Oswaal CBSE Chapterwise & Topicwise Question Bank Class 10 Mathematics Standard Book (For 2022-23 Exam) [18 ed.] 9355951531, 9789355951533
143 6 93MB
English Pages [475] Year 2022
Cover
Title Cover
Copyright Page
Contents
Syllabus
Sample Question Paper, 2021-22
Solved Paper, 2021-22
Topper's Answer 2020
UNIT 1 : Number System
Unit 1 : Number System
Self Assessment Test - 1
UNIT 2 : Algebra
2. Polynomials
3. Pair of Linear Equations in Two Variables
4. Quadratic Equations
5. Arithmetic Progression
Self Assessment Test - 2
UNIT 3 : Coordinate Geometry
6. Lines (in two-dimensions)
Self Assessment Test - 3
UNIT 4 : Geometry
7. Triangles
8. Circles
9. Constructions
Self Assessment Test - 4
UNIT 5 : Trigonometry
10. Introduction to Trigonometry and Trigonometric Identities
11. Heights and Distances
Self Assessment Test - 5
UNIT 6 : Mensuration
12. Areas Related to Circles
13. Surface Areas and Volumes
Self Assessment Test - 6
UNIT 7 : Statistics and Probability
14. Statistics
15. Probability
Self Assessment Test - 7
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18th EDITION YEAR 2022-23
“978-93-5595-153-3”
I SBN
SYLLABUS COVERED
CENTRAL BOARD OF SECONDARY EDUCATION DELHI
PUBLISHED BY
C OPYRIG HT
RESERVED BY THE PUBLISHERS
All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.
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Oswaal Books has exercised due care and caution in collecting all the data before publishing this book. In spite of this, if any omission, inaccuracy or printing error occurs with regard to the data contained in this book, Oswaal Books will not be held responsible or liable. Oswaal Books will be grateful if you could point out any such error or offer your suggestions which will be of great help for other readers. Kindle Edition (2)
TABLE OF CONTENTS Latest CBSE Syllabus released on 21st April 2022 for Academic Year 2022-2023 (CBSE Cir. No. Acad 48/2022) Latest Term-I & Term-II Board Papers 2021-22 - Fully solved
41 - 63
Topper's Answer 2020
l
9 - 12 17 - 40
l
l
In each chapter, for better understanding, questions have been classified with the typology issued by CBSE as : AE - Analyzing & Evaluating
Unit 3 : Coordinate Geometry
Equations in Two Variables
*Cross- Multiplication Method
l
of Linear Equations
(3)
Topic 2. Tangents to a Circle from a point outside it
124 - 124
Self Assessment Test-2
Topic 2. Sum of n terms of an
208 - 229
Segment in a given Ratio
Topic 3. Construction of a Triangle similar to a given Triangle
Self Assessment Test-4
Arithmetic Progression Arithmetic Progression
Topic 1. Division of a Line
98 - 123
th
Topic 1. To find n term of the
186 - 207
Topic 2. Discriminant and Nature of Roots
5. Arithmetic Progression
8. Circles
* 9. Constructions
Topic 1. Solutions of Quadratic Equations
70 - 97
4. Quadratic Equations
Topic 2. Algebraic Methods to solve pair
(Prove) In a triangle, if the square of one side is equal to the sum of the squares on the other two sides, then the angles opposite to the first side is a right angle.
(Prove) In a right triangle, the square on the hypotenuse is equal to the sum of squares of the other two sides.
l
Topic 1. Graphical Solution of Linear
45 - 69
Two Variables
(Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
l
3. Pair of Linear Equations in
*Topic 2. Problems on Polynomials
coefficient of quadratic polynomials
(Motivate) If a perpendicular is drawn from the vertex of a right triangle to the hypotenuse, then the triangles on each side of the perpendicular are similar to each other and to the whole triangle.
230 - 231
l
158 - 185
*
Topic 1. Zeroes of a Polynomial and
7. Triangles
23 - 44
2. Polynomials
157 - 157
Unit 4 : Geometry
Unit 2 : Algebra
Area of Triangle
Self Assessment Test-3
22 - 22
*Topic 2.
Self Assessment Test-1
Non-Terminating Recurring Decimals
Irrational Numbers, Terminating and
*Topic 2.
125 - 156
and section formula
Topic 1. Distance between two points
Topic 1. *Euclid's Division Lemma and Fundamental Theorem of Arithmetic Statements
6. Lines (in two-dimensions)
1 - 21
Unit 1 : Number System 1. Real Numbers
C - Creating
A - Applying
- Understanding,
U
- Remembering,
R
344 - 344
345 - 388
Topic 1. Mean, Median and Mode
*Topic 2. Cumulative Frequency Graph Self Assessment Test-7
389 - 405 406 - 407
15. Probability
involving triangles, simple quadrilaterals and circle"
Self Assessment Test-6
*"Motivate the area of a circle" & "Plane figures
282 - 308
*Topic 3. Frustum of Cone
14. Statistics
281 - 281
Metallic Solid into Another
Unit 7 : Statistics and Probability
258 - 280
Unit 6 : Mensuration 12. Areas Related to Circles
*Topic 2. Conversion of one type of
Topic 2. Trigonometric Identities Self Assessment Test-5
Topic 1. Surface areas and volumes
Complementary Angles
232 - 257
Topic 1. Trigonometric Ratios and
11. Heights and Distances
and Trigonometric Identities
10. Introduction to Trigonometry
309 - 343
13. Surface Areas and Volumes
Unit 5 : Trigonometry
TABLE OF CONTENTS
qq
*Kindly note that this Chapter/Topic has been deleted from the Latest CBSE Syllabus for Academic Year 2022-23. Hence it is optional to study it.
(4)
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WISH TO KNOW WHAT WE HAVE FOR YOU UNDER FREE RESOURCES? HERE YOU GO: Date Sheet
Latest Sample Question Paper
Board Syllabus 2023 Exams
Project Report
Solved Board Paper 2019, 2020, 2021
Lab Activities (Science & Math only)
Toppers’ Answers 2015 to 2020
Flowcharts & Mind Maps
How to decode icons given throughout the book?
Where to report any plausible content or technical error?
Write to us on: [email protected]
Topic-wise & Chapter-wise
and remember to mention the following.
Mnemonics Previous Years’ Board Papers Mind Maps
Full name of the book with its ISBN
Topper’s Answer
Mention the page number and specify the error
Concept Videos
You may also upload an image with the error marked, with a little
Highly Expected Questions
detail of the error.
for the upcoming exam
(6)
LET THE ADVENTURE BEGIN!
The new way of learning; Blended Learning
The pandemic introduced us all to a phenomenon which now seems to be the way forward for learners & teachers alike, it is blended learning. In just a span of a year, we have witnessed a rapid advancement in e-learning. Many researchers say that, in no time e-learning will become mainstream. Oswaal Books identified this as an opportunity and thus we decided to prepare students for this turbulent yet a very useful change with Oswaal 360. Through Oswaal 360 we aim to help the students learn at their own pace. Hence, we wish to make learning adaptive in order to simplify it for every student.
India is currently one of the youngest economies in the world. Hence, it’s imperative for our education system to churn out more learners than ever before. Considering the current learning mindset of students CBSE has recently introduced the curriculum for classes IX to XII containing academic content, syllabus for examinations with learning outcomes, pedagogical practices and assessment guidelines. The Board will conduct the annual scheme of assessment at the end of the Academic session 2022-23 as per the given curriculum.
What Oswaal Question Banks have for you? The structural changes in Education that are being brought in by the latest Syllabus and the new National Educational Policy (NEP) resonate with our approach of focusing on deeper understanding instead of rote learning. We have designed this book to make learning simpler for every student. Our cognitive and innovative exercises make us the leaders of simplified learning. Our study material proves to be so effective because it is based on Bloom’s Taxonomy and also helps retain information for a longer period of time. The last year saw a sudden rise in the need for online classes and self-study. So as blended learning replaces traditional learning methods, Oswaal Books provides the right material to adapt and learn faster. This book takes an intuitive approach for skill development to unleash a student’s maximum potential. Questions like Tabular Based Questions, Passage Based Questions, Picture Based Questions, Fill in the Blanks, Match the Following, etc. have been exclusively developed by the Oswaal Editorial Board to help students master these new Typologies of Questions.
Benefits of studying from Oswaal Question Banks • Previous Years' Questions with Board Marking Scheme Answers • Revision Notes for in-depth study • Modified & Empowered Mind Maps & Mnemonics for quick learning • Chapterwise learning Outcomes & Art integration as per NEP • Include Questions from CBSE official Question Bank released in April 2021 • Unitwise Self-Assessment Tests & Practice Papers • Concept videos for blended learning
Our Heartfelt Gratitude!
We have taken due care in developing this book. There have been a lot of people who have helped us in our journey. We would like to offer heartfelt gratitude to them- our authors, editors, reviewers, and especially students like you who regularly send us suggestions which help us in continuous improvement of this book. Wish you all Happy Learning and a Successful Academic Year 2022-23!! (7)
CBSE CIRCULAR 2022-23 dsUæh; ek/;fed f'k{kk cksMZ
CENTRAL BOARD OF SECONDARY EDUCATION
April 21, 2022
CBSE/Acad/2022
Cir No Acad-48/2022 All Heads of Institutions affiliated to CBSE Subject : CBSE - Secondary and Senior School Curriculum 2022-23
1. CBSE annually provides curriculum for classes IX to XII containing academic content, syllabus for examinations with learning outcomes, pedagogical practices and assessment guidelines.
2. Considering the feedback of stakeholders and other prevailing conditions, the Board will conduct the annual scheme of assessment at the end of the Academic Session 2022-23 and the curriculum has been designed accordingly. Details are available at the link https://cbseacademic.nic.in/ curriculum_2023.html
3. It is important that schools ensure curriculum transaction as per the directions given in the initial pages of the curriculum document. The subjects should be taught as per the curriculum released by the Board with the help of suitable teaching-learning strategies such as Art-Integrated Education, Experiential Learning, and Pedagogical Plans etc. wherever possible.
4. Before making annual pedagogical plan to ensure curriculum transaction for optimal learning, it is desirable that the Head of the School may take a session with all the teachers on the important topics covered in initial pages of the curriculum document as well as the topics covered under subject-wise syllabus.
5. Sample Question Papers with detailed design of the Question Paper will be made available on CBSE’s website in due course of time. 6. Schools are requested to share the curriculum available on https://cbseacademic.nic.in/ curriculum_2023.html including initial pages to all the teachers and students. With Best wishes,
(Dr. Joseph Emmanuel) Director (Academics)
(8)
SYLLABUS
MATHEMATICS STANDARD
Course Structure Class - X (Code No. 041) Latest Syllabus issued by CBSE for Academic Year 2022-23
Unit Name
Marks
I
Number Systems
06
II
Algebra
20
III
Coordinate Geometry
06
IV
Geometry
15
V
Trigonometry
12
VI
Mensuration
10
VII
Statistics Probability
11
Unit No.
Total
80
UNIT I: NUMBER SYSTEMS
(15) Periods
1. REAL NUMBERS
Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality of
2,
3,
5.
UNIT II: ALGEBRA
Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials.
2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
Pair of linear equations in two variables and graphical method of their solution, consistency/ inconsistency.
Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems.
3. QUADRATIC EQUATIONS
2
Standard form of a quadratic equation ax + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots.
(15) Periods
Situational problems based on quadratic equations related to day to day activities to be incorporated.
4. ARITHMETIC PROGRESSIONS
(15) Periods
(8) Periods
1. POLYNOMIALS
th
(10) Periods
Motivation for studying Arithmetic Progression. Derivation of the n term and sum of first n terms of
A.P. and their application in solving daily life problems.
(9)
SYLLABUS UNIT III: COORDINATE GEOMETRY
(15) Periods
1. COORDINATE GEOMETRY
Review : Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).
1. TRIANGLES
UNIT IV: GEOMETRY
Definitions, examples, counter examples of similar triangles.
(15) Periods
1. (Prove) 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.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar.
2. CIRCLES
(10) Periods
5. (Motivate) 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.
Tangent to a circle at, point of contact
1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.
UNIT V: TRIGONOMETRY
Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45° and 60°. Relationships between the ratios.
2. TRIGONOMETRIC IDENTITIES
(10) Periods
1. INTRODUCTION TO TRIGONOMETRY
2
2
(15) Periods
Proof and applications of the identity sin A + cos A = 1. Only simple identities to be given.
(10) Periods
3. HEIGHTS AND DISTANCES: Angle of Elevation, Angle of Depression.
Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.
( 10 )
SYLLABUS UNIT VI : MENSURATION
(12) Periods
1. AREAS RELATED TO CIRCLES
Area of sectors and segments of a circle. 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).
(12) Periods
2. SURFACE AREAS AND VOLUMES
Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.
1. STATISTICS
UNIT VII : STATISTICS AND PROBABILITY
Mean, median and mode of grouped data (bimodal situation to be avoided).
2. PROBABILITY
(18) Periods (10) Periods
Classical definition of probability. Simple problems on finding the probability of an event. qq
( 11 )
SYLLABUS MATHEMATICS-Standard QUESTION PAPER DESIGN CLASS – X (2022-23)
S. No. 1.
Time : 3 Hours
Typology of Questions
3.
Total Marks
% Weightage (approx.)
43
54
19
24
18
22
80
100
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.
2.
Max. Marks : 80
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. 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
10 Marks
Lab Practical (Lab activities to be done from the prescribed books)
05 Marks
qq
05 Marks
( 12 )
Portfolio
20 Marks
Pen Paper Test and Multiple Assessment (5+5)
INTERNAL ASSESSMENT
HOW TO USE THIS BOOK? Chapter Navigation Tools
Chapter Analysis
Syllabus
Mind Maps
Get Concept Clarity
Prescribed by CBSE
Know what’s important from the examination point of view
Latest Typologies of Questions
Topic-wise Segregation
Revision Notes
For Focused Study
Chapter Summary Developed by Oswaal Editorial Board
Passage/ Picture/ Visual & MCQs based questions
Find Exam Oriented Preparation Tools in the Chapter
Mnemonics
QR Codes
Commonly Made Errors
Test
Memorising Better
For Concept Videos
To help you avoid some common errors in your answers
Self Assessment (Unit-wise) Sample Question Papers
Answering Tips
Topper’s Answer
Previous Years’ Board Papers
Academically Important
Tips to Improve your Score
Learn how to write perfect answers
To help you decode the paper pattern
Look out for Highly Expected Questions for the upcoming exams
( 13 )
HEAR IT FROM OUR HAPPY READERS!
This edition is absolutely great for 10th standard student. Full details are given with concept clearty. Must go for it !
This book is a complete guide of Science for class 10 students. It's easily written touching every aspect of each & every topic with a detailed understanding. The book is quite helpful for students from examination point of view.
Rohit S.
Debatrata Mishra
The Question Bank is really helpful for children to understand the subject in a better way. Key notes and questions classified mark wise are some of the best features which I loved about the book.
This book is very useful tool for the 10th standard students. The Mind Maps, Chapter-wise Questions, and the Sample Papers help the students to prepare for the exams thoroughly and also the QR code provided can be referred for further reference.
Simran Bhatia
Aanvi
This book really helps to understand the latest concept that could come up in the latest board exams. A very helpful one. Shivangi Thapa
The book comprises of Revision Notes, Mind Maps and Mnemonics, Chapterwise MCQs and assertion - reasoning. The book is based on latest assessment patterns. Indeed, a must have practice book for students! Aitijhyawearstiara The subject matter of the book exactly sticks to the curriculum and is likely to come in the exam. Very beneficial for the students, I am happy with this purchase This book is great. It's very effective for students to score higher marks. This book is of great help in covid time where the students can study themselves and gets updated with latest syllabus and content. All the topics are covered. Very essential for coming board examination. The answers given are clear and to the point
Kashish Garg If you solve all the papers, you are good to go for your actual exam. You will not even feel the pressure of that one. An excellent book recommended from my side. Ram Kapoor
Sada
( 14 )
I courageuosly move in the direction of my dreams hat I lize w trol a e r n I ot Co d cann the goo t e l and ngs flow thi
I am Sup porte fully by th d e universe
I am wra pp loving en ed in the ergy of universe the I allow m y desires to flow to me now
Affirmations for the new
"YOU"
sly I effortles y m t attrac s e desir
Accept yourself, love yourself, and keep moving forward. If you want to fly, you have to give up what weighs you down.
All of my thoughts are aligned with m y desires
en to I am op riences e p new ex lcome and we ce into n abunda my life
I have the p ower to shift my m indset and see the goo d in everything .
o, I create When I let g mething space for so better.
( 15 )
( 16 )
What are Associations?
To Make clearer and better notes To Concentrate and save time To Plan with ease and ace exams
To Unlock the imagination and come up with ideas To Remember facts and
Associations are one powerful memory aid connecting seemingly unrelated concepts, hence strengthening memory.
move to the subsequent levels of association. This is exactly how the brain functions, therefore these Mind Maps.
level and the chronology continues. The thickest line is the First Level of Association and the lines keep getting thinner as we
from the core concept are the First Level of Association. Then we have a Second Level of Association emitting from the first
It’s a technique connecting the core concept at the Centre to related concepts or ideas. Associations spreading out straight
Learning made simple ‘a winning combination’
Why?
with a blank sheet of paper coloured pens and your creative imagination!
How?
MIND MAP
When?
AN INTERACTIVE MAGICAL TOOL
Result
What?
presenting words and concepts as pictures!!
anytime, as frequency as you like till it becomes a habit!
Learning MaDE SimpLE
MIND MAPS
[ 17
SOLVED PAPER - 2022 (Term-I)
Sample Question Paper, 2021-22 (Issued by CBSE Board on 14th January, 2022) MATHEMATICS STANDARD(Term- II) SOLVED Max. Marks: 40
Time: 2 Hours
General Instructions : 1. 2. 3. 4. 5.
The question paper consists of 14 questions divided into 3 sections A, B, C. All questions are compulsory. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.
Section - A 1. Find the value of a25 - a15 for the AP: 6, 9, 12, 15, ……..
12320 cm². If the radius of its base is 56 cm, then find its height.
OR
If 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term.
5. Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below.
2. Find the value of m so that the quadratic equation mx(5x – 6) + q = 0 has two equal roots.
Marks Obtained
3. From a point P, two tangents PA and PB are drawn to a circle C(0, r).
[2 Marks Each]
Number of Students
If OP = 2r, then find ∠APB. What type of triangle is APB?
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 5
10
–
6
3
6. If Ritu were younger by 5 years than what she really is, then the square of her age would have been 11 more than five times her present age. What is her present age?
OR
Solve for x : 9x² – 6px + (p² – q²) = 0
4. The curved surface area of a right circular cone is
Section - B 7. 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
8. Construct a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Distance (in m) Number of Students
0–1
1–2
2–3
3–4
4–5
40
80
62
38
30
[3 Marks Each]
9. The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Runs scored Number of batsmen
0 – 40 40 – 80 80 – 120 1200 – 160 – 200 160 12
20
35
30
23
18 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
OR
of two poles. (Take
10. Two vertical poles of different heights are standing 20 m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole from the foot of the first pole is 30°. Find the difference between the heights
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take
3 = 1.73)
Section - C
[4 Marks Each] OR
11. The internal and external radii of a spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. Also find the total surface area
3 = 1.73)
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ
22 of the cylinder. ( Take π = ) 7
12. 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 to the centre.
Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower.
(i)
(Lighthouse of Mumbai Harbour. Picture credits Times of India Travel)
(ii)
[2]
A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the
13. Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites.
Case Study-1
After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240( 3 – 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from
boat coming towards the lighthouse and found it to be 30°.
the top of the observation tower?
[2]
Case Study-2
14. Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.
Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour.
[ 19 Form an A.P representing the number of pushups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished? [2]
(i)
Find the total number of push-ups performed by Nitesh up to the day his goal is achieved. [2]
(ii)
But he wants to achieve a target of 3900 push-ups in 1 hour for which he practices regularly. With each day of practice, he is able to make 5 more push-ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved. Keeping the above situation in mind answer the following questions:
CBSE SAMPLE QUESTION PAPER - 2021-22 (Term-II)
nnn
CBSE Marking Scheme 2021-22 (Issued by Board) Section - A
= 6 + (25 – 1) × 3 = 6 + (24) × 3 = 6 + 72 = 78 th 15 term = a15 = 6 + (15 – 1) × 3 = 6 + (14) × 3 = 6 + 42 = 48
Discriminant = 0
⇒ b2 – 4ac = 0 ...(i) Quadratic equation can be written as 5mx2 – 6mx + 9 = 0 Here, a = 5m, b = – 6m, c = 9 Put in (i), (6 m)2 – 4(5 m)(9) = 0 m=0 2 = 36m – 180m = 0 or m = 5 = 36 m(m – 5) = 0 Either m = 0 or m = 5 Since, m = 0 is not possible Hence, m=5
25th term = a25
3.
a25 – a15 = 78 – 48 = 30 OR
Let ∠APO = q sinq =
7(a + 6d) = 5(a + 4d) 7a + 42d = 5a + 20d
(From (i)
= a + 11d
∠APO = q
Radius is always perpendicular to tangent OA ⊥ AP
So,
∠OAP = 90°
=
12 term = a12
Let,
\ OAP is a right angled triangle D
...(i)
2(a + 11d) = 0
Radius of given circle OA = r units
In OAP, D
2a + 22d = 0
=0
(\ PA = PB) ⇒ ∆APB is equilateral [CBSE Marking Scheme, 2021-22]
Detailed Solution:
7a – 5a + 42d – 20d = 0
a + 11d = 0
∠APB = 2q = 60° ∠PAB = ∠PBA = 60°
7a7 = 5a5
⇒ Also
5th term = a5 = a + (5 – 1)d = a + 4d
1 OA = ⇒ q = 30° 2 OP
7th term = a7 = a + (7 – 1)d = a + 6d
nth term = an = a + (n – 1)d
\
Since, mx (5x – 6) + 9 = 0 has equal roots
nth term = an = a + (n – 1)d
According to the Questions,
m = 0, 5 ; rejecting m = 0, we get m = 5 [CBSE Marking Scheme, 2021-22]
Detailed Solution:
Let 'a' and 'd' be the first term and common difference of AP
First term = a = 6 Common Difference = d = 9 – 6 = 3
36m(m – 5) = 0
⇒
t12 = 0 [CBSE Marking Scheme, 2021-22]
6, 9, 12, 15, ....... is the given A.P.
⇒
a + 11d = 0
⇒
(– 6 m)2 – 4 (5 m) (9) = 0
2a + 22d = 0
⇒
for equal roots
7(a + 6d) = 5(a + 4d)
⇒
b2 – 4ac = 0
Since,
a25 – a15 = 30 OR
5mx2 – 6mx + 9 = 0
a25 – a15 = 78 – 48
th
Given
a15 = 6 + 14(3) = 48
2.
a25 = 6 + 24(3) = 78
Detailed Solution:
Hence, 12th term is zero.
a = 6, d = 3 ;
1.
[ 21
sin q =
OA OP
sin q =
1 r = 2r 2
[Length of tangents
∠ABP = ∠BAP = x
Number of students
0 - 20
5
20 - 40
10
40 - 60
x (Say)
60 - 80
6
80 - 100
3
So, from APB, D
PA = PB
[Angle sum property of triangle]
60° + x + x = 180°
⇒
Lower limit of modal class (l) = 40
2
Size of modal class (h) = 20 Frequency corresponding to modal class
Mode = l +
⇒
x − 10 5 = x − 16 2 20
x − 10 × 20 2 x − 16
2 F1 – 16 = 4 F1 – 40
⇒
2 F1 – 4 F1 = – 40 + 16
⇒
x − 10 1 = 2 x − 16 4
f0 = 10, f2 = 6
f 10 45 = 40 + 20 × 1 2 f1 10 6
⇒ ⇒
F1 = 12
– 2 F1 = – 24
h = 20, f1 = = ?,
5 =
⇒
l = 70 cm l2 = h2 + r2 (70)2 = h2 + (56)2 h2 = 4900 – 3136 h2 = 1764 h2 = 422 h = 42 cm
Modal class is 40 – 60, l = 40,
⇒ ⇒ Now, ⇒ ⇒ ⇒ ⇒ ⇒
⇒
12320 × 7 = l 22 × 56
x − 10 × 20 2 x − 10 − 6
22 × 56 × l 7
12320 =
45 = 40 +
⇒
5.
F1 − F0 ×h 2 F1 − F0 − F2
Frequency preceding to modal class (F2) = 6
(F0) = 10
[CBSE Marking Scheme, 2021-22]
Detailed Solution: Let Height of cone = h Radius of cone = 56 cm C.S.A of cone = 12320 cm2 Let slant height of cone = l cm Curved surface area of cone = prl
Frequency preceding to modal class
70 2 − 56 2 = 42 cm
h =
(F1) = x
l = 70 cm
22 × 56 × l = 12320 7
Modal class = 40 – 60
D
CSA (cone) = prl = 12320 cm
\ APB is equilateral triangle.
4.
Mode = 45
x = 60°
or
Marks obtained
∠APB = 2 × ∠APO = 2 × 30° = 60°
∠APB + ∠ABP + ∠BAP = 180°
[CBSE Marking Scheme, 2021-22]
Detailed Solution:
∠APO = 30°
Let,
f1 = 12
q = 30°
from an external point are equal in length]
2f1 – 16 = 4f1 – 40
⇒
⇒
sin q = sin 30°
⇒
f1 − 10 1 = f1 − 16 2 4
⇒
CBSE SAMPLE QUESTION PAPER (TERM-II) 2021-22
22 ]
p+q p−q or 3 3
Now,
[CBSE Marking Scheme, 2021-22]
=
6 p ± 6q −b ± D = 18 2a
x =
= 36q2
= (–6p)2 – 4(9) (p2 – q2)
D = b2 – 4ac
a = 9, b = –6p, c = p2 – q2
2
9x – 6px + (p – q ) = 0
2
=
x =
\
Let present age of Ritu = x years As per question given the following equation can be formed: ⇒ (x –5)2 = 11 + 5x
− ( − 6 p ) ± 36 q 2 −b± D = 2( 9 ) 2a
Detailed Solution:
x=
OR 2
x = 14 years (rejecting x = 1 as in that case Ritu's age 5 years ago will be – ve)
x = 1 or 14
(x – 14) (x – 1) = 0
⇒
x2 – 15x + 14 = 0
(x – 5)2 = 5x + 11
⇒ x2 – 10x + 25 = 11 + 5x 2 ⇒ x – 10x – 5x + 25 – 11 = 0 ⇒ x2 – 15x + 14 = 0 2 ⇒ x – 14x – 1x + 14 = 0 ⇒ x(x – 14) – 1(x – 14) = 0 ⇒ (x – 14) (x – 1) = 0 ⇒ x – 14 = 0, x – 1 = 0 x = 14, x = 1 Since, her age can't be 1, so her present age will be 14 years. OR 9x2 – 6px + (p2 – q2) = 0 In comparing with ax2 + bx + c = 0 a = 9, b = – 6 p, c = p2 – q2 Solve by quadratic formula, D = b2 – 4ac = (– 6p)2 – 4(9)(p2 – q2) = 36p2 – 36p2 + 36q2 = 36q2
Let the present age of Ritu be x years
6.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
6 p ± 6q 18 p+q p−q or 3 3
Section - B
7.
Distance (in m)
0–1
1–2
2–3
3–4
4–5
40
80
62
38
30
40
120
182
220
250
Number of Students
n = 250 2 2 = 125 ⇒ median class is 2 – 3, l = 2, h = 1, cf = 120, f = 62
cf
= 2+
=
50% of students jumped below 2
5 62
5 129 = 2 m or 2.08 m 62 62
5 m and 50% above it. 62
[CBSE Marking Scheme, 2021-22]
n − cf 2 l + ××i h Median = f
[ 23
CBSE SAMPLE QUESTION PAPER (TERM-II) 2021-22
Cumulative frequency
0–1
40
40
1–2
80
120
2–3
62
182
3–4
38
220
4–5
30
250
Draw OA and construct ∠AOB = 120° Draw ∠OBP = ∠OBP = 90° PA and PB are required tangents [CBSE Marking Scheme, 2021-22]
Detailed Solution:
Draw circle of radius 4cm
250 N = = 125 2 2 \ Median class = 2 – 3 Lower limit of median class (l) = 2 Size of median class (h) = 1 Frequency corresponding to median class (f) = 62 Total number of observations Frequency (N) = 250 Cumulative frequency preceding Median class (c.f.) = 120 N − c. f . Median = l + 2 ×h f Take
5 m or 2.08 m 62
\ 50% of students jumped below 2.08 m and 50% of students jumped above 2.08 m.
8.
N = 250
= 2
No. of students (Frequency)
129 62
=
Distance (in m)
Detailed Solution:
9.
Runs Scored
0 – 40
40 – 80
80 – 120
120 – 160
160 – 200
Total
12
20
35
30
23
Sfi =120
xi
20
60
100
140
180
f ix i
240
1200
3500
4200
4140
y – x = 20 3 −
10.
20 40 40 3 = = = 23.06 m 3 3 3
OR
y 20
m
= 88.2 m
3
x =
50
Height of the building = ( 50 3 + 1.7 )m
⇒
20
Let PR be the building and AB be the boy In ∆PQR, tan 60° = PQ ⇒ PQ = 50 3m
x tan30° = 20
In ∆RSQ,
y = 20 3m
⇒
tan60° =
In ∆PQS,
Sfixi =13280
Sfi xi 13280 Mean ( x ) = = = 110.67 runs Sfi 120 [CBSE Marking Scheme, 2021-22]
Number of Batsmen (fi)
5 62
=2+
125 − 120 ×1 62
=2+
Steps of construction: 1. Draw a circle of radius 4 cm. 2. Draw two radii having an angle of 120°. 3. Let the radii intersect circle at A and B. 4. Draw angle of 90° on both A and B. 5. The point where both rays of 90° intersect is P. 6. PA and PB are the required tangents.
[CBSE Marking Scheme, 2021-22]
24 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
Detailed Solution:
= 40 3 . 3 = 23.07m
OR
y 20
In PQA,
∠Q = 90°
3 =
PQ 50
PQ AQ
Total height of the building = PQ + QR
= 86.5 + 1.7 = 88.2 m.
Difference between their heights
50 3 = PQ
x = 20 3 m 3
tan 60° =
=x
3
AQ = 50 m
⇒
D
BR = 50 m
Distance between Boy and building
x 20
=
20
D
tan 30° = RS QS 3
= QR = 1.7 m
∠S = 90°
1
Height of Boy = AB = 1.7 m
3 =
PQ QS
D
tan 60° =
∠Q = 90°
y = 20 3 m
In RSQ,
20 m 3 3
= 20 3 ×2 3
Distance between the poles is QS = 20 m.
In PQS,
Let the heights of two pole be y and x.
y – x = 20 3
14 = 7 cm 2
Let height of cylinder = h cm
2 8 = 2 cm 3 3
According to the question, [When one shape is reshaped into
TSA of cylinder is
Radius of cylinder (R1) =
h=
⇒
4π 3 2 [ 5 − 33 ] = p(7) h 3
⇒
another shape the volumes are
= 2pr(r + h) 22 8 × 7 × 7 + 7 3
same of both shapes] 4 p (R3 – r3) = p(R1)2h 3
29 3
1276 = cm2 or 425.33 cm2 3 [CBSE Marking Scheme, 2021-22]
= 44×
= 2×
Volume of shell = Volume of cylinder
11.
Section - C
Internal radius (r) = 3 cm External radius (R) = 5 cm
Detailed Solution:
4 3 3 (5 – 3 ) = (7)2h 3 4 × (125 – 27) = 49h 3 4 98 =h × 3 49
[ 25
\
OP ⊥ TP[Radius and Tangent are perpendicular to each other]
∠OPT = 90°
⇒
∠TPQ = 90° – ∠OPQ D
In PTQ,
∠TPQ + ∠TQP + ∠PTQ = 180°
[From (i)]
∠TPQ + ∠TPQ + ∠PTQ = 180°
By angle sum property of triangle,
⇒
2(90° – ∠OPQ) + ∠PTQ = 180°
⇒
180 – 2∠OPQ + ∠PTQ = 180°
[From (ii)]
2[TPQ + ∠PTQ = 180°
⇒
∠PTQ = 2∠OPQ
Detailed Solution:
2∠OPQ = ∠PTQ [CBSE Marking Scheme, 2021-22]
...(ii)
∠OPQ + ∠TPQ = 90°
1 ∠OPQ = ∠PTQ 2
θ 2
=
Now, PT is a tangent & OP is radius
θ = 90° − 90° − 2
[Angles opposite to equal sides are equal]
∠OPT = 90° ∠OPQ = ∠OPT – ∠TPQ
θ 2
...(i)
= 90° −
1 (180° – q) 2
∠TQP = ∠TPQ
∠TPQ = ∠TQP =
\
Let ∠PTQ = q TPQ is an isosceles triangle
∠OAP + ∠OBP + ∠APB + ∠AOB = 360° ⇒ 90° + 90° + ∠APB + ∠AOB = 360° (Q Tangent ⊥ radius) ⇒ ∠APB + ∠AOB = 180° OR
Proof: oOAPB is a quadrilateral, OA is radius and PA & PB are tangents. \ ∠OAP = ∠OBP = 90° [Radius is always perpendicular to tangent] By angle sum property of quadrilateral. \ ∠AOB + ∠OBP + ∠APB + ∠OAP = 360° ∠AOB + 90° + ∠APB + 90° = 360° \ ∠APB + ∠AOB = 180° Hence Proved. OR Given: A circle with centre O two tangents TP and TQ to the circle where P and Q are the point of contact. To Prove: ∠PTQ = 2∠OPQ Proof: TP = TQ [length of tangents drawn from same external point to a circle are equal]
∠APB + ∠AOB = 180°
1276 cm2 3
12.
To Prove:
=
29 3
= 44 ×
22 8 =2× × 7 7 7 3
or 425.33 cm2
Given: PA and PB are two tangents to a circle with centre O.
TSA of cylinder = 2pr (h + R1)
8 m or 2 2 cm 3 3
h=
CBSE SAMPLE QUESTION PAPER (TERM-II) 2021-22
Hence Proved.
26 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
13.
Case Study-1 1
(i)
=
240 x
x = 240 3 m.
3
Distance of foot from the foot of observation tower
Distance between boat and light house is reduced by = 24 3 1 m
= 240 3 − 240( 3 − 1)
(ii) After 10 minutes,
(ii) Distance of boat from tower
In ∆PTR,
= 240 3 m. 240 tan 30° = ⇒ x = 240 3m m x
= 240 m Let the angle of depression = q 240 tanq = =1 240
q = 45° [CBSE Marking Scheme, 2021-22]
⇒
= 240 3 – 240 3 1
= 240
3 – 240
3 + 240
= 240 m
Let the point C is the new position of boat,
tan q =
⇒
AB 240 = BC 240
q = 45°
240 x
D
∠R = 90° tan 30° =
In PRT,
(i) Height of light house = 240 m
\ Distance between boat and light house
Detailed Solution:
Case Study-2
nth term = an = a + (n – 1)d
To Find n
d=5 nth term (an) = 3900
Common Difference = d = 3005 – 3000
According to the problem
= 624450 push-ups [CBSE Marking Scheme, 2021-22]
181 × ( 3000 + 3900 ) 2
n (a+ l) 2
Sn =
= 180 days
=
First term = a = 3000
n = 181
Minimum number of days of practice = n – 1
(ii)
3000, 3005, 3010, ...... 3900
900 = 5n – 5 ⇒ 5n = 905
⇒
(i) By the given situation the A.P. to be formed is:
3900 = 3000 + (n – 1)5
3900 = 3000 + (n – 1)5
⇒
Detailed Solution:
an = a + (n – 1)d
14. (i) 3000, 3005, 3010, ..., 3900
[ 27
n = number of days = 181
= 3000 l = last term = 3900
⇒
Put all values in (i) Sn =
...(i)
=
181 [3000 + 3900] 2 181 [6900] 2
= 624450
Total number of Push-ups performed in 181 days
\
n [a + l] Sn = 2
(ii) Total Number of push-ups performed means sum of all push-ups he did in 181 days.
= 180
= 181 – 1 [Excluding the last day]
Minimum number of days he needs to practice before his goal is accomplished
n = 181
or
a = first term of the A.P.
180 + 1 = n
900 =n–1 5
181 days.
900 = (n – 1)5
Sn = Sum of all push-ups in
3900 – 3000 = (n – 1)5
CBSE SAMPLE QUESTION PAPER (TERM-II) 2021-22
= 624450. lll
Solved Paper, 2021-22 MATHEMATICS (STANDARD) Term-I, Set-4 Question Paper
Series : JSK/2
Code No. 030/2/4 Max. Marks : 40
Time allowed : 90 Minutes
General Instructions :
(i) The question paper contains 50 questions out of which 40 questions are to be attenpted. All questions carry equal marks.
(ii) The question paper consists of three sections – Section A, B and C.
(iv) Section–B also contains 20 questions. Attempt any 16 questions from Q. No. 21 to 40.
(vi) There is only one correct option for every Multiple Choice Question (MCQ). Marks will not be awarded for answering more than one option.
(v) Section–C contains of two Case Studies containing 5 questions in each case. Attempt any 4 questions from Q. No. 41 to 45 and another 4 from Q. No. 46 to 50.
(iii) Section–A contains of 20 questions. Attempt any 16 questions from Q. No. 1 to 20.
(vii) There is no negative marking.
SECTION-A Q. No. 1 to 20 are of 1 mark each. Attempt any 16 from Q. No. 1 to 20.
ar(DABC) : ar(DADE) is equal to
In figure, DE || BC, AD = 2 cm and BD = 3 cm, then
(d) – 2
(c) – 3
(b) –3
(a) 2
(a) 2464 m2 (b) 1232 m2 2 2 (c) 616 m (d) 308 m 8. For an event E, P(E) + P( E ) = x, then the value of x3 – 3 is (a) –2 (b) 2 (c) 1 (d) –1 9. 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. 10. In figure, the graph of a polynomial P(x) is shown. The number of zeroes of P(x) is
If A(3, 3 ), B(0, 0) and C(3, k) are the three vertices of an equilateral triangle ABC, then the value of k is
7.
(d) –1, –4
5.
1 2 2 , the value of sec q + cosec q is 3 40 (a) 1 (b) 9 38 1 (c) (d) 5 9 3 The area of a quadrant of a circle where the circumference of circle is 176 m, is If cot θ =
(c) 1, –2
(b) –1, 4
(a) –1, 2
The values of x and y satisfying the two equations 32x + 33y = 34, 33x + 32y = 31 respectively are
(b) 2 : 3 (d) 25 : 4
(d) 5
6.
(c) 3
(b) 2
(a) 1
(a) 4 : 25 (c) 9 : 4
The graph of a polynomial P(x) cuts the x-axis at 3 points and touches it at 2 other points. The number of zeroes of P(x) is
(d) 6
(c) 5
3.
(b) 4
2.
4.
(a) 3
The exponent of 5 in the prime factorisation of 3750 is
1.
[ 29
SOLVED PAPER - 2021-22 (TERM-I)
SECTION-B
23. The diameter of a car wheel is 42 cm. The number of
complete revolutions it will make in moving 132 km is
28.
27.
26.
(a) 44 cm (b) 88 cm (c) 132 cm (d) 176 cm 16. The probability that the drawn card from a pack of 52 cards is neither an ace nor a spade is 35 9 (a) (b) 52 13 19 10 (c) (d) 26 13 17. Three alarm clocks ring their alarms at regular intervals of 20 min., 25 min. and 30 min. respectively. If they first beep together at 12 noon, at what time will they beep again for the first time?
covered by the tip of minute hand from 10:10 am to 10:25 am is
25.
15. The minute hand of a clock is 84 cm long. The distance
24 7
24.
(d)
7 24
(c)
7 14. In DABC right angled at B, sin A = , then the value 25 of cos C is 7 24 (a) (b) 25 25
5
(a) 10 (b) 10 6 3 (c) 10 (d) 10 If q is an acute angle and tan q + cot q = 2, then the 3 3 value of sin q + cos q is 1 (a) 1 (b) 2 2 (c) (d) 2 2 The ratio in which the line 3x + y – 9 = 0 divides the line segment joining the points (1, 3) and (2, 7) is (a) 3 : 2 (b) 2 : 3 (c) 3 : 4 (d) 4 : 3 If x – 1 is a factor of the polynomial p(x) = x3 + ax2 + 2b and a + b = 4, then (a) a = 5, b = –1 (b) a = 9, b = –5 (c) a = 7, b = –3 (d) a = 3, b = 1 If a and b are two coprime numbers, then a3 and b3 are (a) Coprime (b) Not coprime (c) Even (d) Odd The area of a square that can be inscribed in a circle of 1408 2 area cm is 7 (a) 321 cm2 (b) 642 cm2 2 2 (c) 128 cm (d) 256 cm
4
1 13. In DABC and DDEF, ∠F = ∠C, ∠B = ∠E and AB = 2 DE. Then the two triangles are (a) Congruent, but not similar (b) Similar, but not congruent (c) Neither congruent nor similar (d) Congruent as well as similar
Q. No. 21 to 40 are of 1 mark each. Attempt any 16 from Q. 21 to 40. 21. The greatest number which when divides 1251, 9377 and 15628 leaves remainder 1, 2 and 3 respectively is (a) 575 (b) 450 (c) 750 (d) 625 22. Which of the following cannot be the probability of an event? (a) 0.01 (b) 3% 17 16 (c) (d) 16 17
(a) 1 (b) 2 (c) 3 (d) 4 11. Two lines are given to be parallel. The equation of one of the lines is 3x – 2y = 5. The equation of the second line can be (a) 9x + 8y = 7 (b) –12x – 8y = 7 (c) –12x + 8y = 7 (d) 12x + 8y = 7 12. Three vertices of a parallelogram ABCD are A(1, 4), B(–2, 3) and C(5, 8). The ordinate of the fourth vertex D is (a) 8 (b) 9 (c) 7 (d) 6
Y
X
X
(a) 4 : 00 pm (b) 4 : 30 pm (c) 5 : 00 pm (d) 5 : 30 pm 18. A quadratic polynomial, the product and sum of whose zeroes are 5 and 8 respectively is 2 2 (a) k[x – 8x + 5] (b) k[x + 8x + 5] 2 2 (c) k[x – 5x + 8] (d) k[x + 5x + 8] 19. Points A(–1, y) and B(5, 7) lie on a circle with centre O(2, –3y). The values of y are (a) 1, –7 (b) –1, 7 (c) 2, 7 (d) –2, –7 1 + tan θ 20. Given that sec q = 2 , the value of is sin θ (a) 2 2 (b) 2 (c) 3 2 (d) 2
Y
30 ]
(b)
p2 + 1 2p
p2 − 1 2p
(a)
(a) 107° (b) 135° (c) 155° (d) 145° 39. If sec q + tan q = p, then tan q is
p2 − 1
p2 − 1
)
3 , 0 , B(0, 3)
(b)
A(±3 3 , 0), B(3, 0)
(d)
A(– 3 , 0), B(3, 0)
(
(a) A
(c) A(±3 3 , 0), B(0, 3)
p2 + 1
p2 + 1
(d)
40. The base BC of an equilateral DABC lies on the y-axis. The co-ordinates of C are (0, –3). If the origin is the mid-point of the base BC, what are the co-ordinates of A and B?
(c)
Q. No. 41-45 are based on Case Study-I, you have to answer any (4) four questions. Q. No. 46-50 are based on Case Study-II, you have to answer any (4) four questions. Case Study-I A book store shopkeeper gives books on rent for reading. He has variety of books in his store related to fiction, stories and quizzes etc. He takes a fixed charge for the first two days and an additional charge for subsequent day. Amruta paid `22 for a book and kept for 6 days; while Radhika paid `16 for keeping the book for 4 days.
SECTION-C
29. If A(4, –2), B(7, –2) and C(7, 9) are the vertices of a DABC, then DABC is (a) equilateral triangle (b) isosceles triangle (c) right angled triangle (d) isosceles right angled triangle 30. If a, b are the zeroes of the quadratic polynomial 2 p(x) = x – (k +6) x + 2(2k – 1), then the value of k, if 1 a + b = ab, is 2 (a) –7 (b) 7 (c) –3 (d) 3 31. If n is a natural number, then 2(5n + 6n) always ends with (a) 1 (b) 4 (c) 3 (d) 2 32. The line segment joining the points P(–3, 2) and Q(5, 7) is divided by the y-axis in the ratio (a) 3 : 1 (b) 3 : 4 (c) 3 : 2 (d) 3 : 5 33. If a cot q + b cosec q = p and b cot q + a cosec q = q, then p2 – q2 = (a) a2 – b2 (b) b2 – a2 (c) a2 + b2 (d) b – a 34. If the perimeter of a circle is half to that of a square, then the ratio of the area of the circle to the area of the square is (a) 22 : 7 (b) 11 : 7 (c) 7 : 11 (d) 7 : 22 35. A dice is rolled twice. The probability that 5 will not come up either time is 11 1 (a) (b) 36 3 25 13 (c) (d) 36 36 36. The LCM of two numbers is 2400. Which of the
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
following CAN NOT be their HCF? (a) 300 (b) 400 (c) 500 (d) 600 37. In figure, PQ, QB and RC are each perpendicular to AC. If x = 8 cm and z = 6 cm, then y is equal to
o
38. In a DABC, ∠A = x , ∠B = (3x – 2) , ∠C = y . Also ∠C o – ∠B = 9 . The sum of the greatest and the smallest angles of this triangle is
o
24 cm 7
o
7 cm 56
(d)
25 cm 7
(c)
(b)
56 cm 7
(a)
Assume that the fixed charge be `x and additional charge (per day) be `y. Based on the above information, answer any four of the following questions: 41. The situation of amount paid by Radhika, is algebraically represented by (a) x – 4y = 16 (b) x + 4y = 16 (c) x – 2y = 16 (d) x + 2y = 16 42. The situation of amount paid by Amruta, is algebraically represented by (a) x – 2y = 11 (b) x – 2y = 22 (c) x + 4y = 22 (d) x – 4y = 11 43. What are the fixed charges for a book? (a) `9 (b) `13 (c) `10 (d) `15
[ 31
D
49.
50.
O
48.
47.
Case Study-II A farmer has a field in the shape of trapezium, whose map with scale 1 cm = 20 m, is given below: The field is divided into four parts by joining the opposite vertices. 5 cm A B
C 10 cm Based on the above information, answer any four of the following questions: 46. The two triangular regions AOB and COD
(a) Similar by AA criterion (b) Similar by SAS criterion (c) Similar by RHS criterion (d) Not similar The ratio of the area of the DAOB to the area of DCOD, is (a) 4 : 1 (b) 1 : 4 (c) 1 : 2 (d) 2 : 1 If the ratio of the perimeter of DAOB to the perimeter of DCOD of would have been 1 : 4, then (a) AB = 2CD (b) AB = 4CD (c) CD = 2AB (d) CD = 4AB AO AD OD If in Ds AOD and BOC, , then = = BC BO OC (a) DAOD ~ DBOC (b) DAOD ~ DBCO (c) DADO ~ DBCO (d) DODA ~ DOBC If the ratio of areas of two similar triangles AOB and COD is 1 : 4, then which of the following statements is true? (a) The ratio of their perimeters is 3 : 4 (b) The corresponding altitudes have a ratio 1 : 2 (c) The medians have a ratio 1 : 4 (d) The angle bisectors have a ratio 1 : 16
44. What are the additional charges for each subsequent day for a book? (a) `6 (b) `5 (c) `4 (d) `3 45. Which is the total amount paid by both, if both of them have kept the book for 2 more days? (a) `35 (b) `52 (c) `50 (d) `58
SOLVED PAPER - 2021-22 (TERM-I)
Solutions
Explanation: According to the property of the polynomials, Number of zeroes = Number of points at which graph intersects the x-axis. It is mentioned in the question that, the graph intersects x-axis at 3 points and it touches it at 2 further points. This means that the graph intersects the x-axis at 5 different points. Therefore, number of zeroes = 5. (a) –1, 2
Explanation: The given equations are, 32x + 33y = 34 & 33x + 32y = 31 Subtract eq.(ii) from eq.(i) –x + y = 3
3.
...(i) ...(ii)
4.
(d) 5
2.
Explanation: According to the prime factorisation, 3750 can be written as 3750 = 5 × 5 × 5 × 5 × 3 × 2 = 54 × 31 × 21 It is clear from above, that exponent of 5 in the prime factorisation of 3750 is 4.
(b) 4
1.
or y =3+x Put this value of y in (i), we get 32x + 33(3+x) = 34 ⇒ 32x + 99 + 33x = 34 ⇒ 65x = 34 – 99 ⇒ 65x = –65 or x = –1 Also, y =3+x ⇒ y = 3 + (–1) =3–1=2 Hence, the correct solution is x = –1 and y = 2.
SECTION-A
(c) 3 Explanation: A(3, 3 ), B(0, 0) and C(3, k) are the vertices of the triangle ABC.
32 ]
6.
Then, apply distance formula for sides AB and BC.
k =± 3
22 × × 28 × 28 360 o 7 = 616 cm2
(a) –2
8.
90 o
=
Explanation:
Also, in a quadrant q = 90o θ × πr 2 Area of quadrant = 360 o
(d) 25 : 4
1 4 16 = =5 3 3 3
2
k =3
or
2
12 = 9 + k
=4+
9 + k2
or
2
Explanation: It is given that circumference of the circle is 176 cm2 ⇒ 2pr = 176 22 ⇒ 2× × r = 176 7 176 × 7 ⇒ r= =28 cm 2 × 22
12 =
2 2 = (2) + 3
(c) 616 m2
7.
9 + k 2 units
( 3 − 0 )2 + ( k − 0 )2
AB = BC
or
Explanation: Given
10. (c) 3
( 2 + 3 )2 52 25 = 2 = = 4 2 ( 2 )2
AB2 ar ( ∆ABC ) = ar ( ADE ) AD 2
According to the theorem, “Ratio of area of similar triangles is equal to the square of the ratio of corresponding sides.“
Explanation: As the girl needs to walk 95 m and 171 m at the exact number of minutes. So, we have to find HCF of 95 and 171. According to prime factorisation of 95 and 171 95 = 5 × 19 171 = 3 × 3 × 19 HCF(95, 171) = 19 Hence, greatest possible speed is 19 m/min.
By, AA similarity rule ∆ABC ~ ∆ADE
\
(b) 19 m/min.
9.
[Corresponding angles]
∠ADE = ∠ABC
[Common]
...(i)
P(E) + P ( E ) = 1 ...(ii) From (i) and (ii), we get x =1 3 Put value of x in x – 3, we get 3 3 x – 3 = (1) – 3 = 1 – 3 = –2
First we need to prove that ∆ABC & ∆ADE are similar.
∠A = ∠A
P(E) + P ( E ) = x Also, according to the law of probability,
12 units
=
Now,
5.
BC =
9+3 =
2
=
( 3 − 0 )2 + ( 3 − 0 )
⇒ q = 60o Substituting the value of q 2 2 2 o 2 o sec q + cosec q = sec 60 + cosec 60
( x2 − x1 )2 + ( y2 − y1 )2
AB =
Explanation: It is given that 1 cot q = = cot 60o 3
According to the distance formula, d=
1 3
(d) 5
As in the equilateral triangle ABC all sides are equal.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
Explanation: According to the property of the polynomials, Number of zeroes = Number of points at which graph intersects the x-axis.
[ 33
SOLVED PAPER - 2021-22 (TERM-I)
In the question value of ordinate is asked, 3+b 2 12 = 3 + b b =9
X
13. (b) Similar, but not congruent
X
or
6=
Y
Explanation: According to the definition of similarity of two triangles, “Two triangles are similar when their corresponding angles are equal and the sides are in proportion” Y From the figure it is clear that the graph intersects X-axis at three different points. Therefore, the polynomial has 3 zeroes. 11. (c) –12x + 8y = 7
[Given]
14. (a)
AB 1 = DE 2 Which means the triangles are similar but not congruent. Also
7 25
Explanation:
Taking option (C) and applying the above condition on it and in the given equation. 3 −2 5 ≠ or = −12 8 7
Explanation: The given equation is 3x – 2y = 5 According to the condition that if two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel a1 b1 c1 then = ≠ a 2 b2 c 2
According to the question, ∠F = ∠C and ∠B = ∠E 1 Since, AB = DE 2
12. (b) 9
Explanation: Let A(1, 4) B(–2, 3) C(5, 8) and D(a, b) are the vertices of a parallelogram. Midpoint of diagonal AC x + x 2 y1 + y 2 , = 1 2 2
cos q =
Base Hypotenuse
sin A =
BC 7 = AC 25
cos C =
BC 7 = AC 25
and
\
15. (c) 132 cm
−2 + a 3 + b , = 2 2 The diagonals of the parallelogram bisect each other. The diagonals share same mid-point.
Midpoint of diagonal BD x + x 2 y1 + y 2 , = 1 2 2
Perpendicular Hypotenuse
sin q =
6 12 = , = (3, 6) 2 2
1+ 5 4 +8 , = 2 2
3=
−2 + a 3 + b , \ (3, 6) = 2 2 On comparing both sides, we get −2 + a 3+b and 6 = 2 2
Explanation: Length of minute hand = Radius of the quadrant/sector so formed = 84 cm. In 1 minute, minute hand makes an angle of 6o. Therefore, in 15 minutes it makes an angle of o o 15 × 6 = 90
34 ]
Explanation: As points A and B lie on the circle and O is the centre. AO and BO will be the radii of the circle.
π
90 o
=
o
×2×
22 × 84 7
360 = 132 cm.
9 13
( 2 − ( −1) )2 + ( −3 y − y )2
⇒
( 2 − 5 )2 + ( −3 y − 7 )2
( x2 − x1 )2 + ( y2 − y1 )2
⇒ (3)2 + (–4y)2 = (–3)2 + (–3y – 7)2 ⇒ 9 + 16y2 = 9 + 9y2 + 49 + 42y 2 ⇒ 16y – 9y2 – 42y – 49 = 0 ⇒ 7y2 – 42y – 49 = 0 ⇒ 7(y2 – 6y – 49) = 0 2 ⇒ y – 7y + 1y – 49 = 0 ⇒ y(y – 7) + 1(y – 7) = 0 ⇒ (y – 7)(y + 1) = 0 ⇒ y = 7, –1
d=
Explanation: Time when they ring together = LCM (20, 25, 30) According to prime factorisation, 20 = 2 × 2 × 5 25 = 5 × 5 30 = 2 × 3 × 5 LCM (20, 25, 30) = 2 × 2 × 3 × 5 × 5 = 300 Thus, 3 bells ring together after 300 minutes or 5 hours. Since, they rang together first at 12 noon, then they ring together again at 5 pm
=
(Applying distance formula on both AO and BO)
17. (c) 5 : 00 pm
AO = BO
So,
Explanation: Total ace cards = 4 and total spade cards = 13 – 1 = 12 (One card among aces is also a spade) Cards which are neither ace or spade = 52 – 16 = 36 36 9 = Required probability = 52 13
16. (a)
19. (b) –1, 7
Distance covered by the tip of the minute hand = Length of arc θ = ×2 r 360 o
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
20. (a) 2 2
18. (a) k[x2 – 8x + 5]
...(ii)
1 + tan θ 1 + tan 45o = sin θ sin 45o
or b =–8k, a = 1k
...(i)
⇒
−b 8 = a 1
Also, sec 45 = 2 From (i) and (ii), we get o q = 45 1 + tan θ Put value of q in , sin θ
o
−b a
8=
Explanation: It is given that sec q = 2
Explanation: For any quadratic polynomial, ax2 + bx + c −b Sum of zeroes = a
2 2
21. (d) 625
Explanation: First subtract the remainders from their respective numbers, 1251 – 1 = 1250 9377 – 2 = 9375 15628 – 3 = 15625
1+1 1 2
SECTION-B
or
5 c 1 = a
or c = 5k, a = 1k Polynomial whose sum of zeroes or product of zeroes are given, 2 Required Polynomial = ax + bx + c 2 = kx – 8kx + 5k 2 = k(x – 8x + 5)
⇒
c a
5=
c a
Also, product of zeroes =
[ 35
SOLVED PAPER - 2021-22 (TERM-I)
Explanation: Probability of an event is always a proper fraction. Also, 0 ≤ P(E) ≤ 1 17 But >1 16
= 2 2−
17 16
Now taking, 3 3 3 sin q + cos q = (sin q + cos q) – 3sin q cos q (sin q + cos q) 1 3 = ( 2) − 3× × 2 2 2 3 2 = 2 2
25. (c) 3 : 4
22. (d)
According to the prime factorisation, 1250 = 2 × 5 × 5 × 5 × 5 9375 = 3 × 5 × 5 × 5 × 5 × 5 15625 = 5 × 5 × 5 × 5 × 5 × 5 HCF(1250, 9375, 15625) = 5 × 5 × 5 × 5 = 625
Explanation: Let the the point of intersection be M(x, y).
17 Therefore, can never be probability of any 16 event.
23. (b) 105 Explanation: Diameter of wheel = 42 cm 42 Radius of the wheel = = 21 cm 2 Distance in 1 revolution = Circumference of the wheel = 2pr 22 =2× ×3 7 = 132 cm Total distance covered by the wheel = 132 km = 132 × 100000 cm = 13200000 cm Number of revolutions Total distance covered by wheel = Distance covered in 1 revolution
2
(sin q + cos q) = sin q + cos q + 2sinqcosq 1 =1+2× 2 =1+1=2 Therefore, sin q + cos q = 2 ...(ii)
2
...(i)
2
sin q + cos q = 2sin q cos q 1 = 2sin q cos q 1 sin q cos q = 2
Explanation: Given, p(x) = x3 + ax2 + 2b a+b =4 ...(i) x – 1 is a factor of the polynomial P(x), which means x = 1 is a zero of the polynomial p(x). \ p(1) = 0 or (1)3 + a(1)2 + 2b = 0 or 1 + a + 2b = 0 or a + 2b = –1 ...(ii) Subtracting (i) from (ii), we get b = –5 Substituting the value of b in (i), we get a = 9 \ a = 9 & b = –5
26. (b) a = 9, b = –5
sin 2 θ + cos2 θ =2 cosθ sin θ
2
6k + 3 + 7k + 3 – 9(k+1) = 0 4k – 3 = 0 3 k= 4
The ratio is k : 1 or 3 : 4.
or
or
or or
2
or or
2k + 1 7 k + 3 , = k +1 k +1
2k + 1 7 k + 3 Therefore, 3 –9=0 + k +1 k +1
Explanation: tan q + cot q = 2 sin θ cos θ + or =2 cos θ sin θ or
k ( 2 ) + 1 (1) k (7 ) + 1 (3) , = k +1 k + 1
This point M lies on the line l.
2 2
24. (c)
m x + m2 x1 m1 y 2 + m2 y1 , M(x, y) = 1 2 m1 + m2 m1 + m2
13200000 =100000 = 105 132
=
Let the line l divides the line AB in the ratio k : 1. According to the section formula,
36 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
27. (a) Coprime
[7 − 4 ]2 + 9 − ( −2 )
=
32 + 112
=
9 + 121 = 129
2
AC =
Clearly, they are not equilateral or isosceles. Also, AC2 = AB2 + BC2 Which mean it is following Pythagoras theorem. \ DABC is a right angled triangle.
HCF(a3, b3) = HCF(27, 64) = 1
Then,
Explanation: As a and b are co-prime then a3 and b3 are also co-prime. We can understand above situation with the help of an example. Let a= 3 and b = 4 a3 = 33 = 27 and b3 = 43 = 64 Clearly, HCF(a, b) = HCF(3, 4) = 1
30. (b) 7
28. (c) 128 cm2
AB =
+ ( y 2 − y1 )
[7 − 4 ]2 + −2 − ( −2 ) 32 + 0 = 3
2
32. (d) 3 : 5 Explanation: Let the point on y-axis which divides the line PQ is M(0, y) and the ratio be k : 1. According to the section formula, m x + m2 x1 m1 y 2 + m2 y1 , M(x, y)= 1 2 m1 + m2 m1 + m2
5k + ( −3 ) k ( 7 ) + 1 ( 2 ) , M(0, y) = k +1 k +1
0 + 112 = 11
2
=
[7 − 7 ]2 + 9 − ( −2 )
BC =
=
( x2 − x1 )
2
16 256 = =128 cm2 2 2
Explanation: A(4, –2), B(7, –2) and C(7, 9) are the vertices of a triangle. Using distance formula, d=
2
2
Explanation: Let us take an example of different powers of 5. As, 51 = 5 52 = 25 53 = 125 54 = 625 ……. n It is clear from above example that 5 will always end with 5. Similarly, 6n will always end with 6. n n So, 5 + 6 will always end with 5 + 6 = 11 n Also, 2(5 + 6n) always ends with 2 × 11 = 22 i.e., it will always end with 2.
( diagonal of square )2
=
31. (d) 2
29. (c) right angled triangle
1 ab 2
1 2(2k – 1) 2 k + 6 = 2k – 1 –k = –7 k =7
⇒ ⇒ or
2
2 ( 2 k − 1) = 2(2k – 1) 1
k+6 =
1408 7 × 7 22
Area of square =
r2 =
r = 64 = 8 cm Diameter of circle = 2r = 16 cm As square is inscribed in the circle, diameter of circle = diagonal of square = 16 cm
or
or
a+b =
⇒
or
It is given that,
[Given]
1408 pr = 7 22 1408 2 ×r = 7 7 2
or
=
1408 2 cm 7
Area of circle =
Explanation: p(x) = x2 – (k + 6)x + 2(2k – 1) is the given polynomial Here, a = 1, b = –(k + 6) & C = 2(2k – 1) Sum of zeroes = a + b −b = a =k+6 Product of zeroes = ab c = a
Explanation:
[ 37
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36. (c) 500
On comparing, we get 5k − 3 k +1 5k – 3 = 0 3 k= 5
Explanation: According to the property, HCF of two numbers is also a factor of LCM of same two numbers. Out of all the options, only (C) 500 is not a factor of 2400. Therefore, 500 cannot be the HCF.
2
or
or
0=
2
33. (b) b – a
37. (d)
24 cm 7
Explanation:
Explanation: a cot q + b cosec q = p and b cot q + a cosec q = q are the given equations. Taking, p2 – q2 = (a cot q + b cosec q)2 – (b cot q + a cosec q)2 = a2cot2 q + b2cosec2 q + 2ab.cot q.cosec q – b2cot2 q – a2cosec2 q – 2ab.cot q.cosec q 2 2 = a (cot q – cosec2q) + b2 (cot2q – cosec2q) = a2(–1) + b2(–1) = b2 – a2 [using, cosec2q – cot2q = 1]
34. (d) 7 : 22
[Since, DABQ ~ DACR]
(2 2)
(2 3)
(2 4)
(2 5)
(2 6)
y y AC + = x z AC
(3 1)
(3 2)
(3 3)
(3 4)
(3 5)
(3 6)
(4 1)
(4 2)
(4 3)
(4 4)
(4 5)
(4 6)
⇒
y y + =1 x z
(5 1)
(5 2)
(5 3)
(5 4)
(5 5)
(5 6)
(6 1)
(6 2)
(6 3)
(6 4)
(6 5)
(6 6)
⇒
1 1 1 + = x z y
Put x = 8 and z = 6
1 = 1+1 y 8 6
=
⇒
Total events = 36 Out of the events in which 5 will not come up either time are (1, 1) (1, 2) (1, 3) (1, 4) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 6). No. of required events in = 25
⇒
(2 1)
(1 6)
(1 5)
(1 4)
(1 3)
(1 2)
y y AB + BC + = x z AC
⇒
(1 1)
y y CB AB + = + x z AC AC
Explanation: All possible events are written below:
25 36
Adding (i) and (ii), we get
25 35. (d) 36
Required probability =
1 7 or π 22
=
y=
π
2
1
=p×
⇒
y AB = z AC
BQ AB = AP AC
...(i)
y BQ = x CR
⇒
Area of the circle πr 2 = 2 Area of the square a
⇒
In DACR, we have BQ || CR
r 1 = a π
or
BQ CB = AP CA [Since, DCBQ ~ DCAP]
2a r= 2π
⇒
⇒
Explanation: Let radius of the circle be r cm and side of the square is a cm. According to the question, perimeter of the circle is half of perimeter of the square. 1 ⇒ 2pr = (4a) 2
14 7 = 48 24 24 7
38 ]
Using distance formula,
o
o
BC =
( 0 − 0 )2 + ( −3 − 3 )2
or
x = ±3 3
x 2 + 9 = 36 x2 = 27
x2 + 9
(
)
Coordinates of A and B are ± 3 3 , 0 and (0, 3) respectively.
