 2x − 6 )  x +  2  2 = 2x - 6x + x - 3

(1)

and intersects sides BC and AE, by BPT: BD = BA DC AE BD BA (Using 2 AE = AC) = DC AC Hence proved. Income (`) 500-600 600-700 700-800 800-900 900-1000 Total

= x

Frequency (f) 12 14 8 6 10 50

Midpoint (x) 550 650 750 850 950

fx 36300 ∑ = = 726 50 ∑f

fx 6600 9100 6000 5100 9500 36300

Modal class = Class with highest frequency → 600-700 (f = 14)  f1 − f 0  Mode = l +   ×h  2 f1 − f 0 − f 2  Where, l = 600, f1 = 14, f0 = 12, f2 = 8, h = 100   14 − 12 Mode = 600 +   × 100 z 2 14 12 8 − −  ( )   2  600 +  =  × 100  28 − 20  2 Mode = 600 + × 100 = 600 + 25 = 625 8 Mean income: `726 Modal income: `625

Sample Question Paper 1 (Solved) Basic Time Allowed: 3 hours �� Max Marks: 80

General Instructions: I. This Question Paper has 5 Sections A, B, C, D, and E. II. Section A - has 20 MCQs carrying 1 mark each. III. Section B - has 5 questions carrying 2 marks each. IV. Section C - has 6 questions carrying 3 marks each. V. Section D - has 4 questions carrying 4 marks each. VI. Section E - has 3 case-based integrated questions carrying 4 marks each. VII. All questions are compulsory. However, an internal choice is in Qs of 3 marks, 4 marks & 1 Question of 2 marks has been provided. An internal choice has been provided in 3 questions of Section E. 22 VIII. Draw neat figures wherever required. Take π = wherever required if not stated. 7

SECTION A This section consists of 20 questions of 1 mark each. 1. The median class for the data given below is: Class Frequency

(a) 80-100

20-40 10

40-60 12

60-80 14

80-100 13

(b) 60-80

100-120 17

(d) 40-60

2. One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card? 3 3 3 1 (a) (b) (c) (d) 13 14 26 26 3. If θ is the angle (in degrees) of a sector of a circle of radius r, then the area of the sector is: 2π rθ π r 2θ π r 2θ 2π rθ (a) (b) (c) (d) 360 ° 360° 180° 180° 4. A tangent PQ at point of contact P to a circle of radius 12 cm meets the line through centre O to a point Q such that OQ = 20 cm, length of tangent PQ is: P

(a) 15 cm (b) 12 cm

12cm

3 tan⁡2θ - 3 = 0, then θ = ? (a) 30° (b) 60°

20 cm

(d) 16 cm 5. If

(d) 45°

6. There is a small island in the middle of a 50 m wide river. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 60° and 30°, then find the height of the tree. (a) 22.65 m

(b) 23.56 m

(d) 21.65 m

7. In a data, if l = 60, h = 15, f1 = 16, f0 = 6, f2 = 6, then the mode is: (a) 67.5

(b) 72

Examination Time • Sample Question Paper 1 (Solved) Basic

(d) 62 425

8. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for the first time? (a) 12.20 p.m.

(b) 12.12 p.m.

(d) none of these

9. Two vertices of ∆ABC are A(-1, 4) and B(5, -2), and its centroid is G(0, -3). Then, the coordinates of C are: (a) (4, 3)

(b) (4, 15)

(d) (−15, −4)

10. A cylindrical tub of radius 5 cm and height 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and the height of 22 the cone outside the hemisphere is 5 cm, then find the volume of water left in the tub. (Take π = ) 7 (a) 716 cm3 (b) 616 cm3 (c) 600 cm3 (d) 535 cm3 11. In the adjoining figure P and Q are points on the sides AB and AC respectively of ΔABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm, QC = 6 cm and PQ = 4.5 cm. The measure of BC is equal to: A (a) 13.5 cm P (b) 12.5 cm Q (c) 9 cm (d) 15 cm C B

12. A chord of a circle subtends an angle of 60° at the centre. If the length of the chord is 100 cm, find the area of the major segment. (a) 30391.7 cm²

(b) 30720.5 cm²

(d) 31021.42 cm²

13. Raju bought a fish from a shop for his aquarium. The shop keeper takes out one fish from a tank containing 15 male fish and 18 female fish. The probability that the fish taken out is a male fish is 5 6 5 7 (a) (b) (c) (d) 11 11 12 11 2 14. If cos⁡θ = , then 2sec2θ + 2tan2θ - 7 is equal to: 3 (a) 1 (b) 4 (c) 0 (d) 3 15. x2 - 6x + 6 = 0 have

(a) Real and Equal roots

(b) Real roots

(d) Real and Distinct roots

16. In triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B = (a) 75°

(b) 85°

17. If x = 3 is a solution of the equation 3x2 + (k - 1)x + 9 = 0 then = ? (a) 13

(b) -11

18. Choose the HCF: (396, 594, 1188) (a) 198

(b) 99

(d) 65°

(d) -13

(d) 33

Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option: (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false.

(d) Assertion is false, but Reason is true. 19. Assertion: No two positive numbers can have 18 as their HCF and 380 as their LCM. Reason: LCM is always completely divisible by HCF. 20. Assertion: Distance of point (a, b) from origin is b 2 − a 2 Reason: Distance of point (x, y) from origin is 426

(x − 0) + ( y − 0) 2

SECTION B This section has 5 very short answer type questions of 2 marks each. 1 21. If sin⁡ α = and cot β = 3, then find the value of cosec α + cosec β. 2 22. An umbrella has 8 ribs which are equally spaced (see figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

OR Find the area of the segment of a circle of radius 14 cm, if the length of the corresponding arc APB is 22 cm. 23. A quadrilateral ABCD is drawn to circumscribe a circle.

C Q

S A

Prove that: AB + CD = AD + BC

24. Is the pair of linear equations consistent/inconsistent? If consistent, obtain the solution graphically: 2x - 2y - 2 = 0; 4x - 4y - 5 = 0

1 25. In the figure, D and E are points on AB and AC respectively, such that DE || BC. If AD = BD, AE = 4.5 cm, find AC. 3 A D

OR In the given figure, ABCD is a rectangle. P is the mid-point of DC. QB = 7 cm, AD = 9 cm, and DC = 24 cm, then prove that ∠APQ = 90°. Q A

SECTION C This section has 6 short answer type questions of 3 marks each. 26. In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. OR cos3 θ + sin 3 θ cos3 θ − sin 3 θ Prove that: + = 2 cosθ + sin θ cosθ − sin θ 27. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number. Solve the pair of the linear equation obtained by the elimination method. C 28. In a given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.

Examination Time • Sample Question Paper 1 (Solved) Basic

70°

427

29. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting: (a) a king of red colour

(b) a face card

(d) the jack of hearts

(e) a spade

(f) the queen of diamonds

30. Find the zeros of the given quadratic polynomial and verify the relationship between the zeros and the coefficients: 6x2 - 3 - 7x. 31. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number. Solve the pair of linear equations obtained by the elimination method. OR Aditya is walking along the line joining points (1, 4) and (0, 6). Aditi is walking along the line joining points (3, 4) and (1, 0). Represent the graph and find the point where both cross each other.

SECTION D This section has 4 long answer questions of 5 marks each. 32. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: Expenditure (in `) 1000-1500 Frequency 24

1500-2000 40

2000-2500 33

2500-3000 28

3000-3500 30

3500-4000 22

4000-4500 16

4500-5000 7

33. A solid is in the shape of a cone surmounted on a hemisphere with both their diameters being equal to 7 cm and the height of the cone is equal to its radius. Find the volume of the solid. OR A solid consisting of a right cone standing on a hemisphere is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, if:

• •

radius of the cylinder = 60 cm

• height of the cylinder = 180 cm

radius of the hemisphere = 60 cm

• height of the cone = 120 cm

Assume the hemisphere and cone have a common base. 2

 2x   2x  34. Solve for x:   + 5  − 24 = 0, x ≠ 5 x − 5    x −5

The sum of squares of two consecutive multiples of 7 is 637. Find the multiples. 35. If BD and QM are medians of triangles ABC and PQR, respectively, where ∆ABC ~ ∆PQR, prove that

AB BD . = PQ QM

SECTION E This section has 3 case-study based questions of 4 marks each. 36. Two trees are standing on flat ground. The angle of elevation of the top of both the trees from a point X on the ground is 60°. If the horizontal distance between X and the smaller tree is 8 m and the distance to the top of the two trees is 20 m:

(a) Calculate the distance between the point X and the top of the smaller tree.

(b) Calculate the horizontal distance between the two trees. (c) Find the height of the big tree. OR Find the height of the small tree. 428

60° 8m

37. Saving money is a good habit and it should be inculcated in children right from the beginning. Rehan’s mother brought a piggy bank for Rehan and puts one `5 coin of her savings in the piggy bank on the first day. She increases his savings by one `5 coin daily.

Based on the above information, answer the following questions: (i) How many coins were added to the piggy bank on 8th day? (ii) How much money will be there in the piggy bank after 8 days? (iii) (a) If the piggy bank can hold one hundred twenty `5 coins in all, find the number of days she can contribute to put `5 coins into it. OR (b) Find the total money saved, when the piggy bank is full. 38. Karan went to the Lab near to his home for COVID 19 test along with his family members. The seats in the waiting area were as per the norms of distancing during this pandemic (as shown in the figure). 8 7 6 5 4 3 2 1

Neena

Karan

Akash Enter A 1

Binu 4

His family member took their seats surrounded by red circular area.

(a) What is the distance between Neena and Karan? (b) What are the coordinates of seat of Akash?

(c) What will be the coordinates of a point exactly between Akash and Binu where a person can be? OR Find distance between Binu and Karan.