( 0 − x )2 + ( 3 − 0 )2
= 36 BC = AB
=
SECTION-C
2
p −1 2p
Case Study-I 41. (d) x + 2y = 16
39. (b)
AB =
Also,
Explanation: ∠ = x , ∠B = 3x – 2 and ∠C = y Sum of angles in a triangle is 180o. Therefore, x + 3x – 2 + y = 180o or 4x + y = 182 ...(i) o Also, ∠C – ∠B = 90 or y – (3x – 2) = 90o o or y – 3x = 70 ...(ii) Subtracting (ii) from (i), we get 7x = 175 or x = 250 Put x = 25o in (ii), we get y = 82o Therefore, o o ∠A = 25 , ∠B = 3x – 2 = 3(25) – 2 = 73 o And ∠C = y Sum of greatest and smallest angle = 82o + 25o = 107o
o
38. (a) 107o
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
1 + tan 2 θ + tan q = p
2
Explanation: As the fixed charge for two days be `x and additional charge be `y per day It means that Amruta will pay fixed charge for first two days and pays additional charges for next four days. x + 4y = 22
2
1 + tan q = p + tan q – 2p tan q 2
tan q =
p2 − 1 2p
1 – p = –2p tan q
or
2
or
1 = p – 2p(tan q)
or
2
42. (c) x + 4y = 22
1 + tan 2 θ = p – tan q Squaring both sides, we get or
or sec q = 1 + tan 2 θ Put this value in (i), we get
Explanation: Let the fixed charge for two days be `x and additional charge be `y per day. As Radhika has taken book for 4 days. It means that Radhika will pay fixed charge for first two days and pays additional charges for next two days. x + 2y = 16
...(i)
Explanation: sec q + tan q = p is the given equation. 2 2 Since, 1 + tan q = sec q
43. (c) `10 40. (c) A(±3 3 , 0), B(0, 3)
Explanation:
x + 2y = 16 ...(i) x + 4y = 22 ...(ii) Subtracting (ii) from (i), we get y = 3 and put this value of x in (i), we get x = 10. Therefore, fixed charge is x = `10.
Explanation:
44. (d) `3 Explanation: From solution of Q.43, we get y = 3. Therefore, additional charges is y = `3.
45. (c) `50
Explanation: For two more days price charged will be 2y = 2 × 3 = 6 Total money paid by Amruta and Radhika is 22 + 16 + 6 + 6 = `50
O is the midpoint of the base BC i.e., O is the midpoint of B and C(0, –3) Therefore, coordinates of point B is (0, 3) So, BC = 6 units. Let the coordinates of point A be (x, 0).
[ 39
SOLVED PAPER - 2021-22 (TERM-I)
Case Study-II
Explanation:
Perimeter of DAOB AB = Perimeter of DCOD CD 1 AB = 4 CD
⇒
CD = 4AB
⇒
25 =1:4 100
49. (b) DAOD ~ DBCO 50. (c) The medians have a ratio 1 : 4
48. (d) CD = 4AB
=
Perimeter of DAOB 1 = Perimeter of DCOD 4
Also,
ar ( ∆AOB ) ( AB )2 = ar ( COD ) ( CD )2
Explanation: According to the theorem, “Ratio of area of similar triangles is equal to the square of the ratio of corresponding sides“.
46. (b) Similar by SAS criterion 47. (b) 1 : 4
C.B.S.E.
Topper's* Answers
2020
Class-X
Mathematics Standard
Delhi / Outside Delhi Sets *Note : This paper is solely for reference purpose. The pattern of the paper has been changed for the academic year 2022-23.
MM: 80
Maximum Time: 3 hour
General Instructions:
Read the following instructions very carefully and strictly follow them: (i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory. (ii) Section A: Question Numbers 1 to 20 comprises of 20 questions of one mark each. (iii) Section B: Question Numbers 21 to 26 comprises of 6 questions of two marks each. (iv) Section C: Question Numbers 27 to 34 comprises of 8 questions three marks each. (v) Section D: Question Numbers 35 to 40 comprises of 6 questions of four marks each. (vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 questions of one mark, 2 questions of two marks, 3 questions of three marks and 3 questions of four marks. You have to attempt only one of the choices in such questions. (vii) In addition to this, separate instructions are given with each section and question, wherever necessary. (viii) Use of calculators is not permitted.
SECTION - A
(A) 4
(B) ± 4
(C) – 4 (D) 0
Question numbers 1 to 20 carry 1 mark each. Question numbers 1 to 10 are multiple choice questions. Choose the correct option. Sol. 1. The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is
42 ]
Which of the following is not an A.P. ? (A) – 1.2, 0.8, 2.8, ...
2.
4 7 9 12 , , , , ... 3 3 3 3
(D)
−1 −2 −3 , , , ... 5 5 5
3.
Sol.
(C)
(B) 3, 3 + 2 , 3 + 2 2 , 3 + 3 2 , ...
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
In Figure-1, 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
4 cm
Q
O
P R
Figure-1
(A) 3 cm (B) 4 cm (C) 2 cm
(D) 2 2 cm Sol.
m2 + n2
The distance between the points (m, – n) and (– m, n) is (A)
(C) 2 m 2 + n 2
(D) Sol.
(B) m + n
4.
2m2 + 2n2
The degree of polynomial having zeroes – 3 and 4 only is (A) 2 (B) 1 (C) more than 3 (D) 3
5.
TOPPER'S ANSWER - 2020
In Figure-2, ABC is an isosceless triangle, right-angled at C. Therefore
6.
(A) (B) (C) (D)
Sol.
AB2 = 2AC2 BC2 = 2AB2 AC2 = 2AB2 AB2 = 3AC2
Sol.
The point on the x-axis which is equidistant from (– 4, 0) and (10, 0) is (A) (7, 0) (B) (5, 0) (C) (0, 0) (D) (3, 0) OR The centre of a cirele whose end points of a diameter are (– 6, 3) and (6, 4) is
7.
[ 43
44 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
(A) (8, –1) (B) (4, 7)
7 (D) 4 , 2
7 (C) 0 , 2
8.
The pair of linear equations
(B) inconsistent
(C) consistent with one solution
(D) consistent with many solutions
Sol.
3x 5 y + = 7 and 9x + 10y = 14 is 2 3
(A) consistent
Sol.
[ 45
9.
In figure-3, PQ is tangent to the circle with centre at O, at the point B. If ∠AOB = 100°, then ∠ABP is equal to
TOPPER'S ANSWER - 2020
O
10
A
0° Q B P
(A) (B) (C) (D)
Figure-3 50° 40° 60° 80°
The radius of a sphere (in cm) whose volume is 12p cm3, is (A) 3 (B) 3 3 (C) 32/3 (D) 31/3
10.
Sol.
Sol.
Sol.
11.
Fill in the blanks in question numbers 11 to 15. AOBC is a rectangle whose three vertices are A(0, – 3), O (0, 0) and B(4, 0). The length of its diagonal is ........
46 ]
Sol.
Σf u In the formula x = a + i i × h , ui = ................. Σfi
12.
13.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
All concentric circles are .............. to each other.
14.
Sol.
The probability of an event that is sure to happen, is ........................... .
15.
Sol. Simplest form of (1 – cos2 A) (1 + cot2 A) is ................. .
Sol.
16.
Answer the following questions numbers 16 to 20. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, find the other.
17.
Sol.
From a quadratic polynomial, the sum and product of whose zeroes are (– 3) and 2 respectively. OR Can (x – 1) be a remainder while dividing x – 3x2 + 5x – 9 by (x2 + 3) ? Justify your answer with reasons.
2
Sol.
4
[ 47
18.
TOPPER'S ANSWER - 2020
Find the sum of the first 100 natural numbers.
Sol.
19.
Evaluate : 2 sec 30° × tan 60°
20.
Sol.
In figure-4, the angle of elevation of the top of a tower from a point C on the ground. which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. A
30° C
30 m
B
Figure-4 Sol.
SECTION - B 21.
Question number 21 to 26 carry 2 marks each. Find the mode of the following distribution: Marks
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
Number of Students
4
6
7
12
5
6
48 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
22.
In Figure-5, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = BC + AD.
Sol.
A
B
D
C
Figure-5 In Figure-6, find the perimeter of DABC, if AP = 12 cm. AB + CD = BC + AD.
[ 49
TOPPER'S ANSWER - 2020
A
B
D
P
C Q
How many cubes of side 2 cm can be made from a solid cube of side 10 cm ?
24.
Sol.
23.
Sol.
Figure-6
In the given Figure-7, DE||AC and DF || AF. Prove that
BE BF = . FE EC
50 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
A D
E
C
25.
Sol.
F
B
Show that 5 + 2 7 is an irrational number, where
7 is given to be an irrational number.
OR n
Check whether 12 can end with the digit 0 for any natural number n. Sol.
[ 51
26.
Sol.
TOPPER'S ANSWER - 2020
B+C If A, B and C are interior angles of a DABC, then show that cot = tan 2
A 2.
52 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
SECTION - C 27.
Question number 27 to 34 carry 3 marks each. In figure-8, a square OPQR is inscribed in a quadrant OAQB of a circle. If the radius of circle is 6 2 cm, find the area of the shaded region. B Q
R
O
P
Sol.
Figure-8
A
[ 53
28.
TOPPER'S ANSWER - 2020
Construct a DABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3 of the corresponding sides of DABC. 4 OR
Draw a circle of radius 3.5 cm. Take a point P outside the circle at a distance of 7 cm from the centre of the circle and construct a pair of tangents to the circle from that point. Sol.
O
54 ]
29.
Prove that :
2 cos3 θ – cos θ sin θ – 2 sin 3 θ
= cot θ
Sol.
30.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
1 1 when 1 is subtracted from the numberator and it becomes when 8 is added to its 3 4 denominator. Find the fraction. A fraction becomes
OR
The present age of a father is three years 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. Sol.
[ 55
31.
Using Euclid’s Algorithm, find the largest number which divides 870 and 258 leaving remainder 3 in each case.
Sol.
32.
TOPPER'S ANSWER - 2020
Sol.
Fin d 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. OR Show that the points (7, 10), (–2, 5) and (3, – 4) are vertices of an isosceles right triangle.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
Sol.
33.
56 ]
In an A.P. given that the first term (a) = 54, the common difference (d) = – 3 and the nth term (an) = 0, find n and the sum of first n terms (Sn) of the A.P.
[ 57
34.
Read the following passage and answer the questions given at the end : Diwali Fair
TOPPER'S ANSWER - 2020
A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in Figure-9.
Prizes are given, when a black marble is picked. Shweta plays the game once.
1
4
2
10 6
8
Figure-9
(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 balls out of which 6 are black ?
Sol.
What is the probability that she will be allowed to pick a marble from the bag ?
(i)
58 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
SECTION - D 35.
Question number 35 to 40 carry 4 marks each. 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. OR
Sol.
36.
A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
A solid toy is the form of a hemisphere surmounted by a right circular cone of same radius. The height of the cone is 10 cm and the radius of the base is 7 cm. Determine the volume of the toy. Also find the area of the 22 and 149 = 12.2 . coloured sheet required to cover the toy Use π = 7
[ 59
TOPPER'S ANSWER - 2020
37.
Sol.
For the following data, draw a 'less than' ogive and hence find the median of the distribution. Age (in years)
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
Number of Persons
5
15
20
25
15
11
9
OR The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the number of wickets taken. Number of wickets
20 – 60
60 – 100
100 – 140
140 – 180
180 – 220
220 – 260
Number of bowlers
7
5
16
12
2
3
38.
Sol.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
60 ]
From 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.
( Use
3 = 1.73
)
TOPPER'S ANSWER - 2020
[ 61
39.
Sol.
Sol.
Prove that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides.
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
40.
62 ]
Obtain other zeroes of the polynomial p(x) = 2x4 – x3 – 11x2 + 5x + 5 if two of its zeroes are
5 and –
5.
OR 3
What minimum must be added to 2x – 3x + 6x + 7 so that the resulting polynomial will be divisible by x2 – 4x + 8 ?
Sol.
2
[ 63
TOPPER'S ANSWER - 2020
64 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
real numbers
[ 1
UNIT 1: Number System
c h a p ter
1
Real Numbers
Syllabus ¾
¾
Euclid’s division lemma, Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, proofs of irrationality of 2, 3 and 5 . Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals.
Trend Analysis L
2018 ist of Concepts
Delhi
Questions based on HCF & LCM
1 Q (1 M)
Outside Delhi
Delhi
2020
Outside Delhi
Delhi
Outside Delhi
1 Q (3 M) 1 Q (1 M) 1 Q (1 M) 2 Q (1 M) 2 Q (1 M) 3 Q (2 M) 1 Q (2 M) 2 Q (4 M) 2 Q (3 M)
Irrational numbers, Terminating and 1 Q (2 M) non-terminating Recurring Decimals
T
2019
1 Q (1 M) 1 Q (3 M) 2 Q (3 M)
1 Q (4 M)
OPIC - 1
Euclid’s Division Lemma and Fundamental Theorem of Arithmetic
Revision Notes Algorithm: An algorithm is a series of well defined steps which gives a procedure for solving a mathematical problem.
Lemma: A lemma is a proven statement used for proving another statement. Euclid’s Division Lemma: For given positive integers a and b, there exist unique integers q and r, satisfying
a = bq + r where 0 ≤ r < b.
Here, a = Dividend, b = Divisor, q = Quotient and r = Remainder i.e., Dividend = (Divisor × Quotient) + Remainder
TOPIC - 1 Euclid's Division Lemma and Fundamental Theorem of Arithmetic Page No. 2
TOPIC - 2 Irrational Numbers, Terminating and NonTerminating Recurring Decimals Page No. 11
[ 3
real numbers
For example, 1 → Quotient Divisor 3) 5 → Dividend –3 2 → Remainder As per Euclid's Division Lemma, 5 = (3 × 1) + 2
Scan to know more about this topic
Euclid’s division algorithm is applicable for positive integers only but it can be extended for all integers except zero. When ‘a’ and ‘b’ are two positive integers then a = bq + r, where 0 ≤ r < b,
Steps to find the HCF of two positive integers by Euclid’s division algorithm: (i) Let two integers be a and b such that a > b. (ii) Take greater number a as dividend and the smaller number b as divisor. (iii) Now, find whole numbers ‘q’ and ‘r’ as quotient and remainder respectively. ∴ a = bq + r where 0 ≤ r < b. then HCF (a, b) = HCF (b, r). (iv) If r = 0, b is the HCF of a and b. If r ≠ 0, then take r as divisor and b as dividend.
Euclid’s Division Lemma Scan to know more about this topic
Euclid’s Division Algorithm
(v) Repeat step (iii), till the remainder is zero, the divisor thus obtained at last stage is the required HCF.
The Fundamental Theorem of Arithmetic Every composite number can be expressed as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur. Fundamental theorem of arithmetic is also called a Unique Factorization Theorem. Composite number = Product of prime numbers Or Any integer greater than 1 can either be a prime number or can be written as a unique product of prime numbers. e.g., Scan to know (i) 2 × 11 = 22 is the same as 11 × 2 = 22. more about (ii) 6 can be written as 2 × 3 or 3 × 2, where 2 and 3 are prime numbers. this topic (iii) 15 can be written as 3 × 5 or 5 × 3, where 3 and 5 are prime numbers. The prime factorization of a natural number is unique, except to the order of its factors. e.g., 12 detained by multiplying the prime numbers 2, 2 and 3 together, 12 = 2 × 2 × 3 Fundamental theo We would probably write it as 12 = 22 × 3 rem of arithmetic By using Fundamental Theorem of Arithmetic, we shall find the HCF and LCM of given numbers (two or more). This method is also called Prime Factorization Method. Prime Factorization Method to find HCF and LCM: (i) Find all the prime factors of given numbers. (ii) HCF of two or more numbers = Product of the smallest power of each common prime factor, involved in the numbers. (iii) LCM of two or more numbers = Product of the greatest power of each prime factor, involved in the numbers.
Know the Formulae
For two positive integers a and b, we have HCF (a, b) × LCM (a, b) = a × b or
HCF (a, b) =
a×b LCM ( a , b )
and
LCM (a, b) =
a×b HCF ( a , b )
4 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Mnemonics
Euclid's Division Lemma (a = bq + r) Alibaba's best product quotation is assent reward. Concept: a = bq + r Interpretation: Then a = b × q + r.
Alibaba's A = a best's B = b quotation's Q = q assent's A = addition of bq and r reward's R = r
How is it done on the
GREENBOARD?
Q.1. Show that 6n can never end with digit 0 for any natural number n. Solution Step I: Any number which ends in zero must have at least 2 and 5 as its prime factors. Step II: Since, 6 = 2 x 3
Therefore, 6n = (2 × 3)n ⇒ 2n × 3n Hence, prime factors of 6 are 2 and 3. Step III: Since 6n does not contain 5 as a prime factor, hence 6n can never end in 0. Step IV: Since 6n does not contain 5 as a prime factor, hence 6n can never end in 0.
Very Short Answer Type Questions Q. 1. If xy = 180 and HCF (x, y) = 3, then find the LCM (x, y) A [CBSE SQP, 2020-21]
Sol. We know that, “LCM × HCF = Product of two numbers”. (LCM) (3) = 180 ½ LCM = 60. ½ [CBSE Marking Scheme, 2020-21] Detailed Solution:
Product of two numbers, xy = 180 HCF (x, y) = 3 By using fundamental theorem, HCF (x, y) × LCM (x, y) = x × y ⇒ 3 × LCM (x, y) = 180 ⇒ LCM (x, y) = 60.
1 mark each
Q. 2. Find the total number of factors of prime numbers. U [CBSE Delhi Set-I, 2020] Sol. We have only two factors (1 and number itself) of any prime number, such as: 2 = 2 × 1 (2 and 1) 3 = 3 × 1 (3 and 1) ........etc. 1 Q. 3. Find the HCF and the LCM of 12, 21 and 15. U [CBSE Delhi Set-I, 2020] Sol. Prime factors of 12 = 2 × 2 × 3 Prime factors of 21 = 3 × 7 and Prime factors of 15 = 3 × 5 \ HCF of 12, 21 and 15 = 3. ½ and LCM of 12, 21 and 15 = 2 × 2 × 3 × 5 × 7 = 420. ½
[ 5
real numbers
Q. 4. Find the sum of exponents of prime factors in the prime factorisation of 196.
Sol.
U [CBSE OD Set-I, 2020]
Prime factors of 196 = 22 × 72
½
The sum of exponents of prime factors = 2 + 2 = 4.
½
Q. 5. Complete the following statements: Euclid’s Division Lemma states that for two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where .......... .
U [CBSE SQP, 2020]
composite number as 2 as they get confused in prime number and composite number.
ANSWERING TIP Understand the clear difference in prime numbers and composite numbers.
T
Q. 7. Express 429 as a product of its prime factors.
½
Sometimes students assume smallest
Sol. For given positive integers a and b, there exists unique integer q and r satisfying a = bq + r where 0 ≤ r < b. [CBSE Marking Scheme, 2020] Q. 6. Find the LCM of smallest two digit composite number and smallest composite number.
½
COMMONLY MADE ERROR
R [CBSE OD Set-I, 2020]
Sol. Since, the smallest composite number = 4 and smallest 2 digit composite number = 10 \ LCM of 4 and 10. 4 = 2 × 2 10 = 5 × 2 L.C.M. = 2 × 5 × 2 = 20 \ LCM of 4 and 10 = 20
A + R [CBSE Delhi Set-I, 2019]
opper Answer, 2019
Sol.
\ LCM (a, b) = product of greatest power of x and y. = x3y3
Sol. LCM (x3y2, xy3) = x3y3. 1 [CBSE Marking Scheme, 2019]
Sol. Since, HCF × LCM = Product of numbers 6 × LCM = 336 × 54 336×54 LCM = 6 LCM = 56 × 54 LCM = 3024 [CBSE Marking Scheme, 2019] 1
Detailed Solution: Given, positive integers are a = x3y2 and b = xy3 Here, x and y are prime numbers
A Q. 9. If HCF (336, 54) = 6, find LCM (336, 54) [CBSE OD Set-I, II, III, 2019]
Q. 8. Two positive integers a and b can be written as a = x3y2 and b = xy3, where x and y are prime numbers. Find LCM (a, b). A [CBSE Delhi Set-III, 2019]
Q. 10. What is the HCF of smallest prime number and the smallest composite number ?
A + R [CBSE Delhi, OD, 2018]
T
Sol. The required numbers are 2 and 4 and the HCF of 2 and 4 is 2. Detailed Solution:
[CBSE Marking Scheme, 2018] 1
opper Answer, 2018
1
6 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 11. Explain why 13233343563715 is a composite number ? R [Board Term-I, 2016]
Sol. Since, the given number ends in 5. Hence, it is a multiple of 5. Therefore, it is a composite number. [CBSE Marking Scheme, 2016] 1
Short Answer Type Questions-I
Sol. We know that, The three bells again ring together on that time which is the LCM of individual time of each bell
4 = 2 × 2
Q. 1. 3 bells ring at an interval of 4, 7 and 14 minutes. All three bells rang at 6 am, when the three bells will be ring together next ? C + A [CBSE SQP, 2020-21]
2 marks each
7 = 7 × 1
14 = 2 × 7
LCM = 2 × 2 × 7 = 28
The three bells will ring together again at 6: 28 am [CBSE Marking Scheme, 2020-21] 2
T
Q. 2. If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n. A + R [CBSE Delhi, 2019]
opper Answer, 2019
Sol.
Q. 3. Find the HCF of 1260 and 7344 using Euclid’s A [CBSE Delhi Set-I, II, III 2019] algorithm.
COMMONLY MADE ERROR Some students find the HCF directly without using Euclid’s algorithm.
Practice more such questions based
on Euclid’s algorithm as it is the most important topic from examination perspective.
Q. 4. Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer. A [CBSE Delhi Set-I, 2019] Sol. Using Euclid’s Algorithm a = 4q + r, 0 ≤ r < 4 ⇒ a = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3. 1 Now a = 4q and a = 4q + 2 are even numbers. ½ Therefore when a is odd, it is of the form a = 4q + 1 or a = 4q + 3 for some integer q. ½ [CBSE Marking Scheme, 2019]
Sol. Using Euclid’s Algorithm 7344 = 1260 × 5 + 1044 1260 = 1044 × 1 + 216 ½ 1044 = 216 × 4 + 180 216 = 180 × 1 + 36 180 = 36 × 5 + 0 ½ \ HCF of 1260 and 7344 is 36. 1 [CBSE Marking Scheme, 2019]
ANSWERING TIP
[ 7
real numbers
Sol. Smallest number divisible by 306 and 657 = LCM (306, 657) 1 LCM (306, 657) = 22338 1 [CBSE Marking Scheme, 2019] Detailed Solution: The smallest number that is divisible by two numbers is obtained by finding the LCM of these numbers Using Euclid's Algorithm 657 = 306 × 2 + 45 306 = 45 × 6 + 36 45 = 36 × 1 + 9 36 = 9 × 4 + 0 \ HCF(657, 306) = 9 Product of two numbers LCM = HCF( 657 , 306 ) =
657 × 306 = 657 × 34 9
LCM(657, 306) = 22338 Hence, the smallest number which is divisible by 306 and 657 is 22338. Q. 6. The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other number. R [CBSE SQP 2019]
Q. 8. Explain whether 3× 12 × 101 + 4 is a prime number or a composite number. U [Board Term-I 2017, 2015]
Sol. 3 × 12 × 101 + 4 = 4(3 × 3 × 101 + 1) = 4(909 + 1) = 4(910)
1
= 2 × 2 × 2 × 5 × 7 × 13 1 = a composite number [Product of more than two prime factors] [CBSE Marking Scheme, 2015] Q. 9. Find the HCF and LCM of 90 and 144 by the method of prime factorization. U [Board Term-I, 2016] 90 = 2 × 32 × 5
Sol. Since,
144 = 24 × 32
and Hence,
HCF = 2 × 32 = 18
1
4
1
2
and LCM = 2 × 3 × 5 = 720
Q. 10. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. A [Board Term-I, 2016]
Sol. Given, Length = 8 m 50 cm = 850 cm
breadth = 6 m 25 cm = 625 cm
height = 4 m 75 cm = 475 cm
Since, the length of the longest rod is equal to HCF of 850, 625 and 475 625 ) 850 ( 1
625
225 ) 625 ( 2
450
175 ) 225 ( 1
175
50 ) 175 ( 3
150
Sol. Since, HCF × LCM = Product of two numbers Then, 9 × 360 = 45 × 2nd number 1 ( 9 × 360 ) nd 2 number = 45
Sol. Since, LCM (p, q) = a3b3 ½ 2 and HCF (p, q) = a b ½ Hence, LCM (p, q) × HCF (p, q) = a3b3 × a2b = a5b4 ½ = a2b3 × a3b = pq Hence Verified. ½
Detailed Solution: As per Euclid’s Division Lemma Let positive integer be b = 4 If a and b are two positive integers, then a = bq + r, where 0 £ r < b. Hence, a = 4q + r, where 0 £ r 0. and 2a 2a
Know the Formulae
-b
-b
2 2 Roots of ax + bx + c = 0, where a ≠ 0 are 2 a and 2 a , where b – 4ac = 0
Quadratic identities: (i) (a + b)2 = a2 + 2ab + b2 (ii) (a – b)2 = a2 – 2ab + b2 (iii) a2 – b2 = (a + b)(a – b)
Mnemonics
Concept: To Find the roots of quadratic equation,
− b ± b2 − 4ac . 2a
A Negative Boy could not decided if he did or didn't want to go to a Radical party. The Boy was Square so he missed out on 4 Awesome Chicks. This was all over by 2 a.m. Interpretation: A negative boy = (– b) Undecided mean he wanted to go or didn't want to go = (+/–)
To a radical party = (
) 2
Boy was square = (b ) Missed out = (–) 4 Awesome = 4a Chicks = c All over = Divided by 2 a.m. = 2a
How is it done on the
GREENBOARD?
Q.1. Two water taps together can fill Solution Step I: Let time taken by tap of larger 11 hrs. The tap of smaller diameter be x hrs. Then, the time a tank in 2 12 taken by tap of smaller diameter = (x diameter takes 2 hours more than the + 2) hrs. larger one to fill the tank separately. Then, the part of the tank filled by the Find the time in which each tap can tank of larger diameter in 1 hour = 1 x separately fill the tank. 1 and other tap in 1 hour = x +2
74 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Step II: According to question, 1 1 12 + = x x + 2 35
x + 2 + x 12 = x (x + 2) 35
Step III: By cross multiplication,
Step IV: Now factorizing by splitting the middle term. 6x2 – 30x + 7x – 35 = 0 Þ 6x(x – 5) + 7(x – 5) = 0 Þ (x – 5)(6x + 7) = 0 −7 Þ x = 5 or x = 6
35(2x + 2) = 12(x2+2x)
Rejecting x =
Þ 70x + 70 = 12x2 + 24x
be negative). Therefore, larger tap takes 5 hrs and smaller tap takes 7 hrs.
Þ 12x2 – 46x – 70 = 0 Þ 6x2 – 23x – 35 = 0
Very Short Answer Type Questions Sol. ⇒
Sol. x2 + 7x + 10 = 0 x2 + 5x + 2x + 10 = 0 ½ (x + 5)(x + 2) = 0 Either x = – 5 or x = – 2 ½ [CBSE SQP Marking Scheme, 2020-21]
⇒
Alternate Solution: Given, x2 + 7x + 10 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 7 and c = 10 \ ⇒
− b ± b 2 − 4 ac 1 x = − 2a 2 x=
−7 ± (7 )2 − 4 × 1 × 10 2×1
⇒
−7 ± 9 x = 2
⇒
x =
−7 ± 3 2
⇒
x =
−7 + 3 −7 − 3 or 2 2
⇒
x = – 2 or – 5
Hence, the roots of the given equation are – 2 or – 5. ½ Q. 2. Find the roots of the quadratic equation x2 – 0.04 A [CBSE OD Set-I, 2020] = 0.
Q. 1. Find the roots of the equation x2 + 7x + 10 = 0. U [CBSE SQP, 2020-21]
−7 (as time cannot 6
⇒
1 mark each x2 – 0.04 = 0 x2 = 0.04 x = ± 0.04 x = ± 0.2. 1 [CBSE SQP Marking Scheme, 2020] 2
Q. 3. If one root of the equation (k – 1)x – 10x + 3 = 0 is the reciprocal of the other, then find the value of k. R + U [CBSE SQP, 2020] Sol. 4. 1 [CBSE SQP Marking Scheme, 2020] Detailed Solution: Let one root = a 1 and the other root = α
Product of roots = a ×
1 α
½
Given equation is (k – 1) x2 – 10 x + 3 = 0
Product of roots =
3 ( k − 1)
1 3 = a × α k −1 3 = 1 k −1 3 = k – 1 k = 4.
½
[ 75
qUADRATIC eQUATIONS
Q. 4. Find the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each A [CBSE Delhi Set- I, II, III, 2019] other.
½
1 Let one root be a so other root is α
Sol. Roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other ⇒ Product of roots = 1 ½ k ⇒ = 1 ⇒ k = 3 ½ 3 [CBSE Marking Scheme, 2019]
a = 3, b = –10, c = k
Detailed Solution: Given equation: 3x2 – 10x + k = 0 Comparing it with ax2 + bx + c = 0, we get
Now,
\
T
α×
⇒
1 k = α 3
k = 3
Hence, value of k is 3.
Q. 5. Write the discriminant of the quadratic equation (x + 5)2 = 2(5x – 3).
c product of roots = a
½ A [CBSE Bord Term, 2019]
opper Answer, 2019
Sol.
1
2 Q. 6. If x = 3 is one root of the quadratic equation x – 2kx – 6 = 0, then find the value of k. C + U [CBSE Delhi & OD, 2018]
Sol. x = 3 is one root of the equation \ 9 - 6k - 6 = 0 1 Þ k = 2
½
T
[CBSE Marking Scheme, 2018] ½
opper Answer, 2019
76 ]
Q. 7. Find the value of k, for which one root of the quadratic equation kx2 – 14x + 8 = 0 is 2. U [CBSE SQP, 2018] Sol. Try Yourself, Similar to Q. No. 6 (above) of Very Short Answer Type Questions. 3 x 2 + 6 = 9.
Q. 8. Find the positive root of
U [Board Term-II 2015, 2017]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Putting x = −
1 3 - 2
Þ
3x2 + 6 = 81
2
Hence,
3x = 81 – 6 = 75
75 = 25 x2 = 3 \ x = ± 5 Hence, positive root = 5.
1
1 Q. 9. If x = − , is a solution of the quadratic equation 2 3x2 + 2kx – 3 = 0, find the value of k.
C+ U
[Board Term-2 , 2015 & CBSE Delhi Set-I, II, III, 2017]
k =
3 - 12 4
-9 1 4 [CBSE Marking Scheme, 2015]
3 x2 – 2x –
3 = 0.
U [Board Term-II 2015]
2 Sol. Given, 3 x – 2x – 3 = 0 2 Þ 3 x – 3x + x – 3 = 0 Þ 3 x (x – 3 ) +1(x– 3 ) = 0 Þ (x – 3 )( 3 x +1) = 0
\
x =
3 or
-1
1
3
2 marks each
2 x + 9 + x = 13 T
3 –3 4
k =
Short Answer Type Questions-I Q. 1. Solve for x:
k =
Q. 10. Find the roots of the quadratic equation
[CBSE Marking Scheme, 2015]
1 + 2k − – 3 = 0 2
Þ
3x 2 + 6 = 9
Sol.
2
1 in 3x2 + 2kx – 3 = 0 2
A [Board Term-2 OD Set II, 2016]
opper Answer, 2016
Sol.
=0
Q. 2. If x =
2 and x = – 3 are roots of the quadratic 3
equation ax2 + 7x + b = 0, find the values of a and b. C + A [CBSE, Delhi Set I, II, III, 2016] 2 in ax2 + 7x + b = 0 Sol. Substituting x = 3 4 14 a+ + b = 0 9 3
Þ 4a + 42 + 9b = 0
2
Þ 4a + 9b = – 42 ...(i) ½ again, substituting x = – 3 in ax2 + 7x + b = 0 9a – 21 + b = 0 Þ 9a + b = 21 ...(ii) ½ Solving (i) and (ii), we get
a = 3 and b = – 6
1
Q. 3. Solve the following quadratic equation for x:
4x2 – 4a2x + (a4 – b4) = 0 U [Delhi CBSE Term-II, 2015 (Set I, II, 2016)]
[ 77
qUADRATIC eQUATIONS
Then,
x=
−( −4 a 2 ) ± ( ( −4 a 2 )2 − 4 × 4 × ( a 4 − b 4 ) 2×4
4 a 2 ± 16 a 4 - 16 a 4 + 16b 4 Þ = 8 4 a 2 ± 16 b 4 Þ = 8 Þ
x =
\
x =
1
4 a2 ± 4b2 a2 ± b2 = 8 2 2 2 a2 - b2 a 2 +4ba2 - 4 b 1 or x == 8 2 2 [CBSE Marking Scheme, 2015]
Q. 4. 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. A [CBSE, Foreign Set I, 2016] Sol. Let unit's digit and ten's digit of the two digit number be x and y respectively ∴ Number is 10y + x According to question, 10y + x = 4(y + x) ½ Þ 10y + x = 4y + 4x Þ 10y – 4y = 4x – x Þ 6y = 3x Þ 2y = x ...(i) Also, 10y + x = 3xy ...(ii) Þ 10y + 2y = 3(2y)y [From eq (i)]
Sol. On completing the square, x2 – 4x + 4 – 4 – 8 = 0 Þ (x – 2)2 – 8 – 4 = 0 ½ Þ (x – 2)2 – 12 = 0 Þ (x – 2)2 = 12 ½ Þ (x – 2)2 = (2 3 )2
− B ± B2 − 4 AC 2A
Since, x =
Þ 12y = 6y2 Þ 6y2 – 12y = 0 Þ 6y(y – 2) = 0 Þ y = 0 or y = 2 ½ Rejecting y = 0 as tens digit should not be zero for a two digit number. Þ x = 4 ∴ Required number = 10y + x Þ 10 × 2 + 4 = 24. 1 2 Q. 5. Find the roots of x – 4x – 8 = 0 by the method of U completing the square.
Þ Þ
x – 2 = ± 2 3 ½ x = 2 ± 2 3 x = 2 + 2 3 or 2 – 2 3 ½ [CBSE Marking Scheme, 2015]
\
Q. 6. In a cricket match, Harbhajan took three wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Represent the above situation in the form A [Board Term-2, 2015] of quadratic equation. Sol. Let the number of wickets taken by Zahir be x. Then, the number of wickets taken by Harbhajan = 2x – 3 ½ According to question, x(2x – 3) = 20 ½ Þ 2x2 – 3x = 20 \ Required quadratic equation is, 2x2 – 3x – 20 = 0. 1 [CBSE Marking Scheme, 2015]
Sol. Given, 4x2 – 4a2x + (a4 – b4) = 0 Comparing with Ax2 + Bx + C = 0, we get Here, A = 4, B = –4a2 and C = (a4 – b4)
Short Answer Type Questions-II Q. 1. In a flight of 600 km, an aircraft was slowed due to bad wether. 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. C + A [CBSE Delhi Set-I & OD Set-I, 2020] Sol. Let original speed of flight be x km/h, then according to question, 600 600 − = 30 minutes ½ − 200 x x
Distance Q Time = Speed
1 1 30 − = ⇒ 600 60 x − 200 x
3 marks each
⇒ ⇒ ⇒
x − x + 200 1 = x( x − 200 ) 2 × 600
200
2
x − 200 x
=
1 1200
x2 – 200x = 240000
⇒ x2 – 200x – 240000 = 0
Here, a = 1, b = – 200 and c = – 240000 200 ± 40000 + 960000 \ x= 2×1
=
200 ± 1000000 2
=
200 ± 1000 2
½
78 ]
=
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
200 + 1000 200 − 1000 , 2 2
= 600, – 400 Since, speed cannot be negative, therefore original speed = 600 km/h. and original distance = 600 km original distance \ Time = original speed
=
Q. 3. Solve for x: 1
600 km =1h 600 km/hr.
Hence, the original duration of flight is 1 h.
1
Q. 2. A train covers 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. Find the original speed of the train.
A [CBSE Delhi Set-II, 2020]
Sol. Let the speed of the train = x km/h
1 1 11 = , x ≠ – 4, 7. x + 4 x + 7 30
U [CBSE Outside Delhi Set-I, 2020)] 1 1 11 Sol. Given, − = x + 4 x + 7 30 x+7−x−4 11 ⇒ = ( x + 4 )( x + 7 ) 30 3 11 ⇒ = 2 x + 4 x + 7 x + 28 30
3 11 ⇒ = x 2 + 11x + 28 30 ⇒ 11x2 + 121x + 308 = 90 ⇒ 11x2 + 121x + 218 = 0 Comparing with ax2 + bx + c = 0, we get a = 11, b = 121 and c = 218 − b ± b 2 - 4 ac \ x = 2a
Total distance covered by the train = 480 km \ Time taken to cover the distance 480 km
=
480 h x
Then, time taken to cover the distance 480 km
480 h = x −8
1
According to question,
480 480 − = 3 x−8 x
x − x + 8 ⇒ 480 = 3 x( x − 8 ) ⇒
8 x 2 − 8x
=
3 1 = 480 160
⇒ x2 – 8x – 1280 = 0 1 2 Compare with ax + bx + c = 0, we get a = 1, b = – 8 and c = – 1280 \
x = =
8 ± 64 + 4 × 1280 2×1 8 ± 5184 2
8 ± 72 = 2 =
=
8 + 72 8 − 72 , 2 2 80 − 64 , 2 2
= 40, – 32 Since, negative speed cannot be possible. Hence, the original speed of the train = 40 km/h. 1
1
−121 ± 14641 − 9592 22
−121 ± 5049 22 −121 ± 71.06 = 22 −121 + 71.06 −121 − 71.06 ⇒ x = , 22 22
⇒
If the speed has increased 8 km/h, i.e., (x + 8) km/h
=
1
x =
−49.94 −192.06 , 22 22 ⇒ x = – 2.27, – 8.73. 1 Q. 4. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speed of each train. U [CBSE Outside Delhi Set-II, 2020] Sol. Total distance of the journey = 600 km Let speed of fast train = x km/h, then speed of slow train = (x – 10) km/h According to question, 600 600 – = 3 1 x − 10 x
⇒
x =
Distance Q Time = Speed
⇒ ⇒
x − x + 10 600 =3 ( x − 10 )x 6000 2
x − 10 x
= 3
⇒ x2 – 10x – 2000 = 0 ⇒ x2 – 50x + 40x – 2000 = 0 ⇒ x(x – 50) + 40(x – 50) = 0 ⇒ (x – 50)(x + 40) = 0 Either, x = 50 or x = – 40 speed can not be negative.
1
[ 79
qUADRATIC eQUATIONS
COMMONLY MADE ERROR Some students do not know how to frame the equation. Some frame it correctly but fail to solve it.
application
problems
is
Q. 5. Solve for x: x2 + 5x – (a2 + a – 6) = 0 A [CBSE OD Set-II, 2019] [Board Term-2 Foreign Set I, 2015]
Sol. Given, x2 + 5x – (a2 + a – 6) = 0
Then,
x =
-5 ± 25 + 4( a 2 + a - 6 ) 1 2 -b ± b 2 - 4 ac Q x = 2a
-5 ± ( 2 a + 1) 1 = 2 2a - 4 -2 a - 6 = or 2 2 Thus, x = a – 2 or x = – (a + 3) ½+½ [CBSE Marking Scheme, 2019]
Alternate Solution: x2 + 5x – (a2 + a – 6) = 0 2 ⇒ x + 5x – (a2 + 3a – 2a – 6) = 0 ½ 2 ⇒ x + 5x – [a(a + 3) – 2(a + 3)] = 0 ½ ⇒ x2 + 5x – (a + 3)(a – 2) = 0 ½ 2 ⇒ x + [(a + 3) – (a – 2)]x – (a + 3)(a – 2) = 0 ⇒ x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0 ⇒ x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0 ½ ⇒ [x + (a + 3)] [x – (a – 2)] = 0 ⇒ x = – (a + 3) or x = a – 2 Hence, roots of given equation are – (a + 3) and a – 2. 1 Q. 6. Divide 27 into two parts such that the sum of their 3 . reciprocals is 20
A [CBSE Compt. Set I, II, III, 2018]
Sol. Let two parts be x and 27 – x. 1 1 + \ = x 27 − x 27 − x + x Þ = x( 27 − x )
3 20 3 20
Emphasis on solving quadratic equation
Q. 7. A plane left 30 minutes late then its scheduled time and in order to reach of destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
ANSWERING TIP based on necessary.
⇒ x2 – 27x + 180 = 0 ⇒ (x – 15)(x – 12) = 0 ⇒ x = 12 or 15. 3 \ The two parts are 12 and 15 [CBSE Marking Scheme, 2018]
So, the speed of fast train = 50 km/h, and the speed of slow train = 50 – 10 = 40 km/h. 1
A [CBSE Delhi/OD Set-I, II, III, 2016, 2018]
[Board Term-II, 2015] Sol. Let usual speed of the plane be x km/hr.
1500 1500 30 − 1 \ = x x + 100 60 ⇒ x2 + 100x – 300000 = 0 ⇒ x2+ 600x – 500x – 300000 = 0 1 ⇒ (x + 600)(x – 500) = 0 x = – 600 or x = 500 ½ (Rejecting negative value) Speed of plane = 500 km/h ½ [CBSE Marking Scheme, 2018] etailed Solution: D Let the speed of plane be x km/h Time taken to cover 1500 km Distance 1500 = h (t1) = speed x
½ Time taken to cover 1500 km when speed increased by 100 km/h (t2) =
1500 h x + 100
t1 – t2 = 30 minutes =
Given,
½ 1 h 2
Then,
1500 1500 − x x + 100 1500 x + 150000 − 1500 x x( x + 100 )
1 = 2
1
1 = 2
150000
1 = x + 100 x 2 2
x2 + 100x = 30,000 2 x + 100x – 30,000 = 0 x2 + 600x – 500x – 30,000 = 0 x(x + 600) – 500(x + 600) = 0 (x + 600)(x – 500) = 0 Either x + 600 = 0 ⇒ x = – 600, but speed can not be negative. or x – 500 = 0 ⇒ x = 500 \ Speed of the plane = 500 km/h
½ ½
80 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
opper Answer, 2018
T
OR
−2 x + 8
Q. 8. Solve for x:
x+1 x−2 2x + 3 = 4− ; where x ≠ 1, – 2, 2 + x − 1 x 2 x−2 + U [Board Term-2 Delhi Set II, 2016]
Sol. Here,
x 2 + 3x + 2 + x 2 - 3x + 2 2
x +x-2
=
4 x - 8 - 2x - 3 1 x-2
(2x2 + 4)(x – 2) = (2x – 11)(x2 + x – 2) Þ 5x2 + 19x – 30 = 0 1 Þ (5x – 6)(x + 5) = 0 6 Þ x = – 5 or 1 5 [CBSE Marking Scheme, 2016]
2
−7 x−2
x +x−2 –2x2 + 4x + 8x – 16 = – 7x2 – 7x + 14 5x2 + 19x – 30 = 0 2 5x + 25x – 6x –30 = 0 5x(x + 5) – 6(x + 5) = 0 (x + 5)(5x – 6) = 0 if x + 5 = 0 ⇒ x = –5 6 if 5x – 6 = 0 ⇒ x = 5 Q. 9. Solve the following quadratic equation for x: a a + b x2 + + x + 1 = 0 a+b a U [Board Term-2 Delhi Set III, 2016]
Detailed Solution:
=
x+1 x −2 2x + 3 = 4− + x −1 x + 2 x−2 x+1 x−2 2x + 3 −1+ −1 = 2 − x −1 x+2 x−2 x+1−x+1 x −2−x −2 2x − 4 − 2x − 3 + = x −1 x+2 x−2 2 −4 −7 = + x −1 x + 2 x −2 2x + 4 − 4 x + 4 −7 = ( x − 1)( x + 2 ) x −2
Sol. Here, x 2 +
a a+b x+ x+1=0 a+b a
a a+b a Þ x x + 1 + x + = 0 a b a + b a + a a + b Þ x + 1 x + = 0 a a+b -a -( a + b ) 1 Þ x = or a+b a
[CBSE Marking Scheme, 2016]
[ 81
qUADRATIC eQUATIONS
a + 2b 2 a + b ½ , 3 3 Q. 11. 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.
Hence, the roots =
A [Board Term-2 O.D. Set III, 2016] Sol. Let the three consecutive natural numbers be x, x + 1 and x + 2. ½ \ (x + 1)2 = (x + 2)2 – (x)2 + 60 1 2 2 2 Þ x + 2x + 1 = x + 4x + 4 – x + 60 Þ x2 – 2x – 63 = 0 ½ 2 Þ x – 9x + 7x – 63 = 0 Þ x(x – 9) + 7(x – 9) = 0 Þ (x – 9)(x + 7) = 0 Thus, x = 9 or x = – 7 Rejecting – 7, we get x = 9 Hence, three numbers are 9, 10 and 11. 1
Q. 12. P & Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of the circle of radius x cm which touches given circles externally. Given that angle PRQ is 90°. Write an equation in x and solve it. Sol.
Sol.
Sol. Given, (x – 1)2 – 5(x – 1) – 6 = 0 Þ x2 – 2x + 1 – 5x + 5 – 6 = 0
1 1 = 3 − x x−2 x−2−x 3 = x( x − 2 ) 1 3x2 – 6x = – 2 2 3x – 6x + 2 = 0
A [CBSE SQP, 2020]
1 1 1
1
Þ x2 – 7x + 6 – 6 = 0 Þ Þ \
x2 – 7x = 0 x(x – 7) = 0 x = 0 or 7
1 1
5 marks each
6 ± 12 1 6 3+ 3 3− 3 1 = , 3 3 [CBSE SQP Marking Scheme, 2020] Detailed Solution: 1 1 Given, − = 3 − x x 2 x−2−x = 31 ⇒ x( x − 2 ) x =
1 1 = 3, x ≠ 0, 2 x x-2
½ In right DPQR, by Pythagoras theorem PQ2 = PR2 + RQ2 Þ 172 = (x + 9)2 + (x + 2)2 ½ 289 = x2 + 18x + 81 + x2 + 4x + 4 Þ 2x2 + 22x – 204 = 0 Þ x2 + 11x – 102 = 0 ½ 2 Þ x + 17x – 6x – 102 = 0 Þ x(x + 17) – 6(x + 17) = 0 ½ (x – 6)(x + 17) = 0 Þ x = 6 or x – 17 (x can’t be negative) ½ Thus, x = 6 cm [CBSE Marking Scheme, 2016] ½
2 Q. 13. Solve the quadratic equation (x – 1) –5(x – 1) – 6 = 0 A [Board Term-2, 2015]
Long Answer Type Questions Q. 1. Solve the following equation:
A [Board Term-2, SQP, 2016]
Q. 10. Solve the following quadratic equation for x: 9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0 U [Board Term-2 Foreign Set I, 2016] Sol. Given, 9x2 – 9(a + b)x + 2a2 + 5ab + 2b2 = 0 First, we solve, 2a2 + 5ab + 2b2 = 2a2 + 4ab + ab + 2b2 Here, = 2a[a + 2b] + b[a + 2b] = (a + 2b) (2a + b) 1 Hence, the equation becomes 9x2 – 9(a + b)x + (a + 2b)(2a + b) = 0 2 Þ 9x – 3[3a + 3b]x + (a + 2b)(2a + b) = 0 Þ 9x2 – 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a +b) =0 Þ 9x2 – 3(a + 2b)x – 3(2a + b)x +(a + 2b)(2a + b)= 0 Þ 3x[3x – (a + 2b)] – (2a + b) [3x – (a + 2b)] = 0 Þ [3x – (a + 2b)][3x – (2a + b)] = 0 1 Þ 3x – (a + 2b) = 0 or 3x – (2a + b) = 0 a + 2b 2a + b or x = ½ Þ x= 3 3
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
−2
⇒
x 2 − 2x
= 3
⇒ 3x2 – 6x = – 2 ⇒ 3x2 – 6x + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 3, b = – 6 and c = 2 Now,
x =
−( −6 ) ± ( −6 )2 − 4 × 3 × 2 2×3
=
6 ± 36 − 24 6
=
Hence,
=
x =
1 ½
− b ± b 2 − 4 ac 2a
=
=
½
6 ± 12 6
3+ 3 3− 3 , . 3 3
1
Q. 2. A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/h more, it would have taken 48 minutes less for the journey find the original speed of the train. A [CBSE SQP, 2020]
2
⇒ x2 + 5x – 2250 = 0 1 ⇒ (x + 50)(x – 45) = 0 \ x = 45 1 1 Hence original speed of the train = 45 km/h [CBSE SQP Marking Scheme, 2019] Detailed Solution: Let the original speed of the train be x km/h Distance \ Time = Speed Time taken to cover a distance of 360 km, Distance 360 h = t1 = x Speed
1
Time taken to cover a distance of 360 km, when speed is increased by 5 km/h 360 t1 = h 1 x+5 Given, t1 = t2 = 48 minutes 48 4 = h = 60 5 360 360 4 − = x x+5 5 360 x + 1800 − 360 x 4 ⇒ = x( x + 5 ) 5 Then,
450 1 = x 2 + 5x 5 ⇒ x2 + 5x – 2250 = 0 ⇒ x2 + 50x – 45x – 2250 = 0 ⇒ x(x + 50) – 45(x + 50) = 0 ⇒ (x + 50)(x – 45) = 0 If x + 50 = 0 ⇒ x = – 50, but speed can not be negative and if x – 45 = 0 ⇒ x = 45 Hence, the speed of the train = 45 km/h
3± 3 3
4 1800 = x 2 + 5x 5
⇒
1
6±2 3 6
Sol. Let the original speed of the train be x km/h 360 360 48 = \ − + x x 5 60
⇒
1
1 7 Q. 3. Two water taps together can fill a tank in 1 8 hours. 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. A + U [CBSE Delhi Set-I, II, III, 2019, 2017] [OD Set-III, Foreign Set-III, 2016]
Sol. Let the smaller tap fills the tank in x hrs \ the larger tap fills the tank in (x – 2) hrs. 15 h Time taken by both the taps together = 8 1 1 8 = 2 Therefore + − x x 2 15 ⇒ 4x2 – 23x + 15 = 0 1 ⇒ (4x – 3)(x – 5) = 0 3 \ x = 5 1 x ≠ 4 Smaller and larger taps can fill the tank separately in 5 h and 3 h respectively. 1 [CBSE Marking Scheme, 2019]
82 ]
1
Detailed Solution: Let the time taken by the smaller diameter tap = x h. \ Time for larger diameter tap = (x – 2) h. 7 15 h. Total time taken = 1 = 8 8
1
Portion filled in one hour by smaller diameter tap
=
and by larger diameter tap =
According to the problem,
1 x 1 x−2
⇒
1 1 8 + = x − 2 x 15
⇒
x−2+x 8 x( x − 2 ) = 15
⇒ ⇒ ⇒ ⇒
15(2x – 2) = 8x(x – 2) 30x – 30 = 8x2 – 16x 2 8x – 46x + 30 = 0 4x2 – 23x + 15 = 0
1
1
1
[ 83
qUADRATIC eQUATIONS
Hence, the marks in Hindi = 12 and marks in
⇒ 4x2 – 20x – 3x + 30 = 0 ⇒ 4x(x – 5) – 3(x – 5) = 0 ⇒ (4x – 3)(x – 5) = 0
If x =
x=
3 or x = 5 4
−5 3 3 . , then x – 2 = – 2 = 4 4 4
Since, time cannot be negative, we neglect x =
3 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.
1 Q. 4. In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
C + A [CBSE Outside Delhi Set- 1, 2019]
Sol. Let marks in Hindi be x
Then marks in Eng = 30 – x
1
\ (x + 2)(30 – x – 3) = 210
1
⇒
x2 – 25x + 156 = 0 or (x – 13)(x – 12) = 0 1
⇒
x = 13 or x = 12
\
30 – 13 = 17 or 30 – 12 = 18
Sol. Let total length of cloth = l m. 200 l
1
200 2 = 200 ⇒ (l+5) l
1
⇒ l2 + 5l – 500 = 0
1
⇒
⇒
(l + 5)(200 – 2l) = 200l (l + 25)(l – 20) = 0
⇒
l = 20
1
200 Rate per meter = ` 20
= ` 10 per meter 1 [CBSE Marking Scheme, 2019]
1
Detailed Solution:
1
[CBSE Marking Scheme, 2019] Detailed Solution: Let the marks in Hindi be x and the marks in English be y. According to question, x + y = 30
Let length of the cloth = x m Cost of cloth per meter = ` y Given, x × y = 200 200 ...(i) 1 x According to given conditions, 1. If the piece were 5 m longer 2. Each meter of cloth cost ` 2 less i.e., (x + 5)(y – 2) = 200 ⇒ xy – 2x + 5y – 10 = 200 ⇒ xy – 2x + 5y = 210 1 200 200 x − 2x + 5 ⇒ = 210 1 x x [from eq. (i)] 1000 200 − 2 x + ⇒ = 210 x 1000 − 2 x = 10 ⇒ x ⇒ 1000 – 2x2 = 10x 2 ⇒ x + 5x – 500 = 0 1 ⇒ x2 + 25x – 20x – 500 = 0 ⇒ x(x + 25) – 20(x + 25) = 0 ⇒ (x + 25)(x – 20) = 0 ⇒ x = – 25 or x = 20 \ x = 20 (neglecting x = –25) 200 200 = = 10 y = 1 x 20 [From eq. (i)] Hence, length of the piece of cloth is 20 m and rate per meter is ` 10.
⇒
⇒ y = 30 – x ...(i) 1 If he had got 2 marks more in Hindi, then his marks would be = x + 2
and if he had 3 marks less in English, then his marks would be = y – 3 According to question, (x + 2)(y – 3) = 210 ⇒ (x + 2)(30 – x – 3) = 210 ⇒ (x + 2)(27 – x) = 210 ⇒ 27x – x2 + 54 – 2x = 210 ⇒ –x2 + 25x – 156 = 0 ⇒ x2 – 25x + 156 = 0 ⇒ x2 – 13x – 12x + 156 = 0 ⇒ x(x – 13) – 12(x – 13) = 0 ⇒ (x – 12)(x – 13) = 0 ⇒ Either x = 12 or x = 13 when x = 12, then y = 30 – 12 = 18 when x = 13, then y = 30 – 13 = 17
C + A [CBSE OD Set-II, 2019]
Rate per meter = `
\ Marks in Hindi & English are (13, 17) or (12, 18)
English = 18 or the marks in Hindi = 13 and marks in English = 17. 1 Q. 5. The total cost of a certain length of a piece of cloth is ` 200. If the piece was 5 m longer and each metre of cloth costs ` 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre ?
⇒
1
[from eq. (i)]1
1
y =
84 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
T
Q. 6. A shopkeeper buy certain number of books in ` 80. If he buy 4 more books then new cost price of each book is reduced by ` 1. Find the number of books initially he buy. [CBSE Delhi Board Term, 2019]
opper Answer, 2019
Sol.
5
[ 85
qUADRATIC eQUATIONS
x2 – 42x + 3x – 126 = 0 x(x – 42) + 3(x – 42) = 0 (x – 42)(x + 3) = 0 1 If x + 3 = 0 ⇒ x = – 3, speed cannot be negative. If x – 42= 0 ⇒ x = 42 \ Speed of train = 42 km/h
Sol. Let the original average speed of train be x km/hr. 63 72 + Therefore, = 3 1 x x+6 1 x2 – 39x – 126 =0
Q. 8. 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.
(x – 42)(x + 3) = 0 1 x is not equal to – 3. \ x = 42 1 Thus, original speed of train is 42 km/h. 1 [CBSE Marking Scheme, 2018]
C + A [CBSE Delhi/OD Set 2018, 2017]
[Board Term-II, O.D. Set-II, 2016]
T
etailed Solution: D Let the original speed of train be x km/h Distance Time = speed Total time to complete journey = 3 h 63 72 + = 3 x x+6 63x + 378 + 72 x =3 x( x + 6 ) 135 x + 378 =3 x 2 + 6x 3x2 + 18x = 135x + 378 3x2 – 117x – 378 = 0 x2 – 39x – 126 = 0 Detailed Solution:
Sol.
1
1
1
1
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 1 According to the question, 24 24 Þ − = 1 1 18 − x 18 +x 24(18 + x ) − 24(18 − x ) =1 Þ 18 2 − x 2 Þ 432 + 24x – 432 + 24x = 324 – x2 Þ 48x = 324 – x2 2 Þ x + 48x – 324 = 0 1 Þ x2 + 54x – 6x – 324 = 0 Þ x(x + 54) – 6(x + 54) = 0 Þ (x + 54)(x – 6) = 0 Þ x + 54 = 0 or x – 6 = 0 Þ x = – 54 or x = 6 1 Since, speed cannot be negative. Hence, the speed of steam x = 6 km/h. 1 [CBSE Marking Scheme, 2018]
Q. 7. 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/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed ? C + A [CBSE Delhi Set, 2018]
opper Answer, 2018
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
86 ]
5
Q. 9. Solve
1 1 1 1 = + + , where a + b ≠ 0. (a + b + x) a b x A [CBSE SQP, 2018, 2016]
[Board Foreign set II, III 2017] Sol. Given,
Þ
1 1 1 1 = + + a+b+x a b x 1 1 1 1 - = + a+b+x x a b
Þ x(a + b+ x) = – ab Þ x2 + (a + b)x + ab = 0 Þ (x + a)(x + b) = 0 Þ x = – a or x = – b 1 [CBSE Marking Scheme, 2018]
COMMONLY MADE ERROR 1
Candidates do error in simplifying this type of equations.
Þ
x - (a + b + x) a+b = x( a + b + x ) ab
1
Þ
x -a-b-x a+b = x( a + b + x ) ab
1
Þ
-( a + b ) a+b = x( a + b + x ) ab
ANSWERING TIP
1
Adequate
simplifying equations.
practice is necessary for this type of quadratic
T
Q. 10. 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 A [Board OD Set III, 2017] days will B take to finish it ?
Sol.
opper Answer, 2017
[ 87
qUADRATIC eQUATIONS
Q. 11. In a rectangular part of dimensions 50 m× 40 m a rectangular pond is constructed so that the area of grass strip of uniform breadth surrounding the pond would be 1184 m2. Find the length and breadth of the pond.
⇒ x = 12 or x = 13 If x = 12, then marks in Mathematics = 12 and marks in English = 18 If x = 13, then marks in Mathematics = 13 and marks in English = 17.
1
1 1
2
A [Board Foregin Set-I, III 2017] Sol. Let width of grass strip be x m. \ Length of pond = (50 – 2x) m and Breadth of pond = (40 – 2x) m 1 And area of park – area of pond = area of grass strip Þ (50 × 40) – (50 – 2x)(40 – 2x) = 1184 Þ 2000 – 2000 + 180x – 4x2 = 1184 1 Þ x2 – 45x + 296 = 0 Þ x2 – 37x – 8x + 296 = 0 Þ x(x – 37) – 8(x – 37) = 0 Þ x = 8 or 37 1 (37 is rejected, as it gives negatives values for length & breadth) Thus, the length of pond = 50 – 2 × 8 = 34 m 1 and breadth of pond = 40 – 2 × 8 = 24 m. 1 Q. 12. In a class test Raveena got a total of 30 marks in English and Mathematics. Had she got 2 more marks in Mathematics and 3 marks less in English then the product of her marks obtained would have been 210. Find the individual marks obtained A [OD Compt. I, II, III, 2017] in two subjects. Sol. Let marks in mathematics be x. Then marks in English = 30 – x 1 (x + 2)(30 – x – 3) = 210 1 ⇒ (x + 2)(27 – x) = 210 ⇒ 27x + 54 – x2 – 2x = 210 ⇒ –x2 + 25x = 210 – 54 = 156 2 ⇒ x – 25x + 156 = 0 1 ⇒ x2 – 13x – 12x + 156 = 0 ⇒ x(x – 13) – 12(x – 13) = 0 ⇒ (x – 13)(x – 12) = 0
Q. 13. Solve for x:
2x 2x + 5 - 24 = 0, x ≠ 5 x -5 x - 5 U [CBSE S.A.-2, 2016] 2
2x 2x + 5 - 24 = 0 Sol. Given, x - 5 x - 5 2x Let = y ( x - 5) ∴ y2 + 5y – 24 = 0 Þ (y + 8)(y – 3) = 0 y = 3 or – 8
1 1 1
Putting y = 3 we get 2x =3 x−5 2x = 3x – 15 Þ x = 15 1 2x =–8 Again, for y = – 8, x−5 2x = – 8x + 40 10x = 40 x = 4 Hence, x = 15 or 4 1 [CBSE Marking Scheme, 2016]
Q. 14. Find x in terms of a, b and c: a b 2c where x ≠ a, b, c + = x−a x−b x−c
C + U [Board Term-2 Delhi Set 1, 2016]
Sol. a(x – b)(x – c) + b(x – a)(x – c) = 2c(x – a)(x – b) 2 2 Þ x (a + b – 2c) + x(–ab –ac –ab – bc + 2ac + 2bc) = 0 Þ x2(a + b – 2c) + x(–2ab + ac + bc) = 0 2
ac + bc - 2 ab and x = 0 Þ x = − a + b - 2 c
1
[CBSE Marking Scheme, 2016]
88 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 15. 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
Sol.
A [Foreign Set III, 2016]
Sol. Let numerator be x. x 2x + 1
x 2x + 1 58 + = Again, 2 x + 1 x 21
1 1
Þ 21[x2 + (2x + 1)2] = 58 (2x2 + x) Þ 11x2 – 26x – 21 = 0
7 x = 3 or − 11
1
3 1 7 [CBSE Marking Scheme, 2016]
Sol.
x2 + x2 + 14x – 240 = 0
or
x2 + 7x – 120 = 0
1½
x =
-7 ± 49 + 480 2
x =
− 7 ± 23 = 8, – 15 (rejected) 2
∴
x = 8 m and x + 7 = 15 m 1
Hence, distance between two gates = 8 m and 15 m. [CBSE Marking Scheme, 2016] Q. 18. The time taken by a person to cover 150 km was 1 2 hours more than the time taken in the return 2 journey. If he returned at a speed of 10 km/h more than the speed while going, find the speed in km/h in each direction. Sol. Let the speed while going be x km/h.
A [CBSE Delhi Set III, 2016]
Here, a + b + c = 60 and c = 25 a + b = 60 – c a + b = 60 – 25 = 35 Using Pythagoras theorem, a2 + b2 = 625 Using identity, (a + b)2 = a2 + b2 + 2ab 352 = 625 + 2ab or, 1225 – 625 = 2ab Þ ab = 300 1 Hence, area of ∆ABC = ab = 150 cm2. 2
½
x2 + (x + 7)2 = (17)2
fraction =
Q. 16. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle. A [CBSE Delhi Set II, 2016]
1
AP = x + 7
Using Pythagoras Theorem,
(rejected negative value) Hence,
∴
∴
(x – 3)(11x + 7) = 0
Let P be the location of the pole such that its distance from gate B, x metres.
AB is diameter, ∠APB = 90° and AB = 17 m 1
11x2 – 33x + 7x – 21 = 0
1
the fraction =
Then,
1
∴ Speed while returning = (x + 10) km/h
According to question,
1
Þ
150 150 5 = x x + 10 2 x2 + 10x – 600 = 0
Þ (x + 30)(x – 20) = 0 Þ
1 1
[CBSE Marking Scheme, 2016]
Q. 17. A pole has to be erected at a point on the boundary of a circular park of diameter 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. [Foreign Set I, II, 2016]
2
x = 20 or – 30
1
Rejecting negative value, ∴ Speed while going = 20 km/h
1
and speed while returning = 20 + 10 = 30 km/h 1
[CBSE Marking Scheme, 2016]
Q. 19. 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 fraction and original fraction is
29 . Find the 20
original fraction. A [Board Term-2, Delhi, 2015 Set I, III]
[ 89
qUADRATIC eQUATIONS
x -1 1 x +2 x -3 x -1 29 + = Then, x x+2 20 Þ 20[(x – 3)(x + 2) + x(x – 1)] = 29(x2 + 2x) Þ 20(x2 – x – 6 + x2 – x) = 29x2 + 58x 1 Þ 11x2 – 98x – 120 = 0 Þ 11x2 – 110x + 12x – 120 = 0 (11x + 12)(x – 10) = 0 or x = 10 1 7 . [CBSE Marking Scheme, 2015] 1 ∴ The fraction is 10
TOPIC
=
Q. 20. 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. C + A [CBSE OD, Set I, II, III, 2015] Sol. Let the length of shorter side be x m. ∴ Length of diagonal = (x + 16) m 1 and length of longer side = (x + 14) m 1 Using Pythagoras Theorem, x2 + (x + 14)2 = (x + 16)2 1
Sol. Let the denominator be x, then numerators = x – 3 x -3 1 So, the fraction = x By the given condition, x -3+2 new fraction = x+2
x+
16
x
x + 14
Þ x2 – 4x – 60 = 0 2 Þ x + 6x – 10x – 60 = 0 Þ x(x + 6) – 10(x + 6) = 0 x = – 6 or x = 10 Þ x = 10 m 1 ∴ Length of sides are 10 m and 24 m. 1 [CBSE Marking Scheme, 2015]
-2
Discriminant and Nature of Roots Scan to know more about this topic
Revision Notes
For the quadratic equation ax2 + bx + c = 0, the expression b2 – 4ac is known as discriminant i.e., Discriminant D = b2 – 4ac Nature of roots of a quadratic equation: (i) If b2 – 4ac > 0, the quadratic equation has two distinct real roots. (ii) If b2 – 4ac = 0, the quadratic equation has two equal real roots. (iii) If b2 – 4ac < 0, the quadratic equation has no real roots.
Nature of Roots
Know the Formulae Discriminant, D = b2 – 4ac.
How is it done on the
GREENBOARD?
Q.1. Find the value of k for which the equation 4x2 + kx + 25 = 0 has equal roots. Solution: Step 1: 4x2 + kx + 25 = 0 C omparing above equation with ax2 + bx + c = 0 a = 4, b = k and c = 25 Step 2: Condition for equal roots is D=0
i.e., b2 – 4ac = 0 Step 3: Substituting the values of a, b and c in the above condition. (k2) – 4(4) (25) = 0 Þ k2 – 400 = 0 Þ k2 – (20)2 = 0 Þ (k – 20) (k + 20) = 0 Þ k = 20 or – 20
90 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2
Sol. 9x + 6kx + 4 = 0 For equal roots b2 – 4ac = 0 Discriminant D = 0 so, (6k)2 – 4 × 9 × 4 = 0 ½ 36k2 = 144 k2 = 4 k = ± 2 ½ [CBSE Marking Scheme, 2020-21] Detailed Solution: Given, 9x2 + 6kx + 4 = 0. Comparing with ax2 + bx + c = 0, we get a = 9, b = 6k, c = 4 ½ Since, Discriminant, D = b2 – 4ac and for equal roots, b2 – 4ac = 0 [QD = 0] Þ (6k)2 – 4 × 9 × 4 = 0 36k2 – 144 = 0 Þ 36k2 = 144 Þ Þ k2 = 4 k = ± 2 ½ Þ Q. 2. For what value(s) of 'a' quadratic equation 3ax2 – 6x + 1 = 0 has no real roots ? A [CBSE SQP, 2020-21]
Sol. Given that, 3ax2 – 6x + 1 = 0 For no real roots b2 – 4ac < 0 ½ Discriminant D < 0 so, (– 6)2 – 4(3a) (1) < 0 12a > 36 a > 3 ½ [CBSE Marking Scheme, 2020-21] Detailed Solution: Given, 3ax2 – 6x + 1 = 0 On Comparing with ax2 + bx + c = 0, we get a = 3a, b = – 6 and c = 1 Discriminant, D = b2 – 4ac = (– 6)2 – 4 × 3a × 1 = 36 – 12a ½ For condition of 'no real roots', b2 – 4ac < 0 36 – 12a < 0 Þ Þ 12a > 36 Þ a > 3. ½ Q. 3. Find the values (s) of k for which the quadratic equation x2 + 2 2 kx + 18 = 0 has equal roots.
C + A [CBSE SQP, 2020]
⇒
2 D = ( 2 2 k ) – 4 (1)(18) = 0
½
k = ± 3 ½ [CBSE Marking Scheme, 2020]
Detailed Solution:
Since, x2 + 2 2 kx + 18 = 0 has equal roots
D = 0 b2 = 4ac
½
2
( 2 2 k ) = 4 × 1 × 18 4 × 2 × k2 = 72 72 k2 = 8 k2 = 9 ⇒ k =
9
k = ± 3. ½ Q. 4. For what values of k, the roots of the equation x2 + 4x + k = 0 are real ?
A [CBSE Delhi Set-I, II, III, 2019]
Sol. Since roots of the equation x2 + 4x + k = 0 are real ⇒ 16 – 4k ≥ 0 ½ ⇒ k ≤ 4 ½ [CBSE Marking Scheme, 2019]
C + A [CBSE SQP, 2020-21]
Sol.
1 mark each
Detailed Solution: Given quadratic equation is x2 + 4x + k = 0. Comparing the given equation with ax2 + bx + c = 0,
we get a = 1, b = 4, c = k ½ Since, given the equation has real roots ⇒ D ≥ 0 ⇒ b2– 4ac ≥ 0 ⇒ 42 – 4 × 1 × k ≥ 0 ⇒ 4k ≤ 16 ⇒ k ≤ 4 ½ Q. 5. Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0. A [CBSE OD, Set-I, II, III, 2019] Sol. 2x2 – 4x + 3 = 0 ⇒ D = 16 – 24 = – 8 \ Equation has no real roots 1 [CBSE Marking Scheme, 2019]
Q. 1. For what values of k, the given quadratic equation 9x2 + 6kx + 4 = 0 has equal roots ?
Very Short Answer Type Questions
Detailed Solution:
Given: 2x2 – 4x + 3 = 0 On comparing above with ax2 + bx + c = 0, we get, a = 2, b = -4, c = 3 We shall find D = b2 - 4ac So, D = (-4)2 - 4(2) × (3) = -8 < 0 or (-ve) Hence, the given equation has no real roots.
½
½
[ 91
qUADRATIC eQUATIONS
Q. 7. Find the value(s) of k if the quadratic equation 3 x 2 − k 3 x + 4 = 0 has real roots.
COMMONLY MADE ERROR
U [CBSE SQP 2017]
Students often make mistakes in analyzing
Sol. If Discriminant of quadratic equation is equal to zero, or more than zero, then roots are real.
the nature of roots as they get confused with the conditions.
Given,
ANSWERING TIP
3x 2 − k 3x + 4 = 0
Comparing with ax2 + bx + c = 0, we get a = 3, b = −k 3 and c = 4
Understand the different conditions for nature of roots.
Since, Discriminant, D = b2 – 4ac
Q. 6. Find the values (s) of k for which the equation x2 + 5kx + 16 = 0 has real and equal roots. U [CBSE SQP, 2018] 2
Sol. For roots to be real and equal, b – 4ac = 0 ½ ⇒ (5k)2 – 4 × 1 × 16 = 0 ½ 8 k = ± 5 [CBSE Marking Scheme, 2018]
½
and for real roots b2 – 4ac ≥ 0
(
)
2 Þ −k 3 − 4 × 3 × 4 ≥ 0 2 Þ 3k – 48 ≥ 0
Þ
k2 – 16 ≥ 0
Þ (k – 4)(k + 4) ≥ 0 \
Short Answer Type Questions-I
k ≤ – 4 and k ≥ 4.
½
2 marks each
T
Q. 1. For what value of k, the given quadratic equation kx2 – 6x – 1 = 0 has no real roots ? [CBSE Delhi Board Term, 2019]
opper Answer, 2019
Sol.
Q. 2. Find the value of k for which the roots of the quadratic equation 2x2 + kx + 8 = 0 will have the equal roots ? C + A [Board Term-II OD Compt., 2017]
Sol. For equal roots, D = 0
\
Þ
2
b – 4ac = 0 k2 = 4 × 2 × 8
1
k2 = 64
k2 = 64
1
k = ± 64
1 \ k = ± 8 2 Q. 3. If 2 is a root of the equation x + kx + 12 = 0 and the equation x2 + kx + q = 0 has equal roots, find U [CBSE SQP, 2016] the value of q.
92 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Since, 2 is the root of x2 + kx + 12 = 0 Þ (2)2 + 2k + 12 = 0 Þ 2k + 16 = 0 k = – 8 ½ Putting, k = – 8 in x2 + kx + q = 0 Þ x2 –8x + q = 0 ½ For equal roots, ½ (–8)2 – 4(1)q = 0 Þ 64 – 4q = 0 Þ 4q = 64 Þ q = 16 ½ [CBSE Marking Scheme, 2016] 2 Q. 4. Find k so that the quadratic equation (k + 1)x – 2(k + 1)x + 1 = 0 has equal roots. R [Board Term-2 2016] [CBSE Board Term-2, Set- I, III, 2015]
Sol. Since, (k + 1)x2 – 2(k + 1)x + 1 = 0
has equal roots.
D = 0
Þ
b2 = 4ac
Þ
x2 + px + 16 = 0 have equal roots if D = p2 – 4(16)(1) = 0 1 p2 = 64 ⇒ p = ± 8 ½ \ x2 ± 8x + 16 = 0 ⇒ (x ± 4)2 = 0 1 x ± 4 = 0 \ Roots are x = – 4 and x = 4 ½ [CBSE Marking Scheme, 2019] Detailed Solution: Given quadratic equation, x2 + px + 16 = 0 ...(i) If this equation has equal roots, then discriminant value is zero i.e., D = b2 – 4ac = 0 ...(ii) Now, comparing the given quadratic equation with ax2 + bx + c = 0,
4(k + 1) = 4(k + 1) 2
Þ
k + 2k + 1 = k + 1
Þ
k2 + k = 0
Þ
k(k + 1) = 0
Þ
k = – 1 does not satisfy the equation
So,
k = 0 or – 1
1
k = 0 [CBSE Marking Scheme, 2016]
Short Answer Type Questions-II Q. 1. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the A roots of the equation so obtained. [CBSE OD Set-I, II, III, 2019]
1
2
3 marks each
we get a = 1, b = p and c =16 1 \ From eq (ii), p2 – 4 × 1 × 16 = 0 ⇒ þ2 = 64 ⇒ p = ±8 when þ = 8, ½ from eq (i), x2 + 8x +16 = 0 ⇒ x2 + 4x + 4x +16 = 0 ⇒ x(x + 4) + 4(x + 4) = 0 ⇒ (x + 4)(x + 4) = 0 ⇒ x = –4, –4 Hence, roots are –4 and –4. ½ when p = – 8, from eq. (i), x2 – 8x + 16 = 0 ⇒ x2 – 4x – 4x +16 = 0 ⇒ x(x – 4) –4(x – 4) = 0 ⇒ (x – 4)(x – 4) = 0 ⇒ x = 4, 4 Hence, the required roots are either – 4, – 4 or 4, 41
Q. 2. If the roots of the equation (a2 + b2)x2 – 2(ac + bd)x + (c2 + d2) = 0 are equal, prove that
a c = . b d
A [Board OD Set III, 2017] [Delhi Comptt. Set-I, OD Set III, 2017]
Topper Answer, 2017 Sol.
[ 93
qUADRATIC eQUATIONS
3
Q. 3. ad ¹ bc, then prove that the equation. (a2 + b2)x2 + 2(ac + bd)x + (c2 + d2) = 0 has no real U [CBSE OD Set-I 2017]
roots. 2
Sol. We have, A = (a + b ), B = 2(ac + bd) and C = (c2 + d2)
2
½
For no real roots, D < 0 2
i.e., D Þ b – 4ac < 0
b2 – 4ac = [2(ac + bd)]2 – 4(a2 + b2)(c2 + d2)] 1
= 4[a2c2 + 2abcd + b2d2]
= 4[a2c2 + 2abcd + b2d2 – a2c2 – a2d2 – b2c2 – b2d2]
= – 4[a2d2 + b2c2 – 2abcd]
= – 4(ad – bc]2
– 4[a2c2 + a2d2 + b2c2 + b2d2] 1
2
Q. 4. If 2 is a root of the quadratic equation 3x + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0 has equal roots, find k. C + A [O.D. Comptt. Set-II, III, 2017] Sol. Given, 2 is a root of the equation, 3x2 + px – 8 = 0 Putting x = 2 in 3x2 + px – 8 = 0, we get 12 + 2p – 8 = 0 Þ p = – 2 Given,
4x2 – 2px + k = 0 has equal roots,
and
4x2 + 4x + k = 0 has equal roots. 1
\
D Þ b2 – 4ac = 0
Þ (4)2 – 4(4)(k) = 0 Þ
Since,
ad ¹ bc
Therefore,
D < 0
Hence, the equation has no real roots.
1
16 – 16k = 0
Þ 16k = 16 ½
\
1
k = 1.
2
Q. 5. If the roots of the quadratic equation (a – b)x + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.
opper Answer, 2016
T
U [OD, Set -II, 2016]
Sol.
3
94 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Long Answer Type Questions
5 marks each
2 Q. 1. Check whether the equation 5x – 6x – 2 = 0 has real roots and if it has, find them by the method of completing the square. Also, verify that roots obtained satisfy the given equation. U [CBSE SQP, 2017] Sol. Discriminant D = b2 – 4ac. Here, a = 5, b = (– 6) and c = (– 2) Then, b2 – 4ac = (– 6)2 – 4 × 5 × – 2 = 36 + 40 = 76 > 0 1 So the equation has real and two distinct roots. Again, 5x2 – 6x = 2 (dividing both the sides by 5) 1 6 2 2 x − x = 5 5
Þ Þ
Þ a = 0 or a3 + b3 + c3 = 3abc
Þ
On adding (i) and (iii) we get Þ 2r2 = 56 Þ r2 = 28 cm
1 and r1 = 49 – 28 = 21 Hence, radii of smaller circle is 21 cm 1 [CBSE Marking Scheme, 2017]
On adding square of the half of coefficient of x 6 9 2 9 x2 − x + = + Þ 5 25 5 25
Þ
Þ
Verification:
3 + 19 3 + 19 5 −2 − 6 5 5
x−
3 19 = ± 5 5 x=
3 + 19 3 − 19 or 5 5
1
2
=
=
=
1
Q. 3. 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.
U [Board Comptt. I, II, III, 2017]
Sol. We have r2 – r1 = 7 cm, r2 > r1 ...(i) 1 2 2 2 and p(r2 – r1) = 1078 cm ...(ii) Þ p(r2 – r1)(r2 +r1) = 1078 1 r2 + r1 =
1078 = 49 22
...(iii) 1
2
Q. 4. If roots of the quadratic equation x + 2px + mn = 0 are real and equal, show that the roots of the quadratic equation x2 – 2(m + n)x + (m2 + n2 + 2p2) = 0 are A [Foreign Set II, 2016]
also equal. 2
9 + 6 19 + 19 18 + 6 19 − − 2 5 5
1
28 + 6 19 18 + 6 19 − −2 5 5 28 + 6 19 − 18 − 6 19 − 10 5
= 0
Similarly,
3 − 19 3 − 19 5 −2 =0 − 6 5 5
Sol. For equal roots of x + 2px + mn = 0, 4p2 – 4mn = 0 Þ p2 = mn ...(i) 2 For equal roots of x2 – 2 (m + n)x + (m2 + n2 + 2p2) = 0, 4(m + n)2 – 4(m2 + n2 + 2p2) = 0, 1 m2 + n2 + 2mn – m2 – n2 – 2(mn) = 0 (From (i)) \ If roots of x2 + 2px + mn = 0 are equal, then those of x2 – 2a(m + n)x + (m2 + n2 + 2p2) = 0 are also equal. [CBSE Marking Scheme, 2016] 2
2
Hence Verified. 1
2 2 Q. 2. If the roots of the quadratic equation (c – ab)x – 2 2 2 (a – bc)x + b – ac = 0 in x are equal, then show that either a = 0 or a3 + b3 + c3 = 3abc
2
4[a4 + ac3 + ab3 – 3a2bc] = 0 a(a3 + c3 + b3 – 3abc) = 0 1
U [Board OD Set II, III 2017]
Sol. Here, A = (c – ab), B = –2(a2 – bc), C = (b2 – ac)
For real equal roots D Þ B2 – 4AC =0
Þ
Þ 4(a4 + b2c2 – 2a2bc) – 4(b2c2 – c3a – ab3 + a2bc) = 0 ½
Þ 4[a4 + b2c2 – 2a2bc – b2c2 + c3a + ab3 – a2bc) = 0 ½
1
[–2(a2 – bc)]2 – 4(c2 – ab)(b2 – ac) = 0 1
Q. 5. Find the positive values of k for which quadratic equations x2 + kx + 64 = 0 and x2 – 8x + k = 0 both will have the real roots. C + A [Foreign Set I-2016]
Sol. (i) For x2 + kx + 64 = 0 to have real roots k2 – 256 ≥ 0 ...(i) 1½ (ii) For x2 – 8x + k = 0 to have real roots 64 – 4k ≥ 0 ...(ii) 1½ For (i) and (ii) to hold simultaneously k = 16 2 [CBSE Marking Scheme, 2016] 2
Q. 6. If (– 5) is a root of the quadratic equation 2x + px + 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the values of p and k. U [CBSE Delhi Board, Set II, 2015]
[ 95
qUADRATIC eQUATIONS
Sol. Since, (–5) is a root of given quadratic equation 2x2 + px – 15 = 0 \ 2(–5)2 + p(–5) –15 = 0 1 50 – 5p – 15 = 0 5p = 35 Þ p = 7 1 And p(x2 + x) + k = 0 has equal roots ⇒ px2 + px + k = 0 So, (b)2 – 4ac = 0 2 1 (p) – 4p×k = 0 (7)2 – 4 × 7 × k = 0 28k = 49 49 7 k = = 1 28 4
7 . 1 4 [CBSE Marking Scheme, 2015]
Hence, p = 7 and k =
Q. 7. If the roots of the quadratic equation (x – a) (x – b) + (x – b)(x – c) + (x – c)(x – a) = 0 are equal. Then, show that a = b = c. U [CBSE Delhi Board, Set II, 2015] Sol. Given, (x – a)(x – b) + (x – b)(x – c) + (x – c)(x – a) = 0 Þ x2 – ax – bx + ab + x2 – bx – cx + bc + x2 – cx – ax + ac = 0
Þ 1 ( a 2 + b 2 − 2 ab ) + ( b 2 + c 2 − 2bc ) + ( c 2 + a 2 − 2 ac ) = 0 2
Q. 1. 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/h 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]
(i) What will be the distance covered by Ajay’s car in two hours ? (a) 2(x + 5) km (b) (x – 5) km (c) 2(x + 10) km (d) (2x + 5) km Sol. Correct option: (a). Explanation: Speed of Raj's car = x km/hr Speed of Ajay's car = (x + 5) km/hr Distance covered by Ajay in 2 hours = [(x + 5) × 2] km = 2(x + 5) km.
1 ( a − b )2 + ( b − c )2 + ( c − a )2 = 0 2
Þ
Þ (a – b)2 + (b – c)2 + (c – a)2 = 0 if a ¹ b ¹ c
1
(a – b)2 = 0, (b – c)2 = 0 and (c – a)2 = 0 if (a – b)2 = 0 Þ a = b (a – c)2 = 0 Þ b = c (c – a)2 = 0 Þ c = a 1 \ a = b = c Hence Proved.
Visual Case Based Questions ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark
Þ 3x2 – 2ax – 2bx – 2cx + ab + bc + ca = 0 1 Þ 3x2 – 2(a + b + c)x + (ab + bc + ca) = 0 1 For equal roots, B2 – 4AC = 0 Þ {– 2(a + b + c)}2 = 4 × 3(ab + bc + ca) Þ 4(a + b + c)2 – 12(ab + bc + ca) = 0 Þ (a + b + c)2 – 3(ab + bc + ca) = 0 2 2 Þ a + b + c2 + 2ab + 2bc + 2ac – 3ab – 3bc – 3ac =0 Þ a2 + b2 + c2 – ab – ac – bc = 0 1 1 2 a 2 + 2b 2 + 2 c 2 − 2 ab − 2 ac − 2bc = 0 Þ 2
4 marks each
(ii) Which of the following quadratic equation describe the speed of Raj’s car ? (a) x2 – 5x – 500 = 0 (b) x2 + 4x – 400 = 0 (c) x2 + 5x – 500 = 0 (d) x2 – 4x + 400 = 0 Sol. Correct option: (c). (iii) What is the speed of Raj’s car ? (a) 20 km/hour (b) 15 km/hour (c) 25 km/hour (d) 10 km/hour Sol. Correct option: (a). (iv) How much time took Ajay to travel 400 km ? (a) 20 hour (b) 40 hour (c) 25 hour (d) 16 hour Sol. Correct option: (d). Q. 2. The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream. [CBSE QB, 2021]
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(i) If John had x number of marbles, then number of marbles Jivanti had: (a) x – 45 (b) 45 – x (c) 45x (d) x – 5 Sol. Correct option: (b). Explanation: If John had x number of marbles, then Jivanti had (45 – x) marbles, because there are total 45 marbles. 1
As we see in the above figure of right angled triangle playground, the length of the sides are 5x cm and (3x – 1) cm and area of the triangle is 60 cm2. C + AE (i) The value of x is: (a) 8 (b) 3 (c) 4 (d) 5 Sol. Correct option: (b). Explanation: Given, area of triangle = 60 cm2 1 Þ × AB × BC = 0 2
(i) Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be (a) 20 km/hr (b) (20 + x) km/hr (c) (20 – x) km/hr (d) 2 km/hr Sol. Correct option: (c). Explanation: Speed of motorboat in upstream = Speed of motorboat – Speed of stream = (20 – x) km/hr (ii) What is the relation between speed ,distance and time ? (a) speed = (distance )/time (b) distance = (speed )/time (c) time = speed x distance (d) speed = distance x time Sol. Correct option: (b). (iii) Which is the correct quadratic equation for the speed of the current ? (a) x2+ 30x − 200 = 0 (b) x2 + 20x − 400 = 0 (c) x2 + 30x − 400 = 0 (d) x2 − 20x − 400 = 0 Sol. Correct option: (c). (iv) What is the speed of current ? (a) 20 km/hour (b) 10 km/hour (c) 15 km/hour (d) 25 km/hour Sol. Correct option: (b). (v) How much time boat took in downstream? (a) 90 minute (b) 15 minute (c) 30 minute (d) 45 minute Sol. Correct option: (c). Q. 3. John and Jivanti are playing with the marbles. They together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. C
(ii) Number of marbles left with Jivanti, when she lost 5 marbles: (a) x – 45 (b) 40 – x (c) 45 – x (d) x – 40 Sol. Correct option: (b). Explanation: Number of marbles left with Jivanti, when she lost 5 marbles = (45 – x – 5) = (40 – x) 1 (iii) The quadratic equation related to the given problem is: (a) x2 – 45x + 324 = 0 (b) x2 + 45x + 324 = 0 (c) x2 – 45x – 324 = 0 (d) – x2 – 45x + 324 = 0 Sol. Correct option: (a). Explanation: According to question, (x – 5)(40 – x) = 124 Þ – x2 – 200 + 40x + 5x – 124 = 0 Þ x2 – 45x + 324 = 0 1 (iv) Number of marbles John had: (a) 10 (b) 9 (c) 35 (d) 30 Sol. Correct option: (b). Explanation: x2 – 45x + 324 = 0 2 Þ x – 9x – 36x + 324 = 0 Þ x(x – 9) – 36(x – 9) = 0 Þ (x – 9)(x – 36) = 0 Either x = 9 or x = 36. Therefore, the number of marbles John had 9 or 36. 1 (v) If John had 36 marbles, then number of marbles Jivanti had: (a) 10 (b) 9 (c) 36 (d) 35 Sol. Correct option: (b). Explanation: If John had 36 marbles, then Jivanti had (45 – 36) = 9 marbles. 1 Q. 4. There is a triangular playground as shown in the below figure. Many Children and people are playing and walking in the ground.
Þ (5x)(3x – 1) = 120 Þ 3x2 – x – 24 = 0
[ 97
qUADRATIC eQUATIONS
Now, in right angled DABC, AC2 = AB2 + BC2
Þ 3x2 – 9x + 8x – 24 = 0 Þ 3x(x – 3) + 8(x – 3) = 0 Þ (x – 3)(3x + 8) = 0 8 Either x = 3 or x = − 3
O
Since length can't be negative, then x = 3. (ii) The length of AB is: (a) 8 cm (b) 10 cm (c) 15 cm (d) 17 cm Sol. Correct option: (c). Explanation: The length of AB = 5x cm = 5 × 3 cm = 15 cm (iii) The length of AC is: (a) 17 cm (b) 15 cm (c) 21 cm (d) 20 cm Sol. Correct option: (a). Explanation: Q AB = 15 cm and BC = (3x – 1) cm = (3 × 3 – 1) cm = 8 cm
(By using Pythagoras theorem)
= (15)2 + (8)2 = 225 + 64 = 289 = (17)2 1
1
½
Hence, AC = 17 cm. ½ (iv) The perimeter of DABC is: (a) 35 cm (b) 45 cm (c) 30 cm (d) 40 cm Sol. Correct option: (d). Explanation: Here, AB = 15 cm, BC = 8 cm and AC = 17 cm. Then, the perimeter of DABC = (AB + BC + CA) cm = (15 + 8 + 17) cm = 40 cm. 1 (v) The given problem is based on which mathematical concept ? (a) AP (b) Linear equation in one variable (c) Quadratic Equations (d) None of these Sol. Correct option: (c). Explanation: The given problem is based on the concept of quadratic equations.
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
chapter
5
ARITHMETIC PROGRESSIONS
Syllabus ¾
¾
Motivation for studying Arithmetic Progression. Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.
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Sum of nth term of an AP
1 Q (2 M) 1 Q (2 M) 1 Q (4 M) 1 Q (3 M) 1 Q (4 M)
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TOPIC - 1
To Find nth Term of the Arithmetic Progression
Revision Notes
An arithmetic progression is a sequence of numbers in which each term is obtained by adding or subtracting a fixed number d to the preceding term, except the first term. The difference between the two successive terms of an A.P. is called the common difference. Each number in the sequence of arithmetic progression is called a term of an A.P. The arithmetic progression having finite number of terms is called a finite arithmetic progression. The arithmetic progression having infinite number of terms is called an infinite arithmetic progression.
TOPIC - 1 To Find nth Term of the Arithmetic Progression Page No. 99
TOPIC - 2 Sum of n Terms of an Arithmetic Progression Page No. 109
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
A list of numbers a1, a2, a3, …… is an A.P., if the differences a2 – a1, a3 – a2, a4 – a3, … give the same value i.e., ak+1 – ak is same for all different values of k. The general form of an A.P. is a, a + d, a + 2d, a + 3d, …..
Scan to know more about this topic
If the A.P. a, a + d, a + 2d,………, l is reversed to l, l – d, l – 2d, ………, a, the common difference changes to negative of original sequence common difference. nth term of AP
Know the Formulae
The general (nth) term of an A.P. is expressed as: an = a + (n – 1)d. ......... from the starting. where, a is the first term and d is the common difference. The general (nth) term of an A.P. l, l – d, l – 2d,…….., a is given by: an = l + (n – 1)(– d) = l – (n – 1)d .......... from the end. where, l is the last term, d is the common difference and n is the number of terms.
Know the Terms
A sequence is defined as an ordered list of numbers. The first, second and third terms of a sequence are denoted by t1, t2 and t3 respectively. If the terms of sequence are connected with plus (+) or minus (–), the pattern is called a series. Example: 2 + 4 + 6 + 8 + ........ is a series. The sequence of numbers 0, 1, 1, 2, 3, 5, 8, 13,...... was discovered by a famous Italian Mathematician Leonasalo Fibonacci, when he was dealing with the problem of rabbit population. If the terms of a sequence or a series are written under specific conditions, then the sequence or series is called a progression. If a constant is added or subtracted from each term of an A.P., the resulting sequence is also an A.P. If each term of an A.P. is multiplied or divided by a constant, the resulting sequence is also an A.P. If the nth term is in linear form i.e., an + b = an, the sequence is in A.P. If the terms are selected at a regular interval, the given sequence is in A.P. If three consecutive numbers a, b and c are in A.P., the sum two numbers is twice the middle number i.e., 2b = a + c.
How is it done on the
GREENBOARD?
Q.1. Which term of the A.P. 6, 13, 20, 27, ...... is 98 more than its 24th term ? Solution: Step I: The given A.P. is 6, 13, 20, 27, .......... Here first term, a = 6 Common difference, d = 13 – 6 = 7 Step II: The 24th term, a24 = a + (24 – 1)d or, a24 = 6 + 23 × 7 a24 = 6 + 161 a24 = 167
Step III: Now according to question,
a24 + 98 = an
167 + 98 = a + (n – 1)d
265 = 6 + (n – 1)7 259 = (n – 1)7
259 = n – 1 7
or
37 = n – 1 n = 38
Hence, 38th term is the required term.
[ 101
aRITHMETIC pROGRESSIONS
Mnemonics Concept: nth Term of Arithmetic Progressio n = a + (n – 1)d. Nokia Offers Additional Programmers in English To Attract Positive New One Buyer Daily Interpretation: Nokia's 'N' is nth term.
Offer's 'O' is of Additional's 'A' is Arithmetic Programmer's 'P' is Progression In's 'I' is is. English's 'E' is Equal To's 'T' is To Attract's 'A' is a Positive's 'P' is + New's 'N' is n One buyer is – 1 Daily's 'D' is d
Very Short Answer Type Questions Q. 1. Which term of the following A.P. 27, 24, 21, ........ A [CBSE SQP, 2020-21] is zero ?
Sol. We know that an = a + (n – 1)d l = 0 0 = 27 + (n – 1)(– 3) ½ 30 = 3n n = 10 ½ 10th term of the given A.P. is zero. [CBSE Marking Scheme, 2020-21] Detailed Solution: Given A.P. = 27, 24, 21, ........... . Here, a = 27 and d = 24 – 27 = – 3 and, l = 0 = an \ an = a + (n – 1)d ⇒ 0 = 27 + (n – 1)(– 3) ⇒ – 3n + 3 = – 27 ⇒ – 3n = – 27 – 3 = – 30 ⇒ n = 10.
½
½
Q. 2. In an Arithmetic Progression, if d = – 4, n = 7, A [CBSE SQP, 2020-21] an = 4, then find a.
Q. 3. Find the value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P. R [CBSE Delhi, Set-I, 2020] Sol. Q 2x, (x + 10) and (3x + 2) are in A.P. ⇒ (x + 10) – 2x = (3x + 2) – (x + 10) ½ ⇒ – x + 10 = 2x – 8 ⇒ – x – 2x = – 8 – 10 ⇒ – 3x = – 18 Hence, x = 6. ½ Q. 4. If the first term of an A.P. is thp and the common difference is q, then find its 10 term. R [CBSE Delhi, Set-I, 2020] Sol. We have, first term (a) = p, Common difference (d) = q and n = 10 Then, an = a + (n – 1)d ½ ⇒ a10 = p + (10 – 1)q ⇒ a10 = p + 9q. ½ 1 1− p , Q. 5. Find the common difference of the A.P. , p p 1 − 2p , ............... . p
R [CBSE OD Set-I, 2020] 1 1 − p 1 − 2p , ... Sol. Given A.P. = , p p p
Sol. We know that an = a + (n – 1)d 4 = a + 6 × (– 4) ½ a = 28 ½ [CBSE Marking Scheme, 2020-21]
1 mark each
Detailed Solution: We have, d = – 4, n = 7, and an = 4 \ an = a + (n – 1)d ⇒ 4 = a + (7 – 1)(– 4) ⇒ 4 = a + 6 (– 4) = a – 24 ⇒ a = 4 + 24 ⇒ a = 28.
½
Here, let
1 1−p and a2 = p p
\ Common difference = a2 – a1 = =
½
a1 =
=
1−p 1 − p p
1− p −1 p −p p
= – 1.
1
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Numbers are 12, 15, 18, ..., 99 ½ \ 99 = 12 + (n – 1) × 3 ⇒ n = 30 ½ [CBSE Marking Scheme, 2019] Detailed Solution: Numbers divisible by 3 are 3, 6, 9, 12, 15, -------, 96, 99 Lowest two digit number divisible by 3. is 12. and highest two digit number divisible by 3 is 99.
Hence, the sequence start with 12 ends with 99 and common difference is 3.
So, the A.P. will be 12, 15, 18, ----, 96, 99 Here, a = 12, d = 3, l = 99 \ l = a + (n – 1)d \ 99 = 12 + (n – 1)3 ⇒ 99 – 12 = 3(n – 1) 87 3 ⇒ n – 1 = 29 ⇒ n = 30 Therefore, there are 30, two digit numbers divisible by 3. Q. 8. In an A.P., if the common difference (d) = – 4, and the seventh term (a7) is 4, then find the first term. U [Delhi/OD, 2018] ⇒
Q. 6. Find the nth term of the A.P. a, 3a, 5a, ....... . A [CBSE SQP, 2020-21] Sol. Given A.P. = a, 3a, 5a, ... Here first term, a = a and d = 3a – a = 2a ½ \ nth term = a + (n – 1)d = a + (n – 1)2a = a + 2na – 2a = 2na – a = (2n – 1)a. ½ Q. 7. How many two digits numbers are divisible by 3 ? U [CBSE Delhi Set-1, 2019]
n – 1 =
Sol. Since, ⇒
a + 6(– 4) = 4 a = 28 1 [CBSE Marking Scheme, 2018]
Detailed Solution:
Topper Answer, 2018
Q. 9. Which term of the A.P. 8, 14, 20, 26, ....... will be 72 1 \ an = + ( n − 1)1 more than its 41st term. m A [CBSE Outside Delhi Set-II 2017] [CBSE Board Comptt. Set-III, 2017] Hence, an = 1 + n − 1 Sol. Given a = 8 and d = 6. m Let nth term be 72 more than its 41th term. 1 + ( n - 1)m ∴ tn – t41 = 72 = 1 m 8 + (n – 1)6 – (8 + 40 × 6) = 72 8 + (n – 1)6 = 320 Q. 11. If the nth term of the A.P. – 1, 4, 9, 14, .... is 129. (n – 1)6 = 312 Find the value of n. n – 1 = 52 n = 53 1 A [CBSE Delhi Comptt. Set I, II, III, 2017] 1 1 + m 1 + 2m Q. 10. Write the nth term of the A.P. , , , ..... Sol. Given, a = – 1 and d = 4 – (– 1) = 5 m m m 1 1 + m 1 + 2m an = – 1 + (n – 1) × 5 = 129 ½ , ..... , , m m m or, (n – 1)5 = 130
A [CBSE Delhi Comptt. Set-I, II, III, 2017]
Sol. We have,
1 a = m d =
1+ m 1 − =1 m m
(n – 1) = 26
n = 27 th
Hence, 27 term = 129.
½
[CBSE Marking Scheme, 2017]
[ 103
aRITHMETIC pROGRESSIONS
Q. 12. What is the common difference of an A.P. in which a21 – a7 = 84 ? A [CBSE Outside Delhi Set-I, II, III, 2017]
Topper Answer, 2017 Sol.
Q. 13. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P. ? C + A [OD Set II, 2016]
Topper Answer, 2016 Sol.
2 , 8 , 18 , ... .
U [SQP, 2016] [Foreign Set-I, II, III, 2015] Sol. Given sequence is an A.P.
2 , 8 , 18 , ... =
Hence, or,
2 , 2 2 , 3 2 ... a =
2,d=
2 and n = 10
an = a + (n – 1)d a10 =
=
2 + (10 - 1) 2 2 +9 2
= 10 2 Hence,
a10 =
200 .1
Q. 15. Is series
A [CBSE SQP, 2020]
U [CBSE, Term-2, 2015]
Sol. Common difference, d1 = Again,
=
d2 =
6- 3 3
(
2 -1
)
9- 6
= 3 - 6 d3 = 12 - 9 = 2 3 - 3 As common differences are not equal. Hence, the given series is not an A.P. [CBSE Marking Scheme, 2015] 1
Short Answer Type Questions-I
2 marks each
Sol. 110, 120, 130, ........, 990 an = 990 ⇒ 110 + (n – 1) × 10 = 990 1 \ n = 89 1 [CBSE SQP Marking Scheme, 2020]
Q. 1. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
3 , 6 , 9 , 12 , ..... an A.P. ? Give reason.
Q. 14. Find the tenth term of the sequence:
1
104 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Detailed Solution: The number which ends with 0 is divisible by 2 and 5 both. \ Such numbers between 102 and 998 are: 110, 120, 130, .........., 990. Last term, an = 990 a + (n + 1)d = 990 1 110 + (n – 1) × 10 = 990 110 + 10n – 10 = 990 10n + 100 = 990 10n = 990 – 100 10n = 890 890 = 89. 1 n = 10 Q. 2. Show that (a – b)2, (a2 + b2) and (a + b)2 are in A [CBSE Delhi Set-I, 2020] A.P. . 2 2 2 Sol. Given: (a – b) , (a + b ) and (a + b)2 Common difference, d1 = (a2 + b2) – (a – b)2 = a2 + b2 – (a2 + b2 – 2ab) = a2 + b2 – a2 – b2 + 2ab = 2ab ½ and d2 = (a + b)2 – (a2 + b2) = a2 + b2 +2ab – a2 – b2 = 2ab ½ Since, d1 = d2 Hence, (a – b)2, (a2 + b2) and (a + b)2 are is an A.P. 1 Hence Proved. Q. 3. Which term of the A.P. 3, 15, 27, 39,... will be 120 more than its 21st term ? A [CBSE Delhi Set-I, 2019]
Sol. an = a21 + 120 = (3 + 20 × 12) + 120 = 363 1 \ 363 = 3 + (n – 1) × 12 ⇒ n = 31 1 or 31st term is 120 more than a21. [CBSE Marking Scheme, 2019] Detailed Solution: Given A.P. is: 3, 15, 27, 39 Here, first term, a = 3 and common difference, d = 12 Now, 21st term of A.P. is t21 = a + (21 – 1) d[tn= a+(n – 1)d] \ t21 = 3 +20 × 12 = 243 Therefore, 21st term is 243 We need to calculate term which is 120 more than 21st term i.e., it should be 243 +120 = 363 1 Therefore, tn= 363 \ tn = a + (n – 1)d
⇒ 363 = 3+ (n – 1)12 ⇒ 360 = 12(n – 1) ⇒ n – 1 = 30 ⇒ n = 31 So, 31st term is 120 more than 21st term.
Q. 4. Find the 20th term from the last term of the A.P.: 3, 8, 13, ...... 253. A [CBSE SQP, 2018] Sol. 20th term from the end = l – (n – 1)d ½ = 253 – 19 × 5 1 = 158 ½ [CBSE Marking Scheme, 2018] Detailed Solution: Given A.P.: 3, 8, 13, .......... 253 Here, first term (a) = 3, common difference (d) = 8 – 3 = 5 and last term (l) = 253 1 Then, 20th term from the end of the A.P. = l – (n – 1)d = 253 + (20 – 1)5 = 253 – 95 th
Q. 5. If 7 times the 7 term of an A.P. is equal to 11 times its 11th term, then find its 18th term. A [CBSE SQP-2018] [Foreign Set-2017]
[CBSE Board Term-II, 2016]
Sol. 7a7 = 11a11 ⇒ 7(a + 6d) = 11(a + 10d) ⇒
1
a + 17d = 0
\
a18 = 0
1
[CBSE Marking Scheme, 2018] Detailed Solution: Given, Q
7a7 = 11a11 an = a + (n – 1)d
Then, 7[a + (7 – 1)d] = 11[a + (11 – 1)d] ⇒
7(a + 6d) = 11(a + 10d)
⇒
7a + 42d = 11a + 110 d
⇒
11a – 7a = 42d – 110d
⇒
4a = – 68d
⇒ a = – 17d 1 ⇒ a + 17d = 0 i.e., a + (18 – 1)d = 0 Hence, a18 = 0. 1 Q. 6. Find how many integers between 200 and 500 are divisible by 8. A [Board Delhi comptt. Set-I, II, III, 2017] Sol. Integers divisible by 8 are 208, 216, 224, ......, 496. 1 Which is an A.P. Given: a = 208, d = 8 and l = 496 Let the numbers of terms in A.P. be n. \
1
1
= 158.
an = a + (n – 1)d = l 208 + (n – 1)d = 496 (n – 1)8 = 496 – 208 288 n – 1 = 8
½
= 36 n = 36 + 1 = 37 ½ Hence, no. of required integers divisible by 8 = 37.
[ 105
Sol. Here, a5 = a + 4d = 26 ...(i) ½ and a10 = a + 9d = 51 ...(ii) ½ Solving Eqns. (i) and (ii), we get or, 5d = 25 d = 5 ½ and a = 6 Hence, the A.P. is 6, 11, 17 ........ ½ [CBSE Marking Scheme, 2017] Q. 8. How many two digit numbers are divisible by 7 ? A [CBSE SQP, 2016] Sol. Two digit numbers which are divisible by 7 are: 14, 21, 28, ......, 98. ½ It forms an A.P. Here, a = 14, d = 7 and an = 98 ½ Since, an = a + (n – 1)d 98 = 14 + (n – 1)7 ½ 98 – 14 = 7n – 7 84 + 7 = 7n or, 7n = 91 or, n = 13 ½ [CBSE Marking Scheme, 2016]
Q. 9. In a certain A.P. 32th term is twice the 12th term. Prove that 70th term is twice the 31st term. A [Board Term-2, 2015]
Sol. Let the 1st term be a and common difference be d. According to the question, a32 = 2a12 ∴ a + 31d = 2(a + 11d) a + 31d = 2a + 22d a = 9d 1 Again, a70 = a + 69d = 9d + 69d = 78d a31 = a + 30d = 9d + 30d = 39d Hence, a70 = 2a31 Hence Proved. 1 [CBSE Marking Scheme, 2015] Q. 10. The 8th term of an A.P. is zero. Prove that its 38th term is triple of its 18th term. A [CBSE Board Term-2, 2015]
Sol. Given, a8 = 0 or, a + 7d = 0 or, a = – 7d ½ or, a38 = a + 37d or, a38 = – 7d + 37d = 30d ½ And, a18 = a + 17d = – 7d + 17d = 10d ½ or, a38 = 30d = 3 × 10d = 3 × a18 ∴ a38 = 3a18. Hence Proved. ½ [CBSE Marking Scheme, 2015] Q. 11. The fifth term of an A.P. is 20 and the sum of its seventh and eleventh terms is 64. Find the common A [Foreign Set II, 2015] difference. [CBSE Board Term-II, 2015] Sol. Let the first term be a and common difference be d. Then, a +4d = 20 ...(i) ½ and a + 6d + a + 10d = 64 a + 8d = 32 ...(ii) 1 Solving equations (i) and (ii), we get d = 3 Hence, common difference, d = 3 ½ [CBSE Marking Scheme, 2015]
Q. 7. The fifth term of an A.P. is 26 and its 10th term is 51. Find the A.P. A [OD Comptt. Set-II, 2017]
Q. 12. Find the middle term of the A.P. 213, 205, 197, ..... A [CBSE Delhi Board Term, 2015] 37.
aRITHMETIC pROGRESSIONS
Sol. Here, a = 213, d = 205 – 213 = – 8 and l = 37
Let the number of terms be n. ∴ or, or, or,
Let
tn < 0 tn = a + (n – 1) d
½
37 = 213 + (n – 1)(– 8) 37 – 213 = – 8(n – 1) - 176 = 22 -8
½
n = 22 + 1 = 23
½
n – 1 =
23 + 1 = 12th The middle term will be = 2 ∴ a12 = a + (n – 1)d
½
= 213 + (12 – 1)(– 8) = 213 – 88 = 125 Thus, the middle term will be 125.
Short Answer Type Questions-II 1 1 3 Q. 1. Which term of the A.P. 20, 19 , 18 , 17 , ...... 4 2 4 is the first negative term. A [CBSE OD Set-III, 2020] Sol. Here, First term, a = 20 77 3 1 − 20 = − and Common difference, d = 4 4
l = a + (n – 1)d
\ ⇒ ⇒ ⇒
½
[CBSE Marking Scheme, 2015]
3 marks each 3 20 + (n – 1) − < 0 4
½
80 – 3n + 3 < 0 83 – 3n < 0 83 n > 3
⇒ n > 27.6 ⇒ n = 28 Hence, the first negative term is 28.
1
106 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 2. Find the middle term of the A.P. 7, 13, 19, ...., 247. U [CBSE OD Set-III, 2020] Sol. In this A.P., a = 7, d = 13 – 7 = 6 ½ and tn = 247 ½ tn = a + (n – 1)d \ 247 = 7 + (n – 1)6 ⇒ 6(n – 1) = 240 ⇒ n – 1 = 40 ⇒ n = 41 1
Hence,
the middle term =
=
n+1 2 41 + 1 2
42 = 2 = 21.
1
th Q. 3. For what value of n, are the n terms of two A.Ps 63, 65, 67,.... and 3, 10, 17,.... equal ? C + A [CBSE Outside Delhi Set-III, 2017]
Topper Answer, 2017 Sol.
Q. 4. If the 10th term of an A.P. is 52 and the 17th term is 20 more than the 13th term, find A.P.
Sol.
A [CBSE, Outside Delhi Set-I 2017]
a10 = 52
or,
a + 9d = 52
Also
a + 16d – (a + 12d) = 20
4d = 20
d = 5
Substituting, the value of d in (i), we get
Hence,
...(i) 1
a17 – a13 = 20
a = 7 A.P. = 7, 12, 17, 22 .....
½
1 ½
[CBSE Marking Scheme, 2017]
Q. 5. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and A [CBSE SQP, 2016] the common difference.
3
Sol. Let the first term of A.P. be a and common difference be d. Given, a9 = 7a2 or, a + 8d = 7(a + d) ...(i) ½ and a12 = 5a3 + 2 Again, a + 11d = 5(a + 2d) + 2 ...(ii) 1 From (i), a + 8d = 7a + 7d – 6a + d = 0 ...(iii) From (ii), a + 11d = 5a + 10d + 2 – 4a + d = 2 ...(iv) Subtracting (iv) from (iii), we get – 2a = – 2 or, a = 1 1 From (iii), – 6 + d = 0 d = 6 ½ Hence, first term = 1 and common difference = 6 [CBSE Marking Scheme, 2016]
[ 107
aRITHMETIC pROGRESSIONS
Q. 6. The digits of a positive number of three digit number are in A.P. and their sum is 15. The number obtained by A [CBSE Delhi Set II, 2016] reversing the digits is 594 less than the original number. Find the number.
Topper Answer, 2016 Sol.
Q. 7. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes (1st and 4th) to the product of middle (2nd and 3rd) is 5 : 6. U [Foreign Set I, 2016] Sol. Let the four parts be a – 3d, a – d, a + d and a + 3d. ∴ a –3d + a – d + a + d + a + 3d = 56 or, 4a = 56 a = 14 1 Hence, four parts are 14 – 3d, 14 – d, 14 + d and 14 + 3d. Now, according to question, or,
(14 - 3d )(14 + 3d ) 5 = (14 - d )(14 + d ) 6 196 - 9d 2 196 - d 2
=
5 6
or, 6(196 – 9d2) = 5(196 – d2) or, 6 × 196 – 54d2 = 5 × 196 – 5d2 or, 6 × 196 – 5 × 196 = 54d2 – 5d2 or, (6 – 5) × 196 = 49d2 196 or, d2 = =4 49
3
or, d = ± 2 1 ∴ The four parts are {14 – 3(± 2)}, {14 –(± 2)} Hence, first possible division will be 8, 12, 16 and 20. ½ and second possible division will be 20, 16, 12 and 8. ½ th th th Q. 8. The p , q and r terms of an A.P . are a, b and c respectively. Show that a(q – r) + b(r – p) + c(p – q) U [Foreign Set II, 2016] = 0. Sol. Let the first term be a' and the common difference be d. a = a’ + (p – 1)d, b = a’ + (q – 1)d and c = a’ + (r – 1)d 1½ a(q – r) = [a’ + (p – 1)d][q – r] b(r – p) = [a’ + (q – 1)d][r – p] and c(p – q) = [a’ + (r – 1)d][p – q] ½ \ a(q – r) + b(r – p) + c(p – q) = a’[q – r + r – p + p – q] + d[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)] ½ = a’ × 0 + d[pq – pr + qr – pq + pr – qr + (– q + r – r + p – p + q)] = 0 Hence Proved. ½ [CBSE Marking Scheme, 2016]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 9. Prove that the nth term of an A.P. can not be n2 + 1. Justify your answer. [CBSE Board Term-2 2015]
Sol. Let nth term of A.P., an = n2 + 1 Putting the values of n = 1, 2, 3, ......, we get a1 = 12 + 1 = 2 a2 = 22 + 1 = 5 a3 = 32 + 1 = 10
1
The obtained sequence = 2, 5, 10, 17, ......... Their common difference = a2 – a1 = a3 – a2 = a4 – a3 or, 5 – 2 ¹ 10 – 5 ¹ 17 – 10 \ 3 ¹ 5 ¹ 7 1 Since the common difference are not equal. Hence, n2 + 1 is not a form of nth term of an A.P. 1 [CBSE Marking Scheme, 2015]
108 ]
Long Answer Type Questions
7 : 15. Find the numbers. U [CBSE Delhi Set-I, 2020] [CBSE Delhi & OD, 2018]
Sol. Let the four consecutive terms of A.P. be (a – 3d), (a – d), (a + d) and (a + 3d). By given conditions a – 3d + a – d + d + a + 3d = 32 ⇒ 4a = 32
1
⇒
a = 8 1 ( a - 3d )( a + 3d ) 7 1 And = ( a - d )( a + d ) 15 a 2 - 9d 2 7 = 2 2 15 a d d2 = 4 d = ±2 1 Hence, the numbers are 2, 6, 10 and 14 or 14, 10, 6 and 2. 1 [CBSE Marking Scheme, 2018]
Q. 1. The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last term to the product of two middle terms is
5 marks each
Q. 2. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the A.P. is zero. [CBSE Term I, II, III, 2019]
Topper Answer, 2019 Sol.
[ 109
aRITHMETIC pROGRESSIONS
rd
Q. 3. An A.P. consists of 50 terms of which 3 term is 12 and last term is 106. Find the 29th term. U [CBSE SQP, 2018] Sol. Given, n = 50, a3 = 12 and a50 = 106
1
Then
a + 2d = 12
1
and
a + 49d = 106
1 1
On solving, we get d = 2 and a = 8 a29 = a + 28d
1
= 64
[CBSE Marking Scheme, 2018]
Q. 4. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers. A [Delhi Set III, 2016] Sol. Let the three numbers in A.P. be a – d, a and a + d. 1
Then, their sum i.e., 3a = 12 or,
a = 4
Also, (4 – d)3 + 43 + (4 + d)3 = 288 2
3
1 2
3
or, 64 – 48d + 12d – d + 64 + 64 + 48d + 12d + d = 288 or,
24d2 + 192 = 288
Q. 5. Find the value of a, b and c such that the numbers a, 7, b, 23 and c are in A.P.
1
U [CBSE Board Term-2, 2015]
= 8 + 28 × 2
or, d2 = 4 ∴ d = ± 2 1 Hence, the numbers are 2, 4 and 6, or 6, 4 and 2. 1 [CBSE Marking Scheme, 2016]
Sol. Since, a, 7, b, 23 and c are in A.P. Let the common difference be d
∴ a + d = 7 ...(i) ½ and a + 3d = 23 ...(ii) ½ From (i) and (ii), we get a = – 1 and d = 8 1 Again, b = a + 2d b = – 1 + 2 × 8 or, b = – 1 + 16 or, b = 15 1 ∴ c = a + 4d = – 1 + 4 × 8 = – 1 + 32 c = 31 1 ∴ a = – 1, b = 15 and c = 31 1 [CBSE Marking Scheme, 2015]
TOPIC - 2
Sum of n Terms of an Arithmetic Progression
Know the Formulae Sum of n terms of an A.P is given by:
Sn =
5
n [2a + (n – 1)d] 2
where, a is the first term, d is the common difference and n is the total number of terms. Sum of n terms of an A.P. when first and last term is given.
110 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sn =
n [a + l] 2
where, a is the first term and l is the last term. The nth term of an A.P is the difference of the sum of first n terms and the sum to first (n – 1) terms of it. i.e., an = Sn – Sn – 1.
How is it done on the
GREENBOARD?
Q.1. Find the number of terms in the A.P 54, 51, 48, ..... whose sum is 513. Also, give the reason of double answer. Solution: Step I: The given A.P. is 54, 51, 48, ..... Here a = 54, d = 51 – 54 = – 3 Sum required is 513. Step II: Applying the sum formula n Sn = [2a + (n – 1)d] 2
n 513 = [2 × 54 + (n – 1)(– 3)] 2
1026 = n[108 – 3n + 3]
1026 = n[111 – 3n] 1026 = 111n – 3n2 or, 3n2 – 111n + 1026 = 0 or, 3[n2 – 37n + 342] = 0 or, n2 – 37n + 342 = 0 Step III: Factorizing the quadratic equation n2 – 19n – 18n + 342 = 0 n(n –19) – 18(n – 19) = 0 or, (n – 19)(n – 18) = 0 or, n = 18 or 19 Hence, the required number of terms will be 18 or 19. 19th term of A.P. is zero hence double answers are correct.
Very Short Answer Type Questions
1 mark each A [CBSE Board Term, 2019]
Q. 1. Find the sum of the first 10 multiples of 6.
Topper Answer, 2019 Sol.
Q. 2. If nth term of an A.P. is (2n + 1), what is the sum of A [CBSE SQP, 2018] its first three terms ?
Sol. Since, a1 = 3, a2 = 5 and a3 = 7 3 (3 + 7) = 15 2
1 ½ ½
S3 =
[CBSE Marking Scheme, 2018]
[ 111
aRITHMETIC pROGRESSIONS
etailed Solution: D an = (2n + 1) ∴ a1 = 2 × 1 + 1 = 3 l = a3 = 2 × 3 + 1 = 7 n Since, Sn = [a + l] 2
Hence,
S3 =
½
3 [3 + 7] 2
Q. 3. If the first term of an A.P. is – 5 and the common difference is 2, then find the sum of the first 6 terms. R Sol. In the given A.P., a = – 5 and d = 2 n ½ Thus, Sn = [2a + (n – 1)d] 2 \
S3 = 15.
½
S6 =
= 3(– 10 + 10) = 0.
Short Answer Type Questions-I
=
20 [2 × 1 + (20 – 1)3] 2
= 10(2 + 57) = 10 × 59 = 590.
1
Q. 2. The sum of the first 7 terms of an A.P. is 63 and that of its next 7 terms is 161. Find the A.P. . A [CBSE Delhi Set-III, 2020] n Sol. Since, Sn = [2a + (n – 1)d] 2 Given,
S7 = 63 7 So, S7 = [2a +6d] 2
= 63 or, 2a + 6d = 18 ...(i) ½ Now, sum of 14 terms is: S14 = Sfirst 7 terms + Snext 7 terms = 63 + 161 = 224 14 [2a+ 13d] = 224 \ 2 ⇒ 2a+ 13d = 32 On subtracting (i) from (ii), we get (2a +13d) – (2a + 6d) = 32 – 18 ⇒ 7d = 14 ⇒ d = 2 Putting the value of d in (i), we get a = 3 Hence, the A.P. will be: 3, 5, 7, 9, ... .
...(ii) ½
2 marks each
Sol. a1 = S1 = 3 – 4 = –1 ½ a2 = S2 – S1 = [3(2)2 – 4(2)] – (–1) = 5 ½ \ d = a2 – a1 = 6 ½ Hence an = –1 + (n – 1) × 6 = 6n – 7 ½ Alternate method: Sn = 3n2 – 4n \ Sn – 1 = 3(n – 1)2 – 4(n – 1) = 3n2 – 10n + 7 1 Hence an = Sn – Sn – 1 ½ = (3n2 – 4n) – (3n2 – 10n + 7) = 6n – 7 ½ [CBSE Marking Scheme, 2019] Detailed Solution: Given, Sn = 3n2 – 4n Put n = 1, S1 = 3 × 12 – 4 × 1 =–1 ½ So, sum of first term of A.P. is – 1. But sum of first term will be the first term, \ First team, a1 = –1 Put n = 2, S2 = 3 × 22 – 4 × 2 = 4 ½ \ Sum of first two terms is 4. \ First term + Second term = 4 \ –1 + a2 = 4 ⇒ a2 = 5 ½ Hence, Common difference, d = a2 – a1 = 5 – (–1) = 6 \ nth term, an = a1 +(n – 1)d i.e., an = –1 + (n – 1)6 ⇒ an = 6n – 7 Therefore, nth term is 6n – 7. ½
COMMONLY MADE ERROR Some students do not know the basic concepts of arithmetic progression. Many students try to solve with wrong method.
½ ½
Q. 3. If Sn, the sum of first n terms of an A.P. is given by Sn = 3n2 – 4n. Find the nth term. A [CBSE Delhi Set-I, 2019]
½
Q. 1. Find the sum of first 20 terms of the following A.P.: 1, 4, 7, 10, ........ A [CBSE Delhi Set-II, 2020] Sol. Given A.P.: 1, 4, ,7, 10, ... Here, a = 1, d = 4 – 1 = 3 and n = 20 ½ \ The sum of first 20 terms, n S20 = [2a + (n – 1)d] ½ 2
6 [2 × (– 5) + (6 – 1) × 2] 2
ANSWERING TIP Learn
the concept of Arithmetic progression with different examples.
112 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
A [Delhi/OD 2018] [Delhi Comptt. Set-I, 2017]
Q. 4. Find the sum of first 8 multiples of 3.
Sol. Here, S = 3 + 6 + 9 + 12 + ... + 24 = 3(1 + 2 + 3 + ... + 8) 8×9 =3× 2 = 108
1
1 [CBSE Marking Scheme, 2018]
Detailed Solution:
Topper Answer, 2019
9 −11 , − 5 , − ..... 2 2
A [CBSE Delhi comptt. Set-III, 2017]
[CBSE Delhi Set-III, 2016] 11 1 Sol. Given a = – 6 and d = − − ( − 6 ) = 2 2 n Since, Sn = [ 2 a + ( n − 1)d ] 2 Let sum of n terms be zero. \ Sn = 0
n 1 or, 2 × − 6 + ( n − 1) = 0 2 2 or,
n n 1 −12 + − = 0 2 2 2
or,
n n 25 − =0 2 2 2
Sol. Given,
Since, or,
½
a + 42d = 171
½ ...(ii) ½
a = 3 and d = 4
½
[CBSE Marking Scheme, 2017]
Q. 7. Reshma wanted to save at least ` 6,500 for sending her daughter to school next year (after 12 months). She saved ` 450 in the first month and raised her savings by ` 20 every next month. How much will she be able to save in next 12 months ? Will she be able to send her daughter to the school next year ? C [Foreign Set-I, II, III, 2016] [CBSE Delhi Term-II Set-I, II, III, 2015] Sol. Here a = ` 450, d = ` 20, n = 12
Sn =
S12 =
n [2a + (n – 1)d] 2 12 [2 × 450 + 11 × 20] 2
= 6[1120] = 6720 > 6500
n [ 2 a + (n − 1)d ] 2
or, 2a + 9d = 42
15 (2 a + 84 d ) = 2565 2
Hence, given A.P. is 3, 7, 11 .......
S10 = 210
10 (2 a + 9d ) = 210 2
or,
15 (a + 35d + a + 49d ) 2
On solving (i) and (ii), we get
A [Foreign Set-III, 2017]
Sn =
a50 = a + 49d
Sum of last 15 terms =
or,
or, n2 – 25n = 0 1+½ n(n – 25) = 0 n = 25 as n ≠ 0 Hence, terms are needed = 25. Q. 6. In an A.P. of 50 terms, the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.
a36 = a + 35d
Hence,
are needed to give their sum zero.
Since,
and
Q. 5. How many terms of the A.P. −6 ,
1
½ ...(i)
2
\ Reshma will be able to send her daughter to school. [CBSE Marking Scheme, 2016] Q. 8. In an A.P., if S5+ S7 = 167 and S10 = 235, then find the A.P., where Sn denotes the sum of first n terms. A [CBSE Board, Term-2 2015]
[ 113
aRITHMETIC pROGRESSIONS
n [ 2 a + (n − 1)d ] 2
Sn =
Sol. Given,
S5 + S7 = 167 5 7 Hence, ( 2 a + 4 d ) + ( 2 a + 6d ) = 167 2 2
or, or
24a + 62d = 334 12a + 31d = 167
Given,
S10 = 235
or,
5(2a + 9d) = 235
or
2a + 9d = 47
Solving (i) and (ii), wet get Hence
....(i) ½
a = 1 and d = 5 ½ A.P. = 1, 6, 11, .... ½ [CBSE Marking Scheme, 2015]
Short Answer Type Questions-II
Q. 1. Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the last ( a + c )( b + c - 2 a) . term is c is equal to 2( b - a)
A [CBSE OD Set-I, 2020] Sol. Given, first term, A = a and second term = b ⇒ common difference, d = b – a Last term, l = c ⇒ A + (n – 1)d = c [By using, l = a + (n – 1)d] 1 ⇒ a + (n – 1)d = c a + (n – 1)(b – a) = c ⇒ (b – a)(n – 1) = c – a c−a ⇒ n – 1 = b−a
⇒
n =
c−a +1 b−a
=
c−a+b−a b−a
⇒
n =
b + c − 2a b−a
sum =
n [A + l] 2
Now
=
1
(b + c − 2a) [a + c] 2 (b − a) ( a + c )( b + c − 2 a ) 2 (b − a)
1 Hence Proved. Q. 2. Solve the equation: 1 + 4 + 7 + 10 + ... + x = A [CBSE Delhi OD Set-I, 2020] 287. Sol. Given, a = 1 and d = 4 – 1 = 3 ½ Let number of terms is the series be n, then n Sn = [2a + (n – 1)d] ½ 2 =
⇒
n [2 × 1 + (n – 1)3] = 287 2
⇒
n [2 + 3n – 3] = 287 2
⇒ 3n2 – n – 574 = 0
...(ii) ½
½
½
3 marks each
⇒ 3n2 – 42n + 41n – 574 = 0 ⇒ 3n(n – 14) + 41(n – 14) = 0 ⇒ (n – 14)(3n + 41) = 0 41 , it is not possible. Either n = 14 or n = – 3 Thus 14th term is x \ a + (n – 1)d = x ⇒ x = 1 + 13 × 3 = 40. 1 Q. 3. If in an A.P., the sum of first m terms is n and the sum of its first n terms is m, then prove that the sum of its first (m + n) terms is –(m + n). A [CBSE OD Set-II, 2020] Sol. Let 1st term of series be a and common difference be d, then Sm = n(given) m [2a + (m – 1)d] = n ½ ⇒ 2 ⇒ m[2a + (m – 1)d] = 2n...(i) ½ and Sn = m(given) n [2a + (n – 1)d] = m ⇒ 2 ⇒ n[2a + (n – 1)d] = 2m...(ii) ½ On subtracting, 2(n – m) = 2a(m – n) + d[m2 – n2 – (m – n)] ⇒ 2(n – m) = 2a(m – n) + d[(m – n)] [– (m – n) – (m – n)] ⇒ 2(n – m) = (m – n)[2a + d(m + n – 1)] ⇒ – 2 = 2a + d(m + n – 1) ½ m+n [2a + (m + n – 1)d] Now, Sm + n = 2 =
m+n (– 2) 2
= – (m + n) Hence Proved. 1 Q. 4. Find the sum of all 11 terms of an A.P. whose A [CBSE OD Set-II, 2020] middle term is 30. Sol. In an A.P. with 11 terms, 11 + 1 middle term = term 2 = 6th term 1 Now, sixth term i.e., a6 = a + (6 – 1)d i.e., a + 5d = 30 ...(i) [Q middle term i.e., a6 = 30 (given)] 1 Now, the sum of 11 terms,
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
11 [2a + (11 – 1)d] 2 11 [2a + 10d] = 2 11 × 2[a + 5d] = 2 = 11 × 30 [from (i)] = 330. 1 Q. 5. If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its A [CBSE SQP, 2020] first (m + n) terms is zero.
Sol.