Examination Time • Sample Question Paper 1 (Solved) Basic

429

Sample Question Paper 1 SOLUTION 1. (b) 60-80 Explanation: Total frequencies

6. (d) 21.65 m Let the height of the tree be h. In ∆PAT, tan60° =

h h ⇒ 3 = ⇒h = x x

In ∆QAT, tan⁡30°=

h 1 h ⇒ = 50 − x 3 50 − x

(N) = 10 + 12 + 14 + 13 + 17 = 66 66 So, N = = 33 2 2 c.f. just greater than 33 is 36 and the corresponding class is 60-80. Hence, median class = 60-80 2. (c)

3 26

(2 kings + 2 queens + 2 jacks) 6 3 = 52 26

∴ P(getting a face card) = 3. (a) π r θ 360° 2

Area of sector =

360°

π r 2θ

× πr = 360° 2

4. (d) 16 cm

⇒ 3h = 50 −

20 cm

Since OP is perpendicular to PQ, the ∠OPQ = 90° Now, in right angled triangle OPQ, OQ 2 = OP 2 + PQ 2 ⇒ (20)2 = (12)2 + PQ 2 ⇒ PQ = 256 2

⇒ PQ = 16 cm 5. (a) 30° 3 tan 2θ − 3 = 0 3 tan 2θ = 3

⇒ tan 2θ =

⇒ tan 2θ = 3 ⇒ tan 2θ = tan ⁡60° ⇒ 2θ = 60° ⇒ θ = 30°

h 3

50 3  = 21.65 m = x 4 

h   3

7. (a) 67.5  f1 − f 0  Mode = l +  ×h  2 f1 − f 0 − f 2 

16 − 6 = 60 + × 15 2 × 16 − 6 − 6 10 60 + = × 15 32 − 12 10 = 60 + × 15 20 = 60 + 7.5 = 67.5

8. (a) 12.20 p.m.

12 cm

⇒ PQ 2 = 400 - 144

30° 50 - x

P x A

⇒ The height of the tree is 21.65 m

Number of black face cards = 6

430

60°

⇒= h

Total number of cards = 52

⇒

Given: First clock rings every 50 seconds, Second clock rings every 48 seconds We need to find the LCM of 50 and 48 to know after how many seconds both will ring together again. 50 = 2 × 52 48 = 24 × 3 LCM = 24 × 3 × 52 = 16 × 3 × 25 = 1200 So, they will ring together after 1200 seconds. 1200 ÷ 60 = 20 minutes So, if they first beep together at 12.00 p.m. They will beep together again at 12.20 p.m. 9. (c) (-4, -15) Let the vertex C be C(x, y). Then 4+2+ y −1 + 5 + x = 0 and = −3 1 3 ⇒ x + 4 = 0 and 6 + y = -9 ∴ x = -4 and y = -15 So, the coordinates of C are (-4, -15).

10. (b) 616 cm3

= Area of major sector + Area of ∆OAB

( 360° − 60° ) 2 3 2 = × πr + r 360° 4 5 cm 7/2 cm

Volume of water in the cylindrical tub = Volume of the tub  22  3 = π r 2h=  × 5 × 5 × 9.8  cm= 770 cm 3  7  Volume of the solid immersed in the tub = Volume of the hemisphere + Volume of the cone  2 22 7 7 7   1 22 7 7   =  × × × ×  +  × × × × 5   cm 3  3 7 2 2 2   3 7 2 2    539 385   924  3 3 = + 154 cm 3   cm =   cm = 6   6  6 

Volume of water left in the tub = Volume of the tub - Volume of solid immersed = (770 - 154) cm3 = 616 cm3

11. (a) 13.5 cm In ∆ABC, AQ AP 3 3.5 1 ⇒ = ⇒ = ⇒ QC PB 6 7 2 AQ AP Since, , = QC PB Therefore, QP || BC ∴

AQ QP 1 4.5 = ⇒ = ⇒ BC= 13.5 cm AC BC 3 BC

12. (c) 30520.61 cm

In ∆OAB, by angle sum property

300 22 3 2 = × × (100 ) + × 1002 360° 7 4 5 22 2 = × × (100 ) + 3 × 2500 6 7 = 26190.48 + 4330.13 = 30520.61 cm2 5 13. (a) 11 Explanation: Total number of fish = 15 + 18 = 33 Male fish = 15 Number of possible outcomes = 15 Number of total outcomes = 15 + 18 = 33 15 5 Required Probability = = 33 11 14. (c) 0 2 b Given, cos⁡θ= = = k 3 h 2 2 A 2 sec ⁡θ + 2 tan θ - 7 b = 2k, h = 3k In ∆ABC, h2 = p2 + b2

⇒ (3k)2 = p2 + (2k)2 ⇒ 9k2 = p2 + 4k2 ⇒ p2 = 9k2 - 4k2

⇒ p2 = 5k2 ⇒ p = 5k

3k 2 = and tan= θ 2k 3 ⇒ 2sec2θ + 2tan2θ - 7

Then, sec= θ

2  5 3 ⇒ 2   + 2   − 7 2  2 

⇒

9 5 + − 7 2 2

⇒ 2× ⇒

5k = 2k

5 2

9 5 + 2× − 7 4 4

9 + 5 − 14 = 0 2

15. (d) Real and Distinct roots Comparing the given equation to the below equation

O r A

ax2 + bx + c = 0

60° r C 100 cm

a = 1, b = -6, c = 6 B

60° + ∠OAB + ∠OBA = 180° ⇒ 2∠OAB =120° ⇒ ∠OAB = 60° ⇒ ∆OAB is an equilateral triangle. ⇒ r = 100 cm Area of major segment Examination Time • Sample Question Paper 1 (Solved) Basic

D = b2 - 4ac

D = (-6)2 - 4 × 1 × 6 D = 36 - 24 D = 12 D>0

If b2 - 4ac > 0, then the equation has real and distinct roots. Hence Real and Distinct roots. 431

16. (a) 75°

So, Assertion is true.

In ∆ABC and ∆DEF

“LCM is always completely divisible by HCF.” → True

∠A = ∠E = 40°

This is a correct mathematical fact.

AB : ED = AC : EF, ∠F = 65°

⇒

40°

AB AC = ED EF

Since LCM must be divisible by HCF and here 18 does not divide 380, the pair is not possible. 65°

20. (d) Assertion is false, but Reason is true. Distance of point (a, b) from origin is which is incorrect

As, The correct distance is

QIn ∆ABC and ∆EDF AB AC = ED EF

will be

21. cosec= α

40° + 65° + ∠C = 180°

3(3) + (k - 1) 3 + 9 = 0 2

27 + 3k - 3 + 9 = 0

27 + 3k + 6 = 0

3k = -3

k = -11

18. (a) (A) 198 ⇒

594 = 2 × 3 × 11 3

⇒ 1188 = 22 × 33 × 11

HCF = 21 × 32 × 11 = 2 × 9 × 11

= 198 19. (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. HCF × LCM = Product of the two numbers 18 × 380 = 6840 We now need to check if 6840 can be expressed as the product of two numbers whose HCF is 18 and LCM is 380. 380 ÷ 18 ≈ 21.11 (Not an integer) 432

a2 +b2

1 = 1 2

1 + cot 2 β =

1+ 3 =

4= 2

( 2 + 1)

22. Given 8 ribs will divide whole circle in 8 parts. 360° = 45° 8 Area between two consecutive ribs of the umbrella Here, r = 45 cm and θ=

× πr 2 360° 45° 22 22275 = × × 45 × 45= cm 2 360° 7 28

OR 1 = APB = 22 cm

As ⇒ 396 = 22 × 32 × 11

2 + 2 or 2 ∴ cosecα + cosec β =

∴ ∠C = 180° - 105° = 75°

x = 3 is a solution of the equation means it satisfies the equation. Put x = 3, we get

1 = sin α

cosec β =

⇒ 105° + ∠C = 180°

3x2 + (k - 1)x + 9 = 0

Assertion is false, but Reason is true.

∴∠C = ∠F = 65° and ∠B =∠D

17. (b) -11

( a − 0 ) + (b − 0 ) =

Which is correct.

∴ ∆ABC ∼ ∆EDF (SAS criterion)

(Sum of angles of a triangle)

b2 + a2

Now, Distance of point (x, y) from origin is it

∠A = ∠E (each = 40°)

But ∠A + ∠B + ∠C = 180°

b2 − a2

360°

θ=

×2×

22 × 14 = 22 cm 7

O 90°

360° ⇒ θ = 90° 4

Area of the sector

lr 33 × 14 = = = 154 cm 2 2 2 1 Area of triangle AOB = × OA × OB 2 1 = × 14 × 14 2 = 98 cm2 Area of the segment = (154 - 98) cm2

= 56 cm2

14 cm B

P l

23. D

PQ 2 = QC 2 + PC 2

PQ 2 = 16 2 + 122 = 400 ⇒ PQ = 20 cm In right ∆ABQ using Pythagoras theorem

Q S A

AQ 2 = AB 2 + BQ 2

We know that the lengths of tangents drawn from an exterior point to a circle are equal. AP = AS, ... (i) [tangents from A]

⇒ AQ 2 = 242 + 72 = 625 ⇒ AQ = 25 cm In right ∆ADP using Pythagoras theorem AP 2 = AD 2 + DP 2

BP = BQ, ... (ii) [tangents from B]

⇒ AP 2 = 92 + 122

CR = CQ, ... (iii) [tangents from C]

⇒ AP 2 = 81 + 144

DR = DS, ...(iv) [tangents from D]

⇒ AP 2 = 255

AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS) [using (i), (ii), (iii), (iv)] = (AS + DS) + (BQ + CQ)

AP = 15 cm In ∆APQ, AP 2 = 152 = 225

= AD + BC

PQ 2 = 202 = 400

Hence, AB + CD = AD + BC 24. 2x - 2y - 2 = 0

(1)

4x - 4y - 5 = 0

(2)

Here, a1 = 2, b = -2, c1 = -2 a2 = 4, b2 = -4, c2 = -5 a1 b1 c1 We see that = ≠ a 2 b2 c 2

⇒ AP 2 + PQ 2 = 625 Also, AQ 2 = 25 2 = 625 ⇒ AQ 2 = AP 2 + PQ 2 ∴ ∆APQ is a right angled ∆ (using converse of BPT) ∴ ∠APQ = 90° 26. In ∆PQR, by Pythagoras theorem

Hence, the lines represented by the equations (1) and (2) are parallel.