S11 =
Sm = Sn
m n ⇒ [2a + (m – 1)d] = [2a + (n – 1d] 1 2 2 ⇒ 2a(m – n) + d(m2 – m – n2+n) = 0 1 ⇒ (m – n)[2a + (m + n – 1)d] = 0 1 or Sm+n = 0 [CBSE SQP Marking Scheme, 2020] Detailed Solution: Sum of first m terms = Sum of first n terms ⇒ Sm = Sn m n [2a + (m – 1)d] = [2a + (n – 1)d] 2 2
Sol. S4 = 40 ⇒ 2(2a + 3d) = 40 ⇒ 2a + 3d = 20 ½ S14 = 280 ⇒ 7(2a + 13d) = 280 ⇒ 2a + 13d = 40 ½ Solving to get d = 2 ½ and a = 7 ½ n ½ \ Sn = [14 + (n – 1)2] 2
114 ]
= n(n + 6) or (n2 + 6n) ½ [CBSE Marking Scheme, 2019] Detailed Solution: Since, Sum of n terms of an A.P., n ½ Sn = [2a + (n – 1)d] 2 [a be the first term and d be the common difference] According to question, S4 = 40 4 [2a + (4 – 1)d] = 40 ½ ⇒ 2
½
⇒ ⇒ and
½
⇒
m[2a + (m – 1)d] = n[2a + (n – 1)d] m[2a + (m – 1)d] – n[2a + (n – 1)d] = 0 2a(m – n) + [m(m – 1) – n (n – 1)]d = 0 1 2a(m – n) + [m2 – m – n2 + n]d = 0 2a(m – n) + [(m – n)(m + n) – (m – n)]d = 0 (m – n)[2a + (m + n – 1)d] = 0 Here, (m – n) is not equal to zero. So, [2a + (m + n – 1)d] = 0 Hence, Sm + n = 0. 1 Q. 6. If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first A [CBSE Delhi Set-I, 2019] n terms.
14 [2a +(14 – 1)d] = 280 2
⇒ 7 (2a + 13d) = 280 ⇒ 2a + 13d = 40 Solving eq. (i) and (ii), we get a = 7 and d = 2 n \ Sn = [2 × 7 +(n – 1)2] 2 n = [14+ 2n – 2] 2 n = (12 +2n) 2 = 6n+ n2 Hence, Sum of n terms = 6n + n2.
Topper Answer, 2019 Sol.
2[2a+3d] = 40 2a + 3d = 20 S14 = 280
...(i) ½ ½ ...(ii) ½
½
[ 115
aRITHMETIC pROGRESSIONS
3 Q. 7. How many terms of an A.P. 9, 17, 25, .... must be taken to give a sum of 636 ?
A [CBSE OD Set-III, 2017]
Topper Answer, 2017 Sol.
Q. 8. Find the sum of n terms of the series 1 2 3 4 − + 4 − + 4 − + ...... n n n
A [CBSE Delhi Set-I, II, III, 2017]
or, (4 + 4 + 4 + ..... up to n terms) 1 – (1 + 2 + 3 + ..... up to n terms) n
or,
Sol. Let sum of n term be Sn 1 2 3 \ Sn = 4 − + 4 − + 4 − + ...... n n n
up to n terms 1 or, (4 + 4 + 4 + ..... up to n terms) 1 2 3 – + + + ........ up to n terms n n n or, (4 + 4 + 4 + ..... up to n terms) 1 – (1 + 2 + 3 + ..... up to n terms) n
or,
4n −
1 n( n + 1) × n 2 4n −
1½
n+1 7n − 1 = 2 2
Hence, sum of n terms =
3
7n − 1 2
½
[CBSE Marking Scheme, 2017]
Q. 9. If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, find its 20th term.
A [CBSE OD Comptt. Set-III, 2017]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
\
Hence,
S14 =
14 [ 2 × 10 + (14 − 1) d ] 2
1050 = 150 20 + 13d = 7 13d = 130 130 = 10 d = 13
1
an = a + (n – 1)d a20 = 10 + 19 × 10 = 200 1 a20 = 200. [CBSE Marking Scheme, 2017] A [Delhi Comptt. Set-III, 2017]
Sol. Given, 1 + 3 + 5 + 7 + ..... + 49 Let, total odd numbers of terms be n. an = 1 + (n – 1) × 2 = 49 (n – 1) × 2 = 49 – 1 = 48 n – 1 = 24 n = 24 + 1 = 25 25 S25 = (1 + 49 ) 2
1
1
= 25 × 25 = 625 Hence, sum of odd numbers between 0 and 50 = 625 1 [CBSE Marking Scheme, 2017] Q. 11. If mth term of A.P. is
1 1 , find and nth term is n m
the sum of first mn terms.
A [CBSE Set-I, II, 2017]
Sol. Let first term of given A.P. be a and common difference be d. 1 ...(i) ½ \ am = a + (m – 1)d = n 1 ...(ii) ½ and an = a + (n – 1)d = m On subtracting (ii) from (i) we get 1 1 m−n 1 (m – n)d = − = n m mn 1 or, d = mn 1 and a = mn Now
Smn = =
mn 1 1 + ( mn − 1) 2· 2 mn mn mn 2 mn 1 + − 2 mn mn mn
1 [mn + 1] 2
1 [mn + 1] . 1 2
[CBSE Marking Scheme, 2017]
Q. 12. Find the sum of all two digit natural numbers which are divisible by 4. A [Delhi Comptt. Set-II, 2017]
Q. 10. Find the sum of all odd numbers between 0 and 50.
=
mn 1 +1 2 mn
Hence, the sum of first mn terms = ½
= 1050
Smn =
½
Sol. First two digit multiple of 4 is 12 and last is 96 So, a = 12, d = 4 and l = 96 Let nth term be last term = 96 \ an = a + (n – 1)d = l 12 + (n – 1)4 = 96 n – 1 = 21 n = 21 + 1 = 22 22 Now, S22 = [12 + 96 ] 2
Sol. Given, a = 10, and S14 = 1050 Let the common difference of the A.P. be d. Since, Sn = n [ 2 a + ( n − 1)d ] 2
1
1
= 11 × 108 = 1188 1 [CBSE Marking Scheme, 2017] Q. 13. Find the sum of the following series: 5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + .... A [Foreign Set-I, 2017] + (– 5) + 81 + (– 3) Sol. The series can be written as (5 + 9 + 13 + .... + 81) + [(– 41) + (– 39) + (– 37) + (– 35) ... (– 5) + (– 3)] For the series (5 + 9 + 13 + ..... + 81) ½ a = 5 d = 4 and an = 81 Then, an = 5 + (n – 1)4 = 81 or, (n – 1)4 = 76 ½ n = 20 20 ( 5 + 81) Sn = 2
116 ]
= 860 For series (– 41) + (– 39) + (– 37) + .... + (– 5) + (– 3) an = – 3 ½ Then, \
a = – 41 d = 2 an = –41 + (n – 1)(2) n = 20 20 Sn = −41 + ( − 3) 2
= – 440 ½ Hence, the Sum of the series = 860 – 440 = 420 1 [CBSE Marking Scheme, 2017]
[ 117
aRITHMETIC pROGRESSIONS
Now, A.P. is 5, 8, 11, ...... . nth term, an = a + (n – 1)d = 5 + (n – 1)3 = 3n + 2
Sol. Since, and
Hence,
a20 = 3 × 20 + 2
a20 = 62
[CBSE Marking Scheme, 2015]
Also,
S2 =
= and
S3 =
n [2 × 1 + (n – 1)2] 2 n [2n] = n2 2
½
n [2 × 1 + (n – 1)3] 2
=
n( 3n - 1) ½ 2
S1 + S3 =
n( n + 1) n( 3n - 1) + ½ 2 2
Now,
=
n[n + 1 + 3n - 1] 2
n[ 4 n] = 2 = 2n2 = 2S2 Hence Proved. 1 [CBSE Marking Scheme, 2016] Q. 15. If the sum of the first n terms of an A.P. is
1 2
[3n2 + 7n], then find its nth term. Hence write its 20th term. A [CBSE Board Term-2, Set-II 2015] [CBSE SQP-2016] 1 [3n2 + 7n] Sn = 2
1 S1 = [3 × (1)2 + 7(1)] 2
1 = [3 + 7] 2 1 × 10 = 5 = 2
=
Q. 17. If Sn denotes, the sum of the first n terms of an A.P. prove that S12 = 3(S8 – S4). A [CBSE Delhi Board, Set-I, 2015] Sol. Let a be the first term and d be the common difference. Sn =
n [ 2 a + (n − 1)d ] 2
S12 = 6[2a + 11d]
= 12a + 66d ½
Then,
½
S4 = 2[2a + 3d]
= 4a + 6d
...(i) 1
S8 = 4[2a + 7d]
= 8a + 28d and
1 [12 + 14] 2
½
3(S8 – S4) = 3[(8a + 28d) – (4a + 6d)]
= 3[4a + 22d] = 12a + 66d
1 × 26 2
= 13 a1 = S1 = 5 a2 = S2 – S1 = 13 – 5 = 8 d = a2 – a1 = 8 – 5 = 3
12 [2 × 100 + (12 – 1) × 20] ∴ S12 = 2 or, = 6[200 + 11 × 20] or, = 6[200 + 220] or, = 6 × 420 = 2520 3 She will be able to send her daughter to school after 12 weeks. [CBSE Marking Scheme, 2015]
1 S2 = [3 (2)2 + 7 × 2] 2 =
Sol. Here, required money is ` 2500 a = saving in 1st week = ` 100 d = difference in weekly saving = ` 20 A.P. formed by saving, According to the question, Sequence is 100, 120, 140, ..... upto 12 terms n Sn = [2a + (n – 1)d] 2
Since,
Sol.
½
Q. 16. Aditi required ` 2500 after 12 weeks to send her daughter to school. She saved ` 100 in the first week and increased her weekly saving by ` 20 every week. Find whether she will be able to send her daughter after 12 weeks. C [CBSE Board Term-2, Set-I, II, III, 2015]
or,
S1 = 1 + 2 + 3 + .... + n. S2 = 1 + 3 + 5 + ...upto n terms S3 = 1 + 4 + 7 + ...upto n terms n( n + 1) ½ S1 = 2
Q. 14. The sum of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P. is 1 and common differences are 1, 2 and 3 respectively. Prove that A [OD Set III, 2016] S1 + S3 = 2S2.
From equation (i) and (ii), S12 = 3(S8 – S9) ½ ½ ½ ½
1
[CBSE Marking Scheme, 2015]
Q. 18. The 14th term of an A.P. is twice its 8th term. If the 6th term is – 8, then find the sum of its first 20 A [CBSE OD Set-I, 2015] terms. [Foreign Set-I, II, 2015]
118 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Let first term be a and common difference be d. Here, a14 = 2a8 or, a + 13d = 2(a + 7d) a + 13d = 2a + 14d a = – d ...(i) ½ Again, a6 = – 8 or, a + 5d = – 8 ...(ii) ½
Solving (i) and (ii), we get a = 2, d = – 2 ½ 20 [ 2 × 2 + ( 20 - 1)( -2 )] ½ S20 = 2 = 10[4 + 19 × (– 2)] = 10(4 – 38) = 10 × (– 34) = – 340 1 [CBSE Marking Scheme, 2015]
Long Answer Type Questions Q. 1. Solve: 1 + 4 + 7 + 10 + ... + x = 287.
⇒
A [CBSE Delhi Set-I, 2020] Sol. See the solution of Q. 2. from Short Answer Type Question-II. Q. 2. The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the A.P.. [CBSE Delhi Set-II, 2019]
Also
903 =
=
⇒ ⇒ Now,
n (3 + 83) 2 1806 = 86n 1806 n = 86 903 =
n = 21 n Sn = [2a + (n – 1)d] 2 21 [2 × 3 + (21 – 1)d] 903 = 2
⇒ 1806 = 21(6 + 20d) ⇒ 6 + 20d = 86 ⇒ 20d = 80 ⇒ d = 4 Hence, the common difference is 4.
...(i) 1
n [2a + (n – 1)d] 2 n (6 + 80) 2
1 1
1
COMMONLY MADE ERROR
= 43n (using (i)) 1+½ ⇒ n = 21 and d = 4 1+½ [CBSE Marking Scheme, 2019] Detailed Solution: Given: First term, a = 3 Last term, an = 83 Sum of n terms, Sn = 903 n Since, Sn = ( a + an ) 2
⇒
⇒
1
Sol. Here a = 3, an = 83 and Sn = 903 Therefore 83 = 3 + (n – 1)d ⇒ (n – 1)d = 80
5 marks each
Some students fail to find the value of n as they get confused between the nth term and last term.
ANSWERING TIP 1 1
Understand the formulae related to given condition and use them to solve the problems.
Q. 3. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th A [CBSE OD Set III 2017] [CBSE OD Set-I, 2016] terms.
Sol.
Topper Answer, 2017
[ 119
aRITHMETIC pROGRESSIONS
Q. 4. The ratio of the sums of first m and first n terms of an A.P. is m2 : n2. Show that the ratio of its mth and nth terms is (2m – 1) : (2n – 1). [CBSE Delhi Set-I, 2017] Sol. Let first term of given A.P. be a and common difference be d also sum of first m and first n terms be Sm and Sn respectively. Sm m2 \ 1 = 2 Sn n m [ 2 a + (m − 1)d ] m2 1 or, 2 = 2 n n [ 2 a + ( n − 1)d ] 2 2 2 a + ( m − 1)d or, = m ×n =m 1 2 2 a + ( n − 1)d m n n or, m(2a + (n – 1)d) = n[2a + (m – 1)d] 1 d = 2a am a + ( m − 1)d = Now an a + ( n − 1)d a + ( m − 1) × 2 a = a + ( n − 1) × 2 a a + 2ma − 2 a 2ma − a or, = 1 a + 2na − 2 a 2na − a a( 2m − 1) = a( 2n − 1) = 2m – 1 : 2n – 1 Hence Proved. Q. 5. If the pth term of an A.P. is
1 1 and qth term is . q p
Prove that the sum of first pq term of the A.P. is
pq+1 2 .
[CBSE Delhi Set-III, 2017]
Sol. Try yourself similar to Q.No. 11 of VSATQ-II. Q. 6. If the ratio of the 11th term of an A.P. to its 18th term is 2 : 3, find the ratio of the sum of the first five term to the sum of its first 10 terms. [Delhi Comptt. Set-I, II, III, 2017]
5
a11 a + 10d 2 = = a18 a + 17 d 3 or, 2(a + 17d) = 3(a + 10d) 1 a = 4d ...(i) 5 (2 a + 4d ) S5 2 Now, = 1 S10 10 [2 a + 9d ] 2 Putting the value of a = 4d, we get 1 5 (8 d + 4 d ) S5 or, 1 = 2 5 (8 d + 9 d ) S10 6 12d = 1 17 34 d Sol. Since,
Hence, S5 : S10 = 6 : 17. Q. 7. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the A.P. [CBSE SQP, 2017] Sol. Let the middle most terms of the A.P. be (a – d), a and (a + d). Given , a – d + a + a + d = 225 or, 3a = 225 1 or, a = 75 1 37 + 1 = 19th term and the middle term = 2 \ A.P. is (a – 18d), ....., (a – 2d), (a – d), a, (a + d), (a + 2d), .... , (a + 18d) 1 Sum of last three terms (a + 18d) + (a + 17d) + (a + 16d) = 429 or, 3a + 51d = 429 or, 225 + 51d = 429 or, d = 4 1 First term, a1 = a – 18d = 75 – 18 × 4 = 3. a2 = 3 + 4 = 7 Hence, A.P. = 3, 7, 11, ........ , 147. 1 Q. 8. The minimum age of children to be eligible to participate in a painting competition is 8 years. It is observed that the age of youngest boy was 8 years and the ages of rest of participants are having a common difference of 4 months. If the sum of ages
120 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
of all the participants is 168 years, find the age of eldest participant in the painting competition. C [CBSE SQP, 2016] Sol. Here, a = 8, d = 4 months = Sn = 168
1 years and 3
n Sn = [ 2 a + ( n - 1)d ] 2
1
Since
Hence,
1 n2 + 47n – 1008 = 0 2 or, n + 63n – 16n – 1008 = 0 or, (n – 16)(n + 63) = 0 or, n = 16 or n = – 63 n = 16 (n cannot be negative So – 63 rejected) 1 Thus, the age of the eldest participant = a + 15d = 13 years [CBSE Marking Scheme, 2016] 1
168 =
n 1 2( 8 ) + ( n - 1) 2 3
1
Q. 9. 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. C [CBSE Delhi Set I, II, 2016] Sol. Let total time to catch the thief be n minutes. Then, total distance covered by thief = (100n) ½ metres Total distances to be covered by policeman = 100 + 110 + 120 + ... + (n – 1) terms 1 n -1 [200 + (n – 2)10] 1 ∴ 100n = 2 n2 – 3n – 18 = 0 ½ (n – 6)(n + 3) = 0 ½ or, n = 6 ½ Policeman takes 6 minutes to catch the thief. 1 [CBSE Marking Scheme, 2016]
Visual Case Based Questions
Q. 1. India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. [CBSE QB, 2021]
Based on the above information, answer the following questions: (i) Find the production during first year. Sol. ` 5000 Explanation: a6 = 16,000 a + (n + 1)d = 16,000 a + (6 – 1)d = 16,000 a + 5d = 16,000 a9 = 22,600 a + (n – 1)d = 22,600 a + (9 – 1)d = 22,600 a + 8d = 22,600
...(i)
...(ii)
Solving equation (i) and (ii) a + 5d = 16,000 a + 8d = 22,600 – – – – 3d = – 6,600 d = 2,200 Now, putting d = 2,200 in equation (i) a + 5d = 16,000 a + 5 × 2,200 = 16,000 a + 11,000 = 16,000 a = 5,000 (ii) Find the production during 8th year. Sol. Production during 8th year is (a + 7d) = 5000 + 2(2200) = 20400 (iii) Find the production during first 3 years. Sol. Production during first 3 year = 5000 + 7200 + 9400 = 21600 (iv) In which year, the production is ` 29,200. Sol. N = 12 Explanation: an = 29,200 a + (n – 1)d = 29,200 (x – 1)2,900 = 29,200 – 5,000 2,200n – 2,200 = 24,200 2200n = 26,400 26 , 400 n = 2 , 200
ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark
4 marks each
n = 12 (v) Find the difference of the production during 7th year and 4th year. Sol. Difference = 18200 – 11600 = 6600
[ 121
aRITHMETIC pROGRESSIONS
[CBSE QB, 2021]
(i) Which of the following terms are in AP for the given situation (a) 51, 53, 55…. (b) 51, 49, 47…. (c) – 51, – 53, – 55…. (d) 51, 55, 59… Sol. Correct option: (b). Explanation: a = 51 d = – 2 AP = 51, 49, 47 ...... . (ii) What is the minimum number of days he needs to practice till his goal is achieved ? (a) 10 (b) 12 (c) 11 (d) 9 Sol. Correct option: (c). Explanation: Goal = 31 second n = number of days \ an = 31 a + (n – 1)d = 31 51 + (n – 1)(– 2) = 31 51 – 2n + 2 = 31 – 2n = 31 – 53 – 2n = – 22 n = 11 (iii) Which of the following term is not in the AP of the above given situation (a) 41 (b) 30 (c) 37 (d) 39 Sol. Correct option: (b). (iv) If nth term of an AP is given by an = 2n + 3 then common difference of an AP is (a) 2 (b) 3 (c) 5 (d) 1 Sol. Correct option: (a). (v) The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP (a) 6 (b) – 6 (c) 18 (d) – 18 Sol. Correct option: (a). Explanation: Since, 2x, x + 10, 3x + 2 are in AP, this common difference will remain same.
x + 10 – 2x = (3x + 2) – (x + 10) 10 – x = 2x – 8 2x = 18 x = 6 Q. 3. Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ` 1,18,000 by paying every month starting with the first instalment of ` 1000. If he increases the instalment by ` 100 every month, answer the following: [CBSE QB, 2021]
(i) The amount paid by him in 30th installment is (a) 3900 (b) 3500 (c) 3700 (d) 3600 Sol. Correct option: (a). Explanation: a = 1000 d = 100 a80 = a + (n – 1)d = 1000 + (30 – 1)100 = 1000 + 2900 (ii) The amount paid by him in the 30 installments is (a) 37000 (b) 73500 (c) 75300 (d) 75000 Sol. Correct option: (b). Explanation: Sum of 30 installments n = [2a + (n – 1)d] 2
Q. 2. Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
=
30 [2 × 1000 + (30 – 1)100] 2
= 15[2000 + 2900] = 15 × 4900 = 73500 Total Amount paid in 30 installments = ` 73500 (iii) What amount does he still have to pay after 30th installment ? (a) 45500 (b) 49000 (c) 44500 (d) 54000 Sol. Correct option: (c). (iv) If total installments are 40 then amount paid in the last installment ? (a) 4900 (b) 3900 (c) 5900 (d) 9400 Sol. Correct option: (a). Explanation: Amount paid in 40th installment, a40 = a + (n – 1)d = 1000 + (40 – 1)100 = 1000 + 3900 = 5900
122 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(v) The ratio of the 1st installment to the last installment is (a) 1 : 49 (b) 10 : 49 (c) 10 : 39 (d) 39 : 10 Sol. Correct option: (b). Q. 4. Jaspal Singh takes a loan from a bank for his car. Jaspal Singh repays his total loan of ` 118000 by paying every month starting with the first installment of ` 1000. If he increases the installment by ` 100 every month.
(i) If the given problem is based on A.P., then what is the first term and common difference ? (a) 1000, 100 (b) 100, 1000 (c) 100, 100 (d) 1000, 1000 Sol. Correct option: (a). Explanation: The number involved in this case form an A.P. in which first term (a) = 1000 and common difference (d) = 100. (ii) The amount paid by him in 25th installment is: (a) ` 3300 (b) ` 3200 (c) ` 3400 (d) ` 3500 Sol. Correct option: (c). Explanation: The amount paid by him in 25th installment is: T25 = a + 24d = 1000 + 24 × 100 = 1000 + 2400 = ` 3400. (iii) The amount paid by him in 30th installment is (a) ` 3900 (b) ` 3500 (c) ` 3000 (d) ` 3600 Sol. Correct option: (a). Explanation: The amount paid by him in 30th installment, T30 = a + 29d = 1000 + 29 × 100 = 1000 + 2900 = ` 3900.
(v) The difference amount paid by him in 26th and 28th installment is: (a) ` 400 (b) ` 100 (c) ` 500 (d) ` 200 Sol. Correct option: (d). Explanation: The amount paid by him in 26th installment, T26 = a + 25d = 1000 + 25 × 100 = 1000 + 2500 = ` 3500 The amount paid by him in 28th installment, T28 = a + 27d = 1000 + 27 × 100 = 1000 + 2700 = ` 3700 \ The difference amount paid by him in 26th and 28th installment is: = ` (3700 – 3500) = ` 200. Q. 5. A ladder has rungs 25 cm apart. (see the below).
(iv) The total amount paid by him in 25th and 30th installment is: (a) ` 7500
(b) ` 7300
(c) ` 7800 (d) ` 7600 Sol. Correct option: (b). Explanation: Total amount paid by him in 25th and 30th installment = ` (3400 + 3900)
= ` 7300.
25 m
2
1 m 2
25 m 45 m
The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. The top and 1 the bottom rungs are 2 m apart. 2 (i) The top and bottom rungs are apart at a distance: (a) 200 cm (b) 250 cm (c) 300 cm (d) 150 cm Sol. Correct option: (b). Explanation: Since the top and the bottom rungs 1 5 are apart by 2 m = m 2 2 =
5 × 100 cm 2
= 250 cm (ii) Total number of the rungs is: (a) 20 (b) 25 (c) 11 (d) 15 Sol. Correct option: (c). Explanation: The distance between the two rungs is 25 cm.
[ 123
aRITHMETIC pROGRESSIONS
Hence, the total number of rungs =
250 +1 25
= 11. (iii) The given problem is based on A.P. find its first term. (a) 25 (b) 45 (c) 11 (d) 13 Sol. Correct option: (a). Explanation: The length of the rungs increases from 25 to 45 and total number of rungs is 11. Thus, this is in the form of an A.P., whose first term is 25. (iv) What is the last term of A.P. ? (a) 25 (b) 45 (c) 11 (d) 13
Sol. Correct option: (b). Explanation: Total number of terms, n = 11 and the last term, T11 = 45. (v) What is the length of the wood required for the rungs ? (a) 385 (b) 538 (c) 532 (d) 382 Sol. Correct option: (a). Explanation: The required length of the wood, 11 [25 + 45] S11 = 2 11 × 70 = 2 = 385 cm.
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, MATHEMATICS (STANDARD), Class – X
Maximum Time: 1 hour Q. 1. For what value of k, do the equation 3x – y + 8 = 0 and 6x – ky = – 16 represent coincident lines ? R Q. 2. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, then find the product of the other two zeroes. R Q. 3. If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and – 3, then find the value of a. R Q. 4. If in the equation x + 2y = 10, the value of y is 6, then find the value of x. R Q. 5. Find the value of p for which 3x2 – 5x + p = 0 has equal roots. R Q. 6. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time.
MM: 25
(ii) How many flags are left and right to the middle flag ? (a) 14, 12 (b) 13, 13 (c) 13, 14 (d) 14, 13 (iii) How much distance did she cover in completing this job and returning back to collect her books ? (a) 339 m (b) 634 m (c) 364 m (d) 346 m (iv) What is the maximum distance she travelled carrying a flag ? (a) 13 m (b) 52 m (c) 27 m (d) 26 m (v) What is the mathematical concept related to this question ? (a) A.P. (b) Lines (c) Linear equations (d) none of these Q. 7. If p and q are the zeroes of polynomial f(x) = 2x2 – 7x + 3, find the value of p2 + q2. Q. 8. Find the sum of the integers between 100 and 200 that are divisible by 6. A [Board Term-2, 2012] Q. 9. How many three digit numbers are such that when divided by 7, leave a remainder 3 in each case ? [Board Term-2, 2012 Set (1)] x y 2 2 2 2 Q. 10. If (x + y )(a + b ) = (ax + by)2. Prove that = . a b
A [Bord Term-2, 2014]
Q. 11. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth is increased by 3 units. The area is increased by 67 square units if length is increased by 3 units and breadth is increased by 2 units. Find A the perimeter of the rectangle.
(i) What is the position of middle most flag ? (a) 13th (b) 13.5th th (c) 14 (d) 12.5th
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lines (in two-dimensions)
[ 125
UNIT 3: Co-ordinate Geometry
c h a p te r
6
Line (in two-dimensions)
Syllabus ¾
¾
Review : Concepts of co-ordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division). Area of a triangle.
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1 Q (1 M) 1 Q (3 M) 2 Q (1 M) 3 Q (1 M) 1 Q (3 M) 1 Q (3 M) 1 Q (3 M) 1 Q (4 M)
Area of a Triangle
2 Q (3 M)
TOPIC
Distance between two points and 1 Q (2 M) section formula 1 Q (3 M) 1 Q (3 M)
1 Q (1 M) 1 Q (3 M) 1 Q (3 M)
-1
Distance between two points and Section formula Revision Notes
TOPIC - 1 Two perpendicular number lines intersecting at origin are called co-ordinate axes. The horizontal line is the X-axis (denoted by X'OX) and the vertical line is the Y-axis (denoted by Y'OY).
Distance between two points and Section formula Page No. 126
TOPIC - 2 Area of Triangle Page No. 144
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lines (in two-dimensions) Y-axis
X’-axis
X-axis
Y’-axis
The point of intersection of X-axis and Y-axis is called origin and denoted by O. Scan to know Cartesian plane is a plane obtained by putting the co-ordinate axes perpendicular to each more about other in the plane. It is also called co-ordinate plane or XY-plane. this topic The x-co-ordinate of a point is its perpendicular distance from Y-axis. The y-co-ordinate of a point is its perpendicular distance from X-axis. The point where the X-axis and the Y-axis intersect has co-ordinate point (0, 0). The abscissa of a point is the x-co-ordinate of the point. Basic concepts of Co-ordinate The ordinate of a point is the y-co-ordinate of the point. Geometry If the abscissa of a point is x and the ordinate of the point is y, then (x, y) is called the co-ordinates of the point. The axes divide the Cartesian plane into four parts called the quadrants (one fourth part), numbered I, II, III and IV anti-clockwise from OX. The co-ordinates of a point on the X-axis are of the form (x, 0) and that of the point on Y-axis Scan to know are (0, y). more about this topic Sign of co-ordinates depicts the quadrant in which it lies. The co-ordinates of a point are of the form (+, +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant. Three points A, B and C are collinear if the distances AB, BC and CA are such that the sum of Theorem of Cotwo distances is equal to the third. ordinate Geometry Three points A, B and C are the vertices of an equilateral triangle if AB = BC = CA. The points A, B and C are the vertices of an isosceles triangle if AB = BC or BC = CA or CA = AB.
Three points A, B and C are the vertices of a right triangle, if AB2 + BC2 = CA2. C
B
For the given four points A, B, C and D : D
A
A C
B
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
1. If AB = BC = CD = DA; AC = BD, then ABCD is a square. 2. If AB = BC = CD = DA; AC ≠ BD, then ABCD is a rhombus. 3. If AB = CD, BC = DA; AC = BD, then ABCD is a rectangle. 4. If AB = CD, BC = DA; AC ≠ BD, then ABCD is a parallelogram. Diagonals of a square, rhombus, rectangle and parallelogram always bisect each other. Diagonals of rhombus and square bisect each other at right angle. All given points are collinear, if the area of the obtained polygon is zero. Three given points are collinear, if the area of triangle is zero. Centroid is the point of intersection of the three medians of a triangle. In the figure, G is the centroid of a triangle ABC. A
F
E G
B
C
D
Centroid divides each median of a triangle in a ratio of 2 : 1 from vertex to base of the side. If x ≠ y, then (x, y) ≠ (y, x) and if (x, y) = (y, x), then x = y. To plot a point P(3, 4) in the cartesian plane. (i) A distance of 3 units along X-axis. (ii) A distance of 4 units along Y-axis. Y
X'
X
Y'
Know the Formulae The distance between two points i.e., P(x1, y1) and Q(x2, y2) is
d =
The distance of a point P(x, y) from origin is
( x 2 - x1 )2 + ( y 2 - y1 )2
x2 + y2
[ 129
lines (in two-dimensions)
Co-ordinates of point (x, y) which divides the line segment by joining the points (x1, y1) and (x2, y2) in the ratio m : n internally are mx + nx1 my + ny1 and y = 2 x = 2 m + n m + n Co-ordinates of mid-point of the line segment by joining the points (x1, y1) and (x2, y2) are y + y1 x + x1 x = 2 and y = 2 2 2
Know the Facts Co-ordinate geometry is the system of geometry where the position of points on the plane is described using an ordered pair of numbers.
Cartesian plane was discovered by Rene Descartes. The other name of co-ordinate geometry is Analytical Geometry. Co-ordinate Geometry acts as a bridge between the Algebra and Geometry. Medians of a triangle are concurrent. The point of concurrency is called the centroid. Trisection of a line segment means dividing it into 3 equal parts, so 2 points are required. Centroid of a triangle divides its median in the ratio of 2 : 1.
How is it done on the
GREENBOARD?
Q.1. Find the point of trisection of the line segment joining the points (5, –6) and (–7, 5). Solution Step I: Diagrammatic representation. A
x1
B
(5, –6)
x1 =
1(−7) + 2(5) 3 = =1 1+ 2 3
y1 =
1(5) + 2(−6) −7 = 1+ 2 3
−7 \ P(x1, y1) is 1, . 3
Points of trisections divides line Step IV: Finding co-ordinates of Q segment into three equal parts. using section formula i.e., AP = PQ = QB Q B A Step II: The ratio of AP : PQ : QB = 1 : 1:1 2:1 AQ 2 AP 1 = = and or, 2(−7) + 1(5) −9 QB PB 2 1 = = –3 x2 = 2 +1 3 Step III: finding co-ordinates of P 2(5) + 1(−6) 4 using section formula. = y2 = 3 2 +1 P A B 1:2
4 \ Q(x2, y2) is −3, . 3
130 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Very Short Answer Type Questions Q. 1. In which quadrant lies the point which divides the line segment joining the points (8, – 9) and (2, 3) in ratio 1 : 2 internally ? U [CBSE SQP, 2020]
Sol. IV quadrant.
[CBSE SQP Marking Scheme, 2020]
Detailed Solution:
Q. 3. 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.
A [CBSE Delhi Set-I, 2020]
Sol.
½
1 mark each
\
m= 1, n = 2
k =
1( −7 ) + 2( 2 ) 1+ 2 mx 2 + nx1 Q x = m + n
Given, (x1, y1) = (8, – 9)
(x2, y2) = (2, 3)
mx 2 + nx1 my 2 + ny1 , (x, y) = m + n m+n
1 × 2 + 2 × 8 1 × 3 + 2 × ( −9 ) (x, y) = , 1+ 2 1+ 2
2 + 16 3 − 18 (x, y) = , 3 3
18 −15 (x, y) = , 3 3
(x, y) = (6, – 5)
Hence, the point (6, – 5) lies in IV quadrant.
⇒ ½
k =
−7 + 4 3
⇒ 3k = – 3 ⇒ k = – 1.
½
Q. 4. The point P on X-axis equidistant from the points A(–1, 0) and B(5, 0). A [CBSE OD Set-I, 2020] Sol. Let the position of the point P on X-axis be (x, 0), then
PA2 = PB2 ⇒ (x + 1) + (0)2 = (5 – x)2 + (0)2 ½ ⇒ x2 + 2x + 1 = 25 + x2 – 10x ⇒ 2x + 10x = 25 – 1 ⇒ 12x = 24 ⇒ x = 2 Hence, the point P (x, 0) is (2, 0). ½ Q. 5. Find the co-ordinates of the point which is reflection of point (–3, 5) in X-axis. 2
½
Q. 2. Find the distance between the points
(acos q + b sin q, 0) and (0, a sin q – b cos q).
A [CBSE Delhi Set-I, 2020]
Sol. Here, x1 = a cos q + b sin q, y1 = 0 and x2 = 0, y2 = a sin q – b cos q \ Distance = =
( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2
½
A [CBSE OD Set-I, 2020] Sol. By using the graph of coordinate plane, we have the reflection of point (– 3, 5) is x-axis is (– 3, – 5). ½
( 0 − a cos θ − b sin θ)2 + ( a sin θ − b cos θ − 0 )2
Y (–3, 5)
= =
2
2
( −1) ( a cos θ + b sin θ) + ( a sin θ − b cos θ)
5
2
a 2 cos2 θ + b 2 sin 2 θ + 2 ab cos θ sin θ
X
X' –3
O
½
+ a 2 sin 2 θ + b 2 cos2 θ − 2 ab sin θ cos θ
=
a 2 (sin 2 θ + cos2 θ) + b 2 (sin 2 θ + cos2 θ)
=
a2 × 1 + b2 × 1
=
a 2 + b 2 unit
(–3, –5)
½
–5
Y' Q. 6. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y. A [CBSE OD Set-I, 2020]
[ 131
lines (in two-dimensions)
Sol. Here,
x1 = 6, y1 = 5
Detailed Solution: Since, AB is the diameter, center C must be the mid
P(6, 2) 3
A(6, 5) and
1
point of the diameter AB.
B(4, y)
x2 = 4, y2 = y
Then
x =
mx 2 + nx1 my 2 + ny1 and y = ½ m+n m+n
\
2 =
3× y +1× 5 3y + 5 = 3+1 4
⇒ 3y + 5 = 8 ⇒ 3y = 8 – 5 = 3 ⇒
½
Let the co-ordinates of point A be (x, y).
x-coordinate of
C =
⇒
2 =
½
y = 1.
Q. 7. Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, –3) and B is the point (1, 4). A [CBSE Delhi Set-I, II, III, 2019]
Sol. Let the point A be (x, y) x+1 4+y \ = 2 and ½ = −3 2 2 ⇒ x = 3 and y = –10 ½ \ Point A is (3, –10) [CBSE Marking Scheme, 2019]
x+1 2
⇒ ⇒
x+1 2 4 = x + 1 x = 3
and y-coordinate of
C =
y+4 2
y+4 2 ⇒ –6 = y + 4 ⇒ y = –10 Hence, coordinates of point A is (3, –10).
⇒
–3 3= =
½
T
Q. 8. The distance between point A(5, – 3) and B(13, m) is 10 units. Calculate the value of m. [CBSE Delhi Board term, 2019]
opper Answer, 2019
Sol.
1
132 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
a Q. 9. Find the value of a, for which point P , 2 is the 3
AD =
( 5 + 1)2 + (1 - 2 )2 =
37 unit ½ [CBSE Marking Scheme, 2018]
midpoint of the line segment joining the points A [CBSE SQP, 2018]
Q(–5, 4) and R(–1, 0).
-5 + ( -1) 4 + 0 a , = , 2 2 3 2
Sol.
B(1, 5) D A(5, 1)
a -6 = ⇒ a = – 9 3 2
Detailed Solution:
1
[CBSE Marking Scheme, 2018]
C(–3, –1)
Detailed Solution: P
Q
R
Here D is the mid-point of BC.
½
P is the mid-point of QR
or,
a −5 + ( −1) = 3 2
or,
a −6 = 3 2
or,
1 − 3 5 − 1 , = = ( −1, 2 ) ½ 2 2 \ ½
a = – 9.
Q. 10. A(5, 1), B(1, 5) and C(–3, –1) are the vertices of DABC. Find the length of median AD.
Sol. Coordinates of D are (–1, 2)
AD =
( 5 + 1)2 + (1 − 2 )2
=
36 + 1 =
Hence, the length of AD is
U [CBSE Compt. Set-I,II,III, 2018]
Then, the coordinates of D
37
37 unit.
½
Q. 11. If the distance between the points (4, k) and
½
(1, 0) is 5, then what can be the possible values of
A
U [Delhi Set I, II, III, 2017]
k ? Sol. Using distance formula,
( 4 − 1)2 + ( k − 0 )2 = 5
½
or, 32 + k2 = 25 D
B
C
Short Answer Type Questions-I
k = ±4
½
[CBSE Marking Scheme, 2017]
2 marks each
Q. 1. Find the point on X-axis which is equidistant from the points (2, – 2) and (– 4, 2).
Q. 2. P(–2, 5) and Q(3, 2) are two points. Find the coordinates of the point R on PQ such that PR = 2QR.
U [CBSE SQP, 2020-21]
A [CBSE SQP, 2020-21]
Sol. Let P(x, 0) be a point on x-axis
PA = PB
½
PA2 = PB2
½
(x – 2)2 + (0 + 2)2 = (x + 4)2 + (0 – 2)2
2
2
x + 4 – 4x + 4 = x + 16 + 8x + 4 – 4x + 4 = 8x + 16 x = –1
Hence co-ordinate of required point are (–1, 0).
½ ½
Sol.
PR : QR = 2 : 1 mx + nx1 my 2 + ny1 R(x, y) = 2 , m+n m+n
½
1( −2 ) + 2( 3) 1( 5) + 2( 2 ) R , 1 2+1 2 + 1 4 R , 3 . ½ 3 [CBSE Marking Scheme, 2020-21]
[ 133
lines (in two-dimensions)
Detailed Solution:
and also given, PR = 2QR
Given two points : P(– 2, 5) and Q(3, 2) Here,
x1 = – 2, x2 = 3
and
y1 = 5, y2 = 2 Y
PR 2 = QR 1
⇒
PR : QR = 2 : 1
½
mx + nx1 my 2 + ny1 Then R(x, y) = 2 , m+n m+n
P(–2, 5) R(x, y)
2
2 × 3 + 1 × ( −2 ) 2 × 2 + 1 × 5 , = 2+1 2+1
Q(3, 2) 1 X′
X –2
⇒
O
3
½
6 − 2 4 + 5 , = 3 3 4 9 4 = , = , 3 3 3
Y′
3
Q. 3. In parallelogram ABCD, A(3, 1), B(5, 1) C(a, b) and D(4, 3) are the vertices. Find vertex C(a, b).
T
A [CBSE Board Term, 2019]
opper Answer, 2019
Sol.
2
134 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 4. Find the coordinates of the point P which divides the join of A(–2, 5) and B(3, – 5) in the ratio 2 : 3.
A [CBSE SQP, 2018]
Sol. Let
∴
x =
6−6 = 0 5
1
−10 + 15 =1 5 1 [CBSE Marking Scheme, 2018] y =
P A
Since,
and
and y = So, the point is (0, 1).
B
x =
mx2 + nx1 m+n
½
y =
my2 + ny1 m+n
½
2 × 3 + 3 × ( -2 ) = 0 2+3
½
2 × ( -5) + 3 × 5 5 = = 1 ½ 2+3 5
Q. 5. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, – 3). Hence find m. C + A [CBSE Delhi/OD Set-2018]
Detailed Solution:
x =
Sol.
A
B
P Let AP : PB = k : 1 6k + 2 =4 k +1
⇒
Hence
k = 1, ratio is 1 : 1 −3 + 3 = 0 m = 2
½
1 ½
[CBSE Marking Scheme, 2018]
opper Answer, 2018
T
Detailed Solution:
p Q. 6. If 1, is the mid point of the line segment 3
2
2 joining the points (2, 0) and 0 , , then show 9 that the line 5x + 3y + 2 = 0 passes through the point (– 1, 3p).
Sol. Since 1,
p is the mid point of the line segment 3
joining the points (2, 0) and 0 ,
\
A [CBSE SQP, 2017]
2 . 9
2 0+ p 2p 2 9 = = = 3 2 3 9 1 p = 3
1
[ 135
lines (in two-dimensions)
Hence, the line 5x + 3y + 2 = 0, passes through the point (– 1, 1) as 5(– 1) + 3(1) + 2 = 0. 1 [CBSE Marking Scheme, 2017]
Q. 8. If the line segment joining the points A(2, 1) and B(5, – 8) is trisected at the points P and Q, find the coordinates P.
Q. 7. In what ratio does the point P(– 4, 6) divide the
line segment joining the points A(– 6, 10) and A [Delhi Comptt. Set-I, II, III 2017] B(3, – 8) ?
Sol. Let
∴ or, or, Hence,
½
AP : PB = k : 1 3k − 6 = – 4 k +1 3k – 6 = – 4k – 4 7k = 2 2 k = 7
1
AP : PB = 2 : 7 ½ [CBSE Marking Scheme, 2017]
A [Outside Delhi Comptt. Set-I, III, 2017]
Sol. A
(2, 1)
B (5, –8)
Q
P
Let P(x, y) divides AB in the ratio 1 : 2.
\ Using section formula
x =
1× 5+ 2× 2 =3 1+ 2
and
y =
1 × −8 + 2 × 1 = −2 1+ 2
Hence coordinates of P are (3, – 2).
1
1
Q. 9. If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal, then prove that 3x = 2y.
T
U [CBSE OD, Set-II, 2017, 2015]
Sol.
opper Answer, 2017
2
Q. 10. A line intersects the Y-axis and X-axis at the points P and Q respectively. If (2, – 5) is the mid-point of PQ, then find the coordinates of P and Q.
T
U [CBSE OD, Set-III, 2017]
opper Answer, 2017
Sol.
2
136 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 11. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the co-ordinates of P. U [CBSE, Delhi Set I, II, III, 2016]
Sol. Let the point P be (2y, y)
1
Given,
or, ( 2 y − 2 )2 + ( y + 5)2 =
Solving to get y = 8
Hence, coordinates of point P are (16, 8).
PQ = PR ( 2 y + 3 )2 + ( y − 6 )2 ½
½
[CBSE Marking Scheme, 2016] Detailed Solution:
Given, 2
( 2 y − 2 ) + ( y + 5)
2
=
1 2
( 2 y + 3) + ( y − 6 )
2
½
Squaring both sides 4y2 – 8y + 4 + y2 + 25 + 10y = 4y2 + 12y + 9 + y2 + 36 – 12y or, 2y = 16 or y = 8 Hence, coordinates of point P are (16, 8). ½
Q. 12. Find the ratio in which Y-axis divides the line segment joining the points A(5, –6) and B(–1, –4). Also find the co-ordinates of the point of division.
= (y – a – b)2 – (y – b + a)2
or, (x – a – b + x – a + b)(x – a – b – x + a – b)
PQ = PR
A [CBSE, Delhi Set I, II, III, 2016]
Sol. Let the point on y-axis be (0, y) and AP : PB = k : 1. ½ Therefore
Hence, required ratio is 5 : 1
−13 Hence, point on Y-axis is 0 , . 3
y =
= (y – a – b + y – b + a)(y – a – b – y + b – a)
or,
(2x – 2a)(– 2b) = (2y – 2b)(– 2a)
or,
(x – a)b = (y – b)a Hence proved. 1
or,
bx = ay.
[CBSE Marking Scheme, 2016]
Q. 14. Prove that the point (3, 0), (6, 4) and (–1, 3) are the vertices of a right angled isosceles triangle.
U [CBSE, OD Set I, II, III, 2016]
Sol. Let A(3, 0), B(6, 4) and C(–1, 3)
∴ AB2 = (3 – 6)2 + (0 – 4)2 = 9 + 16 = 25
½
BC2 = (6 + 1)2 + (4 –3)2 = 49 + 1 = 50
½
2
2
2
½
AB2 + CA2 = BC2
½
and CA = (–1 –3) + (3 – 0) = 16 + 9 = 25 2
2
AB = CA or, AB = CA
5−k = 0 gives k = 5 k +1
Let the point P be (2y, y)
or,
or, [x – a – b]2 – [x – a + b]2
Squaring, we get
[x – (a + b)]2 + [y – (b – a)]2
∴ Triangle is isosceles.
−4( 5) − 6 −13 = 6 3
½
½
Also,
25 + 25 = 50
½
Q. 13. If the point P(x, y) is equidistant from the points Q(a + b, b – a) and R(a – b, a + b), then prove that bx = ay. [OD, Set I, II, III, 2016]
A [CBSE SQP, 2016]
Sol. |PQ| = |PR|
[ x - ( a + b )]2 + [ y - ( b - a )]2
=
[ x - ( a - b )]2 + [ y - ( b + a )]2
[CBSE Marking Scheme, 2016]
Since, Pythagoras theorem is verified, therefore triangle is a right angled triangle.
[CBSE Marking Scheme, 2016]
Q. 15. If the mid-point of the line segment joining x y + 1 A , and B(x + 1, y – 3) is C(5, – 2), find 2 2
x, y.
U [CBSE Board Term-2, 2016]
[ 137
lines (in two-dimensions)
Sol.
or,
or,
49 + 4 – 4t + t2 = 41 + t2 + 4t + 4
or,
53 – 4t = 45 + 4t
y+1 + y-3 2 = – 2 2
or,
1
x = 6
or,
8t = 8
∴
t =1
[CBSE Marking Scheme, 2015]
Q. 17. Show that the points (a, a), (–a, –a) and ( − 3 a , 3 a)
y + 1 + 2y – 6 = – 8 y = – 1.
Q. 16. If A (5, 2), B (2, – 2) and C (–2, t) are the vertices of a right angled triangle with ∠B = 90°, then find U [CBSE Board, 2015] the value of t.
are the vertices of an equilateral triangle. U [Foreign Set I, II, III, 2015]
1
[CBSE Marking Scheme, 2016]
Sol. AB2 = (2 – 5)2 + (– 2 –2)2 =9 + 16 = 25
Sol. Let A(a, a), B(– a, – a) and C ( - 3 a , 3 a ) Here, AB =
BC =
BC2 = (– 2 – 2)2 + (t + 2)2 = 16 + (t + 2)2 AC2 = (5 + 2)2 + (2 – t)2 = 49 + (2 – t)2
1
a 2 + 3a 2 − 2 3a 2 + a 2 + 2 3a 2 + 3a 2
=
8 a 2 = 2 2 a unit
and AC =
Sol. By using section formula,
3
B(x, y) 4
–1 =
=
½
mx 2 + nx1 m+n 3×x+4×2 3x + 8 = ½ 3+4 7
⇒ 3x + 8 = – 7
( a + 3 a )2 + ( a − 3 a )2
½
a 2 + 3a 2 + 2 3a 2 + a 2 + 3a 2 − 2 3a 2
Since AB = BC = AC, therefore ABC is an equilateral ½ triangle.
= 8a2 = 2 2a unit
[CBSE Marking Scheme, 2015]
Q. 1. If the point C(– 1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
C(–1, 2)
½
Short Answer Type Questions-II
A(2, 5)
( − a + 3 a )2 +( − a − 3 a )2
=
AC2 = AB2 + BC2
∴
( a + a )2 +( a + a )2 = 2 2a unit ½
Since, ∆ABC is a right angled triangle.
1
or,
49 + (2 – t)2 = 25 + 16 + (t + 2)2
x +x+1 2 At mid-point of AB = =5 2
3 marks each
⇒ 3x = – 15 ⇒
x = – 5
and
2 =
=
½
my 2 + ny1 m+n 3×y+4×5 3 y + 20 = ½ 3+4 7
⇒ 3y + 20 = 14 ⇒ 3y = 14 – 20 = – 6 ⇒
y = –2
Hence, the coordinates of B(x, y) is (–5, –2).
½ ½
138 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 2. 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. A + U [CBSE, OD Set-I, 2020] Sol. Since, the mid point of A and B is P, A(3, 4)
P(x, y)
B(k, 6)
3+k = x 2
4+6 10 = = 5 1 and y = 2 2 Also, given, x + y – 10 = 0 3+k and y = 5, we Substituting the value of x = 2 get 3+k + 5 – 10 = 0 1 ⇒ 2 3+k = 5 ⇒ 2 ⇒ 3 + k = 10 ⇒ k = 10 – 3 = 7. Hence, the value of k is 7. 1 Q . 3. The line segment joining the points A(2, 1) and B(5, –8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y+ k = 0, find the value of k. A [CBSE Delhi Set-I, II, III, 2019] Sol.
½
AP : PB = 1 : 2
\ AP = PQ = QB PB = PQ + QB = AP + AP = 2AP AP : PB = AP : 2AP = 1 : 2 \ P divides the line segment AB in the ratio 1 : 2. We know that, coordinates of the point dividing the
line segment joining the points (x1, y1) and (x2, y2) in mx 2 + nx1 my 2 + ny1 , the ratio m : n is given by . m+n m+n
\ Coordinates of P 1 × 5 + 2 × 2 1 × ( −8 ) + 2 × 1 , = 1+ 2 1+ 2 5 + 4 −8 + 2 , = 3 3 9 −6 = , = ( 3, −2 ) 3 3 Point P(3, –2) lies on the given line 2x – y + k = 0 \ 2 × 3 – (–2) + k = 0 ⇒ 6 + 2 +k = 0 ⇒ 8 + k = 0 ⇒ k = –8 Thus, the value of k is –8.
COMMONLY MADE ERROR Some Candidates use mid point formula
instead of section formula to find the coordinates of P. Some candidates also make calculation errors.
½
ANSWERING TIP Understand the concept if a point divides a line segment in the ratio m : n (m ≠ n) then m : n is not equal to n : m.
½+½
Thus point P is (3, –2). Point (3, –2) lies on 2x – y + k = 0 ⇒ 6 + 2 + k = 0 ⇒ k = – 8. 1 [CBSE Marking Scheme, 2019]
Q. 4. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (–2, – 5) and (6, 3). Find the coordinates of the point of A [CBSE OD Set-I, II, III, 2019] intersection.
4+5 2−8 x= = – 2 = 3 and y = 3 3
Sol. Let the line x – 3y = 0 intersect the segment
Detailed Solution:
joining A(–2, –5) and B(6, 3) in the ratio k : 1
6 k − 2 3k − 5 \ Coordinates of P are , k + 1 k + 1 k
The given points are A (2, 1) and B (5, –8). Given, P and Q trisects the line segment AB
P lies on x – 3y = 0 ⇒
½ 1
3k − 5 6k − 2 =3 k + 1 k +1
13 ⇒ k= 3
[ 139
lines (in two-dimensions)
\ Ratio is 13 : 3
13 3 − 5 3 y = 13 +1 3
1
9 3 ⇒ Coordinates of P are , ½ 2 2 [CBSE Marking Scheme, 2019] Detailed Solution: Let the ratio in which line x - 3y = 0 divides the line segment is k : 1
and
=
8×3 16
24 3 = = 16 2
½
Hence, the coordinates of point of intersection 9 3 P(x, y) = , 2 2
½
Q. 5. If A(– 2, 1), B(a, 0), C(4, b) and D(1, 2) are the
k
vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides. A [CBSE Delhi/OD, 2018]
P
Sol. Given, ABCD is a parallelogram and diagonals AC and BD bisect each other Using section formula, we get
x =
k × 6 + 1 × ( −2 ) k +1
6k − 2 = k + 1
and
y =
...(i) ½
k × 3 + 1 × ( −5) k +1
3k − 5 = k + 1
...(ii) ½
The point P(x, y) lies on the line, hence satisfies the equation of the given line.
⇒
⇒ ⇒ ⇒ ⇒
6k − 2 3k − 5 − 3 k + 1 k +1
6k – 2 - 3(3k – 5) = 0
6k – 2 – 9k + 15 = 0
units each. ½+1 [CBSE Marking Scheme, 2018]
We know that diagonals of parallelogram bisect each other.
13 k = 3
\ Mid-point of diagonal AC ½
1 + b −2 + 4 1 + b , = 1, 2 2 2
½
a + 1 0 + 2 a+1 , , 1 = 2 2 2
½
Mid-point of diagonal BD
13 6× −2 3 x = 13 +1 3
Mid point of diagonal AC = mid point of diagonal BD
78 − 6 16
72 = 16 9 = 2
10
Detailed Solution:
–3k + 13 = 0
=
a+1 b+1 = 1 and =1 2 2
⇒ ⇒ a = 1, b = 1. Therefore, length of sides are
= 0
Hence, the required ratio is 13 : 3 Now, substituting value of k in x and y, we get
1 Therefore mid point P of BD is same as mid point of AC ½ a + 1 2 − 2 + 4 1 + b , = , 2 2 2 2
½
a+1 2
1+ b =1 2
1 =
\
2 = a + 1 and 1 + b = 2
Again,
a = 1 AB =
and
and
½
b = 1
( x 2 − x 1 ) + ( y 2 − y1 ) 2
2
½
140 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
=
(1 + 2 )2 + ( 0 − 1)2
AB =
9 + 1 = 10 unit
BC =
( x 2 − x 1 ) + ( y 2 − y1 )
and
=
BC =
COMMONLY MADE ERROR Mostly candidates do not use the mid
2
point formula. Generally they use distance formula.
2
(4 − 1)2 + (1 − 0 )2
ANSWERING TIP
9+1
Candidates should use mid point formula by which they can get correct solution in lesser time.
= 10 unit ABCD is a parallelogram (Given) AB = CD =
10 unit
and
BC = AD =
10 unit
1
Q. 6. The points A(1, – 2), B(2, 3), C(k, 2) and D(– 4, – 3) are the vertices of a parallelogram. Find the value of k. C + U [CBSE SQP, 2018]
Q. 7. If coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, – 5), find coordinates of the other two vertices. U [CBSE Comptt. Set I, II, III, 2018]
\
Sol. Let the coordinates of C and D be (a, b) and (c, d).
Sol.
½ ½
⇒ midpoint of AC = midpoint of BD −4 + 2 -3 + 3 1 + k -2 + 2 , ⇒ = 2 , 2 2 2
⇒
1
and
c+1 =2⇒c=3 2
1+ k −2 = 2 2 k = – 3 [CBSE Marking Scheme, 2018] 24
Diagonals of parallelogram bisect each other
⇒
and
3+a = 2 ⇒ a = 1 2 2+b = – 5 ⇒ b = – 12 2
\
½ ½ ½
d+0 = – 5 ⇒ d = – 10 ½ and 2 Coordinates of C and D are (1, – 12) and (3, – 10) 1 [CBSE Marking Scheme, 2018]
Q. 8. In what ratio does the point 11 , y divide the line segment joining the points P (2, – 2) and Q (3, 7) ? Also find U [CBSE OD Set-I, III, 2017] T
the value of y.
Sol.
opper Answer, 2017
[ 141
lines (in two-dimensions)
3
the line segment joining the points (5, 7) and (8, 10) in 3 equal parts. [Board SQP, 2016] A [CBSE OD Comptt. Set-II, 2017]
Sol. A
B
Q
P
x =
1 (8 ) + 2 ( 5 ) = 6 , 3
and
y =
1 (10 ) + 2 (7 ) =8 3
1
\ P(6, 8) And Q is the mid point of PB.
y1 =
1
8 + 10 =9 2
\ Q(7, 9)
1
[CBSE Marking Scheme, 2017]
( 0 + 4 )2 + ( 4 − 0 ) 2
=
32 ½
= 4 2 units
EF =
( 4 − 0 )2 + ( 0 − 4 ) 2
=
32
and C(2, 0) is similar to DDEF with vertices D(– 4, 0), E(0, 4) and F(4, 0). A [Board Foreign Set-I, II 2017; CBSE Delhi Set-I, II, III, 2017]
FD =
( −4 − 4 )2 + (0 − 0 )2
=
64
= 8 units ½ Since, ratio of the corresponding sides of two similar Ds is equal. AB BC AC ½ i.e., = = DE EF DF 4 1 2 2 2 2 1 = = or, = 8 2 4 2 4 2 \ DABC ~ DDEF Hence Proved. [CBSE Marking Scheme, 2017]
Q. 11. In the given figure DABC is an equilateral triangle
Q. 10. Show that DABC with vertices A(– 2, 0), B(0, 2)
DE =
= 4 2 units
\
and
Let P(x, y) and Q(x1, y1) trisect AB. P divides AB in the ratio 1 : 2
6+8 =7 x1 = 2
Q. 9. Find the co-ordinates of the points which divide
of side 3 units. Find the co-ordinates of the other two vertices. A [Foreign Set-I, II, 2017] [Foreign Set-III, 2015]
Sol. Using distance formula
AB =
( 0 + 2 )2 + ( 2 − 0 ) 2
=
4+4
=
4+4
= 2 2 units BC =
( 2 − 0 )2 + ( 0 − 2 ) 2
½
= 2 2 units
CA =
( −2 − 2 ) + (0 − 0 )
= 4 units
2
2
=
16 Sol. The co-ordinates of B are (5, 0) Let co-ordinates of C be (x, y)
(AB = 3) 1
142 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
AC2 = BC2 (sides of equilateral triangle) (x – 2)2 + (y – 0)2 = (x – 5)2 + (y – 0)2 or, x2 + 4 – 4x + y2 = x2 + 25 – 10x + y2 or, 6x = 21 7 1 x = 2 Since
And (x – 2)2 + (y – 0)2 = 9
A [Foreign Set I, II, III, 2016]
Sol.
y =
my 2 + ny1 m+n
x =
2 3 × 2 + 4 × −2 = − 7 3+4
and
Hence,
A(7, 2), B(9, 10) and C(1, 4). If E and F are the mid-points of AB and AC respectively, prove that 1 A [CBSE Board Term-2, 2015] EF = BC. 2
Sol. Let the mid-points of AB and AC be E(x1, y1) and F(x2, y2)
9 + 7 10 + 2 , ∴ Co-ordinates of point E = 2 2
A
3:4 C
B
(2, 5)
(–1, 2)
(x, y )
3 × −4 + 4 × −2 20 =− ½ 3+4 7 20 2 ½ P = − , − 7 7 [CBSE Marking Scheme, 2015] y =
Q. 14. The co-ordinates of the vertices of ∆ABC are
3 3 27 y2 = or y = ± 2 4 (+ve sign to be taken)
7 3 3 . Hence, C = , 1 2 2 Q. 12. 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 x2 + y2.
mx 2 + nx1 m+n
and
2
7 2 or, − 2 + y = 9 2 9 9 2 or, + y 2 = 9 or, y = 9 − 4 4
or,
x =
½
E(x1, y1) = (8, 6)
AC 3 Given that, = CB 4 Applying section formula for x co-ordinate, 3 x + 4( 2 ) – 1 = 3+4 or, –7 = 3x + 8 or x = – 5 Similarly for y co-ordinate, 3 y + 4( 5 ) 2 = 3+4
1
7 + 1 2 + 4 , Co-ordinates of point F = 2 2
14 = 3y + 20 or y = – 2 ∴ (x, y) is (–5, –2) 1 Hence, x2 + y2 = (–5)2 + (–2)2 = 25 + 4 = 29 1 [CBSE Marking Scheme, 2016] Q. 13. If the co-ordinates of points A and B are (– 2, – 2) and (2, – 4) respectively, find the co-ordinates of 3 P such that AP = AB, where P lies on the line 7 segment AB. U [CBSE OD, 2015, Set I, II, III, 2015] AP =
3 AB or, AP : PB = 3 : 4 1 7
Sol.
½
(x2, y2) = (4, 3) Length of EF = =
(8 − 4 ) + (6 - 3) 2
2
( 4 )2 + (3)2
= 5 units
Length of BC = =
...(i) 1
(9 - 1) + (10 - 4 ) 2
2
(8 )2 + (6 )2
=10 units ...(ii) From equations (i) and (ii), we get 1 EF = BC. Hence proved. 1 2
[CBSE Marking Scheme, 2015]
Q. 15. Find the ratio in which the line segment joining
P (x, y) 3:4 A(–2, –2) ∴ Using the section formula
B(2, –4)
1
the points A(3, –3) and B(–2, 7) is divided by X-axis. Also find the co-ordinates of point of division. A [CBSE Delhi Term-2, 2015]
[ 143
lines (in two-dimensions)
Sol.
P
k
1
(x, 0)
=
B (–2, 7)
Let the ratio be k : 1, Using section formula, we get, 1 ( −3 ) + k ( 7 ) 0 = 1+ k 3 or, k = 7 or, Ratio = 3 : 7 m2 x1 + m1x 2 Also, x = m1 + m2
x = 1
Q. 1. 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. A [CBSE OD Set-III, 2019]
3 7 = 21 − 6 = 3 3 7+3 2 1+ 7
3-2×
5 marks each
Q. 3. The base BC of an equilateral triangle ABC lies on y-axis. The co-ordinates of point C are (0, – 3). The origin is the mid-point of the base. Find the co-ordinates of the point A and B. Also find the co-ordinates of another point D such that A [Foreign Set I, II, 2015] BACD is a rhombus.
1 1
0 =
5k 1 1 ⇒ k= i.e., 1 : 5 k 1 5
1
6 4 6 k 4 -26 -13 = 5 = = 1 k 1 6 3 1 5
1
⇒
⇒
y =
1
[CBSE Marking Scheme, 2014]
Sol. Co-ordinates of point B are (0, 3) ∴ BC = 6 unit Let the co-ordinates of point A be (x, 0).
Sol. Any point on Y-axis is P(0,y) Let P divides AB in k :1
1
1+ k
3 ∴ Co-ordinates of point P are , 0 . 2
Long Answer Type Questions
1 (3) + k ( -2 )
A (3, –3)
½ ½
2
x +9
or,
AB =
∴
AB2 = BC2 x + 9 = 36 2
1
-13 ) [CBSE Marking Scheme, 2019] 1 ⇒ P is (0, 3 Q. 2. If P(9a – 2, – b) divides the line segment joining A(3a + 1, – 3) and B(8a, 5) in the ratio 3 : 1. Find the A [CBSE SQP, 2016] values of a and b.
1
Sol. By section formula
and
From (ii),
9a – 2 = – b =
3( 8 a ) + 1( 3a + 1) 3+1
...(i) 1
3( 5) + 1( -3) 3+1
...(ii) 1
or,
From (i),
15 - 3 =3 – b = 4 b = – 3
1
24 a + 3a + 1 4 1 4(9a – 2) = 27a + 1 36a – 8 = 27a + 1 9a = 9 a = 1 1 [CBSE Marking Scheme, 2016] 9a – 2 =
x2 = 27 or, x = ±3 3
Co-ordinates of point A = ( 3 3 , 0 ) Since ABCD is a rhombus. or, AB = AC = CD = DB
1
∴ Co-ordinates of point D = ( - 3 3 , 0 ).
1
[CBSE Marking Scheme, 2015]
Q. 4. (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. A [CBSE, Term-2, 2015]
144 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
½
= 2 13 unit
Sol.
or,
AB = BC ≠ AC
or, DABC is isosceles. Now, using mid-point formula, the co-ordinates of mid-point of BC are
x =
-5+0 5 = - 2 2
y =
3+4 7 = 2 2
Let the vertices of DABC be A(1, –1), B(0, 4) and C(–5, 3). 1
\ Using distance formula,
AB =
or, 2
(1 - 0 ) + ( -1 - 4 )
2
½
5 7 D(x, y) = - , 2 2
1
\ Length of median, AD =
=
26 unit
BC =
2
=
AC = =
TOPIC
1+ 5
2
½
( -5 - 0 ) + ( 3 - 4 ) 25 + 1 =
½
( -5 - 1) + ( 3 + 1) 36 + 16 =
52
=
-7 9 + 2 2
=
130 = 4
26 unit
2
-5 7 - 1 + + 1 2 2
2
2
=
2
2
∴ Length of median AD is
130 units. 2
Revision Notes If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle, then the co-ordinates of centroid are x1 + x 2 + x3 y1 + y 2 + y3 , G = 3 3
If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle,
Area of DABC =
1 [x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)] 2
If the points are collinear, then the area of triangle is zero.
1
[CBSE Marking Scheme, 2015]
-2
then
2
130 unit 2
Area of Triangle
2
[ 145
lines (in two-dimensions)
How is it done on the
GREENBOARD?
Q.1. Find k, if the point A(2, 3), B(5, K) Step 3: According to question, U and C(7, 9) are collinear. 1 0 = [2(k – 9) + 5(9 – 3) Solution: 2 Step 1: If points A, B, and C are collinear then the area of triangle + 7(3 – k)] formed by them is zero. or, 2k – 18 + 45 – 15 + 21 – 7k = 0 Step 2: Area of triangle formed by the vertices (x1, y1), (x2, y2) and (x3, or, –5k + 33 = 0 y3) is given by or, 5k = 33 1 A = [x1(y2 – y3) + x2(y3 – y1) 33 or, k = 2 5 + x3(y1 – y2)]
Very Short Answer Type Questions Q. 1. If the points A(3, 1), B(5, p) and C(7, – 5) are collinear, then find the value of p. A [CBSE Delhi Set-I, 2020] Sol. Here, x1 = 3, x2 = 5, x3 = 7 and y1 = 1, y2 = p, y3 = – 5 If points are collinear, then area of triangle = 0 1 \ |[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| = 0 ½ 2 ⇒
1 |[3(p + 5) + 5(–5 –1) + 7(1 – p)]| = 0 2
⇒
1 |[3p + 15 – 30 + 7 – 7p]| = 0 2
⇒ ⇒ ⇒
or,
or,
Q. 2. If the points (0, 0), (1, 2) and (x, y) are collinear, A
Sol. The points are collinear, then area of triangle = 0 1 |[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| = 0 ∴ 2
or,
\
1 |[y – 2x]| = 0 2 2x – y = 0 y 1 x = 2 [CBSE Marking Scheme, 2016]
U [Foreign Set-I, II, III, 2015]
Sol. Since, the points are collinear, then Area of triangle = 0 1 |[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| = 0 2
Q. 1. Show that the points A(0, 1), B(2, 3) and C(3, 4) are U [CBSE Term-2, 2016]
½
1 |[x(– 4 + 5) + (– 3) (– 5 – 2) + 7(2 + 4)]| = 0 2
Short Answer Type Questions-I
collinear.
1 |[0(2 – y) + 1(y – 0) + x(0 – 2) ]| = 0 2
Q. 3. If the points A(x, 2), B(– 3, – 4) and C(7, – 5) are collinear, then find the value of x.
– 4p – 8 = 0 – 4p = 8 p = – 2. ½
then find the relation between x and y. [CBSE Term-2, 2015, 2016]
1 mark each
x + 21 + 42 = 0 x = – 63 ½
2 marks each
Sol. Area of the triangle formed by the given points A(0, 1), B(2, 3) and C(3, 4)
146 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
1 |0(3 – 4) + 2(4 – 1) + 3(1 – 3)| 1 2 1 = |0 + (2) (3) + (3) ( – 2)| 2 1 = |6 – 6| 2
1 = (0) 2 =0 ∴ The given points are collinear. 1 [CBSE Marking Scheme, 2016]
Q. 2. Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this triangle. A [Foreign Set I, II, III, 2016]
1 25 sq units. 1 ×5×5= 2 2
[CBSE Marking Scheme, 2016]
Q. 3. Find the relation between x and y, if the points A(x, y), B(– 5, 7) and C(– 4, 5) are collinear.
A [CBSE Term,-2, 2015]
Also,
=
1
Sol. If area covered by the given points is O, the points are collinear. Area of ∆ABC = 0 1 |[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| = 0 ½ 2 1 1 = |[x(7 – 5) – 5(5 – y) – 4(y – 7)]| = 0 2 or, 2x – 25 + 5y – 4y + 28 = 0 ½ or, 2x + y + 3 = 0 [CBSE Marking Scheme, 2015]
Sol. Let the points be A(2, – 2), B(–2, 1) and C(5, 2). Applying distance formula, AB2 = (2 + 2)2 + (–2 –1)2 = 16 + 9 AB2 = 25 or, AB = 5 Similarly BC2 = (–2 –5)2 + (1 – 2)2 = 49 + 1 = 50 or,
= 9 + 16 = 25 or, AC2 = 25 or AC = 5 Clearly AB2 + AC2 = BC2 25 + 25 = 50 Hence, the triangle is right angled, 1 Area of ∆ABC = × Base × Height 2
=
BC2 = 50 or, BC = 5 2 AC2 = (2 – 5)2 + (–2 –2)2
Short Answer Type Questions-II
3 marks each
Q. 1. Find the area of triangle PQR formed by the points P(– 5, 7), Q(– 4, –5) and R(4, 5). A [CBSE Delhi Set-I, 2020] Sol. Here x1 = –5, x2 = – 4, x3 = 4 and y1 = 7, y2 = –5, y3 = 5 ½ 1 \ Area of D PQR = |[x1(y2 – y3) + x2(y3 – y1) 2 + x3 (y1– y2)]| ½
=
1 |[50 + 8 + 48]| 2
½
=
1 × 106 = 53 sq units. 2
½
Q. 2. Find the area of triangle ABC with A(1, –4) and the mid-points of sides through A being (2, –1) and C + A [CBSE OD Set-I, 2020] (0, –1). Sol. Let the coordinates of the points B and C be (x, y) x+1 =2 and (a, b), then 2 y−4 =–1 ⇒ x = 4 – 1 = 3 and 2 ⇒ y = – 2 + 4 = 2 1
½
1 = |[–5 (–5 –5) –4 (5 – 7) 2 +4 (7 +5)]| ½ 1 = |[–5(–10) –4(–2) +4(12)]| 2
a+1 = 0 ⇒ a = – 1 2 b−4 = – 1 ⇒ b = – 2 + 4 = 2 and 2 So, the coordinates of B and C are (3, 2) and (– 1, 2) Here, x1 = 1, y1 = – 4, x2 = 3, y2 = 2 and x3 = – 1, y3 = 2 ½ 1 [x1(y2 – y3) + x2(y3 – y1) \ Area of DABC = 2 + x3(y1 – y2)] 1 [1 (2 – 2) + 3(2 + 4) – 1 (– 4 – 2)] = 2 1 [0 + 18 + 6] = 2 Similarly,
=
1 × 24 2
= 12 square units.
1
[ 147
lines (in two-dimensions)
opper Answer, 2019
T
Detailed Solution:
Q. 3. Two friends Seema and Aditya work in the same office at Delhi. In the Christmas vacations, both decided to go to their hometowns represented by Town A and Town B respectively in the figure given below. Town A and Town B are connected by
3
trains from the same station C (in the given figure) in Delhi. Based on the given situation, answer the following questions :
(i) Who will travel more distance, Seema or Aditya, to reach to their hometown ?
148 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(ii) Seema and Aditya planned to meet at a location D situated at a point D represented by the mid-point of the line joining the points represented by Town A and Town B. Find the coordinates of the point represented by the point D.
(ii) We have, D is the mid-point AB.
(iii) Find the area of the triangle formed by joining the points represented by A, B and C.
1 + 4 7 + 2 , The coordinates of D = 2 2
5 9 = , . 2 2
1
(iii) Here, x1 = 1, y1 = 7, x2 = 4, y2 = 2, x3 = – 4 and y3 = 4. \ Area of DABC 1 = |[ x1 ( y 2 − y3 ) + x 2 ( y3 − y1 ) + x 3 ( y1 − y 2 )]| 2
[CBSE SQP, 2020]
1 = |[1( 2 − 4 ) + 4( 4 − 7 ) + ( −4 )(7 − 2 )]| 2
1 = |[ −2 − 12 − 20]| 2
1 = × | −34 | = | −17 | 2
ar DABC = 17 sq. units
Sol. (i) A(1, 7), B(4, 2) C(– 4, 4)
Distance travelled by Seema =
34 units
Distance travelled by Aditya =
68 units
1
1 + 4 7 + 2 5 9 , (ii) Coordinates of D are = , 1 2 2 2 2 1 (iii) ar (DABC) = |[1 (2 – 4) + 4 (4 – 7) – 4(7 – 2)]| 2
Q. 4. If A(– 5, 7), B(– 4, – 5), C(– 1, – 6) and D(4, 5) are
the vertices of a quadrilateral, find the area of the quadrilateral ABCD. A [CBSE Delhi/OD, 2018]
= 17 sq. units 1 [CBSE SQP Marking Scheme, 2020] Detailed Solution: According to given graph, The coordinates, where the station C is situated = (– 4, 4) The coordinates of Town A = (1, 7) and the coordinates of Town B = (4, 2) (i) Distance travelled by Seema from station C to their home town
A=
(1 + 4 ) 2 + ( 7 − 4 ) 2
=
( 5)2 + ( 3)2 = 34
=
(4 + 4) + (2 − 4)
=
( 8 )2 + ( −2 )2
= 68 unit Hence, Aditya travelled more distance.
Sol. Area of quad. ABCD = arDABD + arDBCD 1 | (– 5) (– 5 – 5) Area of DABD= 2 + (– 4) (5 – 7) + (4) (7 + 5)| = 53 sq. units 1
D
C
B
A
( 5)2 + ( 3)2 == 34 unit Distance travelled by Aditya from C to B 2
[CBSE Term-2, 2015]
Area of ar DBCD =
1 |(–4) (– 6 – 5) + (–1) 2
(5 + 5) + 4 (– 5 + 6)| 1 = |44 – 10 + 4| 2
2
1
= 19 sq. units 1 Hence, area of quad. ABCD = 53 + 19 = 72 sq. units 1 [CBSE Marking Scheme, 2018]
[ 149
lines (in two-dimensions)
COMMONLY MADE ERROR
ANSWERING TIP
Some candidates use correct formula for
finding area of triangle but simplifying in error.
For simplifying, be careful.
opper Answer, 2018
T
Detailed Solution:
Q. 5. Find the value of k for which the points (3k – 1, k – 2) (k, k – 7) and (k – 1, – k – 2) are collinear.
A [CBSE SQP, 2018]
Sol. For collinearity of the points, area of the triangle formed by given points is zero. 1 {(3k – 1) (k – 7 + k + 2) + k(–k –2 – k + 2) 2 + (k – 1)(k – 2 – k + 7)} = 0 1 1 {(3k – 1) (2k – 5) – 2k2 + 5k – 5} = 0 2
4k2 – 12k = 0 1 k = 0, 3 1 If k = 0 then two points are coincide \ k = 3. [CBSE Marking Scheme, 2018]
Q. 6. If the area of triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, find x. U [CBSE SQP, 2018]
Sol. Given,
3
Ar (ABC) = 4
1 |[x(4 – 5) + 4(5 – 3) + 3(3 – 4)]| = 4 1½ 2 (– x + 5) = 8 – x + 5 = 8 x = –3 1 If – (– x + 5) = 8 x = 13 ½ [CBSE Marking Scheme, 2018] Q. 7. If the points A(0, 1), B(6, 3) and C(x, 5) are the vertices of a triangle, find the value of x such that U [CBSE S.A.II 2016] area of ∆ABC = 10. [CBSE Compt. Set-I, II, III-2018] Sol. Given, area of DABC = 10 1 \ |[x1(y2 – y3) + x2(y3 – y1) +x3(y1 – y2)]| = 10 ½ 2 Here, x1 = 0, y1 =1, x2 = 6, y2 = 3, x3 = x and y3 = 5 ½
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
⇒ If
1 |[0(3 – 5) + 6(5 – 1) + x(1 – 3)]| = 10 ½ 2
Q. 10. If a ¹ b ¹ 0, prove that the points (a, a2), (b, b2) and (0, 0) will not be collinear.
1 |[0 + 24 + (–2)x]| = 10 ½ 2
U [CBSE, Delhi Set I, II, III, 2017]
⇒ –(– 2x + 24) = 20 ⇒ 2x = 20 + 24 ⇒
– 2x + 24 = 20 – 2x = – 4
Then,
x = +2. 1
x = 22
Q. 8. Find the area of the triangle formed by joining
the mid-points of the sides of a triangle, whose co-ordinates of vertices are (0, – 1), (2, 1) and A [CBSE OD Comptt. Set I, III, 2017] (0, 3). Sol. Let the vertices of given triangle be A(0, – 1), B(2, 1) and C(0, 3). 1 Then, the coordinates of mid-points are P(1, 0), Q(1, 2) and R(0, 1). 1 Area of DPQR, 1 = x1 ( y 2 − y3 ) + x 2 ( y3 − y1 ) + x3 ( y1 − y 2 ) 2
1 = ( 2 − 1) + 1 (1 − 0 ) + 0 (0 − 2 ) 2
Sol. Given, ar DABC = 5 sq. untis 1 7 or, 2 ( −2 − y ) + 3 ( y − 1) + (1 + 2 ) = 5 2 2
y+
or, If y +
1½
1 21 −4 − 2 y + 3 y − 3 + =5 2 2
or,
(
) (
1 = ab 2 − a 2 b + 0 2
)
2
1 = ab (b − a ) ≠ 0 2
(a ¹ b ¹ 0)
1
Hence, the given points are not collinear.
[CBSE Marking Scheme, 2017]
Q. 11. The points A(4, – 2), B(7, 2), C(0, 9) and D(– 3, 5) form a parallelogram. Find the length of altitude of the parallelogram on the base AB.
7 = 10 2 y = 10 −
Sol. Let the height of parallelogram taking AB as base be h.
\
( 7 − 4 )2 + ( 2 + 2 ) 2
AB =
32 + 4 2 =
=
9 + 16
1
= 5 units.
Ar DABC
1 = x1 ( y 2 − y3 ) + x 2 ( y3 − y 2 ) + x3 ( y1 − y 2 ) 2 1 = 4 ( 2 − 9 ) + 7 (9 + 2 ) + 0 ( −2 − 2 ) 2 1 = | −28 + 77 | 2
1 7 13 = 2 2
7 7 27 = – 10 or y = −10 − = − 2 2 2
1 49 sq. units = × 49 = 2 2
5 × h = 49
Hence, the value of y =
or,
h =
27 13 or − ½ 2 2 [CBSE Marking Scheme, 2017]
1
1 49 Now, × AB × h = 2 2
or,
) (
1 2 a b − 0 + b 0 − a2 + 0 a2 − b2 2
Area =
U [CBSE SQP, 2017]
1 = 1+1+ 0 2 = 1 sq. unit 1 Q. 9. The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, – 2). If the third vertex is 7 , y , find the value of y. 2 A [CBSE, Delhi Set II, 2017]
or,
Sol. If the area covered by the given points is zero, then the points are collinear.
\
150 ]
49 = 9·8 units. 5
1
[CBSE Marking Scheme, 2017]
[ 151
lines (in two-dimensions)
Long Answer Type Questions
5 marks each
opper Answer, 2017
T
Q. 1. If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k. A [CBSE OD, Set-II, 2017]
Q. 2. If the co-ordinates of two points are A(3, 4),
B(5, – 2) and a point P(x, 5) is such that PA = PB, then find the area of DPAB. A [CBSE OD Comptt. Set-I, 2017]
Sol. Given PA = PB or PA2 = PB2, using distance formula, 1 (x – 3)2 + (5 – 4)2 = (x – 5)2 + (5 + 2)2 1 On solving, we get, x = 16 1 \ ar DPAB 1 1 = [16 ( 4 + 2 ) + 3 ( −2 − 5) + 5 ( 5 − 4 )] 2 1 = [ 96 − 21 + 5] = 40 2 Hence, area of triangle = 40 sq. units 1 [CBSE Marking Scheme, 2017] Detailed Solution: The coordinates of P, A and B are (x, 5), (3, 4) and (5, – 2) respectively, then PA = PB (given) ½
( x − 3 )2 + ( 5 − 4 )2 =
5
( x − 5 )2 + ( 5 + 2 )2
( x − 3)2 + 1 = ( x − 5)2 + 49 ⇒ (By using distance formula) ½ Squaring on both sides, we get (x – 3)2 + 1 = (x – 5)2 + 49 1 ⇒ (x – 3)2 – (x – 5)2 = 48 ⇒ [(x – 3) + (x – 5)][(x – 3) – (x – 5)] = 48 [Q a2 – b2 =(a + b)(a – b)] ⇒ (2x – 8)(2) = 48 1 ⇒ 2x – 8 = 24 ⇒ 2x = 32 ⇒ x = 16 1 Now, the point P(x, 5) is P(16, 5) ½ 1 \ Area of DPAB = |16( 4 + 2 ) + 3( −2 − 5) + 5( 5 − 4 )| 2 =
1 | 96 − 21 + 5 | = 40 2
Hence, area of DPAB = 40 sq. units.
½ 1
152 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 3. The co-ordinates of the points A, B and C are (6, 3), (–3, 5) and (4, – 2) respectively. P(x, y) is any point ar ( ∆PBC ) x + y − 2 in the plane. Show that = ar ( ∆ABC ) 7
Sol. Area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is given by,
D =
A [Foreign Set I, 2016]
\
49 2 1 (7 x + 7 y − 14 ) ar(∆PBC ) = 2 49 ar(∆ABC ) 2 7 (x + y − 2 )
=
=
1½
1
49
x+y−2 1 = 7 [CBSE Marking Scheme, 2016]
Q. 4. Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent U [CBSE,Delhi Set I, II, III, 2016] of t. 1 Sol. Area of the triangle = |t(t + 2 – t) + (t + 2) 2 (t – t + 2) + (t + 3)(t – 2 – t –2)| 2 1 = [2t + 2t + 4 – 4t – 12] 1 2 = 4 sq. units. 1 which is independent of t. 1 [CBSE Marking Scheme, 2016] Q. 5. In the given figure, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect sides AB and AC at D and E respectively AD AE 1 such that = = . Calculate the area of AB AC 3 ∆ADE and compare it with area of ∆ABC. A (4, 6)
D
ar DABC =
1 |12 + (–4) + 7| 2
ar DABC =
15 sq. units. 2
In ∆ADE and ∆ABC,
C (7, 2)
A [CBSE, OD Set I, II, III, 2016]
[CBSE Delhi Board, 2015]
½
Hence
∆ADE ∼ ∆ABC
(By AAA) 2
or,
or,
2
1 AD 1 Area ∆ADE = = = 3 9 Area ∆ABC AB
½
Ar ∆ADE 1 = 9 15 2
or, Area ∆ADE =
15 5 = sq. units. 2×9 6
1
Area ADE : Area ABC =
5 15 : = 1 : 9 6 2
1
Q. 6. Find the values of k so that the area of the triangle with vertices (1, – 1), (– 4, 2k) and (– k, – 5) is 24 sq. A [CBSE Board, 2015]
units. Sol. Area of triangle =
1 |x1(y2 – y3) + x2 (y3 – y1) + 2
x3(y1 – y2)|
1 or, 24 = |1(2k + 5) – 4(– 5 + 1) – k(– 1 – 2k)| 2 2k2 + 3k – 27 = 0
or,
(k – 3)(2k + 9) = 0
2
48 = 2k + 5 + 16 + k + 2k2 1
or,
B (1, 5)
AE 1 AD = = AB EC 3
∠DAE = ∠BAC (Common)
or,
1
and
or,
E
½
1 |4(5 – 2) + 1(2 – 6) + 7(6 – 5)| ½ ar DABC = 2
Sol. P(x, y), B(–3, 5), C(4, –2) 1 \ ar (DPBC) = |x(7) + 3(2 + y) + 4(y – 5)| 2 1 = |7x + 7y – 14| 1½ 2 1 ar (DABC) = |6 × 7 – 3 (–5) + 4(3 – 5)| 2
1 |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)| 2
k = 3, k =
1 −9 2
1
[CBSE Marking Scheme, 2015]
[ 153
lines (in two-dimensions)
Visual Case Based Questions
4 marks each
Note: Attempt any four sub parts from each question. Each sub part carries 1 mark each
Q. 1. The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using, l Four clear glass panels, trapezium in shape, all of the same size l One tinted glass panel, half a regular octagon in shape Y A
B J
H
I
B A
D
C
G
E
F B
Top view
A
1 cm P
Q
R
S
Not to scale Front view O
(i) Refer to Top View, find the mid-point of the segment joining the points J(6, 17) and I(9, 16). 3 1 33 15 (b) , (a) , 2 2 2 2 (c)
Sol. Correct option: (a). Explanation: The distance of the point P from the Y-axis = 4. [CBSE Marking Scheme, 2020] 1 (iii) Refer to front view, the distance between the points A and S is (a) 4 (b) 8 (c) 14 (d) 20
Then,
AS =
(15 − 1)2 + ( 8 − 8 )2
=
(14 )2
= 14. 1 [CBSE Marking Scheme, 2020] (iv) Refer to front view, find the co-ordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally. (a) (8.5, 2.0) (b) (2.0, 9.5) (c) (3.0, 7.5) (d) (2.0, 8.5) Sol. Correct option: (d). Explanation: (2.0, 8.5) [CBSE Marking Scheme, 2020] Detailed Solution: The coordinates of A = (1, 8) The coordinates of B = (4, 10) Also, m = 1 and n = 3 1 × 4 + 3 × 1 1 × 10 + 3 × 8 , Then, (x, y) = 1+3 1+3 7 34 = , 4 4 = (1.75, 8.5) 1 (v) Refer to front view, if a point (x, y) is equidistant from the Q(9, 8) and S(17, 8), then (a) x + y = 13 (b) x – 13 = 0 (c) y – 13 = 0 (d) x – y = 13 Sol. Correct option: (b). Explanation: x – 13 = 0. 1 [CBSE Marking Scheme, 2020]
Sol. Correct option: (c). Explanation: A's coordinates = (1, 8) S's coordinates = (15, 8)
A [CBSE SQP, 2020-21]
15 33 1 3 , , (d) 2 2 2 2 Sol. Correct option: (c). Explanation: Mid-point of J(6, 17) and I(9, 16) is 6+9 17 + 16 and y = x = 2 2 15 33 1 x = and y = 2 2 (ii) Refer to front View, the distance of the point P from the y-axis is: (a) 4 (b) 15 (c) 19 (d) 25
X
Scale: 1 cm = 1 m
154 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Detailed Solution: or, (x – 9)2 + (y – 8)2 = (x – 17)2 + (y – 8)2 Let point be P(x, y) or, x – 13 = 0 1 PQ2 = PS2 Q. 2. Ayush Starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km. (13, 14) (5, 8)
(2, 4)
(i) What is the distance between house and bank ? (a) 5 km (b) 10 km (c) 12 km (d) 27 km Sol. Correct option: (b). Explanation: We know that, Distance between two points (x1, y1) and (x2, y2).
d =
( x 2 − x 1 ) 2 + ( y 2 − y1 ) 2
Now, distance between house and bank,
(13, 26)
=
(11)2 + ( 22 )2
=
121 + 484
=
605
= 24.59 = 24.6 km
1
( 5 − 2 )2 + ( 8 − 4 )2
(iv) What is the total distance travelled by Ayush to reach the office ?
=
( 3 )2 + ( 4 )2
(a) 5 km
(b) 10 km
(c) 12 km
(d) 27 km
=
9 + 16
Sol. Correct option: (b).
=
25
Explanation: Distance between daughter's school and office,
=
= 5 km 1 (ii) What is the distance between Daughter's School and bank ? (a) 5 km (b) 10 km (c) 12 km (d) 27 km Sol. Correct option: (b). Explanation: Distance between bank and daughter's school, =
(13 − 5)2 + (14 − 8 )2
=
( 8 )2 + ( 6 )2
=
64 + 36
=
100
1 = 10 km (iii) What is the distance between house and office? (a) 24.6 km (b) 26.4 km (c) 24 km (d) 26 km Sol. Correct option: (b). Explanation: Distance between house to office, =
(13 − 2 )2 + ( 26 − 4 )2
=
(13 − 13)2 + ( 26 − 14 )2
=
0 + (12 )2
= 12 km
½
Total distance (House + Bank + School + Office) travelled = 5 + 10 + 12 = 27 km ½
(v) What is the extra distance travelled by Ayush ? (a) 2 km
(b) 2.2 km
(c) 2.4 km (d) none of these Sol. Correct option: (b). Explanation: Extra distance travelled by Ayush in reaching his office = 27 – 24.6 = 2.4 km. 1 Q. 3. In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1 m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th distance AD on the eighth line and posts a red flag.
[ 155
lines (in two-dimensions)
(v) If Joy has to post a flag at one-fourth distance from green flag, in the line segment joining the green and red flags, then where should he post his flag ? (a) (3.5,24) (b) (0.5,12.5) (c) (2.25,8.5) (d) (25,20) Sol. Correct option: (a).
Explanation: Position of Joy's flag = Mid-point of line segment joining green and blue flags 2 + 5 25 + 22.5 , = 2 2 = [3.5, 23.75] ~ [3.5, 24] Q. 4. The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Find the position of green flag
(a) (2, 25)
(b) (2, 0.25)
(c) (25, 2)
(d) (0, –25)
Sol. Correct option: (a).
(ii) Find the position of red flag
(a) (8, 0)
(b) (20, 8)
(c) (8, 20)
(d) (8, 0.2)
Sol. Correct option: (c). (i) Taking A as origin, find the coordinates of P. (a) (4, 6) (b) (6, 4) (c) (0, 6) (d) (4, 0) Sol. Correct option: (a).
(iii) What is the distance between both the flags ?
(a)
41
(b)
11
(b)
61
(c)
51
Sol. Correct option: (c). Explanation: Position of Green flag = (2, 25)
Position of Red flag = (8, 20)
Distance between both the flags
( 8 − 2 )2 + ( 20 − 25)2 =
6 2 + ( −5)2
=
36 + 25
=
61
(iv) If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag ?
(a) 5, 22.5)
(b) (10, 22)
(c) (2, 8.5)
(d) (2.5, 20)
Sol. Correct option: (a). Explanation: Position of blue flag Mid-point of line segment joining the green and red flags 2 + 8 25 + 20 = , 2 2 = (5, 22.5)
(ii) What will be the coordinates of R, if C is the origin ? (a) (8, 6) (b) (3, 10) (c) (10, 3) (d) (0, 6) Sol. Correct option: (c). (iii) What will be the coordinates of Q, if C is the origin ? (a) (6, 13) (b) (–6, 13) (c) (–13, 6) (d) (13, 6) Sol. Correct option: (d). (iv) Calculate the area of the triangles if A is the origin. (a) 4.5 (b) 6 (c) 8 (d) 6.25 Sol. Correct option: (a). Explanation: Coordinates of P = (4, 6) Coordinates of Q = (3, 2) Coordinates of R = (6, 5) 1 Area of triangle PQR = [x1(y2 – y2) + x2(y3 – y1) 2
+ x3(y1 – y2)] 1 = [4(– 3) + 3(– 1) + 6(4)] 2 =
1 [– 12 + (– 3) + 24] 2
156 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
= =
1 [– 12 + 21] 2 1 [ 9] 2
(v) Calculate the area of the triangles if C is the origin.
(a) 8
(b) 5
(c) 6.25
(d) 4.5
Sol. Correct option: (d).
O
= 4.5 sq. units.
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[ 157
LINES (in two-dimensions)
Maximum Time: 1 hour Q. 1. To locate a point Q on line segment AB such that 5 × AB. What is the ratio of line segment in BQ = 7 which AB is divided ? A [CBSE, Board Term-2, 2013] 1 Q. 2. Find the mid-point of P(– 5, 0) and Q (5, 0). Q. 3. If the distance between the points (4, p) and (1, 0) is 5, then the value of p. R1 Q. 4. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). Find the length of its diagonal. U1 Q. 5. Find the ratio in which the line segment joining the points (6, 4) and (1, – 7) is divided by the x-axis. 1 Q. 6. VISUAL CASE STUDY BASED QUESTIONS: In a room, 4 friends are seated at the points A, B, C and D as shown in figure. Reeta and Meeta walk into the room and after observing for a few minutes Reeta asks Meeta. 10 9 8 B
Rows
7 6 5 C
A 4 3 2 D
1 1
2
3
4
5
6
7
8
9
10
Columns Q. 1. What is the position of A ? (a) (4, 3) (b) (3, 3) (c) (3, 4) (d) None of these
MM: 25
Q. 2. What is the middle position of B and C ? 15 11 (a) , 2 2
(c) 1 , 1 2 2
(b)
2 11 , 15 2
(d) None of these
Q. 3. What is the position of D ?
(a) (6, 0) (c) (6, 1)
(b) (0, 6) (d) (1, 6)
Q. 4. What is the distance between A and B ?
(a) 3 2 unit (b) 2 3 unit
(c) 2 2 unit
(c) 3 3 unit
Q. 5. What is the distance between C and D (a)
2 unit
(b)
2 2 unit
(c) 3 2 unit (c) 4 2 unit Q. 7. If P(2, – 1), Q(3, 4), R(– 2, 3) and S(– 3, – 2) be four points in a plane, show that PQRS is a rhombus A [CBSE Term-2, 2012] 2 but not a square. Q. 8. If (a, b) is the mid-point of the segment joining the points A(10, – 6) and B(k, 4) and a – 2b = 18, find the value of k and the distance AB. C + A [CBSE Term-2, 2012] 2 Q. 9. The co-ordinates of vertices of ∆ABC are A(1, – 1), B(– 4, 6) and C(– 3, – 5). Draw the figure and prove that DABC is a scalene triangle. Find its area also. A [CBSE Term-2, 2014] 3 Q . 10. If (3, 2) and (– 3, 2) are two vertices of an equilateral triangle which contains the origin, find the third A [CBSE, Term-2, 2012] 3 vertex. Q. 11. A(4, – 6), B(3, – 2) and C(5, 2) are the vertices of a DABC and AD is its median. Prove that the median AD divides DABC into two triangles of equal A [CBSE O.D. 2014] 5 areas.
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
UNIT 4: Geometry
c h a pter
7
TRIANGLES
Syllabus Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. 3. (Motivate) If in two triangles, the corresponding angles are equal, then their corresponding sides are proportional and the triangles are similar. 4. (Motivate) If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and the two triangles are similar. 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar. 6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on each side of the perpendicular are similar to each other and to the whole triangle. 7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of squares of the other two sides. 9. (Prove) In a triangle, if the square of one side is equal to the sum of the squares on the other two sides, then the angles opposite to the first side is a right angle.
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Revision Notes A triangle is one of the basic shapes of geometry. It is a polygon with 3 sides and 3 vertices /corners.
Two figures are said to be congruent if they have the same shape and the same size. Those figures which have the same shape but not necessarily the same size are called similar figures. Hence, we can say that all congruent figures are similar but all similar figures are not congruent. Similarity of Triangles: Two triangles are similar, if: (i) their corresponding sides are proportional. (ii) their corresponding angles are equal.
If ∆ABC and ∆DEF are similar, then this similarity can be written as ∆ABC ~ ∆DEF. Scan to know more about this topic
Criteria for Similarity of Triangles: P
L
N Q
M
Criteria of similarity of triangles
R
In ∆LMN and DPQR, if
(a) ∠L = ∠P, ∠M = ∠Q and ∠N = ∠R
then ∆LMN ~ DPQR.
(b)
MN LN LM = = , QR PR PQ
(i) AAA-Criterion: In two triangles, if corresponding angles are equal, then the triangles are similar and hence A D their corresponding sides are in the same ratio. If DABC and DDEF are similar
∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
Then,
CE
F
AB BC AC = = DE EF DF
B
Remark: If two angles of a triangle are respectively equal to the two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal. Therefore, AAA similarity criterion can also be stated as follows: P A AA-Criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. As we know that the sum of all angles in a triangle is 180° so if two angles in DABC and DPQR are same i.e., ∠A = ∠P, ∠B = ∠Q.
R
CQ
B
(ii) SSS-Criterion: In two triangles if the sides of one triangle are proportional to the sides of another triangle, then the two triangles are similar and hence corresponding angles are equal. D A AB BC CA = = If DE EF FD
\ DABC ~ DDEF
then
∠A = ∠D, ∠B = ∠E
and
∠C = ∠F
B
(iii) SAS-Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the two triangles are similar.
If
AB AC and ∠A = ∠D, then DABC ~ DDEF. = DE DF
C E A
B
F D
CE
F
[ 161
TRIANGLES
Some theorems based on similarity of triangles:
A
(i) If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio. It is known as ‘Basic Proportionality Theorem’ or ‘Thales Theorem’.
In DABC, let DE || BC, then
(a)
AD AE AB AC (b) = = DB EC DB EC
(c)
AD AE = . AB AC
D B
A
D
C
B
If
C Scan to know more about this topic
E
(ii) If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. It is the 'Converse of Basic Proportionality Theorem'.
E
Thales theorem
AD AE = , DB EC
then DE||BC (iii) If two triangles are similar, then the ratio of areas of these triangles is equal to the ratio of squares of their corresponding sides. Let DABC ~ DPQR, 2
then
2
2 2 BC ar( ∆ABC ) AB CA AM = = = = QR ar( ∆PQR ) PQ RP PN
A
B
M
P
CQ
N
R
Theorems Based on Right Angled Triangles:
(i) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and B also to each other.
In right DABC, B ^ AC, then DADB ~ DABC DBDC ~ DABC A C D and DADB ~ DBDC (ii) In a right angled triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides. C It is known as Pythagoras Theorem. In right DABC, BC2 = AB2 + AC2. A B Some Important Notes: In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with the twice of the square of the median which bisects the third side. Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the median of the triangle. Three times of the square of any side of an equilateral triangle is equal to four times the square of the altitude.
162 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Mnemonics
Area of Right angled Triangle Audi is the product of half of BMW and Honda Concept: Area of triangle =
1 2
A
× Base × height
(H)
Interpretation: A = Area B = Base
Q
H = Height
How is it done on the A x +3
x +2
Q 3x + 1
B
C
P
Solution: Step I : Given that PQ | | BC Hence by using Basic Proportionality theorem
AP AQ = PB QC x x +3 = x + 2 3x + 1 ⇒ x (3x + 1) = (x + 2) (x + 3) ⇒ 3x2 + x = x2 + 5x + 6 ⇒ 2x2 – 4x – 6 = 0 ⇒ x2 – 2x – 3 = 0 ⇒ (x – 3)(x + 1) = 0 ⇒ x = 3 or x = – 1 Since, side of a triangle is not negative. \ x = 3.
4.5cm
1 mark each
Using basic proportionality theorem AD AE = [By using BPT] ½ BD CE
Q. 1. In the DABC, D and E are points on side AB and AC respectively such that DE || BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE. A [CBSE SQP, 2020-21] Sol. A 2 cm 3 cm E D
B
P
Step II :
Very Short Answer Type Questions
C
D Base (B)
GREENBOARD?
Q.1. In the given triangle, find x if PQ | | BC. x
height
3 2 = 4.5 CE CE = 3 cm ½ [CBSE Marking Scheme, 2020-21] AB 1 = , then Q. 2. Given D ABC ~ D PQR, if PQ 3 ar ( ∆ABC ) . ar ( ∆PQR )
U [CBSE Delhi Set-I, 2020] [CBSE Delhi/OD, 2018] [CBSE SQP, 2017]
[ 163
TRIANGLES
Sol. 1 : 9 Explanation: Since, DABC ~ DPQR, we have 2
2 AB ar( DABC ) 1 = PQ = [\ By using property of similar triangles] ar( DPQR ) 3
=
1 . 9
1 [CBSE Marking Scheme, 2020]
Detailed Solution:
Topper Answer, 2018
Q. 3. In an equilateral triangle of side 2a, find the U [CBSE Delhi Set-I, 2020] length of the altitude. Sol. ABC is an equilateral triangle in which AD ⊥ BC. From DABC,
B
D 2a
a 60°
= 4 : 1.
C
4 1 ½
⇒
C [CBSE Delhi Set-III, 2020] Sol. Let BC be the height of the window above the ground and AC be a ladder. C (window)
cm
10
=
Q. 5. If a ladder 10 m long reaches a window 8 m above the ground, then find the distance of the foot of the ladder from the base of the wall.
AB2 = (AD)2 + (BD)2 ½ (By using Pythagoras Theorem) ⇒ (2a)2 = (AD)2 + (a)2 ⇒ 4a2 – a2 = (AD)2 ⇒ (AD)2 = 3a2
BC 2
2a
60°a
4 BC 2
A
2a
=
8 cm
Q. 4. DABC and DBDE are two equilateral triangles such that D is the mid-point of BC. Find the ratio of the areas of triangles ABC and BDE.
Ground B Here, BC = 8 cm and AC = 10 cm \ In right angled triangle ABC, AC2 = AB2 + BC2
⇒ (10)2 = AB2 + (8)2
⇒
= 36
⇒
Hence, the length of attitude is a 3 .
½
A [CBSE Delhi Set-II, 2020]
Sol.
ar( ∆ABC ) = ar( ∆BDE)
3 ( BC )2 4 3 ( BD)2 4
(By using Pythagoras Theorem) AB2 = 100 – 64 ½
AB = 6 m.
½ B
M
C
D
N
E
=
( BC )2 ( BC ) 2
B 2
½
Q. 6. In fig., MN || BC and AM : MB = 1 : 2, then find ar ( DAMN ) . A [CBSE OD Set-I, 2020] ar ( DABC ) A
A
A
AD = a 3
Sol. Here, and
C
AM = k, MB = 2k. MN || BC
164 ]
\ AB = AM + MB = k + 2k = 3k A k M 2k
[CBSE SQP, 2020]
\
C
ar ( DAMN ) 1 k2 AM 2 = = . = 2 2 9 ar ( DABC ) 9k AB
B
½
N
6 3 = AB 6 \ AB = 12 cm. ½ Q. 10. The perimeters of two similar triangles DABC and DPQR are 35 cm and 45 cm respectively, then find the ratio of the areas of the two triangles.
½
[CBSE SQP Marking Scheme, 2020] 1
Detailed Solution:
Q. 7. In DABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then find ∠B. U [CBSE OD Set-I, 2020] Sol. Given, AB = 6 3 cm, AC = 12 cm and BC = 6 cm.
It can be observed that AB2 = 108 cm, AC2 = 144 cm and BC2 = 36 cm Now AB2 + BC2 = 108 + 36 = 144 cm and AC2 = 144 cm i.e., AB2 + BC2 = AC2, Which is satisfies Pythagoras theorem. So, ∠B = 90°. 1 Q. 8. If two triangles are similar, then find the relation of their corresponding sides. A [CBSE OD Set-I, 2020] Sol. If two triangles are similar, then their corresponding sides are in the same ratio. 1 Q. 9. In the figure,, if ∠ACB = ∠CDA, AC = 6 cm and AD = 3 cm, then find the length of AB C
A
Sol. 49 : 81
Given, Perimeter of DABC and DPQR are 35 cm and 45 cm, Since, the ratio of area of the similar triangles is square of the scalar factor of similarity, Perimeter of triangle ABC Scalar factor = Perimeter of triangle PQR Hence,
7 = 9 =
m 6c
C
A [CBSE Delhi Set-1, 2019]
Sol.
AB = 1 + 2 = 3 cm DABC ~ DADE ½ 2 ar( ABC ) AB 9 \ = = 2 ar( ADE) 1 AD \ ar(DABC) : ar(DADE) = 9 : 1 ½ [CBSE Marking Scheme, 2019] Detailed Solution: Given, AD = 1 cm, BD = 2 cm and DE || BC A D B
B
D
AC AD = AB AC
½
E
B
½
3 cm
2
49 . 81
D
Sol.
A
½
Q. 11. In Figure, DE || BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar (DABC) to the ar (DADE) ? A
[CBSE SQP, 2020] DACB ∼ DADC (AA criterion) ½ AC AB = ⇒ AD AC 6 AB = 3 6 \ AB = 12 cm. ½ [CBSE SQP Marking Scheme, 2020] Detailed Solution: Since ∠CDA = ∠ACB(given) ∠CAD = ∠CAB(common) \ DADC ∼ ∆ACB (AA Similarity) C
35 7 = 45 9
Area of triangle ABC = (scalar factor)2 Area of triangle PQR
B
D
=
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
E C
In DADE and DABC,
∠ADE = ∠ABC (corresponding angles) ∠A = ∠A
(common)
[ 165
TRIANGLES
Therefore, by AA criterion corollary condition
DADE ~ DABC
½
Ratio of areas of similar triangles is equal to the square of the ratio of the corresponding sides,
Q. 13. In Figure, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.
ar( DABC ) AB2 ⇒ = 2 ar( DADE) AD ar( DABC ) ( 3 )2 ⇒ = 2 ar( DADE) (1) [AB = AD + BD = 3] ar( DABC ) 9 ⇒ = ar( DADE) 1
Hence, the ratio of the ar(DABC) to the ar(DADE) is 9 : 1. ½
A [CBSE OD Set-1, 2019]
[CBSE Term-2, 2016] AD AD AE 1.8 = ½ = ⇒ 7.2 BD CE 5.4 7.2 × 1.8 = 2.4 cm. ½ AD = 5.4 [CBSE Marking Scheme, 2019]
Sol.
COMMONLY MADE ERROR
\
Some candidates took the ratio of area 2
AB ∆ABC AB . instead of = AD 2 ∆ADE AD
Detailed Solution: It is given that DE || BC
ANSWERING TIP
Q. 12. In Figure, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.
A [CBSE OD Set-1, 2019]
½
Putting the values, we get
the areas of similar triangles is proportional to the square to the corresponding.
Sol. DABC : Isosceles D ⇒ AC = BC = 4 cm.
AD AE = DB EC
Hence,
It is necessary to explain how the ratio of
½
4 2 + 4 2 = 4 2 cm ½ [CBSE Marking Scheme, 2019]
AB =
AD 1.8 = 7.2 5.4
1.8 × 7.2 5.4
⇒
AD =
⇒
AD =
⇒
AD = 2.4 cm.
12.96 5.4 ½
Q. 14. In figure, if AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm and ∠ADE = 48°. Find ∠ABC. A
Detailed Solution: Given, ABC is an isosceles triangle right angled at C i.e. AC = BC = 4 cm ∠C = 90° ½ Applying pythagoras theorem in DABC, AB2 = BC2 + AC2 = 42 + 42 = 16 + 16 = 32
D
E
C
B
U [CBSE SQP, 2018-19]
Sol.
\
AB = 4 2 cm.
½
\
⇒
AE AD = DB EC DE || BC ∠ADE = ∠ABC = 48°
½ ½
[CBSE Marking Scheme, 2018]
166 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
8
6 cm
Detailed Solution: It is given that, A
cm
D
⇒
3 15 = 2 RP
⇒
30 RP = = 10 cm. 3
Some candidates take the ratio of area ∆ ABC BC BC 2 . = instead of ∆QRP RP RP 2
ANSWERING TIP Candidates should more practice to solve such problems.
Q. 16. In the given figure, ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of DPST to the area of DPRQ.
ar( ∆ ABC) BC = ar( ∆QRP) RP
9 15 = 4 PR
3 S
T
4 R
2
Q
Sol. PS = 3 cm, SR = 4 cm and ST || RQ.
2
⇒ PR=10 cm 1 ⇒ [CBSE Marking Scheme, 2018] Detailed Solution: We have, DABC ~ DQRP AB BC AC = = \ QR RP QP
A [CBSE, SQP, 2017-18]
P
cm, then find PR. [CBSE Comptt. Set I, II, III, 2018] [CBSE, Term-I, 2015] Sol.
= 3 + 4 = 7 cm
PR = PS + SR ar ∆PST PS2 32 9 = = = ar ∆PRQ PR 2 7 2 49
Hence, required ratio = 9 : 49.
D
C R
P
2 ar( ∆ABC ) BC Now, = ar( ∆QRP ) RP [Areas of two similar triangles is equal to the square of the ratio of their corresponding sides]
E
x+ 1
x+ 5
B
C U [CBSE, Term-1, 2016]
15 cm
1
Q. 17. In DABC, DE || BC, find the value of x. A x x+ 3
[corresponding sides of similar triangles] Q A
B
1
COMMONLY MADE ERROR
C B Here, in DABC and DADE, AB = (9 + 6) cm = 15 cm Similarly, AC = AE + EC = (8 + 12) cm = 20 cm ½ AD 6 2 = Now, = AB 15 5 AE 8 2 = and = AC 20 5 AD AE = Then, AB AC i.e., DE || BC \ ∠ABC = ∠ADE (Corresponding angles) Hence, ∠ABC = 48°. ½ ar ( ∆ABC) 9 = , and BC = 15 Q. 15. If ∆ ABC ~∆ QRP, ar ( ∆QRP) 4
2
cm
15 9 = 4 RP
12
9 cm
E
⇒
Sol. As
DE || BC
\
AD AE = DB EC
[ 167
TRIANGLES A
x x+3 = x + 1 x +5 2 2 or, x + 5x = x + 4x + 3 or, x = 3 1 [CBSE Marking Scheme, 2016] Q. 18. In ∆ABC, if X and Y are points on AB and AC AX 3 respectively such that = , AY = 5 and YC XB 4
2x
or,
A
Sol.
Y
X
3
U [CBSE Term-1, 2015, 2016]
x+
= 9, then state whether XY and BC parallel or not.
D
B
Sol.
x C 2x –1 E CD x+3 x CE ½ = or, = AD 2x 2x − 1 BE
3 ½ 5 Q. 20. In the given figure, if ∠A = 90°, ∠B = 90°, OB = 4.5 cm, OA = 6 cm and AP = 4 cm, then find QB. U [CBSE Term-1, 2015]
or,
5x = 3 or, x =
Q
B
C
AX 3 AY 5 = and = XB 4 YC 9
Then,
P
AX AY ≠ XB YC
Hence XY is not parallel to BC.
1
A
Q. 19.
D
C
E
In the figure of DABC, the points D and E are on the sides CA, CB respectively such that DE || AB, AD = 2x, DC = x + 3, BE = 2x – 1 and CE = x. Then, find x. A [CBSE, Term-1, 2016] OR In the figure of ∆ABC, DE || AB. If AD = 2x, DC = x + 3, BE = 2x – 1 and CE = x, then find the value of x. [CBSE, Term-1, 2015]
B
B
O
A
AX 3 = , AY= 5 and YC = 9 (Given) XB 4
Sol. In ∆PAO and ∆QBO, ∠A = ∠B = 90° (Given) ∠POA = ∠QOB (Vertically Opposite Angles) Since, ∆PAO ∼ ∆QBO, (by AA) OA PA = Then, OB QB 6 4 = or, 4.5 QB 4 × 4.5 or, QB = 6 \ QB = 3 cm 1 Q. 21. Are two triangles having corresponding sides equal, similar. R [CBSE Term-1, 2015] Sol. Yes, Two triangles having equal corresponding sides are congruent and all congruent Ds have equal angles, hence they are similar too. 1
Short Answer Type Questions-I Q. 1. In the adjoining figure, DE || AC and DC || AP. BC . BE = Prove that CP EC A
2 marks each
Sol. In D ABP,
DC || AP(Given) BD BC = (From BPT) ....(i) ½ DA CP
\
A
D D
B
E
C
L
A [CBSE Delhi Set-I, 2020]
B
E
C
L
½
168 ] In DABC,
\
From equations (i) and (ii), we have BE BC = EC CP
Hence Proved. ½
Q. 2. In the given figure, ∆ABC and ∆DBC are on the same base BC. AD and BC intersect at O. Prove ar ( ∆ABC ) AO = that ar ( ∆DBC ) DO
AC2 = AD2 + CD2 ...(i) ½ C
D B
A
In right DADB, AB2 = AD2 + BD2 ...(ii) ½ Subtracting eq. (i) from eq. (ii), AB2 – AC2 = BD2 – CD2 Hence, AB2 + CD2 = AC2 + BD2 Hence Proved. 1 Q. 4. 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. C [CBSE SQP, 2020]
BD BE = (From BPT) ....(ii) ½ DA EC
U [CBSE Outside Delhi Set-I, 2020]
DE || AC(Given)
Sol. In right DADC,
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol.
A
[CBSE Term-1, 2016]
A
C O
B
D
ar( ∆ABC ) AO Sol. To prove: = ar( ∆DBC ) DO Construction : Draw AE ⊥ BC and DF ⊥ BC. C
A F O E
B
D
Proof:
½
In ∆AOE and ∆DOF,
∠AEO = ∠DFO = 90° (Construction)
or,
∆AOE ∼ ∆DOF
∴
1 × BC × AE ar( ∆ABC ) Now, = 2 1 ar( ∆DBC ) × BC × DF 2
A
∠AOE = ∠DOF (Vertically opposite angles) (By AA Similarity)
AO AE = ...(i) ½ DO DF
B
=
AE DF
=
AO DO
½
[From equation (i)] ½
D
⇒ 4AB2 = 4AD2 + AC2 ⇒ 4AB2 = 4AD2 + AB2 [\ AB = AC] Hence, 3AB2 = 4AD2. Hence Proved. 1 Q. 5. In the given figure, DEFG is a square and ∠BAC = 90°. Show that FG2 = BG × FC A
Hence Proved. Q. 3. In fig. 6, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2 . A [CBSE OD Set-I, 2020]
D
E
G
F
C
D B
C
In right angled DABD, AB2 = AD2 + BD2 (Using Pythagoras Theorem) BC 2 ⇒ AB2 = AD 2 + 4
C
B
D AD ⊥ BC \ In DABD, AB2 = AD2 + BD2 1 BC 2 2 2 2 2 2 or 4AB = 4AD + BC ⇒ AB = AD + 4 ⇒ 3AB2 = 4AD2 1 [CBSE SQP Marking Scheme, 2020] Detailed Solution: Let ABC be an equilateral triangle, in which AD is the perpendicular bisector on BC. BC \ BD = 2 and AB = BC = CA. 1
B A
C
[CBSE SQP, 2020]
[ 169
TRIANGLES
Detailed Solution: Given, DEFG is a square and A
Sol.
F
C
\
GD BG = FC EF
⇒
GD × EF = BG × FC
A
M N C
Þ
B
DABD,
X
B
TX XN TN = = TC CM TM TX × TM = TC × TN
½ T
...(i) ½
3 3 2
C D
∠D = 90°
2
3 3 (3 3 ) = h + 2 2
2
\
or,
27 = h 2 +
27 4
or,
h2 = 27 −
27 4
or,
h2 =
81 4
9 = 4.5 cm 1 2 Q. 8. In a rectangle ABCD, E is a point on AB such that 2 AB. If AB = 6 km and AD = 3 km, then find AE = 3 DE. A [CBSE, Term-1, 2016]
\
Sol.
FG = BG × FC Hence Proved.
∆TXN ~ ∆TCM
1
h
2
Q. 6. X is a point on the side BC of DABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX2 =TB×TC. [CBSE Comptt. Set-I, II, III, 2018]
Þ
3 3
2
Sol.
TX2 =TC × TB ½ [CBSE Marking Scheme, 2018]
Þ
A
∠BAC = 90° To prove, FG2 = BG × FC Since, DEFG is a square, therefore we can write, DE = EF = FG = GD In DAGE and DGDB, \ ∠A = ∠DGB = 90° ∠ADE = ∠GBD (Corresponding angles) \ By AA similarity, DEAD ∼ DGDB...(i) Now, In DAGF and FCE, ∠A = ∠EFC = 90° ∠AED = ∠FCE (Corresponding angles) \ By AA similarity, DAED ∼ DFCE...(ii) From (i) and (ii), we get DGDB ∼ DFCE Since, corresponding sides of two similar triangles are proportional.
Using (ii) in (i), we get TN TX 2 = = TC × TN Þ TB
3 cm, find the Q. 7. In an equilateral triangle of side 3 length of the altitude. U [CBSE Term-1, 2016, 2015]
E
G
B
...(ii) ½
h =
D
6 km
C
3 km
A
4 km
E
B
D
Again, ∆TBN ~ ∆TXM TB BN TN = = Þ TX XM TM TN × TX TM = Þ TB
Sol. DADE ∼ DGBD and DADE ∼ DFEC ⇒ GBD ∼ FEC (AA Criterion) GD GB = ⇒ FC FE ⇒ GD × FE = GB × FC or FG2 = BG × FG [CBSE SQP Marking Scheme, 2020]
1
2 2 Given, AE = AB = × 6 = 4 km 3 3
In right triangle ADE, DE2 = (3)2 + (4)2 or, DE2 = 25 \ DE = 5 km. 1 [CBSE Marking Scheme, 2016]
Q. 9. In the figure, PQRS is a trapezium in which PQ || RS. On PQ and RS, there are points E and F respectively such that EF intersects SQ at G. Prove that EQ × GS = GQ × FS.
170 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X F
S
A
R
G P
Q
E
B
Sol. In DGEQ and DGFS ∠EGQ = ∠FGS (vert. opp. angles) ∠EQG = ∠FSG (alt. angles) \ DGEQ ~ DGFS(AA similarity) 1 EQ GQ or, = FS GS or, EQ × GS = GQ × FS. 1 [CBSE Marking Scheme, 2016] Q. 10. In DABC, AD ⊥ BC, such that AD2 = BD × CD. Prove that ∆ABC is right angled at A. A [CBSE, Term-1, 2015]
U [CBSE Term-1, 2016]
AD2 = BD × CD AD BD = CD AD
Sol. Given, or, ∴ or,
3 marks each
Hence, the length of the corresponding side of second triangle is 5.4 cm. 1 Q. 2. In an equilateral triangle ABC, D is a point on side 1 BC. Prove that BC such that BD = 3 9AD2 = 7 AB2. [CBSE SQP, 2020-21]
D
A
xc
m
m
9c
B
1
C
E
F
Also given, CA = 9 cm Let FD = x cm AB + BC + CA CA 1 = Then DE + EF + FD FD [Given, DABC ~ DDEF] 25 9 = 15 x ⇒ ⇒ 25x = 135 135 = 5.4 ⇒ x = 25
Sol. DABC ~ DDEF AB + BC + CA AB Perimeter (∆ABC ) 1 = = DE + EF + FD DE Perimeter (∆DEF ) 25 9 ⇒ ½ = 15 x [Using proportionality theorem] ⇒ x = 5.4 cm ½ ⇒ DE = 5.4 cm 1 [CBSE Marking Scheme, 2020-21] Detailed Solution: Given, perimeters of both triangles are 25 cm, 15 cm Let both triangles be ABC and DEF. \ AB + BC + CA = 25 cm and DE + EF + FD = 15 cm
½
DADC ∼ ∆BDA (by SAS; Q∠D = 90°) ∠BAD = ∠ACD; ∠DAC = ∠DBA (Corresponding angles of similar triangles) ½ ∠BAD+ ∠ACD + ∠DAC + ∠DBA = 180° or, 2∠BAD + 2∠DAC = 180° ½ or, ∠BAD + ∠DAC = 90° ∴ ∠A = 90° Hence proved. ½
Short Answer Type Questions-II Q. 1. The perimeter of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm. Find the length of the corresponding side of the second triangle. C + A [CBSE SQP, 2020-21]
C
D
Sol.
A
B
D
C
E
½
Construction: Draw AM ^ BC 1 BD = BC 3
½
1 BC 2
BM =
In DABM,
AB2 = AM2 + BM2 2
½ 2
= AM + (BD + DM)
= AD2 + BD2 + 2BD (BM – BD) ½
= AM2 + DM2 + BD2 + 2BD.DM 2
BC BC BC BC = AD 2 + +2 − 3 3 2 3
= AD 2 + 2
BC 2 9
= AD 2 + 2
AB2 9
Hence, 7AB2 = 9AD2 ½ [CBSE Marking Scheme, 2020-21]
½
[ 171
TRIANGLES
Detailed Solution:
Topper Answer, 2018
3
To prove: ∠B = 90° Construction: Draw DPQR right angled at Q, such that PQ = AB and QR = BC
Q. 3. In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle. A [CBSE OD Set-III, 2019; Delhi Set-I, 2020]
A
Sol. For correct given, to prove construction. and figure 3 × ½ = 1½ For correct Proof 1½ [CBSE Marking Scheme, 2020]
P
B
A
B
C
C
Q
R
Proof: In DPQR, ∠Q = 90° \ By Pythagoras theorem, PR2 = PQ2 + QR2 Since, PQ = AB and QR = BC (By construction)
D
D
D
Detailed Solution: Given: A triangle ABC in which AC2 = AB2 + BC2
172 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
\ PR2 = AB2 + BC2 ...(i) Also, given AC2 = AB2 + BC2 ...(ii) From eq (i) & (ii), PR2 = AC2 ⇒ PR = AC ...(iii) Now, in DABC and DPQR AC =PR [From (iii)] AB =PQ (By construction) BC = QR (By construction) \ DABC @ DPQR (By SSS congruence rule) ⇒ ∠B = ∠Q (By cpct) Since, ∠Q = 90° (By construction) \ ∠B = 90° Hence Proved. Q. 4. In the adjoining figure, ∠D = ∠E and AD = AE , DB EC Prove that DBAC is an isosceles triangle. A
D
E
B
C
A [CBSE Delhi Set-I, 2020]
Sol. Given, DABC ∼ DDEF Then according to question, AB DE = [From BPT] ½ BC EF 2x − 1 18 = ½ ⇒ 2 x + 2 3 x +9 ⇒ (2x – 1)(3x + 9) = 18(2x + 2) ⇒ (2x – 1)(x + 3) = 6(2x + 2) ⇒ 2x2 – x + 6x – 3 = 12x + 12 ⇒ 2x2 + 5x – 12x – 15 = 0 ⇒ 2x2 – 7x – 15 = 0 ½ ⇒ 2x2 – 10x + 3x – 15 = 0 ⇒ 2x(x – 5) + 3(x – 5) = 0 ⇒ (x – 5)(2x + 3) = 0 −3 , which is not possible Either x = 5 or x = 2 So, x = 5 ½ Then in DABC, we have AB = 2x – 1 = 2 × 5 – 1 = 9 BC = 2x + 2 = 2 × 5 + 2 = 12 AC = 3x = 3 × 5 = 15 ½ and in DDEF, we have DE = 18 EF = 3x + 9 = 3 × 5 + 9 = 24 DE = 6x = 6 × 5 = 30. ½ Q. 6. In Figure, ∠ACB = 90° and CD ^ AB, prove that CD2 = BD × AD. C
AD AE Sol. Given: ∠D = ∠E and = DB EC A
A
D
E
C
B
To prove: DBAC is an isosceles triangle. AD AE = Proof: DB EC By converse of BPT, DE || BC 1 \ ∠ADE = ∠ABC (Corresponding angles) ½ and ∠AED = ∠ACB (Corresponding angles) ½ ∠ADE = ∠AED(Given) \ ∠ABC = ∠ACB So, BAC is an isosceles triangle. Hence Proved. 1 Q. 5. In the given figure, if DABC – DEF and their sides of the given figure lengths (in cm) are marked along them, then find the lengths of sides of each triangle. A
D
2x–1 B
3x 2x+2
E
Sol.
Alternate Solution: Given: In ∆ACB, ∠ACB= 90° and CD ⊥ AB C
A 6x
18
C
[CBSE Delhi Set-I, 2019]
DACB ~ DADC (AA similarity) AC AD = ...(1) 1 ⇒ BC CD Also DACB ~ DCDB (AA similarity) AC CD = ...(2) 1 ⇒ BC BD Using equations (1) and (2), AD CD = 1 CD BD ⇒ CD2 = AD × BD [CBSE Marking Scheme, 2019]
D
3x+9
F
C + A [CBSE OD Set-I, 2020]
B
D
B
To prove: CD2 = BD × AD Proof: In ∆CAD, ∠ADC = 90° CA2 = CD2+ AD2 ...(i)
[ 173
TRIANGLES
½ 1
Sol. Correct Figure AQ2 = CQ2 + AC2
AQ2 + BP2 = (AC2 + CQ2) + (PC2 + CB2)1
or or
AQ2 + BP2 = (AC2 + CB2) + (PC2 + CQ2) AQ2 + BP2 = AB2 + PQ2 Hence proved. 1
Commonly Made Error Some candidates make mistakes in applying Pythagoras theorem.
Answering Tip Students should analyse the diagram carefully to write correct conditions to prove the question. Q. 8. Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP. [CBSE OD Set-1, 2019]
and In ∆CDB, ∠CDB = 90° CB2 = CD2 +BD2 ...(ii) 1 On adding eq’s. (i) and (ii), we get CA2 + CB2 = 2CD2 + AD2 + BD2 AB2 = 2CD2 + AD2 + BD2 [Q Given ∠ACB = 90° ⇒AB2 = CA2 + CB2] ⇒ AB2 – AD2 = BD2 + 2CD2 ⇒ (AB + AD)(AB – AD) = BD2 + 2CD2 1 2 2 ⇒ (AB + AD)BD – BD = 2CD ⇒ BD[(AB +AD) – BD] = 2CD2 ⇒ BD[AD +(AB – BD)] = 2CD2 ⇒ BD[AD + AD] = 2CD2 ⇒ BD × 2AD = 2CD2 ⇒ CD2 = BD × AD 1 Hence proved. Q. 7. If P and Q are the points on sides CA and CB respectively of DABC, right angled at C. Prove that (AQ2 +BP2) = (AB2 +PQ2) [CBSE Delhi Set-I, 2019]
DAPB ~ DDPC [AA similarity] 1 AP BP = 1 DP PC
Sol.
⇒
A
B
D
P
Q C
½
AP × PC = BP × DP
A
P
BP2 = CP2 + BC2 ½ \ AQ2 + BP2 = (CQ2 + CP2) + (AC2 + BC2) = PQ2 + AB2. 1 [CBSE Marking Scheme, 2019]
C A ½ [CBSE Marking Scheme, 2019] Detailed Solution: Given: ∠BAC and ∠BDC are 90° each AC and BD intersect at P. A
D
P
Detailed Solution: A
A
C
To prove: AP × PC = BP × DP. Proof: In DBAP and DCDP, P ∠A = ∠D (Given) ∠BPA = ∠DPC B C Q (vertically opposite angles) \ DBAP ~ DCDP (By AA similarity) 1 In right angled triangles ACQ and PCB AP BP = Then AQ2 = AC2 + CQ2 ...(i) ½ DP PC and BP2 = PC2 + CB2 ...(ii) ½ or, AP × PC = BP × DP.Hence proved. 1 Adding eq (i) and eq (ii), we get AD AB Q. 9. In similar triangles, DABC and DPQR, AD and PM are the medians respectively. Prove that = . PM PQ A [CBSE Board Term, 2019]
Topper Answer, 2019
174 ]
3
Sol.
\
Þ
DACF =
F
D
Let the side of the square be 'a' units
Q. 10. Prove that area of the equilateral triangle described on the side of a square is half of the area of the equilateral triangle described on its A [CBSE Delhi Set-2018] diagonal.
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
AC2 = a2 + a2 = 2a2 AC =
Area of equilateral triangle DBCE =
B
3 2 a sq.u ½ 4
Area of equilateral triangle
C
E
A
1
2a units
Area of DBCE =
3 4
( 2a)
2
=
3 2 a sq.u 1 2
1 Area of DACF 2
½
[CBSE Marking Scheme, 2018]
[ 175
TRIANGLES
Detailed Solution:
Topper Answer, 2018
P
S
M
1 2
T
Q
R
N
A [CBSE SQP, 2018-2019]
Sol.