PR 2 = PQ 2 + QR 2

Therefore, equations (1) and (2) have no solution, i.e., the given pair of a linear equation is inconsistent.

[Q PR + QR = 25 cm

25. According to question it is given that D and E are the points on sides AB and AC respectively. 1 Also AD = BD, 3 AE = 4.5 cm, DE || BC 1 BD 4.5 AD AE ∴ = ⇒3 = BD EC BD EC 1 4.5 ⇒ = 3 EC ⇒ EC = 4.5 × 3 cm ⇒ EC = 13.5 cm Now, AC = AE + EC = 4.5 + 13.5 = 18 cm OR According to question it is given that ABCD is a rectangle and p is the midpoint of DC . \ AD = BC = 9 cm QC = BQ + BC = 7 + 9 = 16 cm 1 PC = CD ⇒ PC = 12 cm 2 In right ∆PCQ using Pythagoras theorem Examination Time • Sample Question Paper 1 (Solved) Basic

⇒ (25 - QR)2 = 52 + QR 2 ⇒ PR = 25 - QR] 625 - 50QR + QR 2 = 25 + QR 2 ⇒ 600 - 50QR = 0 600 ⇒ QR = = 12 cm 50 Now, PR + QR = 25 cm

5 cm

⇒ PR = 25 - Q = 25 - 12 = 13 cm QR 12 Hence, sin⁡P = , = PR 13 QR 12 PQ 12 cos⁡P = and, tan⁡P = = = PQ 5 PR 13 OR To prove: = L.H.S

cos3 θ + sin 3 θ cos3 θ − sin 3 θ + = 2 cosθ + sin θ cosθ − sin θ

cos3 θ + sin 3 θ cos3 θ − sin 3 θ + cosθ + sin θ cosθ − sin θ

( cosθ + sinθ ) ( cos2 θ + sin 2 θ − sinθ cosθ ) ( cosθ + sinθ ) ( cosθ − sinθ ) ( cos2 θ + sin 2 θ + sinθ cosθ ) + ( cosθ − sinθ )

433

= (1 - sinθ cos⁡θ) + (1 + sin⁡θ cosθ)

29. Total number of cards in one deck of cards is 52. ∴ Total number of outcomes n = 52

= 1 + 1 - sinθ cosθ + sinθ cosθ

(a) Let E1 = Event of getting a king of red colour.

= 2 R.H.S 27. Let the unit’s digit and the ten’s digit in the twodigit number be x and y respectively. Then the number = 10y + x Also, the number obtained by reversing the order of the digit = 10x + y According to the question, x + y = 9

(1)

9(10y + x) = 2(10x + y)

So number of outcomes favourable to E1 m = 2 m 2 1 So, P(E1= ) = = n 52 26 (b) Let E2 = Event of getting a face card ∴ Numbers of outcomes favourable to E2,

m 12 3 = = n 52 13 (c) Let E3 = Event of getting a red face card m = 12. Hence P(E2= )

∴ Numbers of outcomes favourable to E3 = 6

⇒ 90y + 9x = 20x + 2y ⇒ 11x - 88y = 0 ⇒ x - 8y = 0

(2)

Subtracting equation (2) from equation(1), we get

[Q There are 6 red face cards in a deck] m 6 3 Hence, P(E= ) = = 3 n 52 26 (d) Let E4 = Event of getting a jack of heart

∴ Numbers of outcomes favourable to E4 = 1

9y = 9

9 =1 9 Substituting this value of y in equation (1), we get ⇒y=

x+1=9 ⇒x=9-1=8 Verification: substituting x = 8 and y = 1, we find that both the equations (1) and (2) are satisfied as shown below: x+y=8+1=9 x - 8y = 8 - 8(1) = 0 Hence, the solution is correct. 28. Consider the smaller circle whose centre is given as O. The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle. = 2(70°) ∠AOB = 140°

A 70°

Now consider the larger circle B and the points A, C, B and O along its circumference. AOBC from a cycle quadrilateral. In a cyclic quadrilateral, the opposite angles are supplementary, meaning that the opposite angles add up to 180°. ∠AOB + ∠ACB = 180° ∠ACB = 180° - ∠AOB = 180° - 140° ∠ACB = 40° Therefore, the measure of angle ACB is 40°. 434

m 1 = n 52 (e) Let E5 = Event of getting a spade Hence, (E4= )

∴ Numbers of outcomes favourable to E5 = 13

Hence, the required number is 18.

∠AOB = 2∠APB

[Q There is only one jack of heart in deck of cards.]

[Q There are 13 spade in a deck] m 13 Hence, P(E5= ) = n 52 (f) Let E6 = Event of getting the queen of diamond ∴ Numbers of outcomes favourable to E6 = 1

[Q There is only one queen of diamond in a deck] m 1 Hence, P(E= ) = 6 n 52 2 30. Let p(x) = 6x - 3 - 7x. For zeros of p(x), p(x) = 0 ⇒ 6x2 - 3 - 7x = 0

⇒ 6x2 - 7x - 3 = 0

⇒ 6x2 - 9x + 2x - 3 = 0

⇒ 3x(2x - 3) + (2x - 3) = 0

⇒ (2x - 3)(3x + 1) = 0

⇒ 2x - 3 = 0 or 3x + 1 = 0

3 1 ,− 2 3 3 1 So, the zeros of p(x) are and − 2 3 We observe that sum of its zeros ⇒ x=

3 1 or x = − 2 3

⇒ x=

3  1 3 1 +−  = − 2  3 2 3 9 − 2 7 − ( −7 ) Coefficient of x = = = = 6 6 6 Coefficient of x 2 =

3 1 Product of its zeros=   ×  −  2  3 1 3 Constant term = − = − = 2 6 Coefficient of x 2

31. (a) Let the unit’s digit and the ten’s digit in the two-digit number be x and y respectively.

32. We may observe from the given data that maximum class frequency is 40 belonging to 1500-2000 interval.

Then the number = 10y + x

Class size (h) = 500

Also, the number obtained by reversing the order of the digits = 10x + y

f − f1 Mode = 1+ ×h 2 f − f1 − f 2

According to the question, x + y = 9

(1)

9(10y + x) = 2(10x + y) ⇒ 11x - 88y = 0 ⇒ x - 8y = 0

(2)

Subtracting equation (2) from equation(1), we get ⇒ y=

Frequency (f) of modal class = 40 Frequency (f1) of class preceding modal class = 24

⇒ 90y + 9x = 20x + 2y

9y = 9

Lower limit (l) of modal class = 1500

9 = 1 9

Frequency (f2) of class succeeding modal class = 33 40 − 24 Mode = 1500 + 2 × 40 − 24 − 33 16 1500 + = × 500 80 − 57 = 1500 + 347.826 = 1847.826 ≈ 1847.83

Substituting this value of y in equation (1), we get x+1=9 ⇒x=9-1=8 Hence, the required number is 18. Verification: substituting x = 8 and y = 1, we find that both the equations (1) and (2) are satisfied as shown below: x+y=8+1=9 x - 8y = 8 - 8(1) = 0 OR (b) Let the given points be A(1,4), B(0,6), C(3,4) and D(1,0). On plotting points A and B and joining them, we get the path travelled by Aditya. Similarly, on plotting points C and D and joining them, we get path travelled by Aditi. Y 8 The path travelled 7 by Aditya 6 B(0, 6) 5 A(1, 4) 4 C(3, 4) 3 E(2, 2) 2 1 D(1, 0) 1 2 3 4

1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000

24 40 33 28 30 22 16 7 Σfi = 200

di = xi - 2750

ui fi

1250 1750 2250 2750 = a 3250 3750 4250 4750

-1500 -1000 -500 0 500 1000 1500 2000

-3 -2 -1 0 1 2 3 4

-72 _80 _33 0 30 44 48 28 Σfi di = -35

Σfi di ×h Σf i −35 x = 2750 + × 500 200 = x 2750 − 87.5

Mean x = a+

Hence, the solution is correct.

X' -5 -4 -3 -2 -1 0 -1 -2 -3 Y'

Expenditure Number of (in `) families fi

5 X

The path travelled by Aditi

It is clear from the graph that both of them cross each other at point E(2, 2). Examination Time • Sample Question Paper 1 (Solved) Basic

x = 2662.5 33. Radius of hemisphere = radius of cone =

7 cm 2

7 cm 2 Volume of the solid = Volume of hemisphere + Volume of cone Height of cone =

2 3 1 2 πr + πr h 3 3 1 22 7 7  7 7  = × × × 2× +  3 7 2 2 2 2 539 cm 3 or 134.75 cm 3 = 4

We have radius of cylinder = radius of cone = radius of hemisphere = 60 cm Height of cone = 120 cm 435

∴ Height of cylindrical vessel = 120 + 60 = 180 cm

Putting y = -8 2x = −8 x −5 2x = -8x + 40 10x = 40 x=4 Hence, x is 15, 4.

∴ V = Volume of water that the cylinder contains πr2h = {π × (60)2 × 180} cm3 Let V1 be the volume of the conical part. Then, V 60cm

OR (b) According to the question, let the consecutive multiples of 7 be 7x and 7x + 7 (7x)2 + (7x + 7)2 = 637 or, 49x2 + 49x2 + 49 + 98x = 637 or, 98x2 + 98x - 588 = 0 or, x2 + x - 6 = 0 or, (x + 3)(x - 2) = 0 or, x = -3, 2 Rejecting the value, x = 2 Thus, the required multiples are, 14 and 21.

120 cm O

B 60 cm

60cm

1 V1 = π r 2h1 3 1 ⇒ V1 = × π × 602 × 120 cm 3 3 ={π × 602 × 40} cm 3 For hemispherical part r = Radius = 60 cm

35.