1 2
T
O
M
Q
P S
R
N
∠1 = ∠2 PT = PS ....(i) 1 ∆NSQ ≅ ∆MTR ∠NQS = ∠MRT ∠PQR = ∠PRQ PR = PQ ....(ii) 1 PS PT From (i) and (ii), = PQ PR Also, ∠TPS = ∠RPQ (common) Þ ∆PTS ~ ∆PRQ 1 [CBSE Marking Scheme, 2018]
Q. 11. In figure ∠1 = ∠2 and ∆NSQ ≅ ∆MTR, then prove that ∆PTS~∆PRQ.
3 Þ Þ Þ Þ
Q. 12. In ∆ABC, if AD is the median, then show that AB2+AC2 = 2(AD2+BD2).
176 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
A
From equation (i), we have AB2
B
E
C
D
A [CBSE SQP, 2018-19]
Sol. In DABE, AB2 = AE2 + BE2 (Pythagoras theorem) or, AB2 = AD2 – DE2 + (BD – DE)2 = AD2 – DE2 + BD2 + DE2 – 2BD × DE ∴ AB2 = AD2 + BD2 – 2BD × DE....(i) 1 In DAEC, AC2 = AE2 + EC2 or, AC2 = (AD2 – ED2) + (ED + DC)2 = AD2 – ED2 + ED2 + DC2 +2ED × DC or, AC2 = AD2 + CD2 + 2ED × DC or, AC2 = AD2 + DC2 + 2DC × DE....(ii) 1 Adding equations (i) and (ii), AB2 + AC2 = 2(AD2 + BD2)[ BD = DC] 1 Hence Proved. Q. 13. If the area of two similar triangles are equal, then prove that they are congruent. A [CBSE Delhi/OD Set-2018] Sol. Let,
or, or, and
2
and
ar (∆ABC ) AD 2 . = ar (∆PQR ) PS 2
Proof:
∆ABC ∼ ∆PQR
or,
ar ( ∆ABC )
ar ( ∆PQR )
Also
B
PQ 2
=
BC 2 QR 2
ar ( ∆PQR )
= 1
=
CA 2 RP 2
Q
ar(∆ABC ) AB = 2 ar(∆PQR ) PQ ∠A = ∠P 1 1 ∠A = ∠P 2 2
S
R
1 ...(i)
or,
or,
By eqs. (i) and (ii), ar(∆ABC ) AD 2 Hence Proved. 1 = ar(∆PQR ) PS2 [CBSE Marking Scheme, 2016]
∠BAD = ∠QPS DBAD ~ DQPS (AA similarity) 1 BA AD ...(ii) = QP PS
Q. 15. DABC is a right angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the sides opposite to ∠A, ∠B and ∠C 1 1 1 respectively, then prove that 2 = 2 + 2 . p a b
...(i)
(Given) 1
C
D DABC ~ DPQR ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R 2
(Given)
ar(∆ABC) = ar(∆PQR) ar ( ∆ABC )
or,
=
AB2
P
A
R
U [Board Term-1, 2016]
1
Q
∆ABC ≅ ∆PQR
(SSS) 1
Q. 14. If DABC ~ DPQR and AD and PS are bisectors of corresponding angles A and P, then prove that
P
To prove:
1
criteria for Congruence of triangles.
ar∆ABC = ar∆PQR
CA = 1 RP
Candidates should know about SSS-
2
A
C
=1
ANSWERING TIP
∆ABC ∼ ∆PQR,
B
RP 2
∆ABC ≅ ∆PQR
AB BC AC =1 1 = = 2 2 PR 2 PQ QR Þ AB = PQ, BC= QR and AC= PR 1 Þ Therefore, ∆ABC ≅ ∆PQR (SSS Congruence Rule) [CBSE Marking Scheme, 2018]
PQ 2 QR 2 AB BC = = PQ QR AB = PQ, BC = QR CA = RP ∆ABC ≅ ∆PQR
Most candidates are not able to prove
Þ
Detailed Solution: Given:
CA 2
=
Sol.
ar ( ∆ABC )
2
BC 2
COMMONLY MADE ERROR
∆ABC ~ ∆PQR
AB2 BC 2 AC 2 = = = 2 2 PR 2 QR ar ( ∆PQR ) PQ Given, ar (∆ABC) = ar (∆PQR)
\
=
A [Board Term-1, 2016]
[ 177
TRIANGLES A
Sol.
b
D
c
and
Hence,
\
\
B
or,
or,
1 c = p ab
Squaring on both sides,
(common)
DACB ~ DCDB (by AA Similarity) 1
2
∠B = ∠B b c = p a
p2
=
1
or,
p
2
=
U [Board Term-1, 2016]
∠ACB = ∠CDB = 90°
Sol. Given, in ∆ADB, AB = AD2 + BD2 ...(i) (Pythagoras Theorem) A
1 B
c2 a2b 2 2
a +b
2
2 2
a b
[\ c2= a2+b2]
1 1 1 \ 2 = 2 + 2 Hence Proved. 1 p a b Q. 16. In given figure DABC ~ DDEF. AP bisects ∠CAB and DQ bisects ∠FDE. F
C
2
C
2
In ∆ADC, AC = AD + CD ...(ii) (Pythagoras theorem)
Subtracting eqn. (ii) from eqn. (i),
AB2 – AC2 = BD2 – CD2 2
1 2
3 1 or, = BC – BC 4 4
BC 2 9 1 or, = BC 2 − BC 2 = 2 16 16
∴ 2(AB2 – AC2) = BC2
∴ 2(AB)2 = 2AC2 + BC2. Hence proved. 1
1
Q. 18. In given figure, D is a point on AC such that AD = 2CD, also DE||AB.
Q
P
x
D
3x 2
1
(By AA similarity)
In DACB and DCDB,
DCAP ~ DFDQ
1
Hence Proved.
a
2 ∠3 = 2 ∠4 or, ∠3 = ∠4
Q. 17. The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3CD. Prove that 2(AB)2 = 2(AC)2 + BC2.
p C
∠C = ∠F
A
Prove that: AP AB = (i) DQ DE
D B
(ii) DCAP ~ DFDQ. C Sol.
U [Board Term-1, 2016]
P 3 A
1
(i) Here, \ and or, Also Since,
Hence,
(ii) Q \
F
Q 4 B
D
2
E
DABC ~ DDEF ∠A = ∠D (corresponding angles) 2 ∠1 = 2 ∠2 ∠1 = ∠2 ∠B = ∠E (corresponding angles) DAPB ~ DDQE 1
AP AB = DQ DE DABC ~ DDEF ∠A = ∠D
Find:
E
D
B
A
C
E
ar( ∆DCE) ar( ∆ACB)
Sol. Given
U [Board Term-1, 2015]
AD = 2CD
In ∆CDE and ∆CAB
∠C = ∠C
∠CDE = ∠CAB
∴
∆CDE ∼ ∆CAB (By AA similarity rule)
(Common) (Corresponding angles)
1
Now
CD 2 CD 2 ar( ∆DCE) = = CA 2 ( AD + DC )2 ar( ∆ACB)
or,
CD 2 ar( ∆DCE) 1 = 2 = ar( ∆ACB) 9 3CD ) (
3
[CBSE Marking Scheme, 2015]
178 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 19. If in ∆ABC, AD is median and AE ⊥ BC, then 1 BC2. prove that AB2 + AC2 = 2AD2 + 2 A [Board Term-1, 2015]
Sol. To prove:
AB2 + AC2 = 2AD2 +
1 (BC )2 2
1000
A
2 1 = 2 AD 2 + 2 BC 2 1 1 2 2 = 2 AD + BC (as BD = BC) 1 2 2 Hence proved. Q. 20. From an airport, two aeroplanes start at the same time. If speed of first aeroplane due North is 500 km/h and that of other due East is 650 km/h then find the distance between the two aeroplanes after 2 hours. C [Board Term-1, 2015] N Sol.
E
D
C
Draw AE ⊥ BC In DABE, AB2 = AE2 + BE2 (Pythagoras theorem) or, AB2 = AD2 – DE2 + (BD – DE)2 = AD2 – DE2 + BD2 + DE2 – 2BD × DE ∴ AB2 = AD2 + BD2 – 2BD × DE....(i) 1 In DAEC, AC2 = AE2 + EC2 or, AC2 = (AD2 – ED2) + (ED + DC)2 = AD2 – ED2 + ED2 + DC2 +2ED × DC 2 or, AC = AD2 + CD2 + 2ED × CD or, AC2 = AD2 + DC2 + 2DC × DE....(ii) 1 Adding eqns. (i) & (ii), AB2 + AC2 = 2(AD2 + BD2)( BD = DC)
W
O
1300
E
S Distance covered by first aeroplane due North after two hours = 500 × 2 = 1000 km. 1 Distance covered by second aeroplane due East after two hours = 650 × 2 = 1300 km. 1 Distance between two aeroplanes after 2 hours NE = ON 2 + OE 2
= 2690000 = 1640.12 km
B
Long Answer Type Questions
=
(1000 )2 + (1300 )2
=
1000000 + 1690000 1
5 marks each
Q. 1. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. A [CBSE Delhi/OD Set-2018] [SQP, 2018-19]
Topper Answer, 2019
Sol.
[ 179
TRIANGLES
Topper Answer, 2019
5
Q. 2. Prove that in a right triangle, the square of the hypotenuse is equal to sum of squares of other two sides. Using the above result, prove that, in rhombus ABCD, 4 AB2 = AC2 + BD2.
A [Board Term-1, 2015]
[Sample Question Paper 2017] Sol. We have already proved AB2 + BC2 = AC2 in above Q.1. Given: ABCD is a rhombus. 1 Construction: Draw diagonals AC and BD 1 AC ∴ AO = OC = 2 D
R
X
C
1
Y
Z
Q
U [Board Term-1, 2015]
P
O
AC 2 BD 2 + 4 4
Hence, 4AB2 = AC2 + BD2 1 Hence proved. Q. 3. DPQR is right angled at Q. QX ⊥ PR, XY ⊥ RQ and XZ ⊥ PQ are drawn. Prove that XZ2 = PZ × ZQ.
Sol.
R
X
A
B
1 BD and BO = OD = 2
1
AC ⊥ BD [ ABCD is rhombus] To prove: 4 AB2 = AC2 + BD2 Proof: ∠AOB = 90° (Diagonal of rhombus bisect each other at right angle) AB2 = OA2 + OB2 2
4
or,
Y
32
2
AC BD AB2 = + 2 2
1
P
1 Z
1
Q
Here, RQ ⊥ PQ and XZ ⊥ PQ or, XZ || YQ \ Similarly, XY || ZQ XYQZ is a rectangle. ( ∠PQR = 90°) In DXZQ, ∠1 + ∠2 = 90° and in DPZX, ∠3 + ∠4 = 90° XQ ⊥ PR or, ∠2 + ∠3 = 90° By eqs. (i) and (iii), we get ∠1 = ∠3 By eqs. (ii) and (iii), we get
AB2 =
1 ...(i) ...(ii) ...(iii) ½ ½
180 ]
\
\ Thus,
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
∠2 = ∠4 DPZX ~ DXZQ (AA similarity) 1 PZ XZ = XZ ZQ
Sol. (i) Since, BC and OX bisect each other.
So, BXCO is a parallelogram then BE || XC and BX || CF. In DABX, by B.P.T., AF AO = ...(i) 1 FB OX
2
1 Hence Proved. Q. 4. In ∆ABC, the mid-points of sides BC, CA and AB are D, E and F respectively. Find ratio of ar(∆DEF) to ar(∆ABC). A [Board Term-1, 2015] Sol. XZ = PZ × ZQ
In DAXC, AE = EC Eqn. (i) and (ii) gives, AF = FB
1 In ∆ABC, given that F, E and D are the mid-points of AB, AC and BC respectively. Hence, FE || BC, DE || AB and DF || AC. 1 By mid-point theorem,
If
DE || BA
then
DE || BF
and if
FE || BC
then FE || BD 1 ∴ FEDB is a parallelogram in which DF is diagonal and a diagonal of parallelogram divides it into two equal Areas. Hence ar(∆BDF) = ar(∆DEF) ...(i) Similarly ar(∆CDE) = ar(∆DEF) ...(ii) or (∆AFE) = ar(∆DEF) ...(iii) or (∆DEF) = ar(∆DEF) ...(iv) On adding eqns. (i), (ii), (iii) and (iv), ar(∆BDF) + ar(∆CDE) + ar(∆AFE) + ar(∆DEF) 1
= 4ar(∆DEF)
or,
ar(∆ABC) = 4ar(∆DEF) ar ( ∆DEF )
ar ( ∆ABC )
=
1 4
1
Q. 5. In DABC, AD is a median and O is any point on AD. BO and CO on producing meet AC and AB at E and F respectively. Now AD is produced to X such that OD = DX as shown in figure. A F
E O
B
D
C
Prove that:
(i) EF || BC (ii) AO : AX = AF : AB
A [Board Term-1, 2015]
1
1
OX + OA FB + AE = OA AF AX AB = (from fig.) OA AF or OA : AX = AF : AB Hence Proved. 1 Q. 6. In the right triangle, B is a point on AC such that AB + AD = BC + CD. If AB = x, BC = h and CD = d, then find x (in terms of h and d). C + U [Board Term-1, 2015]
A
x B h C
d
D
Sol. Given, AB + AD = BC + CD AD = BC + CD – AB or, AD = h + d – x 1½ In rt ∆ACD, AD2 = AC2 + DC2 or, (h + d – x)2 = (x + h)2 + d2 or, (h + d – x)2 – (x + h)2 = d2 1½ or, (h + d – x – x – h)(h + d – x + x + h) = d2 or, (d – 2x)(2h + d) = d2 or, 2hd + d2 – 4hx – 2xd = d2 or, 2hd = 4hx + 2xd = 2 (2h + d) x hd 2 or, x = 2 +d h
X
AE EC
So by converse of B.P.T., EF || BC OX FB = (ii) Given OA AF Adding 1 on both sides OX FB +1 = +1 OA AF
AO ...(ii) 1 OX
[CBSE Marking Scheme, 2015]
Q. 7. 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 U [Board Term-1, 2015] to the opposite side.
[ 181
TRIANGLES P
A
Sol.
B
C
D
Q
E
R
1 Given: ∠A = ∠P ∠B = ∠Q, ∠C = ∠R Proof: Let ∠A = ∠P be x In ∆ABC, ∠A + ∠B + ∠C = 180° or, x° + ∠B + ∠B = 180° (given, ∠B = ∠C) or, 2∠B = 180° – x ∠B =
or,
180° - x 2
...(i) 1
Now, in ∆PQR ∠P + ∠Q + ∠R = 180° (Given, ∠Q = ∠R) or, x° + ∠Q + ∠Q = 180° or, 2∠Q = 180° – x 180° - x ...(ii) 1 or, ∠Q = 2
In ∆ABC and ∆PQR, ∠A = ∠P [Given] ∠B = ∠Q [from eqs. (i) and (ii)] ∆ABC ∼ ∆PQR (AA similarity) or,
or,
ar ( ∆ABC )
ar ( ∆PQR )
or,
\
=
AD 2 PE2
1
16 AD 2 = 25 PE2 4 AD = 5 PE AD 4 = . PE 5
1
Given: To prove: Proof: Since,
1 or, or, Let DB be 3x, CD be 2x, so BC = 5x In ∆ADB, ∠D = 90° AB2 = AD2 + DB2 1 or, AB2 = AD2 + (3x)2 or, AB2 = AD2 + 9x2 Now 5AB2 = 5AD2 + 45 x2 or, 5AD2 = 5 AB2 – 45x2 ...(i) 1 and AC2 = AD2 + CD2 or, AC2 = AD2 + (2x)2 or, AC2 = AD2 + 4x2 or, 5AC2 = 5AD2 + 20x2 or, 5AD2 = 5AC2 – 20x2 ...(ii) 1 From equation (i) & (ii), 5AB2 – 45x2 = 5AC2 – 20x2 or, 5AB2 = 5AC2 – 20x2 + 45x2 or, 5AB2 = 5AC2 + 25x2 or, 5AB2 = 5AC2 + (5x)2 \ 5AB2 = 5AC2 + BC2 [Q BC = 5x] Hence proved. 1
Q. 8. In ∆ABC, AD ⊥ BC and point D lies on BC such that 2 DB = 3 CD. Prove that 5AB2 = 5AC2 + BC2. U [Board Term-1, 2015] Sol.
AD ⊥ BC 2DB = 3CD 5 AB2 = 5AC2 + BC2 2DB = 3CD DB 3 = CD 2
Visual Case Based Questions
4 marks each
Note: Attempt any four sub parts from each question. Each sub part carries 1 mark
Q. 1. SCALE FACTOR AND SIMILARITY
Scale factor =
length in image
A scale drawing of an object is of the same shape as the object but of a different size.
The scale of a drawing is a comparison of the length used on a drawing to the length it represents.
The value of scale is written as a ratio.
SIMILAR FIGURES
The ratio of two corresponding sides in similar figures is called the scale factor. If one shape can become another using Resizing then the shapes are Similar.
SCALE FACTOR
corresponding length in object
182 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Rotation or Turn
Hence, two shapes are Similar when one can become the other after a resize, flip, slide or turn. C + A [CBSE SQP, 2020-21] (i) A model of a boat is made on the scale of 1 : 4. The AB BC AC = = \ model is 120 cm long. The full size of the boat has PQ QR PR a width of 60 cm. What is the width of the scale (corresponding sides) model ? A P
Translation or Slide
Reflection or Flip
R C Q E D \ ∠B = ∠Q and ∠ADB = ∠PEQ (each 90°) AD AB a = = (corresponding sides) Now, PE PQ b
B
(a) 20 cm (c) 15 cm
(b) 25 cm (d) 240 cm
Sol. Correct option: (c). Explanation: Width of the scale model = 60/4 = 15 cm. [CBSE SQP Marking Scheme, 2020-21]
[CBSE SQP Marking Scheme, 2020-21] (iv) The shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is
(ii) What will effect the similarity of any two polygons ? (a) They are flipped horizontally (b) They are dilated by a scale factor (c) They are translated down (d) They are not the mirror image of on another.
Sol. Correct option: (d). Explanation: They are not the mirror image of one another. [CBSE SQP Marking Scheme, 2020-21]
(iii) If two similar triangles have a scale factor of a : b, which statement regarding the two triangles is true ? (a) The ratio of their perimeters is 3a : b (b) Their altitudes have a ratio a : b a (c) Their medians have a ratio : b 2
Stick
Shadow
Shadow
(a) 3 m (c) 4. 5 m
(b) 3.5 m (d) 5 m
Sol. Correct option: (d). Explanation: Let shadow of the tree be x.
By the property to similar triangles
we have
(d) Their angle bisectors have a ratio a2 : b2
Sol. Correct option: (b). Explanation: Let ABC and PQR be two similar triangles and AD, PE are two altitudes:
Tree
5 12.5 = 2 x x =
(12.5 × 2 ) = 5 m 5
[CBSE SQP Marking Scheme, 2020-21]
(v) Below you see a student's mathematical model of a farmhouse roof with measurements. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edge of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT, and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m.
[ 183
TRIANGLES T
12m G
H E F D
K A
What is the length of EF, where EF is one of the horizontal edges of the block ? (a) 24 m (b) 3 m (c) 6 m (d) 10 m
12m
L B
Explanation: Given scale factor is 1 : 2 \ A'B' = 2AB Þ A'B' = 2 × 1.5 = 3 cm (iii) What is the sum of angles of quadrilateral A'B'C'D' ? (a) 180° (b) 360° (c) 270° (d) None of these Sol. Correct Option: (b) Explanation: Sum of the angles of quadrilateral A'B'C'D' is 360° (iv) What is the ratio of sides A'B' and A'D' ? (a) 5 : 7 (b) 7 : 5 (c) 1 : 1 (d) 1 : 2 Sol. Correct Option: (a) Explanation: A'B' = 3 cm and A'D' = 2AD = 2 × 2.1 = 4.2 cm A'B' 3 30 = \ = A ' D ' 4.2 42
Q. 2. Seema placed a light bulb at point O on the
=
5 or 5 : 7 7
(v) What is the sum of angles C' and D' ? (a) 105° (b) 100° (c) 155° (d) 140° Sol. Correct Option: (c) Explanation: ÐC' = ÐC = 70° and ÐD' = ÐD = 85° \ ÐC' + ÐD' = 70° + 85° = 155° Q. 3. SIMILAR TRIANGLES 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 if 20 m when Vijay’s house casts a shadow 10 m long on the ground. 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.
ceiling and directly below it placed a table. Now, she put a cardboard of shape ABCD between table and lighted bulb. Then a shadow of ABCD is casted on the table as A'B'C'D' (see figure). Quadrilateral A'B'C'D' in an enlargement of ABCD with scale factor 1 : 2, Also, AB = 1.5 cm, BC = 25 cm, CD = 2.4 cm and AD = 2.1 cm; ÐA = 105°, ÐB C+ A = 100°, ÐC = 70° and ÐD = 85°. O
Sol. Correct option: (c). Explanation: Length of the horizontal edge EF = half of the edge of pyramid 12 = = 6 cm (as E is he mid-point of AT) 2 [CBSE SQP Marking Scheme, 2020-21]
12m
C
M
N
A
B 100°
105° 85°
D D'
A'
B'
70°
C C'
(i) What is the measurement of angle A' ? (a) 105° (b) 100° (c) 70° (d) 80° Sol. Correct Option: (a) Explanation: Quadrilateral A'B'C'D' is similar to ABCD. \ ÐA' = ÐA Þ ÐA' = 105° (ii) What is the length of A'B' ? (a) 1.5 cm (b) 3 cm (c) 5 cm (d) 2.5 cm Sol. Correct Option: (b)
(i) What is the height of the tower? (a) 20 m (b) 50 m (c) 100 m (d) 200 m
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Correct Option: (c) Explanation: When two corresponding angles of two triangles are similar, then ratio of sides are equal. Height of Vijay's house Height of tower = Length of Shadow length of shadow 20 m Height of tower = 10 m 50 m
Height of tower =
20 × 50 1000 = 10 10
= 100 m. (ii) What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12 m? (a) 75 m (b) 50 m (c) 45 m (d) 60 m Sol. Correct Option: (d) (iii) What is the height of Ajay’s house? (a) 30 m (b) 40 m (c) 50 m (d) 20 m Sol. Correct Option: (b) Explanation: d Height of Vijay's house Height of Vijay's house = Length of Shadow Length of Shadow 20 m Height of Vijay's house = 10 m 20 m Height of Ajay's house =
20 m × 20 m 10 m
= 40 m. (iv) When the tower casts a shadow of 40 m, same time what will be the length of the shadow of Ajay’s house? (a) 16 m (b) 32 m (c) 20 m (d) 8 m Sol. Correct Option: (a) (v) When the tower casts a shadow of 40 m, same time what will be the length of the shadow of Vijay’s house? (a) 15 m (b) 32 m (c) 16 m (d) 8 m Sol. Correct Option: (d) Q. 4. Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with another point C are a distance of 12 m, connecting C to point D at a distance of 40 m from point C and the connecting D to the point A which is are a distance of 30 m from D such the ∠ADC = 90° .
(i) Which property of geometry will be used to find the distance AC? (a) Similarity of triangles (b) Thales Theorem (c) Pythagoras Theorem (d) Area of similar triangles Sol. Correct Option: (c) (ii) What is the distance AC? (a) 50 m (b) 12 m (c) 100 m (d) 70 m Sol. Correct Option: (a) Explanation: According to the pythagoras, AC2 = AD2 + CD2 AC2 = (30 m)2 + (40 m)2 AC2 = 900 + 1600 AC2 = 2500 AC = 50 m (iii) Which is the following does not form a Pythagoras triplet? (a) (7, 24, 25) (b) (15,8,17) (c) (5, 12, 13) (d) (21,20,28) Sol. Correct Option: (d) (iv) Find the length AB? (a) 12 m (b) 38 m (c) 50 m (d) 100 m Sol. Correct Option: (b) Explanation: AC = 50 m BC = 12 m AC = AB + BC 50 m = AB + 12 m AB = 50 m – 12 m AB = 38 m (v) Find the length of the rope used. (a) 120 m (b) 70 m (c) 82 m (d) 22 m Sol. Correct Option: (c) Explanation: Length of Rope = BC + CD + DA = 12 m + 40 m + 30 m = 82 m Q. 5. SCALE FACTOR A scale drawing of an object is the same shape at the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor Scale factor= length in image / corresponding length in object
184 ]
If one shape can become another using revising, then the shapes are similar. Hence, two shapes
[ 185
TRIANGLES
Os
(c) Their medians have a ratio 10:4 (d) Their angle bisectors have a ratio 11:5 Sol. Correct Option: (b) (v) The length of AB in the given figure:
(a) 8 cm (b) 6 cm (c) 4 cm (d) 10 cm Sol. Correct Option: (c) Explanation: Since, DABC and DADE are similar, then their ratio of corresponding sides are equal. AB AB + BD = BC DE
are similar when one can become the other after a resize, flip, slide or turn. In the photograph below showing the side view of a train engine. Scale factor is 1:200 This means that a length of 1 cm on the photograph above corresponds to a length of 200 cm or 2 m, of the actual engine. The scale can also be written as the ratio of two lengths. (i) If the length of the model is 11 cm, then the overall length of the engine in the photograph above, including the couplings(mechanism used to connect) is: (a) 22 cm (b) 220 cm (c) 220 m (d) 22 m Sol. Correct Option: (a) (ii) What will affect the similarity of any two polygons? (a) They are flipped horizontally (b) They are dilated by a scale factor (c) They are translated down (d) They are not the mirror image of one another. Sol. Correct Option: (d) (iii) What is the actual width of the door if the width of the door in photograph is 0.35 cm? (a) 0.7 m (b) 0.7 cm (c) 0.07 cm (d) 0.07 m Sol. Correct Option: (a) (iv) If two similar triangles have a scale factor 5:3 which statement regarding the two triangles is true? (a) The ratio of their perimeters is 15:1 (b) Their altitudes have a ratio 25:15
Hence,
( x + 4 ) cm x = 3 cm 6 cm 6x = 3(x + 4) 6x = 3x + 12 6x – 3x = 12 3x = 12 x = 4 AB = 4 cm.
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186 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
c h a p te r
8
Circles
Syllabus Tangent to a circle at point of contact. 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.
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Revision Notes A tangent to a circle is a line that intersects the circle at one point only. The common point of the circle and the tangent is called the point of contact. Secant: Two common points (A and B) between line PQ and circle. A tangent to a circle is a special case of the secant when the two end points of the corresponding chord are coincide. There is no tangent to a circle passing through a point lying inside the circle. At any point on the circle there can be one and only one tangent. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
ce nt
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Tangent Line
Radius Point of Contact
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Secant and Tangent
188 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
There are exactly two tangents to a circle through a point outside the circle. The length of the segment of the tangent from the external point P and the point of contact with the circle is called the length of the tangent. The lengths of the tangents drawn from an external point to a circle are equal. In the figure,
PA = PB. A
O P B
Know the Facts The word 'tangent' comes from the Latin word 'tangere', which means to touch and was introduced by the Danish mathematician Thomas Fincke in 1583.
The line perpendicular to the tangent and passing through the point of contact, is known as the normal. In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
How is it done on the
GREENBOARD?
Q.1. In the given figure if PR = 12 cm, OP = 5 cm and OQ = 4 cm, find RQ. P
12
O 4 cm
5 cm
OR2 = 52 + 122 OR2 = 25 + 144 OR2 = 169
or,
cm R
Q
Solution: Step I: OP ⊥ RP (as radius is ⊥ to tangent at point of contact) In right triangle OPR, OR2 = OR2 + OR2
OR = 169 = 13 cm Step II: Similarly, in right triangle OQR, OR2 = OQ2 + QR2 QR2 = OR2 – OQ2 QR2 = 132 – 42 or, QR2 = 169 – 16 QR2 = 153 or,
QR = 153 cm.
= 3 17 cm
c
[ 189
IRCLES
Very Short Answer Type Questions Q. 1. If PQ = 28 cm, then find the perimeter of ΔPLM. P
1 mark each
Q. 2. PQ is a tangent to a circle with centre O at point P. If DOPQ is an isosceles triangle, then find ÐOQP. A [CBSE SQP, 2020-21] P
N
M
Q
T
Q O
A [CBSE SQP, 2020-21]
Sol. Q PQ = PT PL + LQ = PM + MT PL + LN = PM + MN (LQ = LN, MT = MN) (Tangents to a circle from a common point) Perimeter (DPLM) = PL + LM + PM ½ = PL + LN + MN + PM = 2(PL + LN) = 2(PL + LQ) = 2 × 28 = 56 cm ½ [CBSE SQP Marking Scheme, 2020-21] Detailed Solution: Given, PQ = 28 cm \ PQ = PT (Length of tangents from an external point are equal) i.e., PQ = PT = 28 cm According to figure, Let LQ = x, then PL = (28 – x) cm and let MT = y, then PM = (28 – y) cm and LM = LN + NM = x + y
N
In DOPQ, ÐP+ÐQ+ÐO =180° (ÐO = ÐQ isosceles triangle) 2ÐQ+ÐP = 180° 2ÐQ+90° = 180° 2ÐQ = 90° ÐQ = 45° 1 [CBSE SQP Marking Scheme, 2020-21] Detailed Solution: As we know that ∠OPQ = 90° (Angle between tangent and radius) Let ∠PQO be x°, then ∠QOP = x° Since OPQ is an isosceles triangle. (given) (OP = OQ) P xº Q
½ In DOPQ, ∠OPQ + ∠PQO + ∠QOP = 180° (Property of the sum of angles of a triangle) \ 90° + x° + x° = 180° ⇒ 2x° = 180° – 90° = 90° 90 = 45°. ⇒ x = 2 Hence, ∠OQP is 45° ½ Q. 3. If two tangents inclined at 60˚ are drawn to a circle of radius 3 cm, then find length of each tangent. A [CBSE SQP, 2020-21] Sol. A
3c m
O P
Now, the perimeter of DPLM = PL + LM + PM = (28 – x) + (x + y) + (28 – y) = 28 + 28 = 56 cm. ½
xº O
½
O
L
Sol.
30°
O
30°
B
190 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X A
tan 30° =
AO PA
4 cm
In DPAO,
3 PA
P
PA = 3 3 cm.½ [CBSE SQP Marking Scheme, 2020-21]
Detailed Solution:
PA = PB = ?
Angle between tangents = 60°
(Given)
\ Tangents are equally inclined to each other. A
3 cm
30° 60° 30°
B
Some students were not versed with the
and
∠OAP = 90°
(Angle between tangent and radius)
ANSWERING TIP It is necessary for the students to learn all
In DPAO,
1 3
properties of circle.
Perpendicular OA = Base AP
Q. 5. In the given figure, find the length of PB.
= 3 3 5 cm
O m
AP = 3 3
Hence, the length of each tangent is 3 3 cm.
P
½
Q. 4. In the adjoining figure, if DABC is circumscribing a circle, then find the length of BC.
A
(Using trigonometric Ratios)
3c
⇒
tan 30° =
U [CBSE Delhi Set-I, 2020]
B
U [CBSE OD Set-I, 2020]
properties of circle.
½
∠OPA = ∠OPB = 30°
⇒
C
B
COMMONLY MADE ERROR
⇒
O
Q Similarly, PB and BQ are tangents. \ BP = BQ i.e., BQ = 3 cm ½ Now, CR = AC – AR = 11 – 4 = 7 cm Similarly, CR and CQ are tangents. \ CR = CQ i.e., CQ = 7 cm Now, BC = BQ + CQ = 3 + 7 = 10 cm. Hence, the length of BC is 10 cm. ½
O 3 cm
P
R
3 cm
=
3
cm
1
11
(Using trigonometry) ½
Sol. Since AB is a tangent at P and OP is radius. \ ∠APO = 90°, AO = 5 cm and OP = 3 cm
m O 5 c 3 cm A
Sol. Q AP and AR are tangents to the circle from external point A. \ AP = AR i.e., AR = 4 cm
B
In right angled DOPA, AP2 = AO2 – OP2 (By using Pythagoras theorem) ½ AP2 = (5)2 – (3)2 = 25 – 9 = 16 ⇒ AP = 4 cm Perpendicular from centre to chord bisect the chord ⇒ AP = BP = 4 cm. ½
O
P
c
[ 191
IRCLES
[CBSE SQP, 2020]
Q. 7. If the angle between two tangents drawn from an external point 'P' to a circle of radius 'r' and centre O is 60°, then find the length of OP. [CBSE SQP, 2020] [Foreign Set-I, II, III, 2016] B
Sol.
Sol. Length of Tangent = 2 × 5 2 - 4 2 = 2 × 3 cm = 6 cm
Q. 6. If the radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of one circle which is tangent to the other circle.
½+½ [CBSE SQP Marking Scheme, 2020]
O
CO2 + BC2 = OB2
42 + BC2 = 52
In DOBC
2
16 + BC = 25
BC2 = 25 – 16
BC2 = 9
BC = 3
O 5
4 A
P
A
Detailed Solution:
r
C
½
In OBP,
½
∴ OP = 2r ½ [CBSE Marking Scheme, 2020] Detailed Solution: OA = r PP = ? Angle between tangents = 60° Tangents are equally inclined to each other ⇒ ∠OPA = ∠OPB = 30° In DOPA, ∠POA = 180° – 90° – 30° = 60° ½
B
OB = sin 30° OP
cos 60° =
OA OP
In DOAC,
OC2 + AC2 = OA2
42 + AC2 = 52
AC2 = 9
AC = 3
∴
AB = AC + BC
= 3 + 3
r 1 = 2 OP
T
= 6 cm. ½ ⇒ OP = 2r. ½ Q. 8. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE, Delhi Region, 2019]
opper Answer, 2019
Sol.
1
192 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 9. 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.
R
120°
O
P
O
U [CBSE, OD Set-I, II, III, 2017]
ÐOPQ = ÐOQP (radius of circle) 180° - 70° = 55° = ½ 2 Sol.
\ ∠TPQ = 90° – 55° = 35° ½ [CBSE Marking Scheme, 2017] Detailed Solution: According to the figure, OP = OQ(radii) \ ∠OPQ = ∠OQP (Isosceles triangle property) Now, in DPOQ, ∠OPQ + ∠OQP + ∠POQ = 180° (Angle sum property) ∠OPQ + ∠OPQ + 70° = 180° ∠OPQ = 180° – 70° = 110° ½ ∠OPQ = 55° Since ∠OPT = 90° (Angle between tangent and radius) Hence, ∠TPQ = 90° – ∠OPQ = 90° – 55° = 35°. ½ Q. 10. In the fig. there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS. U [Delhi Comptt. Set-I, II, III, 2017]
Sol. In DOQP
50° C
O
P
B
Here, ∠OAB = 90° – 50° = 40° (Q PA ^ OA) ∠OAB = ∠OBA = 40° (Q OA and OB are radii) \ ∠AOB + 40° + 40° = 180° ∠AOB = 180° – 80° = 100° Hence ∠AOB = 100° 1 Q. 13. In fig., PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure A [CBSE, Delhi Set I, II, III, 2015] of ∠OAB.
70°
Q
∠POR = ÐOQP + ÐOPQ (Exterior angle) ½ \ ∠OPQ = ∠POR – ∠OQP = 120° – 90 = 30° ½ Q. 12. From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, U [Delhi Set-I, II, III, 2016] then find ∠AOB. Sol. A
T
P
Q
A
P
50°
O
B
S
P
Q O R T
Sol. PQ = PR = 5 cm ½ and PQ = QS ½ ∴ PS = 2PQ = 2 × 5 = 10 cm. [CBSE Marking Scheme, 2017] Q. 11. PQ is a tangent drawn from an external point P to a circle with centre O and QOR is the diameter of the circle. If ÐPOR = 120°, What is the measure of U [Foreign Set-I, II, III, 2016, 2017] ÐOPQ ?
∠APB = 50°
Sol. Here,
180° - 50° = 65° ∠PAB = ∠PBA = 2 ∠OAB = 90° – ∠PAB = 90° – 65° = 25° [CBSE Marking Scheme, 2015] 1 Q. 14. In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°, then find ∠OQR. [CBSE Delhi Set-I, II, III, 2015] Q
O
50°
R
P
∠QPR = ∠50° (Given)
Sol.
∠QOR + ∠QPR = 180°
(Supplementary angles) ∴
∠QOR = 180° – 50° = 130°
½
From DOQR,
∠OQR = ∠ORQ =
or,
c
[ 193
IRCLES
C
27 cm
R D
S Q
x cm
A [CBSE SQP, 2020-21]
ÐA = ÐOPA = ÐOSA = 90° ½
Sol.
Hence, Also,
ÐSOP = 90° AP = AS
COMMONLY MADE ERROR
AP = AS = 10 cm
CR = CQ = 27 cm
BQ = BC – CQ
½
Some students does not use appropriate figure to solve the question.
= 38 – 27 = 11 cm
½
ANSWERING TIP
BP = BQ = 11 cm
x = AB = AP + BP
= 10 + 11 = 21 cm
½
Carefully read the question and draw the figure as per the required condition.
[CBSE Marking Scheme, 2020-21]
2 marks each
(Angle between tangent and radius) Finally ∠SOP = 360° – (90° + 90° + 90°) = 90° AP = AS (Tangents from external point A) \ OSAP is a square. AP = AS = SO = 10 cm ½ Q CR = CQ (Tangents from external point C) \ CR = CQ = 27 cm But BC = 38 cm (Given) \ BQ = BC – CQ = (38 – 27) cm BQ = 11 cm ½ BP = BQ (Tangent from external point B) \ BP = 11 cm So, x = AB = AP + PB = (10 + 11) cm = 21 cm Hence, the value of x is 21 cm. ½
Hence, OSAP is a square.
50° = 25° 2
½ [CBSE Marking Scheme, 2015] =
Short Answer Type Questions-I Q. 1. In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ^ AB. If radius of incircle is 10 cm, then find the value of x.
180° - 130° 2
Q. 2. In the given figure, two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.
Detailed Solution: With O as centre, draw a perpendicular OP on AB. Now, in quadrilateral APOS,
P
27 cm
O
T Q
U [CBSE Delhi Set-I, 2020] [CBSE Delhi Set-I, II, III, 2017]
S
Sol. Let ∠ OPQ be q, then ∠ TPQ = 90° – q ½ Since, TP = TQ \ ∠ TQP = 90° – q ½ (opposite angles of equal sides)
and
P ∠SAP = 90° ∠APO = 90° ∠ASO = 90°
(Given) (By construction)
194 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
\
P
θ O T Q
Now, ∠ TPQ + ∠ TQP + ∠ PTQ = 180° ½ (Angle sum property of a Triangle) ⇒ 90° – q + 90° – q + ∠ PTQ = 180° ⇒ ∠ PTQ = 180° – 180° + 2q ⇒ ∠ PTQ = 2q Hence, ∠ PTQ = 2 ∠OPQ ½ Hence Proved. [CBSE Marking Scheme, 2020] Q. 3. In Fig, ABC is a triangle in which ∠B = 90°, BC = 48 cm and AB = 14 cm. A circle is inscribed in the triangle, whose centre is O. Find radius of incircle. C
AC =
½
2500 = 50 cm
Here, ∠OQB = ∠OPB = 90° (Radius is perpendicular to tangent) \ In Quadrilateral OPBQ, ∠POQ = 360° – (OQB + ∠OPB + ∠PBQ) = 360° – (90° + 90° + 90°) = 90° So, OPBQ is a square. Then OP = QB = BP = OQ = r ½ Thus, CQ = BC – QB = 48 – r But CQ = CR (Tangents from external point C) \ CR = 48 – r and AP = AB – BP = 14 – r ½ But AP = AR (Tangents from external point A) \ AR = 14 – r Now AC = 50 cm (proved above) ⇒ AR + RC = 50 ⇒ 14 – r + 48 – r = 50 ⇒ – 2r = 50 – 62 = – 12 ⇒ r = 6 cm. ½ Q. 4. In the fig, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD. A B
O'
O
A [CBSE Comptt. Set I, II, III, 2018]
Sol.
C
AC =
AB2 + BC 2
A [Delhi Comptt. Set-III, 2017]
D
A
B
Sol. Construction : Produce AB and CD to meet at P. A
C
B O'
48 cm
O
Q
D r
r
C
R
O r
B
P
P
14 cm
A
2 2 = 14 + 48 = 2500 = 50 cm ½ ∠OQB = 90° ⇒ OPBQ is a square BQ = r, QC = 48 – r = CR ½ Again, PB = r PA = 14 – r ⇒ RA = 48 – r ½ AR + RC = AC ⇒ 14 – r + 48 – r = 50 r = 6 cm ½ [CBSE Marking Scheme, 2018]
Detailed Solution: In DABC, ∠B = 90° (Given) AC2 = AB2 + BC2 (By using pythagoras theorem) = (14)2 + (48)2 = 196 + 2304 = 2500
Now, PA = PC and PB = PD ½ Tangents to a circle from external point ½ Now, PA – PB = PC – PD ½ Þ AB = CD ½ [CBSE Marking Scheme, 2017] Q. 5. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = DQ = 3 cm, then find PC + PD. A
O
B
C
Q
P
D
A [Delhi Comptt. Set-I, II, III, 2017]
Sol. Here,
AC = CQ (Tangents from external point to a circle) PA = PC + CA = PC + CQ (Q CA = CQ) Þ 12 = PC + 3 Þ PC = 12 – 3 = 9 cm 1
c
[ 195
IRCLES
\
PB = PD + BD PA = PD + DQ 12 – 3 = PD = 9 cm PC + PD = 9 + 9 = 18 cm 1 [CBSE Marking Scheme, 2017]
Q. 6. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord. M
O
N
A [CBSE, OD Set-I, II, III, 2017] [CBSE Delhi Term-II, 2015]
Sol. Q PM = PN (length of tangents are equal)
∠1 = ∠2 (angles opp. to equal sides are equal)
Q
1
180° – ∠1 = 180° – ∠2
(linear pair)
∠3 = ∠4
1 M
P
O
N
[CBSE Marking Scheme, 2015]
T
Detailed Solution:
opper Answer, 2017
Sol.
2
196 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 7. In given figure, O is the centre of the circle and LN is a diameter. If PQ is a tangent to the circle at K and ÐKLN = 30°, find ÐPKL. L
½
PA = PB ∠PAB = ∠PBA = 60°
or,
½
\ DPAB is an equilateral triangle.
½
Hence, AB = PA = 5 cm.
½
[CBSE Marking Scheme, 2016] Q. 9. In the given figure PQ is chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the circle at points P and Q respectively. Find ∠PTQ.
P
O
Sol.
K N Q
U [CBSE OD Comptt. Set-I, II, III, 2017]
P
Sol. Here,
OK = OL (radii) ÐOKL = ÐOLK = 30° (Opposite angles of equal sides) 1 Since ÐOKP = 90° (Tangent) \ ÐPKL = 90° – 30° = 60° 1 [CBSE Marking Scheme, 2017] Q. 8. In Fig., 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
O T Q
U [CBSE S.A.II, 2016] Sol. Here, PQ = 6 cm, OP = OQ = 6 cm
∴
PQ = OP = OQ
∴
∠POQ = 60°
∠OPT = ∠OQT = 90°
∴ ∠PTQ + 90° + 90° + 60° = 360°
(angle of equilateral ∆) ½ B
A
(radius ^ tangent)
O
(angle sum property) ½
U [CBSE Delhi Set I, II, III, 2016]
∠PTQ = 120°
1
T
Q. 10. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA. A [CBSE OD Set-I, II, III, 2016]
opper Answer, 2016
Sol.
2
[ 197
c
IRCLES
Q. 11. In given figure, AB is the diameter of a circle with center O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.
P
U [Board Term-II, 2015 Set-I, II, III] B
O A
B
M
O
Q A
∠AOQ = 58° 1 ∠ABQ = ∠AOQ 2
A [Delhi Board Term-2, 2015]
(Given)
Sol.
T
Sol. Construction : Join OB.
In rt. DOMB, OB2 = 52 + 122 = 132
(Angle on the circumference
of the circle by the same arc) 1 × 58° = 2
∴
OB = 13 cm
Since
OB ^ PB
= 29°
1
∠BAT = 90°
(Q OA ^ AT)
∴
∠ATQ = 90° – 29°
P
O
= 61°
1 (radius ^ tangent)
1
A
[CBSE Marking Scheme, 2015]
B
M
Q. 12. In figure, PQ is a chord of a circle centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ. [CBSE, OD Set-I, II, III, 2015] ∴ In rt. DOBP, Q
OP2 = OB2 + BP2
= 132 + 202 = 569
O R
P
T
Sol. Given, ∠QPT = 60°
∠OPQ = ∠OQP = 90° – 60° = 30°
∠POQ = 180° – (30° + 30°)
or,
OP =
[CBSE Marking Scheme, 2015]
P
1 Reflex ∠POQ 2
1 [Reflex ∠POQ = 360° –120° = 240°]
1 × 240°= 120° 2
[CBSE Marking Scheme, 2015] 1
Q. 13. PB is a tangent to the circle with centre O to B. AB is a chord of length 24 cm at a distance of 5 cm from the centre. If the tangent is of length 20 cm, find the length of PO.
R O
Q
To Prove: OT is the right bisector of line segment PQ. Construction: Join OP and OQ Proof: ∆ OPT and ∆ OTQ PT = PQ (Tangents of the circle) OT = OT (Common side) ∠OPT = ∠OQR = 90°
=
T
1
Q. 14. From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ. A [CBSE Delhi Term-2, 2015 Set-I, II, III] Sol. Given: A circle with centre O. Tangents TP and TQ are drawn from a point T outside a circle.
= 180° – 60° = 120° ∠PRQ =
569 = 23.85 cm.
198 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 15. In figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of DABC is 54 cm2, then find the lengths of sides AB and AC.
\
T
6 cm
Q
U [Foreign Set-I, II, III, 2015]
Sol.
∠TOQ = 180° – 70° = 110° 1 (angle of supplementary )
T C
D
9 cm
R
A [CBSE, OD Set-I, II, III, 2015]
A
Sol.
70° P
O
R
O 3 cm B
AB = 6 + x, AC = 9 + x and BC = 15 ½ 1 ar DABC = [15 + 6 + x + 9 + x].3 = 54 2
45 + 3x = 54 1 or, x = 3 \ AB = 9 cm, AC = 12 cm ½ and BC = 15 cm. [CBSE Marking Scheme, 2015] Q. 16. 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.
A
∴
∆OPT ≅ ∆OQT (R.H.S. Congruency) ∠PTO = ∠QTO (C.P.C.T) DPTR and DTRQ TP = TQ (Tangents of circle) TR = TR (Common) DPTR ≅ DQTR (SAS congruency) ∠PRT = ∠TRQ (c.p.c.t.) PR = QR (c.p.c.t.) ∠PRT + ∠TRQ = 180° \ ∠PRT = ∠TRQ = 90°
x
70°
P
Q
x E
1 Then, ∠TRQ = ∠TOQ 2 (angle at the circumference of the circle by same arc) 1 × 110° = 55°. 1 = 2 [CBSE Marking Scheme, 2015]
F
O
6
6
9
3 cm
D
C
9
Let AF = AE = x.
B
O
Short Answer Type Questions-II
3 marks each
Detailed Solution:
Sol. Let ABCD be the || gm. ∴ AB = CD and AD = BC ...(i) ½ AP + PB + DR + CR = AS + BQ + DS + CQ 1 or, AB + CD = AD + BC ½ From (i), 2AB = 2AD or AB = AD
\ AB = CD and AD = BC
Q. 1. Prove that the parallelogram circumscribing a circle is a rhombus. A [CBSE Delhi Set-II, 2020]
A
P
B
Let ABCD be the parallelogram. ...(i) ½
We know that the tangents drawn to a circle from an exterior point are equal in length. Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS. ½ Adding the above equations. AP + BP + CR + DR = AS + BQ + CQ + DS
½
Q
S
½ R 1 or, ABCD is a rhombus. [CBSE Marking Scheme, 2020]
D
C
c
[ 199
IRCLES
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) ⇒ AB + CD = AD + BC ½ From eq. (i), 2AB = 2AD or, AB = AD Hence, ABCD is a rhombus. Hence Proved. ½ Q. 2. In given Fig., two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q. P
T
From the point B, BQ = BP From the point A, AQ = AR From the point C, CP = CR \ Perimeter of DABC, i.e., AB + BC + CA = 2AQ – BQ + BQ + CR – CR] ⇒ 2AQ = AB + BC + CA ⇒
AQ =
1 (BC + CA + AB) 2 Hence proved. 1
Q. 4. In figure AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.
Q
A O
A [CBSE Delhi Set-III, 2020]
PT = TC (tangents of circle) QT = TC (tangents of circle from extended point) 1 T
Q
A [CBSE Delhi Set-I, 2019, Compt. Set I, II, III 2018] A
Sol. 5c
m
P
O
⇒
½ ½
PT = QT PQ = PT + TQ PQ = PT + PT PQ = 2PT
y P
M B
Given, AB = 8 cm ⇒ AM = 4 cm. \ OM = OA 2 − AM 2
So, Now ⇒ ⇒
P
B
4 cm
Sol. Since, and
8 cm
5 cm
1 PQ = PT 2
(By using Pythagoras theorem)
Hence, the common tangent to the circle at C, bisects the common tangents at P and Q. 1 Q. 3. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, 1 respectively, prove that AQ = (BC + CA + AB). 2 A [CBSE OD Set-I, 2020] Sol. BC + CA + AB = (BP + PC) + (AR – CR) + (AQ – BQ) ½ = AQ + AR – BQ + BP + PC – CR
OM = 52 - 4 2 = 3 cm. Let AP = y cm, PM = x cm. \ DOAP is a right angle triangle. \ OP2 = OA2 + AP2 (Again using Pythagoras theorem) (x + 3)2 = y2 + 25 Þ x2 + 9 + 6x = y2 + 25 ...(i) ½ 2 2 2 Also, x + 4 = y ...(ii) ½ x2 + 6x + 9 = x2 + 16 + 25 6x = 32 32 16 i.e., cm Þ x = 6 3
1
From the same external point, the tangent segments drawn to a circle are equal. ½
y2 = x2 + 16 =
256 + 16 9
400 1 9 20 2 cm or 6 cm. 1 y = 3 3 Q. 5. In the given figure a circle is inscribed in a DABC having sides BC = 8 cm, AB = 10 cm and AC = 12 cm. Find the length BL, CM and AN. =
200 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Alternate method : DOAD ≅ DAOC (By SAS) ⇒ ∠1 = ∠2 1 Similarly, ∠4 = ∠3 ½ But ∠1 + ∠2 + ∠3 + ∠4 = 180° ( PQ || RS) 1 ⇒ ∠2 + ∠3 = ∠1 + ∠4 = (180°) = 90° 2 \ In DAOB, ∠AOB = 180° – (∠2 + ∠3) = 90° ½
A
12 cm M
10 cm N B
C
L 8 cm
A [CBSE Delhi Set-II, 2019]
[Delhi Set-I, II, III, 2016]
Sol. Let
4+
m
Sol. DAOD ≅ DAOC ⇒ ∠1 = ∠2 Similarly ∠4 = ∠3
⇒
1 (180°) 2 ∠2 + ∠3 = 90° or ∠AOB = 90°
In D DOA and D COA
cm 8 – x
10 c
12
circle with centre O and another tangent AB with point of contact C intersecting PQ at A and RS at B. Prove that ∠AOB = 90°. [CBSE Delhi Set-I, II, III, 2017] A [CBSE OD Set-1, 2019]
⇒
Detailed Solution:
x
Q. 6. In Figure PQ and RS are two parallel tangents to a
½ [CBSE Marking Scheme, 2019]
½
8–x
4+
x
BL = x = BN (Tangent from external point B) \ CL = 8 – x = CM (Tangent from external point C) AC = 12 ⇒ AM = 4 + x = AN 1 (Tangent from external point A) Now AB = AN + NB = 10 ⇒ x + 4 + x = 10 ⇒ x = 3 1 \ BL = 3 cm, CM = 5 cm and AN = 7 cm. 1
[SAS] 1 ½ ½
DA = AC (Tangents drawn from common point) ∠ODA = ∠OCA = 90° (angle between tangent and radius) OD = OC (radius of circle) \ D DOA @ DCOA (By SAS) ½ Hence, ∠1= ∠2 i.e., ∠DOA = ∠COA (By cpct) ...(i) Similarly, D BOC @ D BOE (By SAS) ½ \ ∠3 = ∠4 i.e., ∠COB = ∠BOE (By cpct) ...(ii) Now, ∠1+ ∠2 + ∠3 + ∠4 = 180° ½ (angles on a straight line) 2 ∠2 + 2 ∠3 = 180° [from eq. (i) & eq. (ii)] ∠2 + ∠3 = 90° i.e., ∠AOC + ∠BOC = 90° or ∠AOB = 90° Hence Proved. 1
COMMONLY MADE ERROR
∠1 + ∠4 = ∠2 + ∠3 =
½
Some candidates could not apply the appropriate theorem to find out the unknown angles.
ANSWERING TIP
½
Learn circle and related angle properties,
cyclic properties, tangent and secant properties thoroughly.
c
[ 201
IRCLES
Q. 7. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and intersecting the larger circle at P on producing. U [CBSE SQP, 2018-19] Find the length of AP.
Q. 8. Prove that the lengths of two tangents drawn from an external point to a circle are equal. A [CBSE OD Set-I, II, III, 2018] Sol. Given: AP and BP are tangents of circle having centre O. ½
Sol.
A
A
O
P
B
O
D
1 P ÐAPB = 90° (angle in semi-circle) and ÐODB = 90° (radius is perpendicular to tangent) DABP ~ DOBD AB AP = 1 Þ OB OD Þ Hence,
26 AP = 13 8 AP = 16 cm 1 [CBSE Marking Scheme, 2018-19]
To Prove :
B
AP = BP
½ ½
Construction : Join OP, AO and BO ½ Proof : DOAP and DOBP OA = OB (Radius of circle) OP = OP (Common side) ∠OAP = ∠OBP = 90° (Radius – tangent angle) DOAP = DOBP (RHS congruency rule) AP = BP (By cpct) 1 Hence Proved.
T
Detailed Solution:
opper Answer, 2018
3 Q. 9. In the given figure, PA and PB are tangents to a circle from an external point P such that PA = 4 cm and ÐBAC = 135°. Find the length of chord AB. P
A
C
B
U [CBSE OD Set I, II, III, 2017]
202 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol.
PA = PB = 4 cm (Tangents from external point) ½ ÐPAB = 180° – 135° = 45° (Supplementary angles) ÐABP = ÐPAB = 45° (Opposite angles of equal sides) ½ ∴ ÐAPB = 180° – 45° – 45° = 90° So, DABP is an isosceles right angled triangle. Þ AB2 = 2AP2 1 2 Þ AB = 32 1 32 = 4 2 cm
Hence,
AB =
[CBSE Marking Scheme, 2017]
Q. 10. Prove that the tangents drawn at the ends of the diameter of a circle are parallel. A [CBSE Delhi Set I, II, III, 2017]
Sol. Let AB be the diameter of a given circle and let CD and EF be the tangents drawn to the circle at A and B respectively. AB ⊥ CD and AB ⊥ EF 1
C
O
E
B
E A
F
D
8 1 7 O2 6 3 54 G
1
H
C
T
Alternate Method:
Sol.
B
F
∴ ∠CAB = 90° and ∠ABF = 90° ½ ∠CAB = ∠ABF and ∠ABE = ∠BAD ½ ∠CAB and ∠ABF also ∠ABE and ∠BAD are alternate interior angles. 1 ∴ CD EF Hence proved. [CBSE Marking Scheme, 2017] Q. 11. ABC is a triangle. A circle touches sides AB and AC produced and side BC at X, Y and Z respectively. Show that 1 perimeter of ∆ABC. AX = 2 Sol. See Q.3. from SATQ-II.
Long Answer Type Questions Q. 1. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. A [CBSE Delhi Region, 2019] [Foreign Set-I, II, III, 2017] Sol. Given: A circle with centre O is inscribed in a quadrilateral ABCD. In ∆AEO and ∆AFO, OE = OF (radii of circle) ∠OEA = ∠OFA = 90° (radius is ⊥r to tangent) 1
D
A
A [Board Term-2, 2016]
5 marks each The point of contact is perpendicular to the tangent. OA = OA (common side) ∆AEO ≅ ∆AFO (R.H.S. congruency) ∠7 = ∠8 (By cpct) ...(i) 1 Similarly, ∠1 = ∠2 ...(ii) ∠3 = ∠4 ...(iii) ∠5 = ∠6 ...(iv) ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° 1 (angle around a point is 360°) 2 ∠1 + 2 ∠8 + 2 ∠4 + 2 ∠5 = 360° ∠1 + ∠8 + ∠4 + ∠5 = 180° (∠1 + ∠8) + (∠4 + ∠5) = 180° 1 ∠AOB + ∠COD = 180° Hence Proved.
opper Answer, 2018
[ 203
A [CBSE SA-2, 2016]
Sol.
A
N
Þ ∴
AN = AP AC – CN = AB – BP b – r = c – BM b – r = c – (a – r) b – r = c – a + r 2r = a + b – c a+b-c . r = 2
1
1
Hence Proved. [CBSE Marking Scheme, 2016] Q. 3. In Fig. O is the centre of a circle of radius 5 cm. T is a 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 two tangents to the circle.
P b
c O
B 1 a C M Let circle touches CB at M, CA at N and AB at P. Now OM ^ CB and ON ^ AC (radius ^ tangent) OM = ON (radii) CM = CN (Tangents) 1 \ OMCN is a square. Let OM = r = CM = CN 1 AN = AP, CN = CM and BM = BP (tangent from external point)
P A
5
a+b-c . sides of the triangle. Prove that r = 2
Q. 2. a, b and c are the sides of a right triangle, where c is the hypotenuse. A circle, of radius r, touches the
c
IRCLES
5
E
O
13
T
5
B Q
U [CBSE Delhi Set I, II, III, 2016]
204 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. and
PT = 169 - 25 = 12 cm TE = OT – OE = 13 – 5 ½+½
= 8 cm Let
PA = AE = x.
Then,
(Tangents)
TA2 = TE2 + EA2
1
or,
(12 – x)2 = 82 + x2
24x = 80
or, Thus
x = 3.3 cm. (Approx.)
1
AB = 2 × x = 2 × 3·3
= 6.6 cm. (Approx.)
1
1
[CBSE Marking Scheme, 2016]
T
Q. 4. Prove that tangent drawn at any point of a circle is perpendicular to the radius through the point of contact. [CBSE OD Set II, 2016]
opper Answer, 2016
Sol.
Q
5
Q. 5. In the given figure, O is the centre of the circle. Determine ∠APC, if DA and DC are tangents and ∠ADC = 50°. A
Given DA and DC are tangents from point D to a circle with centre O.
D
P
O
C
A [Board Term-2, 2015]
Sol.
or,
∴
C
1
1
∠4 = 130° Reflex ∠4 = 360° – 130° = 230° ∠APC =
1
1 reflex ∠4 2
(angle subtended at centre)
O
∠1 + ∠2 + ∠3 + ∠4 = 360°
or, 90° + 90° + 50° + ∠4 = 360°
P
(radius ^ tangent) 1
A
D
∠1 = ∠2 = 90°
∠APC =
1 × 230° = 115° 2
1
[CBSE Marking Scheme, 2015]
c
[ 205
IRCLES
Visual Case Based Questions
(iv) The distance between Kanabh and Shubhi is: 10 13 (a) (b) m m 3 3
(c)
16 m 3
20 (d) m 3
Chikoo and Shubhi are playing with ball, in which Kanabh and Chikoo are standing on the boundary of the circle. The distance between Kanabh and Chikoo is 8 m. From Shubhi point S, two tangents are drawn as shown in the figure. Give the answer C + AE of the following questions.
Q. 1. There is a circular filed of radius 5 m. Kanabh,
Explanation: ∠SKR + ∠OKR = ∠OKR = 90° (Radius is ^r to tangent) 1
ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark
4 marks each
Sol. Correct Option: (d) Explanation: DSKR and DRKO, ∠RKO = ∠KSR and ∠SRK = ∠ORK \ DKSR ∼ DOKR SK RK = Then KO RO
(By AA Similarity)
SK 4 = 5 3 (RO = 3 m, proved in Q.2.) ⇒ 3SK = 20 20 ⇒ SK = 3
Now,
OR =
(v) What is the mathematical concept related to this question ? (a) Constructions (b) Area (c) Circle (d) none of these Sol. Correct Option: (c) Explanation: The mathematical concept (Circle) is related to this question. 1
Q. 2. ABCD is a playground. Inside the playground a circular track is present such that it touches AB at point P, BC at Q, CD at R and DA at S.
D
OK 2 − KR 2
r
(By using Pythagoras theorem)
=
52 − 4 2 = 25 − 16
r m
52 − 4 2== 25 − 16 9 = 3 m.
(iii) The sum of angles SKR and OKR is: (a) 45° (b) 30° (c) 90° (d) none of these Sol. Correct Option: (c)
See the above figure and give answer of the C + AE following questions: (i) If DR = 5 m, then DS is equal to: (a) 6 m (b) 11 m (c) 5 m (d) 18 m Sol. Correct Option: (c)
1
= 3 m.
C
=
A
m
m
(ii) The length (distance) of OR is: (a) 3 m (b) 4 m (c) 5 m (d) 6 m. Sol. Correct Option: (a) Explanation: In question 1, we have proved SK = SC Then DSKC is an isosceles triangle and SO is the angle bisector of ∠KSC. So, OS ^ KC. \ OS bisects KC, gives KR = RC = 4 cm.
Hence, the distance between Kanabh and Shubhi is 20 1 m. 3
SC ? (a) SK ≠ SC (b) SK = SC (c) SK > SC (d) SK < SC. Sol. Correct Option: (b) Explanation: We know that the lengths of tangents drawn from an external point to a circle are equal. So SK and SC are tangents to a circle with centre O. \ SK = SC 1
(i) What is the relation between the lengths of SK and
⇒
206 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Explanation: DR = 5 m (given) \ DR = DS (Length of tangents are equal) i.e., DS = 5 m. 1 (ii) The length of AS is: (a) 18 m (b) 13 m (c) 14 m (d) 12 m Sol. Correct Option: (a) Explanation: We have AD = 23 m. and DS = 5 m (Proved in Q.1) \ AS = AD – DS = (23 – 5) m = 18 m. 1
(i) In the given figure find ∠ROQ. (a) 60 (b) 100 (c) 150 (d) 90 Sol. Correct option: (c). (ii) Find ∠RQP. (a) 75 (b) 60 (c) 30 (d) 90 Sol. Correct option: (a). (iii) Find ∠RSQ. (a) 60 (b) 75 (c) 100 (d) 30 Sol. Correct option: (b). (iv) Find ∠ORP. (a) 90 (b) 70 (c) 100 (d) 60 Sol. Correct option: (a).
(iii) The length of PB is: (a) 12 m (b) 11 m (c) 13 m (d) 20 m Sol. Correct Option: (b) Explanation: We have, AB = 29 m But AS = AP (lengths of tangents are equal) and AS = 18 m (Proved in Q.2) \ AP = 18 m Now, PB = AB – AP = (29 – 18) m = 11 m. 1 (iv) What is the angle of OQB? (a) 60° (b) 30° (c) 45° (d) 90° Sol. Correct Option: (d) Explanation: ∠OQB = 90° (Radius is ^r to tangent) 1 (v) What is the diameter of given circle? (a) 22 m (b) 33 m (c) 20 m (d) 30 m Sol. Correct Option: (a) Explanation: Q PB = 11 m (proved in Q.3) But PB = BQ (lengths of tangents are equal) \ BQ = 11 m or r = OQ = QB = 11 m Hence, diameter = 2r = 2 × 11 = 22 m. 1 Q. 3. A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a 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.
Explanation: ∠ORP = 90° Because, radius of circle is perpendicular to tangent. Q. 4. 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 ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.
c
[ 207
IRCLES
(iii) Find the length of CF.
(a) 9
(b) 5
(c) 2
(d) 3
Sol. Correct option: (d). (iv) If radius of the circle is 4 cm, find the area of ΔOAB.
(i) Find the length of AD. (a) 7 (c) 5 Sol. Correct option: (a). (ii) Find the Length of BE. (a) 8 (c) 2 Sol. Correct option: (b).
(b) 8 (d) 9 (b) 5 (d) 9
(a) 20
(b) 36
(c) 24
(d) 48
Sol. Correct option: (c).
(v) Find area of ΔABC
(a) 50
(b) 60
(c) 100
(d) 90
Sol. Correct option: (b).
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208 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
c ha pt er
9
CONSTRUCTIONS
Syllabus ¾
¾ Division of a line segment in a given ratio (Internally). ¾¾ Tangents to a circle from a point outside it. ¾¾ Construction of a triangle similar to a given triangle.
Trend Analysis 2018 List of Concepts Division of a line segment and Tangent to circle Construction of Triangles
Delhi 1 Q (5 M)
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2020
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TOPIC - 1
Division of a Line Segment in a Given Ratio
Revision Notes
The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as scale factor. To divide a line segment internally in a given ratio m : n, where both m and n are positive integers. 1st Method: we follow the following steps:
TOPIC - 1 Division of a Line Segment in a Given Ration Page No. 209
TOPIC - 2 Tangents to a Circle from a Point Outside it Page No. 213
TOPIC - 1
Construction of a Triangle Similar to a Given Triangle Page No. 219
210 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Step 1. Draw a line segment AB of given length by using a ruler.
Scan to know more about this topic
Step 2. Draw any ray AX making an acute angle with AB. Step 3. Along AX mark off (m + n) points A1, A2,………,Am, Am+1,………, Am+n, such that
AA1 = A1A2 = Am+n–1 Am+n.