Let V2 be the volume of the hemisphere. Then,

2  V 2  π × 603  cm 3 = 3 

⇒ V 2 = {2π × 20 × 60 } cm = {40π ⋅ 60 } cm 2

A 3

Let V3 the volume of the water left-out in the cylinder. Then, V 3 =V − V1 − V 2

V 3 ={π × 602 × 180 − π × 602 × 40 − 40π × 602 } cm 3 V 3 =π × 602 × {180 − 40 − 40} cm 3

22 V 3 = × 3600 × 100 cm 3 7 22 × 360000 22 × 360000 3 cm 3 m = ⇒ V3 = 3 7 7 × (100 ) =

22 × 36 3 m = 1.1314 m 3 700 2

34. (a) We have given,  2x  + 5  2x  − 24 = 0  x −5  x −5 2x Let be y ( x − 5) ∴ y2 + 5y - 24 = 0

Now factorise, y2 + 8y - 3y - 24 = 0 y(y + 8) - 3(y + 8) = 0 Putting y = 3 2x =3 x −5 2x = 3x - 15 x = 15 436

(y + 8)(y - 3) = 0 y = 3, -8

Given: ∆ABC ∼ ∆PQR and BD, QM are medians AB BD To prove: = PQ QM Proof: ∆ABC ∼ ∆PQR (given) AB AC ∴ = PQ PR AB 2 AD ⇒ = PQ 2PM AB AD ⇒ = PQ PM AB 2 AD ⇒ = ( BD and QM are medians ) PQ 2PM In ∆ABD and ∆PQM AB AD = PQ PM

( proved above )

∠A = ∠P (∆ABC ∼ ∆PQR) ∴∆ABD ∼ ∆PQM (SAS criteria) AB BD ∴ (Congruent sides of congruent traingles) = PQ QM 36. (a) In ∆DCX

DC tan60° = CX DC 3= 8 DC = 8 3 m

= DX

DC 2 + CX 2

(8 3 ) + 8 = 2

192 + 64

= 256 = 16 m Hence, distance between X and top of smaller tree is 16 m.

(b) In ∆BAX

AX BX 1 AC + 8 = 2 36

cos60° =

36 = 2AC + 16

n 120 2 (1) + (n − 1)1 = 2 n[2 + n - 1] = 240

n2 + n = 240

n2 + n - 240 = 0

= n(n + 16) - 15(n + 16) = 0

=(n + 16)(n - 15) = 0

⇒ n = -16 and n = 15

We get, n = 15 as n ≠ -16

∴ Number of days = 15

20 = 2AC 20 = 10 AC 2 AC = 10

(b) For amount 5, 10, 15…

∴ Horizontal distance between both trees is 10 m. (c) Height of big tree = AB ∴ In ∆BAX AB AB = ° = tan60 AX 18 AB = 18 3 m OR Height of small tree = CD

a = 5, d = 10 - 5 = 5, n = 15 n S= 2a + (n − 1)d  n 2 15 S= 2 × 5 + (15 − 1) 5 15 2  15 = [10 + 70] 2 15 = × 80 = 600 2 Total money saved = 120 × 5 = `600

38. (a) Position of Neena = (3, 6) Position of Karan =(6, 5)

In ∆CDX

Distance between Neena and Karan

CD tan60° = CX CD 3= 8 CD = 8 3 m 37. (i) Daily 1 coin

( 6 − 3) + ( 5 − 6 )

9 + ( −1)

= 10 (b) Co-ordinate of seat of Akash = 2, 3

This in 8 days = 8 coins (ii) Money in the piggy bank day wise 5, 10, 15, 20… a = 5, d = 10 - 5 = 5 8 S8 = [2 × 5 +(8 - 1)5] 2 = 4[10 + 7 × 5] = 4[10 + 35]

(c)

Akash

Middle point

Binu

(2, 3)

(5, 2)

 2+5 3+2 Co-ordinate of middle point =  ,  2   2 = (3.5, 2.5) OR Binu = (5, 5);

= 4 × 45

Karan = (6, 5)

=180 Money after 8 days = `180

Distance =

( 6 − 5) + ( 5 − 2 ) = 2

1+ 9 =

(iii) (a) We can have at most 120 coins.

Sn ≤ 120 number of coins = 1, 2, 3, 4….

a = 1, d = 1

where S= n

n 2a + (n − 1)d  2

Examination Time • Sample Question Paper 1 (Solved) Basic

437

Sample Question Paper 2 (Solved) Standard Time Allowed: 3 hours �� Max Marks: 80

SECTION A This section consists of 20 questions of 1 mark each. 1. 7 × 11 × 13 + 13 is a/an:

(a) odd number but not composite

(b) square number

(d) composite number

2. Find the discriminant of the equation: 3x2 - 2x + 8 = 0 (a) -94

(b) 94

(d) -92

3. The graph of y = p(x) is given in the following figure for some polynomial p(x). Find the number of zeros of p(x). Y

(a) 2

(b) 1

(d) 3

4. The values of k for which the quadratic equation 16x + 4kx + 9 = 0 has real and equal roots are: 3 3 1 (a) 6, -6 (b) ,− (c) 36, -36 (d) 6, − 4 4 6 2

5. For three terms p, s, q to be in AP, (a) 2p = p + s

(b) p > s > q

(d) 2s = p + q

6. Find the coordinates of the point equidistant from three given points A(5, 3), B(5, -5) and C(1, -5). (a) (2,-1)

438

(b) (3, -1)

(d) (5, -1)

7. If A(1, 3), B(-1, 2), C(2, 5) and D(x, y) are the vertices of a parallelogram ABCD, then the value of (x, y) is: 3  (a) (1, 2) (b) (3, 4) (c)  , 4  (d) (4, 6) 2  8. The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm, respectively. If PQ = 12 cm, then find AB: (a) 24 cm

(b) 32 cm

(d) 16 cm

9. In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. Q

4c m

If ∠QPR = 90°, then length of PQ is:

(a) 3 cm

(b) 4 cm (c) 2 cm

P R

(d) 2 2 cm

10. Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is: (a) 6 cm

(b) 4 cm

11. Find the value of θ if tan⁡θ = cotθ (a) 30°

(b) 60°

12. If 2sin2θ - cos2θ = 2, then find the value of θ. (a) 100°

(b) 70°

(d) 3 cm

(d) 90°

(d) 80°

13. A pole 6 m high casts a shadow 2 3 m long on the ground, then the sun’s elevation is: (a) 30°

(b) 60°

14. Find the area of a sector with radius 7 cm and central angle 90°. (a) 38 cm²

(b) 39 cm²

(d) 90° (d) 37.5 cm²

15. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (a) 28.6 cm²

(b) 26.6 cm²

16. A coin is tossed thrice. The probability of getting at least two tails is: 4 2 (a) (b) (c) 1 5 3 17. Find the modal class from the following table: Size Frequency

(a) 75-85

45-55 7

55-65

65-75

(b) 85-95

75-85

85-95

95-105 105-115 6 10

(d) 22.6 cm²

(d)

1 2

(d) 105-115

18. Netra has a total of 361 songs in her playlist out of which 165 are Hindi, 87 are Punjabi and 109 are English. She starts listening to music by choosing the first song from Hindi category. She will continue listening if the next song which will be played automatically is Punjabi. What is the probability that Netra will continue listening to music? 87 87 109 109 (a) (b) (c) (d) 361 360 361 360 Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option: (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false. (d) Assertion is false, but Reason is true. 19. Assertion: A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. 22 Hence the quantity of water it can hold is 4939 cm³. Use π = . 7 Reason: Quantity of water it can hold = volume of spherical glass vessel + volume of cylindrical neck Examination Time • Sample Question Paper 2 (Solved) Standard

439

20. Assertion:

7 4 ,a ,2 are three consecutive terms of an AP only if a = 5 5

Reason: If p, q and r are in AP, then q - p = r - q

SECTION B This section has 5 very short answer type questions of 2 marks each. 21. Find by prime factorisation the LCM of the numbers 18180 and 7575. Also, find the HCF of the two numbers. 22. Evaluate: 2sin 60° − tan 30° . sec2 45° 2

23. Two tangent segments BC and BD are drawn to a circle with centre O such that ∠CBD = 120°. Prove that OB = 2BC.

B D

24. Prove the trigonometric identity: 1 − cos A Prove that: = 1 + cos A

1 1 + = 2sec2 θ 1 − sin θ 1 + sin θ OR

( cot A − cosecA )

25. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°. OR Find the area of a quadrant of a circle whose circumference is 22 cm.

SECTION C This section has 6 short answer type questions of 3 marks each. 26. In a seminar, the number of participants in Hindi, English and Mathematics is 60, 84 and 108, respectively. Find the minimum number of rooms required, if in each room the same number of participants are to be seated and all of them being in the same subject. 27. If α and β are zeros of the quadratic polynomial 4x2 + x + 1, then form a quadratic polynomial whose zeros are 2α and 2β. 28. If the 3rd and the 9th terms of an AP are 4 and -8 respectively. Which term of this AP is zero? OR A man invested an amount at 12% per annum simple interest and another amount at 10% per annum simple interest. He received an annual interest of `1145. But if he had interchanged the amounts invested, he would have received `90 less. What amounts did he invest at the different rates? 29. Prove that

3 is an irrational number.

30. If sin⁡θ + cos⁡θ = p⁡ and sec⁡θ + cosecθ = q, show that q(p2 - 1) = 2p 31. A box contains 20 balls bearing numbers 1, 2, 3, ..., 20, respectively. A ball is taken out at random from the box. What is the probability that the number on the ball is:

(a) an odd number?

(b) divisible by 2 or 3? (c) a prime number?

(d) not divisible by 10?

440

SECTION D This section has 4 long answer questions of 5 marks each. 32. Solve the following quadratic equation: mnx2 - (m + n)2x + (m + n)2 = 0 OR A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it with an area of 120 m². Find the width of the path. 33. In the figure, AB || PQ || CD,

AB = x units

CD = y units

PQ = z units 1 1 1 Prove that: + = x y z

z B

34. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 12 cm and its base is of diameter 14 cm, find the total 22 surface area of the wooden article. Also, find the volume of the wood left in the article. Use π = . 7 OR A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of `600 per m². 35. 100 surnames were randomly picked up from a local telephone directory, and the distribution of the number of letters of the English alphabet in the surnames was obtained as follows: Number of Letters

1-4

4-7

7-10

10-13

13-16

Number of Surnames

16-19

Determine the median and mean number of letters in the surname. Also, find the modal size of surnames.

SECTION E This section has 3 case-study based questions of 4 marks each. 36. Deepa has to buy a scooty. She can buy the scooty either making cash down payment of `25,000 or by making 15 monthly installments as: 1st month: `3425, 2nd month: `3225, 3rd month: `3025, 4th month: `2825 and so on. (a) Find the amount of 6th installment.

(b) Find the total amount paid in 15 installments. OR (c) If Deepa pays `2625, then find the number of the installment.

(d) Deepa paid 10th and 11th installment together, find the amount paid that month. 37. One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60°. (a) At what distance from the foot of the tree was he observing the bird sitting on the tree?

Examination Time • Sample Question Paper 2 (Solved) Standard

Ball

A Bird

D Bird

Tree 80 m

Height 80 m

° 45 30°

Ground level

441

(b) How far did the bird fly in the mentioned time?

After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?

( 3 + 1) m?

38. Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women’s World Cup’s history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane.

J Full-Back I G

Central Midfielder

Centre-Back Goal keeper H

A Defensive Midfielder

Forward E

B Attacking Midfielder

Striker

Sweeper K

Wing-Back

Winger

(a) At an instance, the midfielders and forward formed a parallelogram. Find the position of the central midfielder (D) if the position of other players who formed the parallelogram are: -A(1, 2), B(4, 3) and C(6, 6). (b) Check if the Goal keeper G(-3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line. OR Check if the Full-back J(5, -3) and centre-back I(-4,6) are equidistant from forward C(0, 1) and if C is the mid-point of IJ. (c) If Defensive midfielder A(1, 4), Attacking midfielder B(2, -3) and Striker E(a, b) lie on the same straight line and B is equidistant from A and E, find the position of E.

442

Sample Question Paper 2 SOLUTION 1. (d) Composite number We have 7 × 11 × 13 + 13 = 13(77 + 1) = 13 × 78. Since the given number has more than two factors other than 1 and itself, therefore it is a composite number. 2. (d) -92 3x2 - 2x + 8 = 0 D = b - 4ac 2

where a = 3, b = -2, and c = 8 Substitute the values: D = (-2)2 - 4(3)(8) = 4 - 96 = -92 3. (b) 1 To find the number of zeros of the polynomial p(x), we count the number of times the graph intersects the x-axis, since the zeros of a polynomial are the x-values where y = 0. From the image provided, the graph intersects the x-axis at 1 point. 4. (a) 6, -6 Given equation is: 16k2 + 4kx + 9 = 0 Here a = 16, b = 4k, c = 9 Now, D = b2 - 4ac = (4k)2 - 4 ⋅ 16⋅ 9 = 16k2 - 576 Roots are real and equal if D = 0 ⇒ 16k2 - 576 = 0 ⇒ 16k2 = 576 ⇒ k2 = 36 = (±6)2 ⇒ k = ±6 5. (d) 2s = q + p If p, s, q are in AP q-s=s-p ⇒q+p=s+s ⇒ 2s = q + p 6. (b) (3, -1) Given: A(5, 3), B(5,-5), and C(1, -5) are equidistance from a point. Let the required point be P(x, y). Then AP = BP = CP ⇒ (AP)2 = (BP)2 = (CP)2 This means (AP)2 = (BP)2 Examination Time • Sample Question Paper 2 (Solved) Standard

⇒ (x - 5)2 + (y - 3)2 = (x - 5)2 + (y + 5)2 ⇒ x2 - 10x + 25 + y2 - 6y + 9

= x2 - 10x + 25 + y2 + 10y + 25

⇒ -10x + 34 - 6y = -10x + 50 + 10y ⇒ -16y = 16 ⇒ y = -1

And (BP)2 = (CP)2

⇒ (x - 5)2 + (y + 5)2 = (x - 1)2 + (y + 5)2 ⇒ x2 - 10x + 25 + y2 + 10y + 25 = x2 - 2x + 1 + y2 + 10y + 25

⇒ -10x + 10y + 50 = -2x + 10y + 26 ⇒ -8x = -24 ⇒ x = 3

Hence, the required point is (3, -1).

7. (d) (4, 6)

Parallelogram ABCD,

A(1, 3), B(-1, 2), C(2, 5), D(x, y)

Using midpoint property:

Diagonals bisect each other

⇒ Midpoint of AC = Midpoint of BD

1+ 2 3 + 5   3  Midpoint = of AC = ,   , 4 2  2   2  −1 + x 2 + y  Midpoint of BD =  , 2   2 Equation we get: −1 + x = 3 2 2 2+ y ⇒ x = 4 and =4 ⇒y=6 2 8. (d) 16 cm Use ratio of perimeters = ratio of corresponding sides: 4 AB 32 4 = = ⇒ AB = × 12 =16 cm PQ 24 3 3 9. (b) 4 cm Given: radius = 4 cm, ∠QPR = 90° Draw OR OQ = OR = 4 cm ∠OQP = ∠ORP = 90° (Angle made by radius and tangent) ⇒A ll angles are 90°, and pair of adjacent sides are 4 cm, thus it is a square with side 4. Thus, PQ = 4 cm 10. (d) 3 cm In a circumscribed quadrilateral: AB + CD = BC + AD ⇒ 6 + 4 = 7 + AD ⇒ AD = 3 cm 443

11. (c) 45° If tan θ = cotθ 1 tan θ = ⇒ tan 2 θ =1 ⇒ θ =45° tan θ 12. (c) 90°

19. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion. 4 cm

7 cm

2sin2θ - cos⁡2θ = 2

21 cm

⇒ 2sin2θ - (1 - sin2θ) = 2

22 2 ×2 ×7 7 = 88 cm3 4 4 22 Volume of sphere: π r 3 = × × (10.5 )3 3 3 7 = 4851 cm3 Volume of cylinder: π r 2h =

⇒ 2sin θ - 1 + sin θ = 2 2

⇒ 3sin2θ = 3 ⇒ sin2θ = 1 ⇒ θ = 90° 13. (b) 60°

height 6 3 tan θ = = = = shadow 2 3 3

14. (c) 38.5 cm2

Area of sector = 15. (a) 28.6 cm2

3 ⇒ θ = 60°

θ 90 22 2 × πr 2 = × ×7 360 360 7 1 22 1078 = × × 49 = =38.5 cm 2 4 7 28

P A

O Q

Area of segment = Sector area - Triangle area Sector area = 90 × π r 2 =1 × 22 × 100 360 4 7 2200 = 78.6 28 1 1 Triangle area = × OA × OB = × 102 =50 2 2 Area of segment = 78.6 - 50 = 28.6 cm2

16. (d)

1 2 Total outcomes when coin tossed thrice = 8

Favourable: (TTH, THT, HTT, TTT) = 4 cases 4 1 ⇒ P(at least two tails)= = 5 2 17. (b) 85-95 Modal class = class with highest frequency

444

Assertion is correct. Reason is also correct, and Reason explains Assertion. 20. (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion. 4 Given: , a , 2 are in AP 5 4 4 14 7 a − = 2 − a ⇒ 2a = 2 + = ⇒ a = 5 5 5 5 Assertion is correct. Reason is also correct, and Reason explains Assertion. 21. 18180 = 22 × 32 × 5 × 101 7575 = 3 × 52 × 101 LCM = 22 × 32 × 52 × 101 = 90900 HCF = 3 × 5 × 101 = 1515 22. 2sin 60° − tan 30° sec2 45° 2

 3  1  3 1 2   −  2  −     3  2 =  4 3 7 = 2 2 12 2

( )

23. Given: ∠CBD = 120° To prove: OB = 2BC In ∆OBC: ∠OBC = ∠OBD = 60°

B D

(Since line joining centre and point of contact bisects the angle between tangents)

⇒ Class is 85-95

∠OCB = 90° (tangent at point C is perpendicular to radius OC)

87 360 First song = Hindi (164 left), total now = 360 87 P(next Punjabi) = 360

Now, applying sine rule in ∆BOC: BC 1 = sin= 30° OB 2 ⇒ OB = 2BC

Max frequency = 32 18. (b)

Total = 88 + 4851 = 4939 cm3

Thus, ∠BOC = 180° - 60° - 90° = 30°

24. Start with the Left Hand Side (L.H.S): 1 1 (1 + sinθ ) + (1 − sinθ ) 2 + = = 1 − sin θ 1 + sin θ (1 − sinθ )(1 + sinθ ) 1 − sinθ 1 - sin2θ = cos2θ 2 = = 2sec2 θ cos2 θ Hence, L.H.S = R.H.S

(1 − cos A )

= 1 − cos2 A

(1 − cos A )

Maths: 108 ÷ 12 = 9 rooms Minimum number of rooms = 5 + 7 + 9 = 21 rooms 27. Let the given polynomial be p(x) = 4x2 + 4x + 1

sin 2 A

sin 2 A 2 cos A   1 =  −   sin A sin A 

= (cosec⁡A - cotA)2

Thus, (cot⁡A - cosecA)2 1 − cos A 2 Hence, = ( cot A − cosecA ) 1 + cos A 25. (a) Given: Radius r = 6 cm, Angle θ = 60° Area of sector =

Now divide each subject’s participants by 12: English: 84 ÷ 12 = 7 rooms

1 − cos A (1 − cos A ) × (1 − cos A ) = 1 + cos A (1 + cos A ) × (1 − cos A ) 2 (1 − cos A ) = (1 + cos A )(1 − cos A )