Step 4. Join BAm+n. Step 5. Through the point Am draw a line parallel to Am+nB by making an angle equal to ∠AAm+nB at Am. i.e., ÐAAmP.
Division of line segment
This line meets AB at point P. The point P is the required point which divides AB internally in the ratio m : n. 2nd Method: 1. Draw an ray AX making an acute angle with line segment AB B
A
X
2. Draw ray BY || AX Y B
A
X
3. Locate A1, A2, A3 (m = 3) on AX and B1, B2 (n = 2) on By join A3B2, intersecting AB at C, AC : BC = 3 : 2. Y B A A
How is it done on the
B
C A
B A
X
GREENBOARD?
Q.1. Draw a line segment AB = 6 cm and divide it in the ratio of 2 : 3. Solution:
Steps of construction: (i) Draw a line segment AB = 6 cm. (ii) At A draw an acute angle BAC. (iii) Mark A1, A2, A3, A4, A5 at equal distance. (iv) Join A5 to B. (v) Draw a line A2P parallel to A5B. (vi) P is the required point on AB which divides it in the ratio 2 : 3.
Very Short Answer Type Questions Q. 1. In the figure, if B1, B2, B3,…... and A1, A2, A3,….. points have been marked at equal distances. On
1 mark each
lines X and Y in what ratio does C divides AB?
U [CBSE SQP, 2020-21]
c
[ 211
ONSTRUCTIONS
Y
B5
B4
B3
B2
Sol. The line segment AB is divided in the ratio AP : PB = 2 : (5 – 2) = 2 : 3
B1
P
B
C A A1
A2
A3
A4
A5
A6
A7
A8
X
Sol. C divides AB internally in the ratio 8 : 5. 1 Q. 2. To find a point P on the line segment AB = 6 cm, such AP 2 = , in which ratio the line segment AB that AB 5 A is divided.
Q. 3. A line Segment AB is divided at point P such that PB 3 A = , then find the radio AP : PB. AB 7 Sol. Here, AB = 7 , PB = 3 ∴
AP = AB – PB = 7 – 3 = 4
∴
AP : PB = 4 : 3
1
2 marks each 2. Draw any ray AX making an acute angle down ward with AB.
3. Mark the points A1, A2, A3,....., A8 on AX such that AA1 = A1A2 = A2A3 = ...., A7 A8.
4. Join BA8.
5. Through the point A3, draw a line parallel to BA8. To meet AB on P. 1
Hence AP : PB = 3 : 5
[CBSE Marking Scheme, 2015]
Q. 3. Draw a line segment of length 5 cm and divide it in U [Board Term-2, 2015] the ratio 3 : 7. Sol.
½ Q. 2. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3 : 5. U [Board Term-2, 2015]
cm
Short Answer Type Questions-I Q. 1. Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5. U [CBSE Board, Delhi Region, 2017] Sol. Steps of Construction: (i) Draw a line segment AB = 8 cm and draw a ray AX making an acute angle with AB at A. (ii) Locate (4 + 5) = 9 points A1, A2, A3, ..... A9 on AX such that AA1 = A1A2 = A2A3 ............ = A8A9. (iii) Join BA9 (iv) Through the point A4 draw a line parallel to BA9 intersecting AB at C. (v) Therefore, C is the point which divides the AB in the ratio 4 : 5. ½
1
B
A
Sol. Steps of construction: 1. Draw a line segment AB = 7 cm.
Steps of Construction :
1
1. Draw a line segment AB = 5 cm.
2. Draw any ray AX making an acute angle down ward with AB.
3. Mark the points A1, A2, A3, .... A10 on AX such that AA1 = A1A2 = .... = A9A10.
4. Join BA10.
5. Through the point A3 draw a line parallel to BA10. To meet AB at P. 1
Hence AP : PB = 3 : 7.
1
[CBSE Marking Scheme, 2015]
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Short Answer Type Questions-II Q. 1. Draw a line segment of length 6 cm and divide it internally in the ratio 4 : 3. Prove your assertion.
C + AE
Sol. Steps of Construction: (i) Draw a line segment AB = 6 cm. (ii) Draw a ray AX, making an acute ∠BAX. (iii) Along AX, mark (4 + 3) = 7 points A1, A2, A3, ...., A7 such that AA1 = A1A2 = ..... = A6A7.
1 (iv) Join BA7. (v) Through the point A4 draw a line parallel to BA7. To meet AB at C. Hence, AC : CB = 4 : 3.
3 marks each
(iv) Join BA9. (v) Through the points A2 and A5, draw BA9 || CA2 and BA9 || DA5. To meet AB at C and D respectively. (vi) On measuring the three parts, we get AC = 3 cm, CD = 4.5 cm and DB = 6 cm. Verification: AC = 2x, CD = 3x and DB = 5x \ 2x + 3x + 5x = 13.5 ⇒ 9x = 13.5 ⇒ x = 1.5 Sum of ratio = 2 + 3 + 4 = 9 2x × 13.5 = 3 cm \ AC = 9x
CD =
3x × 13.5 = 4.5 cm 9x
and
DB =
4x × 13.5 = 6 cm 9x
Hence, AC + CD + DB = (3 + 4.5 + 6) cm = 13.5 cm. 1 Q. 3. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8 and verify it. Sol.
Proof : Suppose AA1 = A1A2 = ..... = A6A7 = x In DBAA7, we have CA4 || BA7 AA4 AC 4x 4 = = = CB A4 A7 3x 3
Hence,
[By using Thale's theorem] 1
AC : CB = 4 : 3.
Q. 2. Draw a line segment of length 13.5 cm and divide it internally in the ratio 2 : 3 : 4. Measure each part and verify it. C + AE Sol. Steps of Construction:
1
(i) Draw a line segment AB = 13.5 cm. (ii) Draw a ray AX, making an acute ∠BAX.
(iii) Along AX, make (2 + 3 + 4) = 9 points A1, A2, A3, ......, A9 such that AA1 = A1A2 = A2A3 = ....... = A8A9.
1
\
1
Steps of Construction: 1 (i) Draw a line segment AB = 7.6 cm. (ii) Draw a ray AX, making an acute ∠BAX. (iii) Along AX, make (5 + 8) = 13 points A1, A2, A3, ......, A13 such that AA1 = A1A2 = A2A3 = ....... = A12A13. (iv) Join BA13. (v) Through the point A5 draw a line parallel to BA13. To meet AB at C. Hence, AC : CB = 5 : 8. (vi) On measuring the two parts, we get AC = 4.7 cm and BC = 2.9 cm. Justification: In DACX5 and DABX13, we have. BA13 || CA5. AA5 AC 5 = \ = CB 8 A5 A13 ⇒
AC : CB = 5 : 8.
1
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Long Answer Type Questions
Sol.
C
1½
Steps of Construction: (i) Draw a line segment AB = 8 cm. (ii) Draw AX || BY such that ∠A and ∠B are acute. (iii) Divide AX and BY in 3 and 5 parts equally by compass and mark A1, A2, A3, B1, B2, B3, B4 and B5 respectively. (iv) Join A3B5 which intersect AB at P and divides AP : PB = 3 : 5. 1½ Measurement of each part: AP = 3 cm and PB = 5 cm. 1 Justification: In DAA3P and DBB5P, AX || BY (By construction) ∠A = ∠B (Alternate angles) ∠A3PA = ∠B5PB (Vertically opposite angles) \ DAA3P ~ DBB5P (By AA criterion of similarity)
AA3 AP = BB5 BP (Let each equal part be x cm) ⇒ AP : BP = 3 : 5. 1 Q. 2. Draw a line segment 12.6 cm long with ruler and compass divide it into three line segments in the ratio 2 : 3 : 5. Measure each of the three parts. C ⇒
Sol.
2
Q. 1. Draw a line segment of length 8 cm. Find a point P on it which divides it in the ratio 3 : 5. Write the measurement of each part and give Justification.
5 marks each
Steps of Construction: 2 (i) Draw a line segment AB = 12.6 cm. (ii) At A, draw an acute angle BAX. (iii) On AX mark 10 (2 + 3 + 5) points A1, A2, A3, ....., A10 such that AA1 = A1 A2 = A2 A3 = ..... = A9A10. (iv) Join A10B. (v) Through A2 and A5 draw two line parallel to A10B intersecting AB at points P and Q respectively, which divide AB in the ratio 2 : 3 : 5. i.e., AP : PQ : QB = 2 : 3 : 5. Measurement of each part: AP = 2.6 cm, PQ = 3.5 cm and QB = 6.5 cm. 1 Q. 3. Draw a line segment of length 11 cm. Find a point R on it which divides it in the ratio 4 : 5. Write the measurements of each part and give Justification. R+ C Sol. Try yourself similar to Q.No. 1 of LATQ.
TOPIC - 2
Tangents to a Circle from a Point Outside It Scan to know more about this topic
Revision Notes To draw the tangent to a circle at a given point on it, when the centre of the circle is known. Given: A circle with centre O and a point P on it. To construct: The tangent to the circle at the point P. Steps of construction: (i) Join OP. (ii) Draw a line segment AB ^ OP at the point P. APB is the required tangent at P.
To draw the tangent to a circle at given point Scan to know more about this topic
O
A P
B
Tangents to a circle from a point outside it
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
To draw the tangent to a circle at a given point on it, when the centre is not known. Q Given: P is a point on the circle. To construct: Draw a tangent from the point P. Steps of construction: (i) Draw any chord PQ and join P and Q to a point R. A P (ii) Draw ∠QPA equal to ∠PRQ on opposite side of chord PQ. The line segment BPA is the tangent to the circle at P. To draw the tangent to a circle from an external point when its centre A is known. Given: A circle with centre O and a point P outside it. To construct : The tangents to the circle from P. P Steps of construction: O M (i) Join OP and bisect it. Let M be the mid-point of OP. (ii) Taking M as centre and MO as radius, draw a circle to intersect B C (O, r) in two points, say A and B. (iii) Join PA and PB. These are the required tangents from P to the circle. To draw tangents to a circle from a point outside it when its centre is not known. Given: P is a point outside the circle. To construct: To draw tangents from the point P. Steps of construction: (i) Draw a circle of given radius. (ii) Through P draw a secant PAB to meet the circle at A and B. (iii) Produce AP to C such that PC = PA. Bisect CB at Q. (iv) With CB as diameter and centre as Q, draw a semi-circle. (v) Draw PD ⊥ CB, to meet semi-circle at the point D. (vi) Taking P as centre and PD as radius draw an arc to interest the circle at T and T’. (vii) Join P to T and T'. Hence, PT and PT’ are the required tangents.
How is it done on the
GREENBOARD?
Q.1. Draw a pair of tangents to a circle of any convenient radius, which are inclined to the line joining the centre of the circle and the point at which A they intersect at an angle of 45°. OR Draw a circle of radius 3.5 cm. Draw two tangents to the circle which are perpendicular to each other. Solution: Steps of construction: Step I: Draw a circle of any convenient radius with O as centre. Step II: Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Step III: Draw a radius OB, making an angle of 90° with OA. Step IV: Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. Step V: Join OP. PA and PB are the required tangents, which make an angle of 45° with OP.
R
B
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Short Answer Type Questions-I
Sol.
(iii) Bisect the line AB. Let mid-point of AB be C. (iv) Taking C as centre draw a circle of radius AC which intersects the two circles at point P, Q, R and S. (v) Join BP, BQ, AS and AR.
Q. 1. Draw a line segment AB of length 9 cm. With A and B as centres, draw circles of radius 5 cm and 3 cm respectively. Construct tangents to each circle from the centre of the other circle.
2 marks each
[CBSE Marking Scheme, 2020-21] 2
Detailed Solution:
BP, BQ and AR, AS are the required tangents. 2 [CBSE Marking Scheme, 2015] Q. 3. Draw a circle of radius 1·5 cm. Take a point P outside it. Without using the centre draw two
Steps of Construction : (i) Draw a line segment AB of 9 cm. (ii) Taking A and B as centres draw two circles of radii 5 cm and 3 cm respectively. (iii) Bisect the line AB. Let mid-point of AB be C. (iv) Taking C as centre draw a circle of radius AC which intersects the two circles at point P, Q, R and S. (v) Join BP, BQ, AS and AR. BP, BQ and AR, AS are the required tangents. Q. 2. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle. A [Delhi CBSE Term-2, 2015]
Sol. Steps of Construction: (i) Draw a line segment AB of 7 cm. (ii) Taking A and B as centres draw two circles of radii 3 cm and 2 cm respectively. (iii) Bisect the line AB. Let mid-point of AB be C.
A tangents to the circle from the point P. Sol. Steps of construction : (i) Draw a circle of radius 1·5 cm. Take a point P outside it. (ii) Through P draw a secant PAB to meet the circle at A and B. (iii) Produce AP to C such that PC = PA. Bisect CB at Q. (iv) With CB as diameter and centre as Q, draw a semicircle. (v) Draw PD ⊥ CB, to meet semi-circle at the point D. (vi) Intersect P as centre and PD as radius draw an arc to intersect the circle at T and T’. 1 PT and PT’ are the required tangents. (vii) Join P to T and T' Hence, PT and PT’ are the required tangents.
Short Answer Type Questions-II Q. 1. Draw a circle of radius of 3 cm. Take two points P and Q one of its diameter extended on both sides, each at a distance of 7 cm on opposite
1
3 marks each sides of its centre. Draw tangents to the circle from these two points. A [Foreign Set-III, 2017]
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Sol. Steps of Construction : (i) Draw a circle with centre O and radius 3 cm. (ii) Draw its diameter MON and extend it to both the sides to P and Q. Such that OP = OQ = 7 cm. (iii) Taking diameters as OP and OQ draw two circles each of which intersects the first circle at the points A, B and C, D respectively. (iv) Join PA, PB, QC and QO to get the required tangents. 1
2
Q. 2. Construct a pair of tangents PQ and PR to a circle of radius 4 cm from a point P outside the circle 8 cm away from the centre. Measure PQ and PR. A [Board Term-2, 2015] Sol. Steps of construction : (i) Draw a line segment OP = 8 cm (ii) Taking O as centre and radius 4 cm, draw a circle. (iii) Taking OP as diameter draw another circle which intersects the first circle at Q and R. (iv) Join P to Q and P to R.
On measuring, we get
PQ = PR = 6.92 cm (Approx.)
1
2
Q. 4. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C and D is drawn. Construct the tangents from A to this A circle. Sol. Steps of construction :
(i) Draw a line segment BC = 8 cm.
(ii) Make a right angle at the point B i.e., ÐCBX = 90° (iii) Draw a arc of radius 6 cm as centre B which intersect BX at the point A. (iv) Join AC \ ABC is required right angle triangle. (v) Draw an arc taking centre B which intersects AC at the point K and L respectively taking K and L as centre draw two arcs of same radius which intersect at the point M. (vi) Join BM \ ÐBDC = 90° (vii) Draw perpendicular bisector of BC.
2
Q. 3. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius A [Board Term-2, 2013] 6 cm. Sol. Steps of construction : (i) Draw a circle of radius 4 cm with centre O. (ii) Draw another circle of radius 6 cm with same centre O. (iii) Take a point P on second circle and join OP. (iv) Draw perpendicular bisector of OP which intersect OP at O’. (v) Draw a circle with centre O’ which intersects the inner circle at points A and B.
Long Answer Type Questions Q. 1. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm.
1 (viii) Draw a circle taking radius equal to OB and centre O which passes through B, D and C. (ix) Draw an arc taking centre A and radius equal to AB to intersect the circle at point P \ AP and AB are the tangents. 1
5 marks each Construct tangents to each circle from the centre A [CBSE Delhi Set-I, 2020] of the other circle.
Sol. Steps of construction : (i) Draw a line segment AB = 7 cm. ½ (ii) With A as centre and radius 3 cm draw a circle. ½ (iii) With B as centre and radius 2 cm draw another circle. ½ (iv) Taking AB as diameter draw another circle, which intersects first two circles at P and Q, R and S. 1 (v) Join B to P and Q, A to R and S. Hence, BP, BQ, AR and AS are the required tangents. 1
cm
1½
2½
Q. 4. Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre, draw two tangents to the circle. A [CBSE OD Set-III, 2020] Sol. Steps of construction : 2½ (i) Draw a line segment OP = 6 cm. (ii) From the point O, draw a circle of radius = 3.5 cm. (iii) Draw a perpendicular bisector of OP. Let M be the mid point of OP. (iv) Taking M as centre and OM as radius draw a circle. (v) This circle intersects the given circle at Q and R. (vi) Join PQ and PR, which are tangents to the circle.
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Q. 2. Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm. From P, draw two tangents to the A [CBSE OD Set-I, 2020] circle. Sol. Steps of construction : 2½ (i) Draw a line segment OP = 6.5 cm.
2½
Q. 5. Construct a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of A [CBSE SQP, 2020] 60°. Sol. Correct construction of given circle 2 Correct construction of two tangents 3 [CBSE SQP Marking Scheme, 2020] Detailed Solution: (i) Draw a circle of 3.0 cm.
2½
cm
A [CBSE OD Set-II, 2020] Sol. Steps of construction : 2½ (i) Draw a circle of radius 4 cm with O as centre. (ii) Draw two radii OA and OB inclined to each other at an angle of 120°. (iii) Draw AP ⊥ OA at A and BP ⊥ OB at B. Which meet at P. (iv) PA and PB are the required tangents inclined to each other an angle of 60°.
(ii) Join O to R. (iii) Construct 90° angle at point R as ∠ORP (iv) Construct 120° angle at point O which meet at point Q. (v) Construct 90° angle at point Q as ∠OQP
(ii) Taking O as centre and radius 2 cm, draw a circle. (iii) Taking OP as diameter draw another circle which intersects the first circle at Q and R. (iv) Join P to Q and P to R. Hence PQ and PR are two tangents. Q. 3. Draw two tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Hence, PR & PQ are the required tangents.
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Q. 6. Drawn two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. A [CBSE OD, Set-1, 2019] Measure PA.
Steps of construction : (i) Draw a circle with radius 3 cm and centre O.
1½
(ii) Draw another circle with radius 5 cm and same centre O.
(iii) Take a point P on the circumference of larger circle and join O to P.
Detailed Solution:
Sol. Constructing two concentric circle of radii 2 cm and 5 cm 1½ Drawing two tangents PA and PB 2 PA = 4.5 cm (approx) 1½ [CBSE Marking Scheme, 2019]
(iv) Taking OP as diameter draw another circle which intersects the smaller circle at A and B. 1½ (v) Join A and B to P.
Hence AP and BP are the required tangents.
Measure PA =
52 - 22 =
21 =4.6 cm (Approx.)
2
Q. 7. Construct a pair of tangents to a circle of radius 4 cm from an external point at a distance 6 cm from the centre A [CBSE Delhi Region, 2019] of the circle.
Topper Answer, 2019
Sol.
Q. 8. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm, and taking B as centre draw another circle of radius 3 cm. Construct tangents to each circle from the centre A [Foreign Set II, 2017] of the other circle.
5 Sol. Steps of construction : (i) Construct a line segment AB = 8 cm. 1½ (ii) With A as centre and radius 4 cm draw a circle. (iii) With B as centre and radius 3 cm draw another circle.
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(iv) Taking AB as diameter draw third circle. Which intersects first two circles at P and Q, and R and S.1
2½
(v) Join B to P, B to Q, A to R and A to S. Hence BP, BQ, AR and AS are the required tangents. Q. 9. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other. A [O.D. Set I, 2016] [Foreign Set I, II, III 2015]
Sol. Steps of construction : (i) Draw a circle of radius 4 cm with O as centre. (ii) Draw two radii OA and OB inclined to each other at an angle of 120°. (iii) Draw AP ⊥ OA at A and BP ⊥ OB at B. Which meet at P. (iv) PA and PB are the required tangents inclined to each other an angle of 60°. 2
3
TOPIC - 3
Construction of a Triangle Similar to a Given Triangle
Revision Notes Construction of triangles similar to a given triangle: (a) Steps of construction (when m < n): Step I. Construct the given triangle ABC by using the given data. Step II. Take any one of the three sides of the given triangle as base. Let AB be the base of the given triangle. Step III. At one end, say A, of base AB. Construct an acute ∠BAX below the base AB. Step IV. Along AX mark off n points A1, A2, A3,……, An such that AA1 = A1A2 = ……… = An–1 An Step V. Join AnB. C
C'
A
B'
B
A A
A A
Step VI. Draw A3B’ parallel to AnB which meets AB at B’. Step VII. From B’ draw B’C’||CB meeting AC at C’.
th
m Triangle AB’C’ is the required triangle each of whose sides is of the n corresponding sides of ∆ABC. (b) Steps of construction (when m > n): Step I. Construct the given triangle by using the given data. Step II. Take any one of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle. Step III. At one end, say A, of base AB. Construct an acute angle ∠BAX below the base AB i.e., on the opposite side of the vertex C. Step IV. Along AX mark off m (larger of m and n) points A1, A2, A3,………Am such that AA1 = A1A2 = ………= Am–1Am.
Scan to know more about this topic
Construction of a triangle similar to a given triangle
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Step V. Join An to B and draw a line through Am parallel to AnB, intersecting the extended line segment AB at B’. Step VI. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. DAB’C’ so obtained is the required triangle.
How is it done on the
GREENBOARD?
Q.1. Construct a right triangle whose hypotenuse and one side measure 5 cm and 4 cm respectively. Then construct another 3 times triangle whose sides are 5 of the corresponding sides of this triangle. Solution:
Steps of Construction: Step I: (i) Draw a line segment BC = 4 cm. (ii) Draw an angle of 90° at B. (iii) Taking C as centre and radius equal to hypotenuse = 5 cm draw an arc to intersect at A. Join A to C. Step II: Draw a line making an acute angle with side BC and divide it in five equal sections and mark B1, B2, B3, B4, B5. Step III: Join B5 to C, and draw a parallel line B3C' to B5C which meets BC at C'. Step IV: D raw a parallel line A'C' to AC. Hence, A'BC' is the required triangle.
5c
m
3 cm 4 cm
Very Short Answer Type Questions
1 mark each
Q. 1. When construction of a triangle similar to a given triangle in the scale factor
5 , then what is the nature of given 3
triangle ? R Sol. Given triangle is smaller than the constructed triangle. 1 Q. 2. In figure, DADE is constructed similar to ∆ABC, write down the scale factor. R C
E
A
D
B
A A A
Sol. Scale factor =
A
3 4
Q. 3. When are the two triangles said to be similar ?
Sol. Two triangles are said to be similar when their corresponding sides are proportional and angles are equal.
1 R
1
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Short Answer Type Questions-I
2 marks each
Q. 1. Construct a triangle similar to a given equilateral ∆PQR with side 5 cm such that each of its side is 6 A of the corresponding sides of ∆PQR. 7
(iii) Cut BA = 5 cm.
Sol. Steps of construction : (i) Draw a line segment QR = 5 cm. (ii) With Q as centre and radius = PQ = 5 cm, draw an arc. (iii) With R as centre and radius = PR = 5 cm, draw another arc meeting the arc drawn in step 2 at the point P. (iv) Join PQ and PR to obtain DPQR. (v) Below QR, construct an acute ∠RQX. (vi) Along QX, mark off seven points Q1, Q2, …… Q7 such that QQ1 = Q1Q2 = Q2Q3 = …… = Q6Q7. (vii) Join Q7R. (viii) Draw Q6R’ || Q7R. (ix) From R’ draw R’P’ || RP. Hence, P’QR’ is the required triangle. 1
(vi) Divide BY in 3 equal segments by marking arc at same distance at B1, B2, and B3.
(iv) Join AC. ΔABC is the given triangle. (v) Draw an acute ∠CBY such that A and Y are in opposite direction with respect to BC.
(vii) Join B3C. (viii) Draw B2C’ || B2C by making equal alternate angles at B2 and B3. (ix) From point C’, draw C’A’ || CA by making equal alternate angles at C and C’.
ΔA’BC’ is the required triangle of scale factor
2 . 3 1
This triangle is also a right triangle.
Q. 3. Two line-segments AB and AC include an angle of 60°, where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC respectively such that AP = 3 1 AB and AQ = AC. Join P and Q and measure 4 4 A
the length PQ. Sol.
cm
1 cm
1
cm
Q. 2. Draw a right angled ΔABC in which BC = 12 cm, AB = 5 cm, and ∠B = 90°. Construct a triangle 2 similar to it and of scale factor . Is the new 3
A
triangle also a right triangle ?
2 Sol. Here, scale factor or ratio factor is < 1, So, triangle 3 to be constructed will be smaller than given ΔABC.
Steps of construction :
(i) Draw ∠BAC = 60° such that AB = 5 cm and AC = 7 cm. (ii) Draw acute angle CAX and mark X1, X2, X3, and X4 equally spaced.
(iii) Join X4C. 1
(iv) Draw X1Q || X4C.
(v) Similarly, draw ∠BAY and divide AY in 4 equal parts, i.e., Y1, Y2, Y3 and Y4.
Steps of construction :
(vi) Join Y4B and draw Y3P || Y4B.
(i) Draw BC = 12 cm.
(vii) Join PQ and measure it.
(ii) Draw ∠CBA = 90° with scale and compass.
(viii) PQ is equal to 3.3 cm.
1
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Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Q. 1. Construct a right-angled triangle whose base is 5 cm and sum of its hypotenuse and other side is 10 cm. Construct another triangle whose sides are 1·4 times the corresponding sides of the previously drawn triangle. Give the justification of the construction. A
Sol. Let us assume that DABC is right-angled at B, with base BC = 5 cm and AC + AB = 10 cm. A DA'BC' whose sides are 1.4 =
Steps of construction :
7 times of DABC, 5
(i) Draw a line segment BC of length 5 cm. (ii) At B, draw ∠XBC = 90°. Taking B as centre and radius as 10 cm, draw an arc that intersects the ray BX at Y. (iii) Join CY and draw its perpendicular bisector to intersect BY at A. Join AC. (iv) Draw a ray BZ making an acute angle with line segment BC.
Short Answer Type Questions-II
3 marks each
Q. 2. Construct a triangle whose perimeter is 13·5 cm
A and the ratio of the three sides is 2 : 3 : 4. Sol. Steps of construction: (i) Draw a line segment PR of length 13·5 cm. (ii) At the point P draw a ray PQ making an acute angle RPQ with PR. (iii) On PQ mark (2 + 3 + 4) 9 points P1, P2, P3, P4, P5, P6, P7, P8, P9 such that PP1 = P1P2 = P2P3 = P3P4 = P4P5 = P 5 P 6 = P 6 P 7 = P 7 P 8 = P 8 P9 . (iv) Join P9R. (v) Through P2 and P5 draw lines P2A and P5B respectively parallel to P9R intersecting PR at A and B respectively.
(vi) With A as centre and radius AP draw an arc. 2 (vii) With B as centre and radius BR draw another arc to intersect first arc. (viii) Join A to C and B to C.
ABC is the required triangle.
(v) Locate 7 points B1, B2, B3, B4, B5, B6 and B7 on BZ such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7. (vi) Join CB5 and draw a line C'B7 parallel to CB5 to intersect extended line segment BC at point C'. (vii) Draw a line through C' parallel to AC intersecting the ray BX at A'. DA'BC' is the required triangle.
2
1 Q. 3. Construct a rhombus ABCD in which AB = 4 cm and ∠ABC = 60°. Divide it into two triangles ABC and ADC. Construct the triangle 2 AB'C' similar to DABC with scale factor . Draw 3
a line segment C'D' parallel to CD, where D' lies on A AD. Is AB'C'D' a rhombus ? Give reasons. Sol. Steps of construction:
(i) The rhombus ABCD is drawn in which AB = 4 cm and ∠ABC = 60°. (ii) Join AC. ABCD is divided into two triangles ABC and ADC. 1
(iii) At the point A draw a ray AX making an acute angle BAX with AB. (iv) Along AX mark off three points A1, A2, A3 such that AA1 = A1A2 = A2A3. (v) Join A3B. from A2 draw A2B'|| A3B. (vi) From B' draw B'C'||BC.
(vii) From C draw C'D'||CD.
2 . 3
(viii) So, D AB'C' similar to D ABC with scale factor
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2
Thus, AB'C'D' is a rhombus.
It can be observed that : AB' 2 AC' = = AB 3 AC C'D' AC' = CD AC
=
AD' 2 = AD 3
Therefore, AB' = B'C' = C'D' = AD' =
Also,
2 AB. 3
Long Answer Type Questions Q. 1. Construct a ∆ABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then construct another triangle 2 times the corresponding sides whose sides are 3 of ∆ABC. Give the justification of the construction. A [CBSE, Delhi Set-II, 2020]
Sol. Steps of construction: (i) Draw a line segment BC = 5 cm. (ii) With B as centre and radius = AB = 4 cm, draw an arc. (iii) With C as centre and radius = AC = 6 cm, draw another arc, intersecting the arc drawn in step (ii) at the point A. (iv) Join AB and AC to obtain DABC. (v) Below BC, make an acute angle ∠CBX. (vi) Along BX mark off three points B1, B2, B3 such that BB1 = B1B2 = B2B3. (vii) Join B3C. (viii) From B2, draw B2C’ || B3C. (ix) From C’, draw C’A’ || CA, meeting BA at the point A’. Then A’BC’ is the required triangle. 3 A
A'
C'
B
C
B B B
[CBSE Marking Scheme 2020] 2 X
1
5 marks each
Justification: Consider, BB3 3 BC = ...(i) = BB2 2 BC ' Also, as A'C'||AC, we can say that ∠A'C'B = ∠ACB (Corresponding angles) ...(ii) Now, in DA'B'C' and DABC, ∠B = ∠B(Common) ∠A'C'B = ∠ACB [From (ii)] \ DA'BC' = ∠ABC (By cpct) AC BC AB So, = = A ' C ' BC ' A ' B From eq. (i), we have BC 3 = BC ' 2 AC BC = A ' C ' BC ' AB 3 = = . A'B 2 \
1
Q. 2. Draw a DABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct another triangle whose 3 times the corresponding sides of sides are 4 DABC. Give the justification of the construction. A [CBSE, Delhi Set-III, 2020] Sol. In D ABC, ∠A + ∠B + ∠C = 180° [Angle sum property of a triangle] ⇒ 105° + 45° +∠C = 180° ⇒ ∠C = 30° 1 Steps of Construction: (i) Draw BC = 7 cm. (ii) Construct ∠CBY = 45° and ∠BCZ = 30°. (iii) Rays BY and CZ intersect at A.
224 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
(iv) D ABC is given. (v) From B, draw a ray BX below BC making acute angle with BC. (vi) Along it mark 4 points B1, B2, B3, B4 such that BB1 = B 1 B 2 = B 2 B3 = B 3 B 4 . (vii) Join B4C. Make ∠BB4C at B3 such that the ray intersects BC at C'. \ ∠BB4C = ∠BB3C'. So, B4C || B3C'.
(iv) Below BC make an acute angle CBX. 3 (v) Mark B1, B2, B3, B4 on BX such that BB1 = B1B2 = B 2 B3 = B 3 B 4 . (vi) Join B4 to C. (vii) Draw a line segment B3C' || B4C to meet BC at C'. (viii) Draw line segment C'A' || CA to meet AB at A'. Hence, A' BC' is the required triangle.
(viii) From C', make ∠BC'A' = ∠BCA so that C'A' || CA. Thus, A'BC' is the required triangle. 1
Justification: Here,
2
\
B3C' || B4C BC 3 = BC ' 1
Now,
BC BC '+ C ' C = BC ' BC '
=
\ and So, Hence,
(By Construction)
(i) Draw a line segment BC = 5 cm. (ii) With B as centre and radius 3 cm, draw an arc.
(iv) Join AB and AC to obtain DABC.
1 4 = 3 3
(v) Below BC make an acute angle ∠CBX.
2
(vi) Along BC mark off five points B1, B2, B3, B4, B5 such as BB1 = B1B2 = B2B3 = B3B4 = B4B5. (vii) Join B5C.
C'A' || CA DBC'A' ~ DBCA BC ' BA ' = BC BA C' A' 3 = . CA 4
A [CBSE OD Set-II, 2020]
(iii) With C as a centre and radius 4 cm, draw another arc meeting the previous arc at the point A.
BC ' 3 = BC 4
=
Sol. Steps of construction:
BC ' C ' C + BC ' BC '
= 1 +
Q. 4. Construct a triangle ABC with sides 3 cm, 4 cm and 5 cm. Now, construct another triangle whose sides 4 times the corresponding sides of DABC. are 5
(viii) From B4, draw B4C' || B5C. (ix) From C', draw CA' || CA meeting BA at the point A'. Hence, A' BC' is the required triangle. 1
Q. 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then construct another triangle whose 3 times the corresponding sides of the sides are 4
A [CBSE OD Set-I, 2020] first triangle. Sol. Steps of construction: (i) Draw a line segment BC = 5 cm. (ii) With B and C as centres and radii 7 cm and 6 cm draw two arcs, which intersects at A. (iii) Join BA and CA to obtain DABC.
2
c
[ 225
ONSTRUCTIONS
Justification: Here, \
B4C' || B5C BC ' 4 = BC 5
Now,
BC BC '+ C ' C = BC ' BC '
=
BC ' 4 \ = BC 5
Hence,
BC BC '+ C ' C = BC ' BC '
Now,
=
BC ' C ' C + BC ' BC '
= 1+
BC ' 2 \ = BC 3
BC ' C ' C + BC ' BC '
= 1+
and So,
(By Construction)
1 5 = 4 4
and So, Hence,
C'A' || CA (Given) DBC'A ~ DBCA BC ' BA ' C' A' 4 1 = . = = BC BA CA 5
sides of the DABC.
A [CBSE SQP 2020]
Sol. Correct construction of given triangle. 2 Correct construction of similar D with scale factor 3. [CBSE Marking Scheme, 2020] 3 4
Detailed Solution: See the solution of Q.2. From LATQ. Q. 7. Construct a DABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°. Then construct a triangle whose 3 of the corresponding sides of DABC. sides are 5 A [CBSE Delhi Set-I, 2019]
A [CBSE OD Set-III, 2020] of DABC. Sol. Steps of Construction: (i) Draw a line segment BC = 5 cm. (ii) At point B, draw a line By making an angle of 60°. (iii) With B as centre mark an arc of 6 cm which cuts BY at A. (iv) Join CA. (v) Draw a ray BX making an acute angle with BC. (vi) Locate three points B1, B2, B3 on the line segment BX at equal distance. (vii) Join B3C. (viii) Draw through B2 a line parallel to B3C intersecting line segment BC at C'. (ix) Through C' draw a line parallel to AC intersecting line segment AB at A. 2 Hence, DA'BC' is the required triangle.
C'A' || CA (Given) DBC'A' ~ DBCA BC ' BA ' C' A' 3 1 = . = = BC BA CA 2
Q. 6. Draw a triangle ABC with side BC = 6.5 cm, ÐB = 30°, ÐA = 105°, Then construct another tri3 times the corresponding angle whose sides are 4
Q. 5. Construct a DABC with AB = 6 cm, BC = 5 cm
and ∠B = 60°. Now construct another triangle 2 times the corresponding sides whose sides are 3
1 3 = 2 2
Sol. Correct construction of DABC. 2 Correct construction of triangle similar to triangle ABC. [CBSE Marking Scheme, 2019] 3
Detailed Solution: Steps of Construction: (i) Draw a line segment AB = 5 cm. (ii) At A make ÐBAY = 45° (iii) Take A as a centre and radius AC = 6 cm, draw an arc cutting AY at C.
(iv) Join BC to obtain the triangle ABC.
2
(v) Draw any ray AX making an acute angle with AB on the side apposite to the vertex C. (vi) Mark off 5 points
Say A1, A2, A3, A4, A5 on AX, so that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
(vii) Join A5B (viii) Draw A3B parallel to A5B which meets AB at B'. (ix) From B', draw B'C' || CB meeting AC at C'.
Justification: Here, \
B2C' || B3C BC ' 2 = BC 3
(By Construction)
Triangle AB'C' is the required triangle, each of 3 whose sides is 5 DABC.
th
of the corresponding sides of 3
226 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
(vii) Join B3C (viii) Form B2C' || B3C (ix) Form C', draw C'A' || CA.
Hence A'BC' is the required triangle.
COMMONLY MADE ERROR
Some candidates follow incorrect methods for construction.
2
ANSWERING TIP
of DABC.
A [CBSE Delhi Set-II, 2019]
Sol. Correct construction of DABC 2 Correct construction of triangle similar to DABC 3 [CBSE Marking Scheme, 2019]
Detailed Solution: See the Detailed Solution of Q.2 from LATQ. Q. 9. Construct an equilateral DABC with each side 5 cm. Then construct another triangle whose sides 2 are times the corresponding sides of DABC. 3
Read the construction based questions carefully and requirement.
solve
them
as
per
Q. 10. Draw a DABC with sides 6 cm, 8 cm and 9 cm and then construct a triangle similar to DABC whose 3 of the corresponding sides of DABC. sides are 5 A [CBSE SQP, 2018] [CBSE Comptt. Set-I, II, III, 2018]
Q. 8. Construct a triangle ABC with side BC = 6 cm, ÐB = 45°, ÐA = 105°. Then construct another triangle 3 times the corresponding sides whose sides are 4
Sol. Correct construction of DABC 2 Correct construction of similar triangle. 3 [CBSE Marking Scheme, 2018]
2
2
A [CBSE OD Set-I, 2019]
Sol. Construction of an equilateral triangle of side 5 cm 2 Construction another similar D with scale factor 2 . [CBSE Marking Scheme, 2019] 3 3
(iv) Join AB and AC to obtain DABC. (v) Below BC, construct an acute angle ∠CBX. (vi) Along BX mark off 3 points B1, B2, B3 such that BB1 = B1B2 = B2B3.
Detailed Solution: Steps of Construction: (i) Draw a line segment BC = 5 cm. (ii) With B as a centre and radius BC = 5 cm, draw another arc. 1½ (iii) With C as a centre and radius AC = 5 cm, draw another arc meeting the arc drawn in step 2 at the point A.
Detailed Solution: Steps of Construction: (i) Draw a line segment BC = 8 cm. (ii) With B as a centre and radius 6 cm, draw an arc. (iii) With C as a centre and radius 9 cm, draw another arc meeting the arc drawn in step 2 at the point A. (iv) Join AB and AC to obtain DABC. (v) Below BC make an acute angle ∠CBX. 3 (vi) Along BX mark off five points B1, B2, B3, B4, B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5. (vii) Join B5C. (viii) From B3 draw B3C' || B5C (ix) From C', draw C'A' || CA meeting BA at the point A'. Hence A'BC' in the required triangle.
1½
c
[ 227
ONSTRUCTIONS
Q. 11. Draw a DABC in which BC = 6 cm, AB = 5 cm and ÐABC = 60°. Then construct another triangle whose sides are 3 of the corresponding sides of DABC. A [CBSE Delhi/OD Set-I, 2018] [Delhi Comptt. Set-I, 2017] 4 [Delhi CBSE Board Term-2, 2015]
Sol. Correct construction of DABC Correct construction of D similar to DABC
2 [CBSE Marking Scheme, 2018] 3
Detailed Solution:
Topper Answer, 2017
Q. 12. Draw a right triangle in which sides (other than hypotenuse) are 8 cm and 6 cm. Then construct 3 another triangle whose sides are times the 4
Alternate Method: Steps of Construction: (i) Construct a triangle ABC in which BC = 6 cm. AB = 5 cm and ∠ABC = 60°. (ii) Draw a ray BX such that ∠CBX is an acute angle. (iii) Locate 4 points B1, B2, B3 and B4 on the ray BX such that BB1 = B1B2 = B2B3 = B3B4. (iv) Join B4C. (v) Through B3 draw a line parallel to B4C which meet BC at C'. (vi) Through C' draw a line parallel to AC which meet AB at A'. Hence DA'BC' is the required triangle.
5
(corresponding) sides of given triangle. A [Delhi Comptt. Set-I, 2017] Sol.
X A A'
B
C'
C
B1 B2 B3
B4
Y 2 Steps of Construction: (i) Draw a line segment BC = 8 cm. (ii) Draw line segment BX making an angle of 90° at the point B of BC.
228 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, mathematics (STANDARD), Class – X
Q. 14. Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle 3 with sides are times the corresponding sides of 4
(iii) From B mark an arc on BX at a distance of 6 cm, Let it is A.
(iv) Join A to C. (v) Making an acute angle draw a line segment BY from B.
A [Delhi CBSE Board Term-2, 2015]
(vi) Mark B1, B2, B3 , B4 on BY such that BB1 = B1B2 = B 2 B 3 = B 3 B4 .
the isosceles triangle. Sol. Steps of Construction:
(vii) Join B4 to C.
(i) Draw a line segment BC = 6 cm.
(viii) Draw a line segment B3C' || to B4C to meet BC at C'.
(ii) Draw a perpendicular bisector of BC which intersects the line BC at Q.
(ix) Draw line segment C'A' ||to CA to meet AB at A' A'BC' is the required triangle. 3 [CBSE Marking Scheme, 2017] Q. 13. Construct a right triangle whose hypotenuse and one side measures 10 cm and 8 cm respectively. Then construct another triangle whose sides are 4 times the corresponding sides of this triangle. 5 A [Board Term-2, 2015]
2
Sol. Steps of construction :
(i) Draw a line segment BC = 8 cm. (iii) Taking C as centre and radius as 10 cm, draw an arc that intersects the ray BM at A.
(ii) Construct AM⊥ BC.
(iv) Join CA to obtain ∆ ABC.
(iii) Mark A on the line such that QA = 4 cm.
(v) Below BC, make an acute angle CBX.
(iv) Join A to B and C.
(vi) Along BX mark off 5 points B1, B2, B3, B4, B5 such that BB1 = B1B2 = B2 B3 = ......... = B4B5.
(v) Draw a ray BX making an acute angle with BC. (vi) Mark four points B1, B2, B3 and B4 on the ray BX. Such that BB1 = B1B2 = B2B3 = B3B4. (vii) Join B4C. (viii) Draw a line parallel to B4C through B3 intersecting line segment BC at C’. (ix) Draw C'A' || CA from point C' Hence ∆ A’BC’ is the required triangle.
3
[CBSE Marking Scheme, 2015]
Q. 15. Draw triangle ABC such that BC = 5 cm, ∠ABC = 60°, ∠ACB = 30°. Now construct ∆A'BC' corresponds to ∆ABC with A'B : AB = 3 : 2. A [Board Term-2, 2015]
Sol.
2
(vii) Join B5C. (viii) From B4, Draw B4C’ || B5C.
Hence A’BC’ is the required triangle.
3
[CBSE Marking Scheme, 2015]
(ix) From the point C’ draw C’A’ || CA meeting BA at point A’.
2
Steps of Construction: (i) Draw a line segment BC of length 5 cm. (ii) Draw the angles of 60° and 30° on the points B and C respectively, which intersect each other at A. (iii) ∆ABC is the given triangle. (iv) Draw a ray BX making an acute angle with BC. (v) Locate three points B1, B2, and B3 on line segment BX. Such that BB1 = B1B2 = B2B3.
(vi) Join B2C.
c
[ 229
ONSTRUCTIONS
(vii) Draw B3C' || B2C to intersect the extended line BC at C'. (viii) Through C' draw a line parallel to AC intersecting extended line segment BA at A'.
[CBSE Marking Scheme, 2015]
Visual Case Based Questions Then,
∠ED7D = ∠PD3D
(Corresponding angles)
\
Q. 1. A school conducted Annual Sports Day on a triangular playground. On the ground, parallel lines have been drawn with chalk powder at a distance of 1 m. 7 flower pots have been placed at a distance of 1 m from each other along DM as shown in the figure.
4 marks each
ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark
3
∆A'BC' is the required triangle.
∠ED7D = 82°.
(iii) The ratio in which P divides DE, is: (a) 3 : 4
(b) 7 : 3
(c) 3 : 7
(d) 2 : 5
Sol. Correct Option: (a)
Explanation: P divides DE in the ratio 3 : 4. (iv) The ratio of DE to DP will be: (a) 2 : 5
(b) 3 : 4
(c) 3 : 7
(d) 7 : 3
Sol. Correct Option: (d) Explanation:
DE = 7 m
[Q DD1 = D1D2 = D2D3 = D3D4 Now answer the following questions:
DP = 3 m [Q DD1 = D1D2 = D2D3]
(ii) If ∠PD3D = 82°, then the measure of ∠ED7D is: (a) 98° (b) 82° (c) 90° (d) 45° Sol. Correct Option: (b) Explanation: We have, PD3 || ED7
7m 7 DE = \ = 3m 3 DP Hence, the ratio of DE to DP is 7 : 3.
(i) PD3 is parallel to: (a) PD (b) PE (c) ED7 (d) None of these. Sol. Correct Option: (c) Explanation: In DED7D, PD3 || ED7.
= D4D5 = D5D6 = D6D7] and
(v) The total distance used for putting 7 flower pots is: (a) 6 m
(b) 7 m
(c) 5 m
(d) 8 m.
Sol. Correct Option: (b) Explanation: Since, 7 flower pots have been placed at a distance of 1 m from each other, then total distance = 7 m.
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230 ]
Oswaal CBSE Chapterwise & Topicwise Question Bank, MATHEMATICS (STANDARD), Class – X
MM: 25
Q. 5. If a line segment AB is divided at point P such that PB 3 = , then find the ratio of AP : PB. AB 7
VISUAL CASE BASED QUESTIONS Q. 6. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour.
Q. 2. If the sides of two similar triangles are in the ratio 4 : 9, then find the ratio of the areas of the two triangles.
U
Q. 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
VERY SHORT ANSWER TYPE QUESTIONS: (1 mark each) Q. 1. In the figure given below, 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, find ∠PBA.
Maximum Time: 1 hour
A + U
that OQ = 12 cm. Find length of PQ.
A+ U
(b) 1200
(c) 1500
(d) 1800
(ii) What is the distance travelled by aeroplane towards west after 1½ hours? (a) 1000
(b) 1200
(c) 1500
(d) 1800
(iii) ÐAOB is (a) 90°
(b) 45°
(c) 30°
(d) 60°
(iv) How far apart will be the two planes after 1 2 hours ? (a)
22 , 50 , 000
(c) 54 , 90 , 000
110°, then find ∠ PTQ.
(a) 1000
1
Q. 4. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ =
(i) What is the distance travelled by aeroplane towards north after 1½ hours?
(b)
32 , 40 , 000
(d) none of these
(v) The given problem is based on which concept? (a) Triangles (b) Co-ordinate geometry (c) Height and Distance (d) None of these Q. 7. In Figure, common tangents AB and CD to the two circles with centres O1 and O2 intersect at E. Prove A [CBSE OD, 2014] that AB = CD.
E
2
Q. 8. In the given figure, CB || QR and CA || PR. If AQ = 12 cm, AR = 20 cm, PB = CQ = 15 cm, calculate A [Board Term-I, 2012] PC and BR.
A [Board Term-2, 2014]
Q. 10. In the given figure, D and E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2.
c
[ 231
ONSTRUCTIONS
cm
cm
U [Board Term-1, 2013, LK-59] Q. 11. Draw an isosceles ΔABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ΔABC in which PQ = 8 cm. Also justify the construction.5
Q. 9. In the given figure, OP is equal to the diameter of a circle with centre O and PA and PB are tangents. Prove that ABP is an equilateral triangle.
cm
cm
Finished Solving the Paper ? Time to evaluate yourself !
OR SCAN THE CODE
For elaborate Solutions
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
UNIT 5: Trigonometry
c h a p te r
10
introduction to trigonometry & trigonometric identities
Syllabus Introduction to Trigonometry : Trigonometric ratios of an acute angle of a right-angled
triangle. Proof of their existence (well defined) motivate the ratios, which are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45° and 60°. Relationships between the ratios. Trigonometric Identities : Proof and applications of the identity, sin2 A + cos2 A = 1. Only simple identities to be given. Trigonometric ratios of complementary angles.
Trend Analysis 2018 List of Concepts
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Trigonometric Ratios Complementary Angles
and 3 Q (1 M) 1 Q (3 M)
Trigonometric Identities
TOPIC
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1 Q (4 M)
2019
2020
Delhi
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Trigonometric Ratios and Complementary Angles Revision Notes
TOPIC - 1 In fig., a right triangle ABC right angled at B is given and ∠BAC = q is an acute angle. Here side AB which is adjacent to ∠A is base, side BC opposite to ∠A is perpendicular and the side AC is hypotenuse which is opposite to the right angle B. C
Trigonometric Ratios and Complementary Angles Page No. 233
TOPIC - 2 Trigonometric Identities Page No. 245 A
B
234 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Know the Formulae The trigonometric ratios of ∠A in right triangle ABC are defined as BC Perpendicular or opposite side sine of ∠A = sin θ = = AC Hypotenuse
Scan to know more about this topic
cosine of ∠A = cos θ =
tangent of ∠A = tan θ =
cotangent of ∠A = cot θ =
AB Base or adjecent side = AC Hypotenuse Perpendicular or opposite side BC = Base adjacent side AB
Trigonometric ratios
Base or adjacent side AB 1 = = Perpendicular or oppsite side BC tan θ
Scan to know more about this topic
Hypotenuse AC 1 = = secant of ∠A = sec θ = Base or adjacent side AB cosθ cosecant of ∠A = cosec θ =
Hypotenuse AC 1 = = Perpendicular or opposite side BC sin θ C
Trigonometric ratios with Angles
B
A
It is clear from the above ratios that cosecant, secant and cotangent are the reciprocals of sine, cosine and tangent respectively. sin θ Also, tan θ = cos θ and
cot θ =
cos θ sin θ
The trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and length of its sides. The value of trigonometric ratio of an angle does not depend on the size of the triangle but depends on the angle only. Complementary Angles: Two angles are said to be complementary if their sum is 90°. Thus, (in fig.) ∠A and ∠C are complementary angles. A Scan to know more about this topic
C B Trigonometric Ratios of Complementary Angles: We have, BC = Base, AB = Perpendicular, and AC = Hypotenuse, with respect to q. AB BC AB , cos θ = , tan θ = ∴ sin θ = AC AC BC
and
cosec θ =
AC AC BC , sec q = , cot θ = . AB BC AB
Again, with respect to the angle (90° – q), BC = Perpendicular, AB = Base and AC = Hypotenuse BC = cos θ ∴ sin (90° – θ) = AC
Complementary angles
[ 235
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
AB AC BC tan (90° – θ) = AB AB cot (90° – θ) = BC AC sec (90° – θ) = AB cos (90° – θ) =
cosec (90° – θ) =
= sin θ = cot θ = tan θ = cosec θ
AC = sec θ BC
∠A
0°
30°
45°
60°
90°
sin A
0
1 2
1
3 2
1
cos A
1
3 2
1
1 2
0
tan A
0
cot A
Not defined (∞)
sec A
1
cosec A
Not defined (∞)
2
1 3
3 2
2
1 1
3
2
2
2
Not defined (∞)
3 1
0
3
2
Not defined (∞)
2
1
3
C
Mnemonics
oncept
The relation of Trigonometric Ratios C
In right angled DABC, we have
sin q =
cot q =
BC BA BC , cos q = , tan q = , AC AC AB AC AB AC , cosec q = , sec q = BA BC BC c
Interpretation: Here,
sin q =
Pandit P Perpendicular BC = = = Har H Hypotenuse AC
cos q =
Badri B Base BA = = = Har H Hypotenuse AC
tan q =
Prasad P Perpendicular BC = = = Bhole B Base AB
B
A
236 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Bhole B Base AB = = = Prasad P Perpendicular BC
sec q =
Har H Hypotenuse AC = = = Badri B Base BA
cosec q =
cot q =
T
Har H Hypotenuse AC = = = . Pandit P Perpendicular BC
rigonometric Ratios
Hints: We learn these ratios in following ways:
P H B "Curly Brown Hair" cos q = H
"Some people have" sin q =
"Through proper Brushing" tan q = (i) sin q =
C
Some ↓ sin q
People ↓ Perpendicular
H
A
Curly ↓ cos q
Brown ↓ Base
P
B
B
Hair ↓ Hypotenuse
BC P = AB B Through ↓ tan q
Interpretation:
How is it done on the Q.1. If
Proper ↓ Perpendicular
2 sin q = 1, find the value of
= ( 2)2 – ( 2)2
2 sin q = 1
sin q =
q = 45° Step I: Now, sec2 q - cosec2 q. Step II:
...(ii)
= (sec2 45°)2 – (cosec2 45°)2
Solution Step I: Given
Brushing ↓ Base
GREENBOARD?
sec2 q - cosec2 q.
or,
Have ↓ Hypotenuse
AB B = AC H
Interpretation:
(ii) tan q =
BC P = AC H
Interpretation:
(ii) Cos q =
P . B
1 2
= sin 45°
...(i)
= 2 – 2 = 0
[ 237
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
Very Short Answer Type Questions Q. 1. If sin A + cos B = 1, A = 30° and B is an acute angle, then find the value of B.
R [CBSE SQP, 2020-21]
sin 30° + cos B = 1
1 + cos B = 1 2
Sol.
\ cos B = 1 −
½ 1 1 = 2 2 1 Q cos 60° = 2
∠B = 60°.
½
[CBSE Marking Scheme, 2020-21] cos 80° + cos 59° cosec 31°. Q. 2. Find the value of sin 10° R [CBSE Delhi Set-I, 2020] cos 80 o + cos 59° cosec 31° Sol. o sin 10 1 cos ( 90° − 10° ) + cos (90° – 31°) × ½ = sin 31o sin 10° sin 10 o 1 = + sin 31° × sin 10 o sin 31o 1 [Q cos(90° – q) = sin q and cosec q = ] sin θ = 1 + 1 = 2. ½ 2 2 sin 35° cos 43° + − 2 cos 60°. Q. 3. Find the value of cos 55° sin 47° – 2 cos 60°.
2
R [CBSE Delhi Set-II, 2020]
sin 35° cos 43° Sol. + – 2 cos 60° cos 55° sin 47° 2 cos( 90° − 47°) sin(90° − 55°) + = – 2 cos 60° sin 47° cos 55° 2 2 cos 55° sin 47° – 2 cos 60° = ½ + cos 55° sin 47° [Q sin (90° – q) = cos q and cos (90° – q = sin q)] 1 1 = (1)2 + (1)2 – 2 × Q cos 60° = 2 2
Sol.
2 cos 67° tan 40° – cos 0°. − sin 23° cot 50° R [CBSE Delhi Set-III, 2020]
2 cos 67° tan 40° – cos 0° sin 23° cot 50°
=
2 sin 23° cot 50° – cos 0° sin 23° cot 50°
2 cos( 90° - 23°) tan( 90° - 50°) – cos 0° sin 23° cot 50°
[ cos 0° = 1] ½
= 0.
Q. 5. Find the value of (tan 1° tan 2° tan 3°... tan 89°). U [CBSE OD Delhi Set-I, 2020] Sol. (tan 1° tan 2° tan 3° ... tan 89°) = (tan 1° tan 89°)(tan 2° tan 88°)(tan 3° tan 87°) (tan 44° tan 46°) ... (tan 45°) = [tan 1° tan (90° – 1)][tan 2° tan (90° – 2)] [tan 3° tan (90° – 3)] ... [tan 45° tan (90° – 45°)] = tan 1° cot 1° tan 2° cot 2° tan 3° cot 3° (tan 44° cot 44°) ... tan 45° ½ tan 44° 1 1 1 ... tan 45° tan 2°. tan 3°. = tan 1° × tan 1° tan 2° tan 3° tan 44°
1 1 1 tan 44° tan 1° × tan 2°. tan 3°. ... tan 45° tan 1° tan 2° tan 3° tan 44° = 1.1.1.1....1.1 = 1. ½ Q. 6. If tan A = cot B, then find the value of (A + B). R + U [CBSE OD Set-II, 2020] Sol. tan A = cot B (Given) ⇒ tan A = tan(90° – B) [ tan (90° – q) = cot q] ⇒ A = 90° – B Hence, A + B = 90°. 1 tan 35° cot 78° + . Q. 7. Find the value of cot 55° tan 12° R [CBSE OD Set-III, 2020]
tan( 90° - 55°) cot( 90° - 12°) tan 35° cot 78° Sol. + + = cot 55° tan 12° cot 55° tan 12°
=
cot 55° tan 12° + cot 55° tan 12°
[tan (90° – q) = cot q and cot (90° – q) = tan q] ½
½
= 1 + 1 = 2. Q. 8. If sin a =
3 and cos b = 0, then find the value of 2
b – a.
½
= 1 + 1 – 1 = 1. Q. 4. Find the value of
=
= 2 – 1 – 1
i.e., cos B = cos 60° Hence,
[Q cos (90° – q) = sin q and tan (90° – q) = cot q] ½
1 mark each
U [CBSE SQP, 2020]
Sol. 30°
[CBSE SQP Marking Scheme, 2020] 1
Detailed Solution: 3 2
sin a =
sin a = sin 60° [Q sin 60° =
3 ] 2
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
a = 60° and cos b = 0 cos b = cos 90° [Q cos 90° = 0] ½ b = 90° Now, b – a = 90° – 60° = 30°. ½ Q. 9. Find A, if tan 2A = cot (A – 24°). A [CBSE Delhi Set-I, II, III, 2019] [CBSE Delhi/OD 2018] [Board Term-I, 2016] Sol.
tan 2A = cot(90° – 2A) ½ 90° – 2A = A – 24° ½ A = 38° [CBSE Marking Scheme, 2019]
⇒ Detailed Solution: Given, tan 2A = cot (A – 24°) ⇒ cot (90° – 2A) = cot (A – 24°) ½ [∴ tan q = cot (90° – q)] On comparing angles, we get 90° – 2A = A – 24°
∴ 3A = 90° + 24° ∴ 3A = 114° 114 o = 38° ½ ∴ A = 3 Hence, angle A = 38°. Q. 10. Evaluate: sin2 60° + 2 tan 45° – cos2 30° A [CBSE OD Set-I, II, 2019]
238 ]
Sol. sin2 60° + 2 tan 45° – cos2 30° 2
æ 3ö æ 3ö = ç 2 ÷ + 2(1) - ç 2 ÷ è ø è ø
= 2
2
[For any two correct values] ½ [CBSE Marking Scheme, 2019] ½
2 2 Q. 11. What is the value of (cos 67° – sin 23°) ? U [CBSE Delhi/OD, 2018]
Sol. Q cos2 67° = cos2 (90° – 23°) = sin2 23° \ sin2 23° – sin2 23° = 0 [CBSE Marking Scheme, 2018] 1
opper Answer, 2018
T
Detailed Solution:
1
Q. 12. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
C + U [CBSE Delhi/OD, 2018]
½ ½ [CBSE Marking Scheme, 2018]
opper Answer, 2018
T
Sol. tan 2A = cot (A – 18°) ⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = 36° Detailed Solution:
1
[ 239
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
Q. 14. If sec q.sin q = 0, then find the value of q. R [Board Term-1, 2016]
COMMONLY MADE ERROR
or, or, \
not done and the angles are equated and simplified incorrectly.
ANSWERING TIP convert the tan to cot before equating the angles.
Q. 13. If sin q = cos q, then find the value of 2 tan q U [CBSE SQP, 2018] + cos2 q.
sin q = cos q q = 45° 1 5 = 2tan q + cos2 q = 2 + 2 2 [CBSE Marking Scheme, 2018] 1
Detailed Solution: We have, sin q = cos q We know that, cos q = sin (90° – q) \ sin q = sin (90° – q) ⇒ q = 90° – q ⇒ 2q = 90° ⇒ q = 45°. Now, 2 tan q + cos2 q = 2 tan 45° + cos2 45° 1 = 2 × 1 + 2 = 2 +
1 5 = . 2 2
tan q = 0 = tan 0° q = 0° 1 [CBSE Marking Scheme, 2016]
sin 25° tan 23° Q. 15. Find the value of cos 65° + cot 67° . U [Board Term-1, 2015]
The candidates should remember to
Sol. Given,
sec q.sin q = 0 sin θ = 0 cos θ
Sol. Given,
Generally conversion from tan to cot is
½
2
sin 25° tan 23° sin 25° tan 23° = + + cos ( 90 ° 25 ° ) cot ( 90° - 23°) cos65° cot 67° = 1 + 1 = 2 1 [CBSE Marking Scheme, 2015] Q. 16. If cos 2A = sin (A – 15°), find A. U [Board Term-1, 2015] Sol.
Sol. sin (90° – 2A) = sin(A – 15°) or, 90° – 2A = A – 15° or, 3A = 105° \ A = 35° 1 [CBSE Marking Scheme, 2015] Q. 17. If tan (3x + 30°) = 1, then find the value of x. U [Board Term-1, 2015] Sol. or,
tan(3x + 30°) = 1 = tan 45° 3x + 30° = 45° or, x = 5° 1 [CBSE Marking Scheme, 2015]
Q. 18. What happens to value of cos q when q increases A [Board Term-1, 2015] from 0° to 90° ? ½
Sol. cos q decreases from 1 to 0. 1 [CBSE Marking Scheme, 2015]
Short Answer Type Questions-I Q. 1. If tan A =
3 1 1 , find the value of + . 4 sin A cos A
R [CBSE SQP, 2020-21]
Sol. Given that,
3 3k = 4 4k 3k 3 = sin A = 5k 5
tan A =
4k 4 = 5k 5 1 1 5 5 + = + 3 4 sin A cos A 20 + 15 = 12
cos A =
Detailed Solution: We have,
tan A =
3 4 Perpendicular Base
½
=
½
i.e., perpendicular = 3k and base = 4k. Let ABC be a right angled triangle, then BC = 3k and AB = 4k C
½
35 ½ 12 [CBSE Marking Scheme, 2020-21] =
2 marks each
3k
A
Now
4k B AC2 = AB2 + BC2 (By using Pythagoras theorem)
240 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
= (4k2) + (3k)2 = 25k2 ⇒
½
AC = 5k
Now,
sin A =
BC 3k 3 = = AC 5k 5
and
cos A =
AB 4k 4 = = ½ AC 5k 5
1 1 1 1 5 5 + = + + = 3 4 sin A cos A 3/5 4/5 20 + 15 35 = . = 12 12
½
Hence,
CD CD + AD AC
3 3 41 + = . 2.6 1.5 13
3 sin θ = cos θ
or,
sin θ 1 = cos θ 3
U [CBSE Comptt. Set-I, II, III, 2018]
Sol.
or,
tan θ =
\
3
θ = 30°.
A+B C = 90° – 2 2 cosec æç è
1
Cö A + Bö C æ 1 = cosec ç 90° - ÷ = sec è 2ø 2 ÷ø 2 [CBSE Marking Scheme, 2018]
Detailed Solution: As we know that the sum of interior angles of a triangle is 180°.
1
= tan 30°
A + B + C = 180°
1
½
æ A + Bö C cosec ç = sec è 2 ÷ø 2
3 sin θ – cos θ = 0 and 0° < θ < 90°
or,
=
½
Q. 4. A, B, C are interior angles of DABC. Prove that
½
AC 1.5 15 = = AD 2.6 26
(ii) sec q + cosec q =
Q. 2. If 3 sin θ – cos θ = 0 and 0° < θ < 90°, find the U [Board SQP, 2020-21] value of θ. Sol. Here
tan q =
(i)
= 16k2 + 9k2
\ sin q ù é êQtan q = cos q ú ë û
1
Q. 3. The rod AC of TV disc antenna is fixed at right angles to wall AB and a rod CD is supporting the disc as shown in figure. If AC=1.5 m long and CD = 3 m, find (i) tan q and (ii) sec q + cosec q.
½
A + B + C = 180°
⇒
A + B = 180° – C
Dividing by 2 on both sides, A+B 180 − C = 2 2
= 90° −
C 2
½
Multiplying by cosec on both sides,
C A + B cosec = cosec 90° − 2 2
⇒
C A + B cosec = sec . 2 2
Hence Proved 1
Sol. Given, and
C + R [CBSE OD Set-II, 2020] AC = 1.5 m CD = 3 m 1.5 cm C A
3 tan 2 30° + tan 2 60° + cos ec 30° − tan 45° cot 2 45°
U [Board Term-1, 2016]
Q. 5. Evaluate:
2 2 Sol. 3 tan 30° + tan 60° + co sec 30° - tan 45° cot 2 45° 2
3 cm
D ½ In right angled triangle CAD, AD2 + AC2 = DC2 (Using Pythagoras theorem) ⇒ AD2 + (1.5)2 = (3)2 ⇒ AD2 = 9 – 2.25 = 6.75
⇒
AD =
6.75 = 2.6 m
(Approx) ½
1 3× + ( 3 )2 + 2 − 1 3 = 1 (1)2 3×
1 +3+ 2 −1 3 1
= = 1 + 3 + 2 – 1 = 5 1 [CBSE Marking Scheme, 2016]
[ 241
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
Sometimes students get confused with
1 1 3 3 × + × 2 2 2 2
1
=
ANSWERING TIP
Q. 8. Find cosec 30° and cos 60° geometrically. U [Board Term-1, 2015]
Memorize the values of trigonometric
angles properly and practice more such problems to not to get confused.