HCF = 22 × 3 = 4 × 3 = 12 Hindi: 60 ÷ 12 = 5 rooms

L.H.S :

108 = 22 × 33

× πr 2 360° 60 22 2 = × ×6 360 7 1 22 792 = × × 36 = 6 7 42 = 18.86 cm2 (approx.) OR

(b) Circumference = 22 cm 22 22 22 × 7 2π r = 22 ⇒ r = = = = 3.5 cm 2π 2 × 22 44 7 1 1 22 Area of quadrant: = × π r 2 = × × ( 3.5 )2 4 4 7 1 22 = × × 12.25 4 7 1 269.5 269.5 = × = 4 7 28 2 = 9.625 cm 26. All participants in a room are from the same subject. Each room has equal number of participants Find HCF of 60, 84, and 108 60 = 22 × 3 × 5 84 = 22 × 3 × 7 Examination Time • Sample Question Paper 2 (Solved) Standard

Since, α, β are zeros of p(x), b −4 α +β = − = = −1 a 4 c 1 Also, αβ= = a 4 Now, a quadratic polynomial whose zeros are 2α and 2β: x2 - (sum of zeros)x + product of zeros = x2 - (2α + 2β)x + 2α ⋅ 2β = x2 - 2(α + β)x + 4(αβ) 1 = x2 - 2(-1)x + 4 ⋅ 4 = x2 + 2x + 1 28. a + 2d = 4

(1)

a + 8d = -8 (2) Subtract (1) from (2): (a + 8d) - (a + 2d) = -8 -4 ⇒ 6d = -12 ⇒ d = -2 Substitute d = -2 in equation (1): a + 2(-2) = 4 ⇒a-4=4 ⇒a=8 Let the nth term be zero: a + (n - 1)d = 0 ⇒ 8 + (n - 1)(-2) = 0 ⇒ 8 - 2(n - 1) = 0 ⇒ 2(n - 1) = 8 ⇒n-1=4 ⇒n=5 Thus, 5th term is zero. OR Let the amount invested at 12% be x, and at 10% be y. Given: He receives annual interest 1145 12% of x = 0.12x, 10% of y = 0.12x + 0.10y = 1145

(1) 445

q(p2 - 1) = (secθ + cosecθ)(1 + 2sinθ cosθ - 1)

Interchanging the investments: 0.10x + 0.12y = 1055

(2)

[1145 - 90 = 1055]

1   1 =  +  2sin θ cosθ  cosθ sin θ   sin θ + cosθ  =  2sin θ cosθ  sin θ cosθ  = 2(sinθ + cosθ) = 2p

Multiply both equations to eliminate decimals: 12x + 10y = 114500

Equation (1) × 100:

10x + 12y = 105500

Equation (2) × 100:

Hence, proved: q(p2 - 1) = 2p

Now solve:

31. A box contains 20 balls. Total outcomes (n) = 20

Multiply (1) by 6: 72x + 60y = 687000

(a) Probability of getting an odd number

Multiply (2) by 5:

Odd numbers from 1 to 20:

50x + 60y = 527500

1, 3, 5, 7, 9, 11, 13, 15, 17, 19 ⇒ 10 numbers 10 1 P(odd number) = = 20 2 (b) Probability of number divisible by 2 or 3

Now subtract: (72x + 60y) - (50x + 60y) = 687000 - 527500 ⇒ 22x = 159500 ⇒ x = 7250

Multiples of 2 between 1-20:

Substitute in (1):

2, 4, 6, 8, 10, 12, 14, 16, 18, 20 ⇒ 10 numbers

0.12(7250) + 0.10y = 1145

Multiples of 3 between 1-20:

⇒ 870 + 0.10y = 1145

3, 6, 9, 12, 15, 18 ⇒ 6 numbers

⇒ 0.10y = 275

Common multiples of 2 and 3 (i.e., LCM 6):

⇒ y = 2750

6, 12, 18 ⇒ 3 numbers

He invested `7250 at 12% and `2750 at 10%. 29. Let ∴

So, numbers divisible by 2 or 3 = 10 + 6 - 3 = 13 13 P(divisible by 2 or 3) = 20 (c) Probability of getting a prime number

3 be a rational number. 3=

p , where q ≠ 0 q

Prime numbers between 1 and 20:

Let p and q be co-prime. ⇒ 3q2 = p2 ⇒ p2 is divisible by 3 ⇒ p is divisible by 3

(i)

2, 3, 5, 7, 11, 13, 17, 19 ⇒ 8 numbers 8 2 P(prime number) = = 20 5 (d) Probability of a number not divisible by 10

⇒ p = 3a, where ‘a’ is some integer

Only one number from 1 to 20 is divisible by 10: → 10

Using 3q = p and p = 3a 2

So, not divisible by 10 = 20 - 1 = 19 numbers

9a2 = 3q2 ⇒ q2 = 3a2 ⇒ q2 is divisible by 3 ⇒ q is divisible by 3

(ii)

(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are co-prime. Thus

3 is irrational number.

30. To prove: q(p2 - 1) = 2p q = sec⁡θ + cosecθ (sin⁡θ + cos⁡θ) = sin θ + cos θ + 2sinθ cosθ

446

(m + n ) ± ( − (m + n ) ) − 4 (mn )(m + n )

= 1 + 2sin⁡θ cosθ

2 2

2mn

(m + n ) ± (m + n ) − 4mn (m + n ) 2

L.H.S: q(p2 - 1) = (sec⁡θ + cosecθ)((sinθ + cosθ)2 -1)

−b ± b 2 − 4ac 2a a = mn, b = - (m + n)2, c = (m + n)2 x=

Given: p = sinθ + cosθ

19 20 2 2 32. Given: mnx - (m + n) x + (m + n)2 = 0 P(not divisible by 10) =

2mn

(m + n ) ± (m + n ) (m + n ) − 4mn 2

2mn

(m + n ) ± (m + n ) (m 2 + n 2 + 2mn − 4mn ) 2

x= = x= x=

2mn

(m + n ) ( (m + n ) ± m 2 + n 2 − 2mn ) 2mn

(

(m + n ) (m + n ) ± (m − n )

)

2mn (m + n ) ( (m + n ) ± (m − n ) ) 2mn

(m + n )( 2m ) 2mn

(m + n ) n

(m + n )( 2n )

(m + n ) m

2mn

Let the width of the path be x m. Length = 16 + 2x, Width = 10 + 2x Area of the entire outer rectangle = (16 + 2x)(10 + 2x) Area of the inner field = 16 × 10 = 160 m2 Area of the path = Outer area - Inner area (16 + 2x)(10 + 2x) - 160 = 120 (16 + 2x)(10 + 2x) - 160 = 120 160 + 32x + 20x + 4x2 = 280 ⇒ 4x2 + 52x + 160 = 280 ⇒ 4x2 + 52x - 120 = 0 ⇒ x2 + 13x - 30 = 0 ⇒ (x + 15)(x - 2) = 0 ⇒ x = -15 or x = 2 x=2m Path is 2 m wide. 33.

A x

P z B

z z DB + = = 1 x y DB

1 z z  1 ⇒  + = zx y  z 1 1 1 + = x y z Hence proved. 34. Given: Diameter = 14 cm → r = 7 cm, Height = 12 cm Since hemispheres are scooped out, the circular ends are hollow. TSA = Curved surface area of cylinder + Curved surface area of 2 hemispheres 22 CSA of cylinder = 2πrh = 2 × × 7 × 12 = 528 cm2 7 CSA of 2 hemispheres = 2 × 2πr2 = 4πr2

22 2 ×7 7 22 =4 × × 49 7 = 616 cm2 =4 ×

TSA = 528 + 616 = 1144 cm2 Now, Volume = Volume of cylinder - Volume of 2 hemispheres 22 Volume of cylinder = πr2h = × 72 × 12 7 22 = × 49 × 12 7 = 1848 cm3 2 3 Volume of 2 hemispheres = 2 × πr 3 4 22 = × × 73 3 7 4 22 = × × 343 3 7 = 628.57 cm3 (approx) Volume left = 1848 - 628.57 = 1219.43 cm2

Total surface area = 1144 cm2,

Volume of wood left = 1219.43 cm2

PQ QD AB||PQ ⇒ = AB DB

(1) (By thales theorem)

PQ BQ = CD BD

(2) (By thales theorem)

PQ||CD ⇒

⇒

Adding (1) and (2) we get, PQ PQ QD BQ + = + AB CD DB BD z z ( DQ + QB ) ⇒ + = x y DB Examination Time • Sample Question Paper 2 (Solved) Standard

Canvas Area = CSA of cone + CSA of cylinder 22 CSA of linder = 2πrh = 2 × × 2 × 2.1 7 88 × 2.1) = ( = 26.4 m2 7 22 CSA of cone = πrl = × 2 × 2.8 = 17.6 m2 7 Total canvas area = 26.4 + 17.6 = 44 m2 Cost = 44 × 600 = `26,400 447

35. Let the assumed mean A be 8.5. Class interval h = 3. Frequency Class interval h = 3.MidClass 1-4 4-7 7-10 10-13 13-16 16-19

value xi

6 30 40 16 8 4

2.5

xi − A h

fi ui

C.F.

-2

-12 -30 0 16 16 12

6 36 76 92 96 100

ui =

5.5 8.5 = A 11.5 14.5 17.5

N = Total frequency = 100

-1 0 1 2 3

 Σf u  Mean= A + h  i i   N   −6   −6  = 8.5 + 3  8.5 + 3  =  100    100  18 = 8.5 − 100 Mean = 8.5 - 0.18 = 8.32 N Now, = 50; 2 Cumulative frequency just after 50 is 76. ∴ Median class is 7 - 10. l = 7, h = 3, N = 100, f = 40, cf = 36 N   2 − cf  Median = l + h    f    21  50 − 36  = = 7 + 3 7+ 7 + 1.05 = 8.05 = 20  40  Now, from the table, it can be observed that the maximum class frequency is 40, belonging to class interval 7-10.

(c) Let an = a + (n - 1)d, given an = `2625 2625 = 3425 + (n - 1) × (-200) ⇒ -800 = (n - 1)(-200) ⇒ n-1=4 ⇒ n=5 So in 5th installment, she pays `2625. (d) an + an + 1 = a + (n - 1)d + a +(n)d

= 2a + 2nd - d

where n = 10

= 2 × 3425 + 2 × 10 × (-200) - 200

= 6850 - 4200 = 2650

⇒ Total = `2650 37. (a) In ∆ABC, tan 45= °

AB 80 = ⇒ CB = 80 cm CB CB

(b) In ∆DEC,

DE 80 = CE CE

tan 30 = ° ⇒

80 = CE

⇒ CE = 80 3 Distance the bird flew = AD = BE = CE - CB = 80 3 − 80 = 80 In ∆FGC,

FG 80 = CG CG

tan60 = °

 f1 − f 0  Mode = l+  ×h  2 f1 − f 0 − f 2 

80 ⇒ 3= CG 80 ⇒ CG = 3

Mode = 7.88 36. (a) Installments are: 3425, 3225, 3025, ... are in AP, with a = 3425, d = 3225 - 3425 = -200 Now 6th installment = a + 5d = 3425 + (5 × - 200) (b) Total amount paid = S= n

= `2425

n 2a + (n − 1)d  2

15 2 ( 3425 ) + 14 ( −2000 )  2  15 = ( 6850 − 2800 ) 2 15 = ( 4050 ) = `30,375 2

( 3 − 1) m OR

h = 3, l = 7, f1 = 40, f0 = 30, f2 = 16

 40 − 30   10  Mode = 7 +   × 3 = 7 +  34  × 3 80 − 30 − 16     = 7 + 0.88 = 7.88

448

Distance the ball travelled after hitting the tree = FA = GB = CB - CG GB =80 −

1   =80  1 − m 3 3 

(

)

20 3 + 1 Distance m/sec = Time taken 2 20

( 3 + 1) × 60 m/min

= 600

( 3 + 1) m/min

38. (a) Let D be (a, b), then

Mid point of AC = Midpoint of BD 1+ 6 2 + 6   4 + a 3 +b  , ,  =  2   2 2   2 4+a 7 3+b 8 = =

a 3= b 5 = Central midfielder is at (3, 5) (b) GH =

( −3 − 3) + ( 5 − 1) 2

36 + 16

= 52 = 2 13 GK =

( 0 + 3) + ( 3 − 5)

9+4

( 3 − 0 ) + (1 − 3) 2

9+4

( 0 − 5) + (1 + 3 )

25 + 16

= 41 CI =

( 0 + 4 ) + (1 − 6 )

16 + 25

= 41

Full-back J(5, -3) and centre-back I(-4, 6) are equidistant from forward C(0, 1) 5 − 4 −3 + 6   1 3  Mid-point of IJ = , =   ,  2  2 2  2 C is NOT the mid-point of IJ. (c) A, B and E lie on the same straight line and B is equidistant from A and E.

= 13 HK =

CJ =

= 13 GK + HK = GH ⇒ G, H and K lie on a same straight line

Examination Time • Sample Question Paper 2 (Solved) Standard

⇒ B is the mid-point of AE 1+ a 4 +b  , =   ( 2, −3 ) 2   2 4 + b = -6; b = -10E is (3, -10). 1 + a = 4; a = 3

449

Sample Question Paper 3 (Unsolved) Standard Time Allowed: 3 hours �� Max Marks: 80

SECTION A This section has 19 multiple choice type questions of 1 marks each. 1. Find LCM of 336 and 54. (a) 3042

(b) 3024

(d) 3044

(1 + 3 ) − (1 − 3 ) is: 2

(a) a positive rational number.

(b) a negative integer.

1 , −1 as the sum and product of its zeros respectively. 4 (b) 4x2 + x - 4 = 0

(d) 4x2 + x + 4 = 0

4. The lines 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0 are: (a) parallel

(d) a negative irrational number.

(b) intersecting

5. If the sum of n terms of an AP is 3n2 + 5n, then which of its term is 164? (a) 27th

(b) none of these

(d) none of these (d) 26th

6. If a point (c, d) lies in the 3rd quadrant, which of the following is true? (a) c is positive and d is negative (b) both c and d are positive

(d) c is negative and d is positive

7 7. If sum of the zeros of a cubic polynomial ax3 + (-7x2) + (-13x) + (d) is and product of zeros is 1, then the values 5 of ‘d’ and ‘a’ are: (a) 5, 5

(b) -5, 5

(d) -5, 5

8. The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm, respectively. If PQ = 12 cm, then find AB: (a) 24 cm

450

(b) 32 cm

(d) 16 cm

9. In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, then the radius of the circle with centre A is:

(a) 3 cm

(b) 1.5 cm

(d) 2.5 cm

r3 Q

r1 A r1 P

r2 r2

10. If ∆ABC ~ ∆DEF such that 2AB = DE and BC = 6 cm, find EF: (a) 10 cm

11. If cot θ =

(b) 12 cm

(d) 4 cm

7 , then evaluate tan2θ. 8

8 49 (b) 7 8 12. The value of (1 + tan2θ)(1 - sinθ)(1 + sinθ) is: (a)

(c)

49 64

(d)

64 49

(a)

(b) 1 (c) 2 (d) 0 2 13. Ravi and Raju are standing at an equal distance from a pole, but on the opposite sides. Both are looking at the top of the pole. The angle of elevation of Ravi’s eye sight from the horizontal is 45°, and that of Raju is 30°. If so, then which of the following statements is true?

(a) Raju is taller than Ravi. (b) Ravi is taller than Raju.

(d) Data is insufficient to compare the heights of Raju and Ravi. 14. Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 60°. (a)

132 cm 2 3

(b)

132 cm 2 2

(c)

15. The total surface area of a right circular cylinder is given by (a) 2πr(r + h)

(b) 2πr(r - h)

132 cm 2 7

(d)

132 cm 2 5

(d) πr(r + h)

16. Time taken by a group of swimmers for different range of ages is being recorded and maintained in a table. Which formula could be used to find the middle-most age? n   2 − cf   f1 − f 0  Σfx Σfd (a) (b) a + (c) l +  (d) l +   ×h  ×h Σf Σf f   2 f1 − f 0 − f 2      17. If a letter of English alphabet is chosen at random, then the probability of this letter to be a consonant is: 5 10 21 (a) 11 (b) (c) (d) 26 13 26 13 18. The median class for the data given below is: Class Frequency

20-40 10

40-60

60-80

80-100

100-120

(a) 80-100 (b) 60-80 (c) 20-40 (d) 40-60 Directions: In questions 19 and 20, a statement of Assertion is followed by a statement of Reason. Choose the correct option: (a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false. (d) Assertion is false, but Reason is true. Examination Time • Sample Question Paper 3 (Unsolved) Standard

451

19. Assertion: From a solid cylinder whose height and radius are 8 cm and 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Hence the volume of the remaining solid will be 192 cm3. Reason: Volume of remaining solid = Volume of the cylinder - Volume of the cone removed 20. Assertion: If the solution of system of equations x - y = 4 and x + y = 6 is x = p and y = 2q then p = 5. Reason: A pair of linear equations in two variables can be algebraically solved by elimination method.

SECTION B This section has 5 very short answer type questions of 2 marks each. 21. Prove that 5 2 is irrational.

22. In ∆ABC, D and E are the points on the sides AB and AC respectively such that DE || BC.

If AD = 6x - 7, DB = 4x - 3, AE = 3x - 3 and EC = 2x - 1, find the value of x. OR In figure check whether AD is the bisector of ∠A of ∆ABC if AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm.

23. The tangent at a point A of a circle with centre O intersects the diameter PQ of the circle (when extended) at the A point B. If ∠BAQ = 105°, find ∠APQ. B

24. Show that the area of a sector is half the product of the length of an arc and radius. OR A chord of a circle of radius 8 cm subtends a right angle at the centre. Find the area of the corresponding:

(a) Major sector

(b) Minor segment

25 Given 15 cot⁡ A = 8, find sin⁡ A and sec ⁡A.

OR sin θ 1 + cosθ Prove the trigonometric identity: + = 2cosecθ 1 + cosθ sin θ

SECTION C This section has 6 short answer type questions of 3 marks each. 26. On morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? 27. If the 7th term of an AP is 27 and the 13th term is 16 more than the 9th term, find the AP. 28. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number. Solve the pair of the linear equation obtained by the elimination method. OR The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number. 29. In the given figure, PA and PB are tangents to a circle centred at O. Prove that (i) OP bisects ∠APB (ii) OP is the right bisector of AB. A O

B 452

OR Prove that the lengths of tangents drawn from an external point to a circle are equal. 30. Prove that: (cosecA - sinA)(sec⁡A - cosA) =

1 tan A + cot A

31. Find the mean using the step deviation method. Class Frequency

0-10

10-20

20-30

30-40

40-50

SECTION D This section has 4 long answer questions of 5 marks each. 32. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other, they meet in one hour. Find the speed of the two cars. OR Ved travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and the rest by car. Find the speed of the train and the car. 33. ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD. 34. A solid is in the shape of a cone surmounted on a hemisphere with both their diameters being equal to 7 cm and the height of the cone is equal to its radius. Find the volume of the solid. OR A solid consisting of a right cone standing on a hemisphere is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm, the radius of the hemisphere is 60 cm and height of the cone is 120 cm, assuming that the hemisphere and the cone have common base. 35. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is `18. Find the missing frequency f. Class Interval

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequency fi

SECTION E This section has 3 case-study based questions of 4 marks each. 36. Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of `118,000 by paying every month starting with the first installment of `1000. If he increases the installment by `100 every month, then answer the following questions: (i) Find the amount paid by him in 30th installment. (ii) Find the amount to be paid in the 40th installment. (iii) (a) What is the ratio of the 19th instalment to the 28th installment? OR (b) Find the total amount paid by him in 30 installments. Examination Time • Sample Question Paper 3 (Unsolved) Standard

453

37. Rinku was very happy to receive a fancy jumbo pencil from his best friend Rohan on his birthday. Pencil is a basic writing tool, when sharpened its shape is a combination of cylinder and cone as given in the picture. Cylindrical pencil with conical head is a common shape worldwide since ages. Commonly pencils are made up of wood and plastic but we should promote pencils made up of eco-friendly material (many options available in the market these days) to save environment. The dimensions of Rinku’s pencil are given as follows: Length of cylindrical portion is 21cm. Diameter of the base is 1 cm and height of the conical portion is 1.2 cm. Based on the above information, answer the following questions: (i) Find the slant height of the sharpened part.

(ii) Find curved surface area of sharpened part (in terms of π).

(iii) (a) Find the total surface area of the pencil (in terms of π). OR (b) The pencil’s total height decreases by 8.2 cm after sharpening it many times, what is the volume of the cylindrical part of the shortened pencil (in terms of π)? 38. Vijay lives in a flat in a multi-storey building. Initially, his driving was rough so his father keeps eye on his driving. Once he drives from his house to Faridabad. His father was standing on the top of the building at point A as shown in the figure. At point C, the angle of depression of a car from the building was 60°. After accelerating 20 m from point C, Vijay stops at point D to buy ice cream and the angle of depression changed to 30°. (a) Find the value of x.

(b) Find the height of the building AB.

454

Sample Question Paper 4 (Unsolved) Standard Time Allowed: 3 hours �� Max Marks: 80

SECTION A This section consists of 20 questions of 1 mark each. 1. The LCM of 12, 15 and 21 is (a) 410

(b) 420

(d) 450

2. The sum and the product of the zeros of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is: (a) x2 - 3x + 10

(b) x2 - 3x - 10

(d) x2 + 3x + 10

3. The graph of y = p(x) is given in the following figure for some polynomial p(x). Find the number of zeros of p(x). Y

(a) 2

(b) 1

4. If the mean of 6, 7, x, 8, y, 14 is 9, then _______. (a) x + y = 21

(b) x + y = 19

(d) 3

(d) x - y = 21

5. The sum of two numbers is 8. If their sum is four times their difference, then the numbers are: (a) None of these

(b) 7 and 1

6. Find the distance between the points (0, 0) and (36, 15). (a) 29

(b) 39

Examination Time • Sample Question Paper 4 (Unsolved) Standard

(d) 5 and 3

(d) 59

455

7. If x = a cosθ and y = b sinθ, then b2x2 + a2y2 = (a) a2 + b2

(b) ab

(d) a 2b 2

8. In the following figure,in ∆PQR, seg RS is the bisector of ∠PRQ. If PS = 6, SQ = 8, PR = 15, find QR. P

(a) 14 (b) 15

(d) 20

9. Two dice are rolled simultaneously. The probability that they get different numbers on both dices is: (a)

5 6

(b)

1 6

(c)

1 3

(d)

2 3

10. If 3 tan 2θ − 3 =, 0 then θ = ? (a) 30°

(b) 60°

(d) 45°

11. Find the area of the sector of a circle having radius 6 cm and of angle 30°. (a) 8.42 cm2

(b) 9 cm2

(d) 9.52 cm2

12. If the midpoint of the line segment joining the points (a, b - 2) and (-2, 4) is (2, -3), then the values of a and b are: (a) 6, 8

(b) 6, -8

(d) -6, 8

13. A pole 6 m high casts a shadow 2 3 m long on the ground, then the Sun’s elevation is: (a) 60°

(b) 45°

(d) 30°

14. The coordinates of the point which divides the join of (−6, 10) and (3, −8) in the ratio 2 : 7 is: (a) (4, -6)

(b) (-4, 6)

(d) (-1, 3)

15. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (a) 28.6 cm2

(b) 26.6 cm2

(d) 22.6 cm2

16. In a family of two children, the probability of having at least one girl is: (a)

1 2

(b)

2 5

(c)

3 4

(d)

1 4

17. The first three terms of an AP respectively are 3y - 1, 3y + 5 and 5y + 1. Then y equals: (a) -3

(b) 4

(d) 2

18. The volume of a right circular cone whose area of the base is 156 cm and the vertical height is 8 cm,is 2

(a) 2496 cm3

(b) 1248 cm3

(d) 416 cm3

(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion. (c) Assertion is true, but Reason is false. (d) Assertion is false, but Reason is true.

19. Assertion: In a solid hemisphere of radius 10 cm, a right cone of same radius is removed out. The surface area of the remaining solid is 570.74 cm2 [Take π = 3.14 and 2 = 1.4 ] Reason: Expression used here to calculate surface area of remaining solid = Curved surface area of hemisphere + Curved surface area of cone PA PB = 20. Assertion: In a triangle PQR, A and B are points on sides PQ and PR such that then AB || QR. AQ BR Reason: By converse of BPT. 456

SECTION B This section has 5 very short answer type questions of 2 marks each. 21. Find LCM and HCF of 120 and 144 by the fundamental theorem of arithmetic. 22. In ∆ABC, D and E are the points on the sides AB and AC respectively such that DE || BC.

If AD = 6x − 7, DB = 4x − 3, AE = 3x − 3, and EC = 2x − 1, find the value of x. OR In figure check whether AD is the bisector of ∠A of ∆ABC if AB = 6 cm, AC = 8 cm, BD = 1.5 cm, and CD = 2 cm.

23. Complete the factor-tree and find the composite number M. M 2

N 2

8190 2

4095 O

1365 3

455 5

91 P

1 − cosθ 24. Prove that: ( cosecθ − cot θ )2 = 1 + cosθ OR 1 + sec A sin 2 A Prove that: = sec A 1 − cos A

22  25. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.  Use π =  7   OR In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the sector formed by 22  the arc.  Use π =  7  

SECTION C

This section has 6 short answer type questions of 3 marks each. 26. National Art convention got registrations from students from all parts of the country, of which 60 are interested in music, 84 are interested in dance and 108 students are interested in handicrafts. For optimum cultural exchange, organisers wish to keep them in minimum number of groups such that each group consists of students interested in the same artform and the number of students in each group is the same. Find the number of students in each group. Find the number of groups in each art form. How many rooms are required if each group will be allotted a room? 27. Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are: y = x, y = 2x, and y + x = 6 21 28. If α and β are the zeros of the polynomial p(x) = 2x2 + 5x + k satisfying the relation α 2 + β 2 + αβ = , then find 4 the value of k. OR The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves `3000 per month, then find the monthly income of each. 29. A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size (in inches) of the tile required that has to be cut and how many such tiles are required? Examination Time • Sample Question Paper 4 (Unsolved) Standard

457

30. A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. OR Form a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. 31. Prove:

Prove:

5 mm 14 mm

sin θ − cosθ + 1 1 = using identity sec2θ = 1 + tan⁡2θ sin θ + cosθ − 1 secθ − tan θ OR 1 1 − cosecA = cosecA − ( cot A )( sec A ) − cot A ( cot A )( sec A ) + cot A

SECTION D This section has 4 long answer questions of 5 marks each. 32. Solve for x:

1 1 1 1 2a − b = + + ; x ≠ 0, x ≠ , a, b ≠ 0 2a + b + 2x 2a b 2x 2 OR

If the roots of the quadratic equation (c2 - ab)x2 - 2(a2 - bc)x + b2 - ac = 0 in x are equal, then show that either a = 0 or a3 + b3 + c3 = 3abc. 33. In the following figure, ∆FEC ≅ ∆GBD and ∠1 = ∠2. Prove that ∆ADE ≅ ∆ABC A D

E 4

34. A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is 14 cm, find the rise in the water level. OR A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of `400 per m2. 35. Calculate the mode of the following frequency distribution table: Marks

Number of Students

25 or more than 25 35 or more than 35 45 or more than 45 55 or more than 55 65 or more than 65 75 or more than 75 85 or more than 85

52 47 37 17 8 2 0

SECTION E This section has 3 case-study based questions of 4 marks each. 36. Rukhsar is celebrating her birthday. She invited her friends. She bought a packet of chocolates which contains 120 chocolates. She arranges the chocolates such that in the first row there are 3 chocolates, 5 chocolates in second row, 7 chocolates in third row and so on. 458

(i)

(a) Find the total number of rows of chocolates. OR (b) If Rukhsar decides to make 15 rows, then how many total chocolates will be placed by her with the same arrangement?

(ii) How many chocolates are placed in last row?

(iii) Find the difference in number of chocolates placed in 7th and 3rd row. 37. Basant Kumar is a farmer in a remote village of Rajasthan. He has a small square farmland. He wants to do fencing of the land so that stray animals may not enter his farmland. For this, he wants to get the perimeter of the land. There is a pole at one corner of his field. He wants to hang an effigy on the top of it to keep birds away. He is standing in one corner of his square field and observes that the angle subtended by the pole in the corner just diagonally opposite to this corner is 60°. When he retires 80 m from the corner, along the same straight line, he finds the angle to be 30°.

A Pole D

30° 80 m

60° Square Field

(a) Find the length of his square field so that he can buy material to do the fencing work accordingly.

(b) Find the height of the pole too so that he can arrange a ladder accordingly to put an effigy on the pole. (c) Find the distance from Farmer at position C and top of the pole? OR Find the distance from Farmer at position D and top of the pole? 38. Some students were asked to list their favourite colour. The measure of each colour is shown by the central angle of a pie chart given below: (a) If a student is chosen at random, then find the probability of his/her favourite colour being white. (b) What is the probability of his/her favourite colour being blue or green?

Green 60°

60° 30°

120° White

Red

90° Yellow

OR What is the probability of the favourite colour being red or blue?

Examination Time • Sample Question Paper 4 (Unsolved) Standard

459

Mathematics (Subject Code–041)

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