A
Sol.
1 ,0≤A+B 2
≤ 90° and A > B, then find A and B. U [Board Term-1, 2016] Sol. Here, or,
sin (A + B) = 1 = sin 90° A + B = 90° 1 = sin 30° sin (A – B) = 2
B
C D Let a triangle ABC with each side equal to 2a. ½ ∠A = ∠B = ∠C = 60° Draw AD perpendicular to BC DBDA ≅ DCDA (by RHS) ½ BD = CD ∠BAD = ∠CAD = 30° (by c.p.c.t) AB 2 a ½ In DBDA, cosec 30° = = =2 BD a BD a 1 = = ½ and cos 60° = AB 2 a 2 [CBSE Marking Scheme, 2015]
...(i)
1
or, A – B = 30° ...(ii) Solving eq. (i) and (ii), A = 60° and B = 30° 1 [CBSE Marking Scheme, 2016]
=
1 3 + 4 4 4 = 1 1 = 4 It is equal to sin 90° = 1 but not equal to cos 90° as cos 90° = 0. [CBSE Marking Scheme, 2016]
the values of trigonometric angles. They substitute wrong values which leads to the wrong result.
Q. 6. If sin (A + B) = 1 and sin (A – B) =
Sol. sin 30° cos 60° + cos 30° sin 60°
COMMONLY MADE ERROR
Q. 7. Find the value of : sin 30°. cos 60° + cos 30°. sin 60° Is it equal to sin 90° or cos 90° ? U [Board Term-1, 2016]
Short Answer Type Questions-II Q. 1. Evaluate: 2
o
2
cos ( 45 + q) + cos ( 45° − q) o
o
tan(60 + q) × tan(30 − q) + (cos 30° + sin 90°) × (tan 60° – sec 0°) A [CBSE SQP 2020]
cos2 ( 45o + q) + cos2 ( 45° − q)
3 marks each
=
cos2 ( 45o + q) + {cos[90° − ( 45° + q)}2 tan( 60 o + q) × tan[90° − ( 60 o + q)] + (cot 30° + sin 90°) × (tan 60° – sec 0°) 1
2
o
cos ( 45 + q) + sin 2 ( 45° + q) = tan( 60 o + q) × cot( 60 o + q)
= 1 + 2 = 3 [CBSE SQP Marking Scheme, 2020] 1
+ (cot 30° + sin 90°) × (tan 60° – sec 0°) ½ 1 = 1 tan( 60° + θ) × tan( 60° + θ) + (cot 30° + sin 90°) × (tan 60° – sec 0°) [Q sin2 q + cos2 q = 1] ½
Detailed Solution:
Sol.
tan( 60 o + q) × tan( 30 o − q)
+ (cot 30° + sin 90°) × (tan 60° – sec 0°) cos ( 45 + q) + sin 2 ( 45° + q) = + ( 3 + 1) ´ ( 3 - 1) 2 tan( 60 o + q) × cot( 60 o + q) 2
o
2
o
2
cos ( 45 + q) + cos ( 45° − q) Given, tan( 60 o + q) × tan( 30 o − q) + (cot 30° + sin 90°) × (tan 60° – sec 0°)
= 1 + ( 3 + 1) × ( 3 − 1)
= 1 + 3 – 1 = 3.
[Q (a + b)(a – b) = a2 – b2] ½ ½
242 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
1
Q. 2. If tan (A + B) = 1 and tan (A – B) =
3
, 0° < A + B < 90°, A > B, then find the values of A and B.
T
R [CBSE Delhi Region, 2019]
opper Answer, 2017
Sol.
Detailed Solution: LHS =
Q. 3. Evaluate: o 2
3 sin 43 cos 37 cosec 53 o o tan 5 tan 25o tan 45° tan 65o tan 85o cos 47 o
o
2
A [CBSE OD Set-I, III, 2019]
æ 3 sin 43o ö cos 37 o cosec 53 o – Sol. ç o ÷ o tan 5 tan 25o tan 45o tan 65o tan 85o è cos 47 ø
æ 3 sin 43o ö = ç cos( 90 o - 43o ) ÷ è ø
3
–
tan 5o tan 25o (1)tan( 90 o - 25o )tan( 90 o - 5o )
1
2
1 = 8 1
2
cos 37 o cosec (90 o - 37 o ) o
tan 5 tan 25o ´ 1 ´ tan( 90 o - 25o )tan( 90 o - 5o )
1
2
cos 37 o cosec (90 o - 37 o )
= 9 -
æ 3 sin 43o ö = ç ÷ è cos( 90 o - 43o ) ø
–
2
æ 3 sin 43o ö cos 37 o sec 37° – = ç o ÷ o tan 5 tan 25o (1)cot 25o cot 5o è sin 43 ø
æ 3 sin 43o ö cos 37 o cosec 53o ç ÷ tan 5o tan 25o tan 45o tan 65o tan 85o è cos 47 o ø
1
[CBSE Marking Scheme, 2019] 1
æ 3 sin 43o ö cos 37 o ´ sec37 o = ç o ÷ o tan 5 tan 25o cot 25o cot 5o è sin 43 ø 1 cos 37 o =9– 1 1 tan 5o tan 25o ´ ´ tan 25o tan 5o 1 = 9- =9–1 1
1
cos 37 o ´
= 8.
½
½
[ 243
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
Alternative Method: Given 4 tan q = 3
4 sin q - cos q + 1 ö Q. 4. If 4 tan q = 3, evaluate æç è 4 sin q + cos q - 1 ÷ø U [CBSE Delhi/OD, 2018]
4 tan q = 3 3 ⇒ tan q = 4 3 4 ⇒ sin q = and cos q = 5 5 3 4 4 ´ - +1 4 sin q - cos q + 1 ö 5 5 \ æç = ÷ 3 4 è 4 sin q + cos q - 1 ø 4 ´ + -1 5 5
tan q =
3 4
tan2 q =
9 16
sec2 q = 1 + tan2 q = 1 +
Sol. Given,
½
1
13 1 11 [CBSE Marking Scheme, 2018] =
Detailed Solution: We have
tan q =
We have,
3 4 Perpendicular Base
=
4 tan θ − 1 + sec θ 4 tan θ + 1 − sec θ
5 8+5 13 4 = = = 5 16 − 5 11 4− 4
Base = 4x
Divide by cos q
2+
Also let ABC be a right angled D. AC =
4 sin θ − cos θ + 1 4 sin θ + cos θ − 1
3−1+
Let perpendicular = 3x and
25 5 = 16 4
5 4 = 5 3+1− 4
4 tan q = 3
⇒ tan q =
sec q =
9 16
COMMONLY MADE ERROR
AB2 + BC 2
(By using Pythagoras theorem) C
Mostly candidates do not find the values of sine and cosine. Some candidates do the wrong calculation.
ANSWERING TIP B
A
AC =
16 x 2 + 9 x 2
=
25x 2 = 5x
\
Then
BC 3x 3 sin q = = = AC 5x 5
and
cos q =
AB 4x 4 = = AC 5x 5
Candidates should find the value of sin q and cos q by using Pythagoras theorem.
Q. 5. If sin (A + 2B) =
1
sin (A + 2B) =
3 ⇒ A + 2B = 60° 1 2
⇒ cos (A + 4B) = 0, ⇒ A + 4B = 90° 1 Solving, we get A = 30° and B = 15° ½+½ [CBSE Marking Scheme, 2018]
8 13 +1 5 = = 5 16 11 −1 5 5 13 . 11
and A + 4B ≤ 90°, then find A and B. C + U [CBSE Comptt. Set-I, II, III, 2018] Sol. Given,
3 4 4× − +1 4 sin θ − cos θ + 1 5 5 \ = 4 cos θ + cos θ − 1 4 × 3 + 4 − 1 5 5
=
3 and cos(A + 4B) = 0, A > B, 2
Detailed Solution: We have sin (A + 2B) = 1
\ ⇒
3 2
sin (A + 2B) = sin 60° A + 2B = 60°
½ ...(i) ½
244 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
and cos (A + 4B) = 0 \ cos (A+ 4B) = cos 90° [Q cos 90° = 0] ½ ⇒ A + 4B = 90° ...(ii) ½ Solving eq. (i) and (ii), we get A = 30° and B = 15°. 1 Q. 6. If in a triangle ABC right angled at B, AB = 6 units and BC = 8 units, then find the value of sin A.cos C U [Board Term-1, 2016] + cos A.sin C. 2
Sol. Here,
2
2
2
AC = (8k) + (6k) = 100k
or,
R Q In the given ∆PQR, right–angled at Q, QR = 9 cm and PR – PQ = 1 cm. Determine the value of
AC = 10k C
P
Q. 7.
U [Board Term-1, 2015]
sin R + cos R.
Sol.
8k
A
B
6k
8k 6 , cos A = 10 k 10
1
6k 8 sin C = , cos C = 10 k 10
1
\ sin A = and
P
Q
PQ2 + QR2 = PR2 (By Pythagoras theorem) or, PQ2 + 92 = PR2 or, PQ2 + 81 = (PQ + 1)2 or, PQ2 + 81 = PQ2 + 1 + 2PQ or, PQ = 40 PR – PQ = 1 (Given) or, PR = 1 + 40 or, PR = 41 40 9 49 + = 3 ∴ sin R + cos R = 41 41 41 [CBSE Marking Scheme, 2015]
\ sin A cos C + cos A sin C 8 8 6 6 × + × 10 10 10 10 64 36 = + 100 100 =
100 = 1. 100
1 [CBSE Marking Scheme, 2016] =
R
Long Answer Type Questions
Q. 1. Evaluate: tan2 30° sin 30° + cos 60° sin2 90° tan2 60° – 2 tan 45° cos2 0° sin 90° R [Board Term-1, 2015] Sol. tan2 30° sin 30° + cos 60° sin2 90° tan2 60 – 2 tan 45° cos2 0° sin 90°
2
1 1 1 2 × + × (1) × = 3 2 2 1
1
( 3)
2
− 2 × 1 × 12 × 1
5 marks each
Sol. Let cot θ = x, then
3 cot2 θ – 4 cot θ +
3 x2 – 3x – x +
or,
(x –
3 =0
3 )( 3 x – 1) = 0
x =
1
= 3 ´ 2 + 2 × 1 × 3 – 2 × 1 × 1 × 1.
or,
1 + 9 - 12 1 3 = + -2 = 6 6 2
∴ If
2 1 = - =6 3
[CBSE Marking Scheme, 2015] 5
2 Q. 2. If 3 cot θ – 4cot θ + 2 of cot θ + tan2 θ.
3 = 0, then find the value U
1
3 x – 4x + 3 = 0
or,
∴
3 = 0 becomes
2
cot θ =
3 or
1 3
1
1
3 or cot θ =
3
θ = 30° or θ = 60° θ = 30°, then
1 cot2 30° + tan2 30° = ( 3 )2 + 3 1 10 = 3 + = 3 3
1
2
1
[ 245
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
If θ = 60°, then
or, 2
Adding eqns. (i) and (ii), we get 2B = 75°
1 cot2 60° + tan2 60° = + ( 3 )2 3 1 10 = +3= 3 3
or,
1
sin (A + B – C ) =
or,
A – C = – 7.5°
...(iii) 1
A + B + C = 180° A + C = 142.5°
...(iv)
Adding eqns. (iii) and (iv),
1 = sin 30° 2
2A = 135°
or,
...(i) 1
A + B – C = 30° 1 and cos ( B + C – A ) = = cos 45° 2
TOPIC
2(A – C) = – 15°
or,
A
or,
1
B = 37.5°
Now subtracting eqn. (ii) from eqn. (i),
Q. 3. In an acute angled triangle ABC, if sin (A + B – C) 1 1 and cos (B + C – A ) = , find ∠A, ∠B = 2 2 and ∠C. Sol. We have
...(ii) 1
B + C – A = 45°
and Hence,
A = 67.5° C = 75° ∠A = 67.5°, ∠B = 37.5° and ∠C = 75° 1
-2
Trigonometric Identities
Revision Notes A
An equation is called an identity if it is true for all values of the variable(s) involved.
An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle.
In ∆ ABC, right-angled at B, By Pythagoras Theorem, AB2 + BC2 = AC2
...(i)
2
Dividing each term of (i) by AC , AB2
AC 2
+
2
or or or
BC 2 AC 2
=
AC 2 AC 2
2
AB BC AC + = AC AC AC
2
Scan to know more about this topic
(cos A)2 + (sin A)2 = 1 cos2 A + sin2 A = 1
...(ii)
This is true for all values of A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity. Now divide eqn.(i) by AB2. AB2
AB2
+
2
or or
BC 2 AB2
=
B
C
AC 2
Trigonometric identities
AB2
2
AB BC AC + = AB AB AB
2
1 + tan2 A = sec2 A...(iii)
Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and sec A are not defined for A = 90°. So, eqn. (iii) is true for all values of A such that 0° ≤ A < 90°. Again dividing eqn. (i) by BC2. AB2 BC
2
+
BC 2 2
BC
=
AC 2 BC 2
246 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X 2
AB BC + BC BC
or or
2
AC = BC
2
cot A + 1 = cosec2 A ...(iv) 2
Note that cosec A and cot A are not defined for all A = 0°. Therefore eqn. (iv) is true for all value of A such that 0° < A ≤ 90°. Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can determine the values of other trigonometric ratios.
How is it done on the Q.1. Prove that.
GREENBOARD? sec θ + tan θ = sec2 θ − tan2 θ
1 sec θ − tan θ =
1 + sin θ cos θ
Solution Step I: L.H.S. =
[ (a + b)(a – b) = a2 – b2] = sec q + tan q [ sec2 q – tan2 q = 1]
1 sec θ − tan θ
Multiplying with sec q + tan q 1 sec θ + tan θ × L.H.S. = sec θ − tan θ sec θ + tan θ
1 sin θ + = cos θ cos θ 1+ sin θ = R.H.S. = cos θ
Very Short Answer Type Questions Q. 1. If x = 2 sin2 q and y = 2 cos2 q + 1, then find the value of x + y. C [CBSE SQP, 2020-21] 2
2
= sin2 q + cos2 q [Q
Sol. x + y = 2 sin q + 2 cos q + 1 ½ = 2(sin2 q + cos2 q) + 1 (As sin2 x + cos2 x = 1) = 3. ½ [CBSE Marking Scheme, 2020-21] Detailed Solution: We have x = 2 sin2 q and y = 2 cos2 q + 1 Then, x + y = 2 sin2 q + 2 cos2 q + 1 ½ = 2(sin2 q + cos2 q) + 1 = 2 × 1 + 1 [Q sin2 q + cos2 q = 1] = 2 + 1 = 3. ½ 1 Q. 2. Find the value of sin 2 θ + 1 + tan 2 θ
U [CBSE Delhi Set-I, 2020]
1 1 = sin2 q + Sol. sin2 q + 2 2 1 + tan q sec q 2 [Q 1 + tan q = sec2 q] ½
1 mark each
1 = cos q] secθ
= 1. [Q sin2 q + cos2 q = 1] ½ Q. 3. Find the value of (1 + tan2 q)(1 – sin q)(1 + sin q). U [CBSE Delhi Set-I, 2020]
Sol. (1 + tan2 q)(1 – sin q)(1+ sin q) [Q 1 + tan2 q = sec2 q] = sec2 q(1 – sin q)(1 + sin q) = sec2 q(1 – sin2 q) [ (a – b)(a + b) = a2 – b2] ½ = sec2 q × cos2 q[Q 1 – sin2 q = cos2 q] 1 1 × cos2 q [Q sec q = ] = cos2 q cosθ = 1. ½ [CBSE Marking Scheme, 2020] Q. 4. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A). C + U [CBSE OD Set-I, 2020] Sol. Given, sin A + sin2 A = 1 ⇒ sin A = 1 – sin2 A = cos2 A ½
[ 247
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
= sin2 32° + cos2 32° = 1. Q. 7. If sin q + cos q =
[sin2 q + cos2 q = 1] ½
2 cos q, (q ¹ 90°) then the value
of tan q is : A [CBSE SQP, 2020] Sol. 2 - 1 [CBSE SQP Marking Scheme, 2020]
On squaring both sides, we get sin2 A = cos4 A ⇒ 1 – cos2 A = cos4 A 2 ⇒ cos A + cos4 A = 1. ½ Q. 5. Find the value of sin 23° cos 67° + cos 23° sin 67°. R [CBSE OD Set-II, 2020] Sol. sin 23° cos 67° + cos 23° sin 67° = sin 23° cos (90° – 23°) + cos 23° sin (90° – 23°) = sin 23° sin 23° + cos 23° cos 23° [Q cos (90° – q) = sin q and sin (90° – q) = cos q] ½ = sin2 23° + cos2 23° = 1. [sin2A + cos2A = 1] ½ Q. 6. Find the value of sin 32° cos 58° + cos 32° sin 58°. R [CBSE OD Set-III, 2020] Sol. sin 32° cos 58° + cos 32° sin 58° = sin 32° cos (90° – 32°) + cos 32° sin (90° – 32°) = sin 32° sin 32° + cos 32° cos 32° [Q cos (90° – q) = sin q and sin (90° – q) = cos q] ½ 5 , find the value of sec a. Q. 8. If tan a = 12
Detailed Solution:
sin q + cos q =
2 cos q
or,
sin q + cos q = cos q
2
or,
sin q cos q + = cos q cos q
2
or,
tan q + 1 =
2
or,
tan q =
½
½
2 – 1.
[CBSE Delhi Region, 2019]
Topper Answer, 2019
Sol.
2 Q. 9. Write the value of cot q –
1 2
sin q U [CBSE SQP, 2018]
cot2 q –
1
= cot2 q – cosec2 q sin 2 q = cot2 q – 1 – cot2 q = –1 1 [CBSE Marking Scheme, 2018] Sol.
1
2 Q. 10. If k + 1 = sec q(1 + sin q)(1 – sin q), then find the value of k. C + U [Board Term-1, 2015]
k + 1 = sec2 q(1 + sin q)(1 – sin q) k + 1 = sec2 q(1 – sin2 q) k + 1 = sec2 q.cos2 q [Q sin2 q + cos2 q = 1] 1 or, k + 1 = sec2 q × sec 2 θ or, k + 1 = 1 ½ or, k = 1 – 1 ½ ∴ k = 0. [CBSE Marking Scheme, 2015] Sol. or, or,
248 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Short Answer Type Questions-I Q. 1. Prove that: 1 +
cot 2 α = cosec a. 1 + cosec α A [CBSE OD Set-I, 2020]
Sol.
L.H.S = 1 +
= 1 +
Q. 4. Prove that: (sin 4 θ + cos 4 θ) = 1 1 - 2 sin 2 θ cos2 θ
cot 2 α 1 + cosec α
Sol.
2
cosec α − 1 1 + cosec α
(1+cosec α )(cosec α − 1) = 1 + 1 1 + cosec α = 1 + cosec a – 1 = cosec a = R.H.S. \ L.H.S = R.H.S. Hence Proved. 1 Q. 2. Show that tan4 q + tan2 q = sec4 q – sec2 q A [CBSE OD Set-I, 2020] Sol. L.H.S. = tan4 q + tan2 q = tan2 q(1 + tan2 q) = tan2 q × sec2 q 1 = (sec2 q – 1)sec2 q = sec4 q – sec2 q = R.H.S. Q L.H.S = R.H.S. Hence Proved. 1 Q. 3. Express the trigonometric ratio of sec A and tan A U [Board Term-1, 2015] in terms of sin A. 1 1 = 1 cos A 1 - sin 2 A sin A sin A = 1 tan A = cos A 1 - sin 2 A [CBSE Marking Scheme, 2015]
Sol.
LHS =
= = =
1 - 2 sin 2 θ cos2 θ (sin 2 θ)2 + (cos2 θ)2 1 - 2 sin 2 θ cos2 θ (sin 2 θ + cos2 θ)2 - 2 sin 2 θ cos2 θ 1 - 2 sin 2 θ cos2 θ 1 - 2 sin 2 θ cos2 θ 1 - 2 sin 2 θ cos2 θ
3 , then prove that
tan q + cot q = 1
Sol. sin q + cos q =
A [CBSE SQP 2020] [CBSE Delhi Set-I, 2020]
2 3 ⇒ (sin q + cos q) = 3
1
Some students make mistakes to prove the sum and become confused.
ANSWERING TIP Follow step by step simplification to avoid errors.
Given
sin q + cos q =
∴
2 (sin q + cos q)2 = ( 3 )
sin2 q + cos2 q + 2 sin q cos q = 3 1 + 2 sin q cos q = 3 2 sin q cos q = 2 sin q cos q = 1
tan q + cot q = =
sin q cos q + cos q sin q
sin 2 q + cos2 q cos q sin q
=
1 cos q sin q
=
1 = 1 1
3
[From equation (i)] 1 Hence Proved
Q. 2. Prove that:
Squaring on both sides,
3 marks each
Detailed Solution:
2
COMMONLY MADE ERROR
⇒ 1 + 2 sin q cos q = 3 ⇒ sin q cos q = 1 1 sin q cos q ∴ tan q + cot q = + = 1 1 cos q sin q [CBSE SQP Marking Scheme, 2020]
= 1 = RHS
Hence Proved [CBSE Marking Scheme, 2015]
Short Answer Type Questions-II Q. 1. sin q + cos q =
A [Board Term-1, 2015]
(sin 4 θ + cos4 θ)
sec A =
and
2 marks each
1
... (i) 1
2(sin6 q + cos6 q) – 3(sin4 q + cos4 q) + 1 = 0.
A [CBSE Delhi Set-II, 2020] Sol. L.H.S. = 2 (sin6 q + cos6 q) – 3 (sin4 q + cos4 q) + 1 = 2 [(sin2 q)3 + (cos2 q)3] – 3 (sin4 q + cos4 q) + 1 ½
[ 249
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
= 2 [(sin2 q + cos2 q)(sin4 q – sin2 q cos2 q + cos4 q] – 3 (sin4 q + cos4 q) + 1 ½ 3 [ a + b3 = (a + b)(a2 – ab + b2)] = 2 (sin4 q – sin2 q cos2 q + cos4 q) – 3 (sin4 q + cos4 q) + 1 [ sin2 q + cos2 q = 1] 1 = – sin4 q – cos4 q – 2sin2 q cos2 q + 1 = – (sin4 q + cos4 q + 2sin2 q cos2 q) + 1 = – (sin2 q + cos2 q)2 + 1 [ (a + b)2 = a2 + b2 + 2ab] ½ = – 1 + 1 = 0 = R.H.S. Hence Proved. ½ 1 + cos θ cot θ + cosec θ − 1 Q. 3. Prove that: = . sin θ cot θ - cosec θ + 1 A [CBSE Delhi Set-III, 2020] cot q + co sec q - 1 Sol. L.H.S. = cot q - co sec q + 1 cos q 1 + -1 sin q sin q = cos q 1 +1 sin q sin q
½
=
sin q(cos q - sin q +1) sin q(cos q + sin q - 1)
=
sin q cos q - sin 2 q + sinq sin q(cos q + sin q - 1)
=
sin q cos q + sin q - (1 - cos2 q) sin q(cos q + sin q - 1)
½
1
sin q(cos q + 1) - [(1 - cos q)(1 + cos q)] ½ = sin q(cos q + sin q - 1) (1 + cos q)(sin q - 1 + cos q) = sin q(cos q + sin q - 1) 1 + cos q sin q Hence Proved. ½
Q. 4. If sin q + cos q =
2 , prove that tan q + cot q = 2.
=
cos2 ( 45° + θ) + cos2 {90° − ( 45° + θ)} tan( 60° + θ)cot{90° − ( 30° − θ)} cos2 ( 45° + θ) + sin 2 ( 45° + θ) tan( 60° + θ)cot( 60° + θ)
1 1
[ cos2q + sin2q = 1 and tan q =
1 ] cot θ
1 1 = 1 = R.H.S. Hence, Proved. 1 Q. 7. Prove that: (sin q + cosec q)2 + (cos q + sec q)2 = 7+tan2 q + cot2 q A [CBSE Delhi Set-I, II, III, 2019]
A [CBSE OD Set-I, 2020] [CBSE SQP, 2017] [Board Term-I, 2015]
Sol. LHS = sin2 q + cosec2 q +2sin q cosec q + cos2 q + sec2 q + 2 cos q sec q 1 2 2 = (sin q + cos q) + cosec2 q + sec2 q 2 sin q cos q +2 + sin q cos q
=
=
= R.H.S.
⇒ cosec2 q + 1 = 3 cot q ⇒ 1 + cot2 q + 1 = 3 cot q 2 ⇒ cot q – 3 cot q + 2 = 0 1 ⇒ cot2 q – 2 cot q – cot q + 2 = 0 ⇒ cot q(cot q – 2) – 1(cot q – 2) = 0 ⇒ (cot q – 2)(cot q – 1) = 0 If cot q = 1 or 2 1 Then, tan q = 1 or . 2 Hence proved. 1 cos2 ( 45° + q) + cos2 ( 45° - q) = 1. Q. 6. Show that : tan(60° + q)tan(30° - q) U [CBSE OD Set-III, 2020] cos2 ( 45° + θ) + cos2 ( 45° − θ) Sol. L.H.S. tan( 60° + θ)tan( 30° − θ)
=
Q. 5. If 1 + sin2 q = 3 sin q cos q, prove that tan q = 1 1 or . A [CBSE OD Set-II, 2020] 2 Sol. Given, 1 + sin2 q = 3 sin q cos q On dividing by sin2 q on both sides, we get 1 + 1 = 3 cot q 1 2 sin θ cos q ù é êQ cot q = sin q ú û ë
Sol. Given,
sin q + cos q =
2
On squaring both the sides, we get
(sin q + cos q)2 = ( 2 )2
⇒ sin2 q + cos2 q + 2 sin q cos q = 2 ⇒ 1 + 2 sin q cos q = 2 ⇒ 2 sin q cos q = 2 – 1 = 1 1 ⇒ =2 ...(i) 1 sin θ cos θ Now,
tan q + cot q =
Q. 8. If sec q = x +
sin θ cos θ + cos θ sin θ
sin 2 q + cos2 q 1 = cos q sin q cos θ sin θ
From (i) and (ii) we get tan q + cot q = 2
= 1 + 1 + cot2 q + 1 + tan2 q + 2 + 2 1½ = 7 + cot2 q + tan2 q = RHS Hence Proved ½ [CBSE Marking Scheme, 2019]
1 . 2x ...(ii) 1
1
1 , prove that sec q + tan q = 2x or 4x A [CBSE Delhi Set-III, 2019]
1 4x
Sol.
sec q = x +
2 sec2 q = x +
1 1 + 2·x 4x 16 x 2
250 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
1 1 + 16 x 2 2
2 1 + tan2 q = x +
2 tan2 q = x +
2 tan2 q = x +
2 tan2 q = x +
1ö æ tan2 q = ç x - ÷ è 4x ø
1 16 x
2
1 16 x 2
+
1 -1 2
-
1 2
æ sin 2 q + cos2 q ö = (cos q + sin q) ç ÷ è cos q sin q ø
1 1 − 2·x· 4x 16 x 2
1 tan q = ± x − 4x
If
tan q = x −
1 4x
Given, Now,
sec q = x +
1 4x
Q. 9. Prove that: cot q – tan q =
cos θ + sin θ cos3 θ - sin 3 θ + cos θ + sin θ cos θ - sin θ (cos θ + sin θ)(cos2 θ + sin 2 θ - sin θ cos θ) (cos θ + sin θ) (cos θ - sin θ)(cos2 θ + sin 2 θ + sin θ cos θ) 1 (cos θ - sin θ) 1
= 2 = RHS Hence proved. Q . 12. If bcos q = a, then prove that cosec q + cot q b+a . b−a
=
U [CBSE Board Term-1, 2015]
A
Sol.
2 cos2 q - 1 sin q cos q
LHS = cot q – tan q cos q sin q = sin q cos q Sol.
cos2 q - sin 2 q = sin q cos q
C
B
1
Given,
½
1
cos q =
a b
AC2 = AB2 – BC2
AC =
b2 − a2 k
cosec q =
cosec q + cot q =
2
cos q - 1 + cos q sin q cos q
½
Q. 10. Prove that: sin q (1 + tan q) + cos q (1 + cot q) = sec q + cosecq U [CBSE SQP, 2018] Sol. LHS = sin q(1 + tan q) + cos q(1 + cot q)
1
sin q ö cos q ö æ æ + cos q ç 1 + = sin q ç 1 + è è cos q ÷ø sin q ÷ø
æ cos q + sin q ö æ sin q + cos q ö + cos q ç = sin q ç 1 ÷ è ø è ø÷ cos q sin q
b -a
2
b+a
, cot q = =
a 2
b - a2
2
b -a
2
b+a b-a
1 − cos θ Q. 13. Prove that: (cot q – cosec q)2 = 1 + cos θ U [CBSE Board Term-1, 2015]
2 cos q - 1 = = RHS Hence Proved sin q cos q [CBSE Marking Scheme, 2018]
b 2
3 [CBSE Marking Scheme, 2015]
2
3
= 2 – sin q cos q + sin q cos q1
U [CBSE SQP, 2018]
=
3
= (1 – sin q cos q) + (1 + sin q cos q)
1 1 1 1 + = 4 x 4 x 2x Hence Proved.
2
=
+
tan q + sec q = 2x
sec q + tan q =
Sol. LHS =
1 1 If tan q = − x − = − x + 1 4x 4x 1 Given, sec q = x + 4x Now,
U [CBSE Board Term-1, 2015]
1
Taking square root on both sides
cos q + sin q = cosec q + sec q = RHS 1 cos q sin q Hence Proved [CBSE Marking Scheme, 2018]
3 3 3 3 Q. 11. Prove that: cos θ + sin θ + cos θ - sin θ = 2. cos θ + sin θ cos θ - sin θ
2
=
Sol.
LHS = (cot q – cosec q)2
1 cos θ = θ θ sin sin cos θ - 1 = sin θ
2
2
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INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
=
=
(1 - cos θ)2
COMMONLY MADE ERROR
(Q sin2 θ + cos2 θ = 1)
sin 2 θ
Some students tried to prove the identity
(1 - cos θ)2
by getting
2
(1 - cos θ)
=
1 - cos θ 1 + cos θ Hence proved.
Students must be advised not to change the form of a given identity.
[CBSE Marking Scheme, 2015] 3
Long Answer Type Questions sin A - cos A + 1 1 = sin A + cos A - 1 sec A - tan A A [CBSE Delhi Set-I, 2019]
Sol. LHS =
sin A − cos A + 1 sin A + cos A − 1
Dividing num. & deno. by cos A, tan A − 1 + sec A = tan A + 1 − sec A =
2
2
(tan A − sec A ) + (sec A − tan A ) tan A − 1 + sec A = (tan A − sec A ) + (1 − sec A − tan A )
= = = =
=
=
1 = RHS sec A - tan A
tan 2 A
1
2
Detailed Answer : sin A - cos A + 1 LHS = sin A + cos A - 1
sec 2 A - tan 2 A sec A - tan A
2
tan A - 1
+
c osec 2 A
1
sin A + cos A − 1 + sin 2 A + cos A sin A − sin A
(sin A − cos A + 1)(1 + sin A ) 2
−1 + cos A + (1 − cos A ) + sin A cos A
1
(sin A − cos A + 1)(1 + sin A ) cos A − cos A + sin A cos A
1 + sin A 1 sin A = + cos A cos A cos A
2
=
1 1 - 2 cos2 A
Sol.
sin 2 A 1 2 2 cos A sin A LHS = + 2 1 1 sin A − −1 cos2 A sin 2 A cos2 A sin 2 A sin 2 A − cos2 A 1
+
cos2 A sin 2 A − cos2 A
1
1
1½ sin 2 A − cos2 A 1 Hence Proved 1½ = 1 − 2 cos2 A [CBSE Marking Scheme, 2019] Q. 3. Prove that:
sin θ sin θ =2+ cot θ + cosec θ cot θ − cosec θ A [CBSE OD, Set-I, 2019]
2
(sin A − cos A + 1)(1 + sin A ) cos A(1 − cos A + sin A )
Hence Proved. 1
A [CBSE Delhi Set-II, 2019]
=
(sin A − cos A + 1)(1 + sin A )
2
sec A - cosec A
=
sin A - cos A + 1 1 + sin A ´ sin A + cos A - 1 1 + sin A
1
Q. 2. Prove that:
1 −1 = RHS 1 = = tan A − sec A sec A − tan A Hence Proved [CBSE Marking Scheme, 2019]
=
=
1
tan A − 1 + sec A
5 marks each
= sec A + tan A (sec A + tan A ) ´ (sec A - tan A ) = (sec A - tan A )
Q. 1. Prove that
1 − cos θ 1 + cos θ
ANSWERING TIP
= RHS
cot q – cosec q =
(1 - cos θ)(1 - cos θ) = (1 - cos θ)(1 + cos θ)
1
Sol. No sequence sin θ sin θ − = cosec θ + cot θ cot θ − cosec θ sin θ sin θ + = cosec θ + cot θ cosec θ − cot θ
252 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
=
=
sin θ[cosec θ − cot θ + cosec θ + cot θ] cosec 2 θ − cot 2 θ
1+1
sin θ ( 2cosec θ) 1
= 2
1
[CBSE Marking Scheme, 2019]
Q. 4. Prove that: 1 1 1 1 + + + =2 2 2 2 2 1 + sin θ 1 + cos θ 1 + sec θ 1 + cosec θ T
2
sin θ sin θ Hence =2+ cot − cosec θ θ cosec θ + cot θ
A [CBSE Delhi, Region, 2019]
opper Answer, 2019
Sol.
Q. 5. Prove that:
sin A - 2 sin 3 A 3
2 cos A - cos A
= tan A.
=
U [CBSE Delhi/OD, 2018] [Board Term-I, 2015]
Sol.
LHS = =
3
sin A - 2 sin A 2 cos3 A - cos A
cos A( 2 cos2 A - 1)
1
cos A( 2 cos2 A - 1)
1
2
= tanA = RHS
sin A(1 - 2 sin 2 A )
T
sin A(1 - 2(1 - cos2 A ))
= tan A ( 2 cos A - 1) ( 2 cos2 A - 1)
Detailed Solution:
Sol.
5
opper Answer, 2018
1½
Hence Proved 1½
[CBSE Marking Scheme, 2018]
[ 253
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
5
(1 + p2)sin2 q + 2 sin q + (1 – p2) = 0 D = 4 – 4(1 + p2)(1 – p2) = 4 – 4(1 – p4) = 4p4
COMMONLY MADE ERROR Some common errors observed were:
(i) Working with both sides together. (ii) Skipping of necessary steps so as to get the answer. (iii) Some opened the LHS expression but failed to simplify and come to the RHS.
ANSWERING TIP Ensure that while proving identities
students proceed with either LHS or RHS but must not work with both sides simultaneously.
Q. 6. If sec q + tan q = p, then find the value of cosec q. U [CBSE SQP-2018]
Sol. sec q + tan q = p
1 sin q + = cos q cos q = 1 + sin q = pcos q
(1 + sin q)2 = p2(1 – sin2 q)
½
2
2
2
2
1 + sin q + 2 sin q = p – p sin q
=
p2 - 1 p2 + 1
1
½
,–1
p2 + 1 , – 1 Hence Proved 1 \ cosec q = 2 p -1 [CBSE Marking Scheme, 2018] Alternative Method: Given sec q + tan q = p...(i) we know sec2 q – tan2 q = 1 (sec q + tan q)(sec q – tan q) = 1 p(sec q – tan q) = 1 1 \ sec q – tan q = ...(ii) p Add (i) and (ii) we get
1
-2 ± 4 p -1 ± p 2 = sin q = 2 2(1 + p ) (1 + p 2 )
p 1 - sin 2 q =
1
4
2 sec q = p + sec q =
1 p2 + 1 = p p
p2 + 1 2p
Subtract (ii) from (i) we get
2 tan q = p −
1 p2 − 1 = p p
254 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
p2 − 1 2p
1 1 θ cos = cosec q = sin θ sin θ cos θ
cosec q =
=
\ cosec q = Q. 7. Prove that:
=
p2 − 1 2p .
cos θ × sin θ sin θ
cosec q + cot q = p...(ii) from eq. (i) and (ii)
cos θ − sin θ + 1 = cosec θ + cot θ cos θ + sin θ − 1
cosec q =
p2 + 1 2p
and
cot q =
p2 − 1 2p
cos q =
cos θ × sin θ sin θ
[CBSE SQP, 2017-18] cos θ − sin θ + 1 Sol. LHS = cos θ + sin θ − 1 = =
sin θ(cos θ − sin θ + 1) sin θ(cos θ + sin θ − 1)
1
= cot θ ×
2
sin θ cos θ − sin θ + sin θ sin θ(cos θ + sin θ − 1)
sin θ cos θ + sin θ − (1 − cos2 θ) = sin θ(cos θ + sin θ − 1)
1
sin θ(cos θ + 1) − [(1 − cos θ)(1 + cos θ)] = sin θ(cos θ + sin θ − 1)
=
(1 + cos θ)(sin θ − 1 + cos θ) sin θ(cos θ + sin θ − 1)
=
(1 + cos θ)(cos θ + sin θ − 1) 1 + cos θ = sin θ(cos θ + sin θ − 1) sin θ
=
1 cos θ + sin θ sin θ
1
Hence Proved 1
cos θ =
p2 - 1 . p2 + 1
U [Board Term-1, 2016]
p2 - 1 Sol. RHS = 2 p +1
( cosec θ + cot θ)2 - 1 = 2 ( cosec θ + cot θ) + 1 cosec 2 θ + cot 2 θ + 2 cosec θ cot θ − 1 = cosec 2 θ + cot 2 θ + 2 cosec θ cot θ + 1
p2 − 1 2p × 2 2p p +1 p2 − 1 p2 + 1
Hence, find the values of cos q and sin q. A [Board Term-1, 2015] 1 1 (sec θ - tan θ) 1 = × p sec θ + tan θ sec θ - tan θ 1 sec θ - tan θ = sec q – tan q 1 = 2 p sec θ - tan 2 θ 1 Solving, sec q + tan q = p and sec q – tan q = p Sol.
We get sec q = 1 and
tan q =
1
1 + cot 2 θ + cot 2 θ + 2 cosec θ cot θ − 1 = 1 cosec 2 θ + cosec 2 θ − 1 + 2 cosec θ cot θ + 1
1 cosec θ
Hence Proved 1 Q . 9. If sec q + tan q = p, show that sec q – tan q = p .
Q. 8. If cosec θ + cot θ = p, then prove that
= \ cos q =
1
= cosec q + cot q = RHS
1
Alternative Method: cosec q + cot q = p(given) We know cosec2 q – cot2 q = 1 (cosec q + cot q)(cosec q – cot q) = 1 p(cosec q – cot q) = 1 1 \ cosec q – cot q = ...(i) p
p2 + 1 2p
p2 − 1
2 cot θ (cot θ + cosec θ) 2 cosec θ ( cosec θ + cot θ)
= cos θ = LHS. Hence proved. 1 [CBSE Marking Scheme, 2016]
sec θ tan θ
p2 + 1
=
\ tan q =
\ cos q =
1 1 p2 + 1 p + = 2 p 2p
1
1 1 p2 - 1 p - = 2 p 2p
1
2p
and sin q =
p2 - 1
1 [CBSE Marking Scheme, 2015] 2
p +1
p2 + 1
[ 255
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES 2 2 2 2 2 2 Q. 10. Prove that b x – a y = a b , if : (i) x = a sec q, y = b tan q, or
(ii) x = a cosec q, y = b cot q. U [Board Term-1, 2015]
x2
Sol. (i)
Q. 11. If cosec q – cot q = 2 cot q, then prove that cosec q + cot q = 2 cosec q.
(ii)
or,
a2 x2
\
a
2
-
or,
\
b2
y2
x
2
a2 x2 a
2
-
y2 b2
= tan2 q
b2
½
= sec2 q – tan2 q = 1.
b2x2 – a2y2 = a2b2.
y2
= sec2 q,
Hence Proved. 2
= cosec2 q,
y2 b2
= cot2 q
½
= cosec2 q – cot2 q = 1
b2x2 – a2y2 = a2b2
Hence Proved. 2
U [Board Term-1, 2015]
Sol. Given, cosec q – cot q =
Squaring both the sides, cosec2 θ + cot2 q – 2 cosec q cot q = 2 cot2 q 1½ or, cosec2 θ – cot2 θ = 2 cosec θ cot θ [ a2 – b2 = (a + b)(a – b)] or, (cosec q + cot q)(cosec q – cot q) = 2 cosec θ cot θ 1 (cosec q – cot q = 2 cot q) Given : 2 cosec θ cot θ 1 or, cosec q + cot q = 2 cot θ
cosec q + cot q =
1½ 2 cosec q Hence Proved. [CBSE Marking Scheme, 2015]
Visual Case Based Questions Note: Attempt any four sub parts from each question. Each sub part carries 1 mark Q. 1. 'Skysails' is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively. Based on the following figure related to sky sailing C + AE answer the questions:
A Rope
C
B
(i) In the given figure, if tan q = cot (30° + q), where q and 30° + q are acute angles, then the value of q is: (a) 45° (b) 30° (c) 60° (d) None of these Sol. Correct option: (b). Explanation: Given, tan q = cot(30° + q) = tan[90° – (30° + q)] = tan(90° – 30° – q) ⇒ tan q = tan(60° – q) ⇒ q = 60° – q
2 cot q
4 marks each ⇒ 2q = 60° ⇒ q = 30°. (ii) The value of tan 30°. cot 60° is: 1 (b) (a) 3 3 1 (c) 1 (d) 3
1
Sol. Correct option: (d). Explanation:
1
tan 30° × cot 60° =
=
3
×
1 3
1 . 3
1
(iii) What should be the length of the rope of the kite sail in order to pull the ship at the angle q and be at a vertical height of 200 m ? (a) 400 m (b) 300 m (c) 100 m (d) 200 m Sol. Correct option: (a). Explanation: In DABC, we have q = 30°, AB = 200 m Perpendicular AB Then, sin 30° = = Hypotenuse AC ⇒ ⇒
1 200 = 2 AC AC = 400 m.
1 2 (iv) If cos A = , then the value of 9 cot A – 1 is : 2 (a) 1 (c) 2 Sol. Correct option: (c).
(b) 3 (d) 4
1
256 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Explanation: Given, cos A =
1 2
⇒ cos A = cos 60° ⇒
A = 60°
Then,
9 cot2 A – 1 = 9(cot 60°)2 – 1
⇒ AB = 8 m. 2 2 (ii) The value of sin 30° + cos 60° is: 1 1 (b) (a) 4 2
2
1 = 9 −1 3 = 9 ×
3 4 Sol. Correct option: (b). Explanation:
1 −1 = 3 – 1 3 1
= 2
(v) In the given figure, the value of (sin C + cos A) is: (a) 1
(b) 2
(c) 3
(d) 4
Sol. Correct option: (a). Explanation: We have,
AB = 200 m and AC = 400 m
[Proved in Q.3]
Then, sin C + cos A =
AB AB + AC AC
= 2 ×
AB AC
= 2 ×
200 = 1 400
Based on the following figure related to the slide C + AE answer the questions: A
1 1 sin2 30° + cos2 60° = + 2 2
=
1 1 + 4 4
=
2 1 = . 4 2
(iii) If cos A =
B
C
(a) 8 m
(b) 6 m
(c) 5 m
(d) 10 m
Sol. Correct option: (a). Explanation: We have, ∠B = 30° and
Then,
1
1 , then the value of 12 cot2 A – 2 is: 2 (b) 4 (d) 2 1 2
⇒ cos A = cos 60° ⇒ A = 60° 2 Then 12 cot A – 2 = 12(cot 60°) – 2 2
1 = 12 −2 3 = 12 ×
1 −2 3
= 4 – 2 = 2. 1 (iv) In the given figure, the value of (sin C × cos A) is: 1 1 (b) (a) 3 2 (c)
sin C × cos A = sin 90° ×
(i) The distance of AB is :
2
1 1 (d) 4 5 Sol. Correct option: (b). Explanation: Since, AC ^ BC, then ∠C = 90°
30°
3 2
(d)
Explanation: since, cos A = 1
1
2
(c)
(a) 5 (c) 3 Sol. Correct option: (d).
Q. 2. Authority wants to construct a slide in a city park for children. The slide was to be constructed for children below the age of 12 years. Authority prefers the top of the slide at a height of 4 m above the ground and inclined at an angle of 30° to the ground.
1 4 = 2 AB
⇒
AC = 4 m sin 30° =
AC AB
= 1 ×
AC AB
4 8
1 = 1 2 (v) In the given figure, if AB + BC = 25 cm and AC = 5 cm, then the value of BC is: (a) 25 cm (b) 15 cm (c) 10 cm (d) 12 cm Sol. Correct option: (d).
[ 257
INTRODUCTION TO TRIGONOMETRY AND TRIGONOMETRIC IDENTITIES
O
Explanation: We have, ∠C = 90° AB + BC = 25 cm and AC = 5 cm Let BC be x cm, then AB = (25 – x) cm By using Pythagoras theorem, AB2 = BC2 + AC2 ⇒ (25 – x2) = x2 + (5)2
⇒ 625 – 50x + x2 = x2 + 25 ⇒ 50x = 600 ⇒ Hence,
x =
600 = 12 50
BC = 12 cm.
1
swaal Learning Tool
To learn from Oswaal Concept Video Visit : http:qrgo.page.link OR Scan this code
258 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
c h a p te r
11
Heights and distances
Syllabus Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression 30°, 45°, 60° only.
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Revision Notes
The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of elevation of a point on the object being viewed is the angle formed by the line of sight with the horizontal when it is above the horizontal level, i.e., the case when we raise our head to look at a point on the object. Now, we may identify line of sight, angles and altitude (height). B (Object)
Scan to know more about this topic
Line of sight
Angle of elevation
O(Observer)
Horizontal
A
Height and Distances
(i) ∠AOB is the angle of elevation.
(ii) The height AB, it means object is at point B from the ground at point A. (iii) AO is the distance between the observer from the point A. The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when it is below the horizontal level, i.e., the case when we lower our head to look at a point on the object.
260 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X (Observer)
Horizontal Angle of depression
Lin
eo
f si
ght (Object)
The height of object above the water surface is equal to the depth of its image below the water surface. The values of the trigonometric ratios of an angle do not vary with the length of the sides of the triangle, if the angles remain the same.
How is it done on the
GREENBOARD?
Q.1. The angle of elevation of the top of a tower from a point 50 m away from the base of the tower is 45°. The angle of elevation of the top of the flag mounted on the tower is 60°. U Find the height of the flag. Solution: Step I: Let BC be the tower, CD be the flag and A is the point of observation. D
C
ABC,
ÐB = 90°
BC 50 BC ⇒ 1 = 50 ⇒ BC = 50 metre Step III: Calculation of height BD. In DABD, ÐB = 90° BD tan 60° = 50 BD ⇒ 3 = 50
tan 45° =
⇒ BD = 50 3 metre Step IV: Calculation for height of flag Height of flag CD = BD – BC = 50 3 − 50
(
)
⇒ CD = 50 3 − 1 B A 50 m = 50 (1.732 – 1) Step II: First we will calculate the = 50 × 0.732 height of tower BC using triangle ⇒ Height of flag = 36.6 metre 60°
45°
Very Short Answer Type Questions Q. 1. The ratio of the length of a vertical rod and the length its shadow is 1 : 3 . Find the angle of elevation of the sun at that moment ? A [CBSE Delhi Set-I, 2020] Sol. Let AB be a vertical rod and BC be its shadow. From the figure, ∠ACB = q. In DABC,
1 mark each A
C
B
[ 261
Heights and Distances
AB BC
½
1 AB (Given) = ∵ 3 3 BC ⇒ tan q = tan 30° ⇒ q = 30° Hence, the angle of elevation of the sun is 30°. ½
⇒ tan q =
1
\ ∠AO1X = 90° – 60° = 30° and ∠AO2X = ∠BAO2 = 45°. ½ Q. 3. 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 ? U [CBSE Term-2, 2016] [CBSE Delhi Set-I, II, 2017]
tan q =
Sol.
C
Q. 2. In the given figure, find the angles of depressions from the observing positions O1 and O2 respectively of the object A. A [CBSE OD Set-I, 2020] O1
O2
q
60°
A
B
45°
A
Sol.
Let the height of tower be AB and its shadow be BC. BC = tan q \ AB
C
B O2
X
O1 60°
A
45° C
B
AC || O1X
Draw
3 1 = tan 60° Hence, the angle of elevation of Sun = 60°. 1 [CBSE Marking Scheme, 2017] =
½
Q. 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 ? U [CBSE OD & Comptt. OD Set-I, II, III 2017] [Foreign Set-I, II, III, 2015]
Topper Answer, 2017 Sol.
1 Q. 5. In the given figure, AB is a 6 m high pole and DC is a ladder inclined at an angle of 60° to the horizontal and reaches up to point D of pole. If AD = 2.54 m, find the length of the ladder. (use 3 = 1.73)
Sol. Given, \ In DBCD,
A
D
\ La
dd
AD = 2.54 m DB = 6 – 2.54 = 3.46 m ÐB = 90° BD sin 60° = DC 3 3.46 = 2 DC 3.46 × 2 3.46 × 2 DC = = =4m 1.73 3
\ Length of ladder = 4 m.
er
1
Q. 6. An observer, 1.7 m tall, is 20 3 m away from a
B
C
C + A [CBSE Delhi Set I, II, III, 2016]
tower. The angle of elevation from the eye of observer to the top of tower is 30°. Find the height A [Foreign Set I, II, III, 2016] of the tower.
262 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
A
Sol.
(h – 1.7)
E
1.7 m B
20 3
Let height of the tower AB be h metre ⇒ AE = h – 1.7
BC = DE = 20 3 m. (Given) ÐE = 90° h - 1.7 tan 30° = 20 3
In ∆ADE,
1
⇒
tan q =
⇒
C
tan q =
h
30°
D
Sol. Let the ∠ACB be q, ÐB = 90° AB BC 20 20 3
=
1 3
= tan 30° 1
q = 30°
Thus, the sun's altitude is 30°. [CBSE Marking Scheme, 2015]
Q. 8 If the length of the ladder placed against a wall is twice the distance between the foot of the ladder and the wall. Find the angle made by the ladder A [CBSE S.A.-2, 2015]
with the horizontal.
Sol. Let the distance between the foot of the ladder and the wall, AB be x.
h - 1.7
then AC, the length of the ladder = 2x C
= 20 3 3 ⇒ h – 1.7 = 20 or h = 20 + 1.7 = 21.7 m 1 Q. 7. In figure, a tower AB is 20 m high and BC, its
2x
er
Wall
La
dd
shadow on the ground, is 20 3 m long. Find the sun's altitude. A
C
B 20 3 C + A [CBSE OD Set-I, II, III, 2015]
A
B
x
ÐB = 90°
In ∆ABC,
cos A =
x 2x
⇒
cos A =
1 = cos 60° ⇒ A = 60° 1 2
Short Answer Type Questions-I
2 marks each
Based on the following figure related to sky sailing, answer the questions : (i) In the given figure, if sin q = cos (3q – 30°), where q and 3q – 30° are acute angles, then find the value of q.
Q. 1. 'Sky sails' is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The 'Sky sails' technology allows the towing kite to gain a height of anything between 100 metres – 300 metres. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of 'telescopic mast' that enables the kite to be raised properly and effectively. (ii) What should be the length of the rope of the kite sail in order to pull the ship at the angle q (calculated above) and be at a vertical height of 200 m ? C [CBSE SQP, 2020-21]
[ 263
Heights and Distances
Sol.
Sol. (i) cos (90° – q) = cos (3q – 30°) ⇒ 90° – q = 3q – 30° ⇒ q = 30° 1 AB (ii) = sin 30° AC 200 1 = AC 2 \ Length of rope = AC = 400 m 1 [CBSE SQP Marking Scheme, 2020] Q. 2. From the top of a 7 m high building the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. C + A [CBSE Delhi Set-I, II, III, 2017] Sol. Let AB be building = 7 m C
h
A
X
E
Tower
7m
7m
Building
B
D
X
CD be the height of tower = (7 + h) m Let the distance BD be x m. Then BD = AE = x m AB = tan 45° Now in DABD BD
h = x
½
D z
z
M
Let M be the centre of the line joining their feet. Let BM = MD = z perpendicular \ tan θ = base In DABM, x \ = tan 30° z 1 ⇒ x = z × ...(i) ½ 3 In DCDM, y = tan 60° z y = z × 3 ...(ii) ½ 1 z× x 3 = From (i) and (ii), y z× 3
x 1 = \ y 3 x : y = 1 : 3 1 [CBSE Marking Scheme, 2015] Q. 4. From the top of light house, 40 m above the water, the angle of depression of a small boat is 60°. Find how far the boat is from the base of the light U [CBSE Board Term-2, 2015] house.
Sol. Let AB be the light house and C be the position of the boat. Since, ∠PAC = 60° ∴ ∠ACB = 60° 1 Let BC be x m. AB In ∆ABC, = tan 60° BC 40 ⇒ = 3 x
7 = 1 Þ x Þ x = 7 m CE = tan 60° In DCEA AE
½
y x
½
3
Þ h = x 3 Substituting the value of x we get
h = 7 3
Ligh thouse
CD = CE + ED
(
)
= 7 + 7 3 m
(
)
Hence, the height of tower = 7 1 + 3 m.
½
[CBSE Marking Scheme, 2017] Q. 3. The tops of two towers of height x and y, standing on the ground, subtend the angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y.
A [CBSE Delhi Set-I, II, III, 2015]
⇒
x =
40 3
×
3 3
40 3 m = 3 40 3 Hence, the boat is m away from the foot of 3 the light house. [CBSE Marking Scheme, 2015] 1
264 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Short Answer Type Questions-II
Q. 1.
Sol.
3 marks each
½ In DABD, ÐADB = ÐDBY = 45° (alternate angles) and in DABC, ÐBCA = ÐXBC = 60° AB = tan 45° AD
If the angles of elevation of the top of the candle from two coins distant ‘a’ cm and ‘b’ cm (a > b) from its base and in the same straight line from it are 30˚ and 60˚, then find the height of the candle. C + A [CBSE SQP, 2020-21]
Candle m
½
Let AB = candle C and D are two Points. AB h = In DACB, tan 60° = BC b h 3 = b In DADB,
½
...(i) ½
h = b 3 tan 30° = 1
=
3
h =
120 =1 AD
AD = 120 m ...(i) 1 AB = tan 60° Now, CA 120 = 3 CA 120 = 40 3 m ½ Þ CA = 3 CD = AD + CA = 120 + 40 3 = 120 + 40 × 1·732 = 120 + 69·28 = 189·28 m Hence the distance between two cars = 189·28 m. 1 [CBSE Marking Scheme, 2017] Q. 3. 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. A [Foreign Set-I, II, III, 2017] Sol.
AB BD h a a 3
Tower
...(ii) ½
Multiplying (i) and (ii), we get 2
Þ
Sol.
h = b 3 ×
a 3
In DABC, ½
2
h = ba h = ab m
Hence, the height of the candle is
½ ab m.
Q. 2. 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. A [Delhi/OD Compt. Set-I, II, III, 2017]
Þ In DABD, Þ \
½ AB = tan 60° AC h = 3 x h= x 3 1 AB = tan q AD h = tan q 3x 1 x 3 = tan 30° 1 = 3 3x q = 30° ½ [CBSE Marking Scheme, 2017]
[ 265
Heights and Distances
T
Q. 4. On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then A [CBSE OD Set-I, II, III, 2017] find the height of the tower.
opper Answer, 2017
Q. 5. The angles of depression of the top and bottom of
A
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) A [CBSE Delhi Set I, II, III, 2016] h − 50 x x = h – 50 ½ h tan 60° = ½ x
Sol.
tan 45° =
⇒ ⇒
x =
Hence
h – 50 =
h
3
½
3h − 50 3 = h 3h − h = 50 3 h
(
)
B
3 − 1 = 50 3
D
x
50 3 h = 3 −1
h =
3 h
C
hE
Tower
3
Building
Sol.
h =
50 3
(
(
½
)
3 +1
3−1
50 3 + 3
)
2
⇒ h = 75 + 25 3 = 75 + 43.25 = 118.25 m. 1 [CBSE Marking Scheme, 2016]
266 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
T
Q. 6. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the A [CBSE OD, Set-II, 2016] ship and the height of the hill.
opper Answer, 2016
Sol.
3
Q. 7. 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 upto one place of decimal. (Use 3 = 1.73) A [Foreign Set II, 2016] A
Sol.
⇒ In DACD, ⇒
x+7 y = tan 60° =
y 3 = x + 7
x 3 = x + 7 7 = ( 3 − 1)x x =
Tower
3
Putting y = x, then ⇒
B
1
x = y
x
1+½
7( 3 + 1) 2
=
7( 2.73) 2
=
19.21 = 9.60 2
= 9.6 m
C
In DBCD,
i.e.,
y
BC = tan 45° CD x = tan 45° = 1 y
D
½
[CBSE Marking Scheme, 2016] Q. 8. At a point A, 20 metre above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud A [CBSE OD Set-I, II, III, 2015] from A ?
[ 267
Heights and Distances
C
Sol.
⇒
A
B
x
h
From, DABC,
D
1 h = tan 30° = x 3
⇒ From DABD,
...(i) ½
x = 3h. 40 + h = tan 60° = x
3
∴
3 h =
A [CBSE SQP, 2020-21]
A
C
h
h
1 80 m Let BD = width of river = 80 m ½ AB = CD = height of both palm trees = h BO = x ½ OD = 80 – x In DABO, h ½ tan 60° = x h 3 = ...(i) ½ x
B
h =
3x
h tan 30° = ( 80 − x ) 1
3
⇒ 3h = 40 + h m.
⇒
h = 20 m.
∴
x = 20 3 m
∴
AC =
=
1
( BC )2 + ( AB)2 ( 20 )2 + ( 20 3 )2
= 400 + 1200 1
Hence, the distance of the cloud = 40 m.
= 40 m.
[CBSE Marking Scheme, 2015]
5 marks each
h =
3x [From eqn. (i)] = 1.73 × 20 = 34.6 The height of the trees = h = 34.6 m ½ BO = x = 20 m DO = 80 – x = 80 – 20 = 60 m ½ \ The distances of the point O from the trees are 20 m and 60 m respectively. Q. 2. The angles of depression of the top and bottom of a building 50 meters high as observed from the top of a tower are 30˚ and 60˚ respectively. Find the height of the tower, and also the horizontal distance between the building and the tower. A [CBSE SQP, 2020-21] X R Sol. 30° 60° (h–50) m
A
30° xm
S hm
Bulding 50 m 50 m Tower
½
h = ...(ii) ½ ( 80 − x) 3
Solving (i) and (ii), we get x = 20
40 + h
60° B
T
1
xm
In DCDO,
...(ii) ½
Q. 1. The two palm trees are of equal heights and are standing opposite to each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees. (use 3 = 1.73) Sol.
3
From (i) and (ii),
Long Answer Type Questions
40 + h
h
x =
Let, height of building AB = 50 m
268 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
and height of tower, RT = h m BT = AS = x m AB = ST = 50 m RS = TR – TS = (h – 50) m RS In DARS, tan 30° = AS 1 h − 50 = x 3 In DRBT,
½ ½
...(i) ½
RT BT h 3 = x
tan 60° =
½
Hence, height of the tower = 75 m Distance between the building and the tower = 25 3
⇒
h ( 3 − 1) = 6 6
×
⇒
h =
⇒
h =
⇒
h = 3 ( 3 + 1)
3 −1
3 +1 3 +1
½
6( 3 + 1) 2
½
⇒
h = 3 × 2.73
⇒
h = 8.19 m.
\ The height of the tower is 8.19 m.
1
½
Q. 4. From a point on the ground, the angles of elevation of the bottom and the top of a tower are 45° and 60° respectively. Find the height of the tower.
½
A [CBSE OD, Set-I, 2020] Sol. Let the height of the tower be BD
In DPAB,
= 25 × 1.732 = 43.30 m½ Q. 3. A vertical tower stands on horizontal plane and is surmounted by a vertical flag-staff of height 6 m. The angles at a point on the bottom and top of the flag-staff with the ground are 30° and 45° respectively. Find the height of the tower. (Take
h 3 – h = 6
= 3(1.73 + 1) ½
Solving (i) and (ii), we get h = 75 h 75 From (ii), x = = 3 3 = 25 3
⇒
⇒ ⇒
tan 45° = 1 =
AB AP
1
20 AP
AP = 20 m
U [CBSE Delhi Set-I, 2020]
3 = 1.73)
Sol. According to question,
1
AD is a flagstaff and BD is a tower.
1
In DABC,
tan 45° =
AB BC
1 =
h+6 BC
⇒ ⇒
BC = h + 6
½
...(i) ½
⇒ ⇒
tan 30° =
DB BC
1
½
h = 3 h+6
[from (i)]
h 3 = h + 6
½
tan 60° =
AD 20 + BD = AP 20
1
20 + BD 20
⇒
3 =
⇒
20 + BD = 20 3
⇒
In DDBC,
In DPAD,
BD = 20 3 – 20
= 20 ( 3 − 1)
1
= 20 (1.732 – 1) = 20 × 0.732 = 14.64 m. Hence, the height of the tower is 14.64 m.
1
Q. 5. The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of a tower from the foot of the building is 60°. If the tower is 50 m high, then find the height A [CBSE OD, Set-II, 2020] of the building.
[ 269
Heights and Distances
Sol. According to question,
Detailed Solution: ∠AED = 60° and ∠BEC = 30°
AD = BC = 3000 3 m Let the speed of the aeroplane = x m/s
Then,
AB = DC = 30 × x = 30x m
In ∆AED,
ÐD = 90°
tan 60° =
3 =
DE = 3000 m
In ∆BEC,
ÐC = 90°
50 m
⇒ ⇒ Now in DBDC,
⇒
1
½
CD tan 30° = BD 1 h h 3 = = 50 3 50 3
1 ½
⇒ 3h = 50 50 ⇒ h = = 16.67 3 Q. 6. The angle of elevation of an air plane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the air plane is flying at a constant height of 3000 3 m, find the speed of the aeroplane.
A [CBSE SQP, 2020]
1
Sol.
⇒
In DDAC,
3000 3 DE ...(ii) 1
BC EC
3000 3 = 3 DE + CD DE + CD = 3000 × 3 3000 + 30x = 9000 (from (i) and (ii)) 30x = 6000 x = 200 m/s Hence, the speed of plane = 200 m/s. 1 18 = 200 × = 720 km/h1 5
Sol.
1 1
AB = 3000 m DC = tan 30° 1 AC
⇒
AC = 9000 m BC = AC – AB = 6000 m 1 6000 m = 200 m/sec. 1 \ Speed of aeroplane = 30 s
AD DE
Q. 7. 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 in between them on the road, the angles of elevation of the top of poles are 60° and 30° respectively. Find the height of the poles and the distances of the point P from the A [CBSE Delhi, Set-I, 2019] poles.
Correct figure BE = tan 60° In DABE, AB
tan 30° =
...(i) 1
1
Hence, the height of the building is 16.67 m.
1
AB tan 60° = BD 50 3 = BD 50 BD = 3
[CBSE SQP Marking Scheme, 2020]
1
In DABD
1
h = tan 60˚= 3 ...(i) ½ x 1 h ...(ii) ½ In ∆CDP, = tan 30˚= 3 80 - x In
∆ABP,
270 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
⇒
80 - x 3 = x 1
or
3x = 80 – x 4x = 80
⇒
x = 20 m.
1
and
h = 20 3 m.
1
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. Given, AP = 200 m and FD = 50 m 1 In DPQA, ∠Q = 90° PQ ½ sin 30° = PA
Dividing (i) by (ii), we get
⇒
Height of poles is 20 3 m
⇒
and the distances P are 20 m and 60 m from poles. 1
Q. 8. 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 and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak. A [CBSE OD, Set-I, 2019]
[CBSE Marking Scheme, 2019]
Sol. x
PQ 1 = 2 200 200 PQ = = 100 m 2
½
\ PR = PQ – RQ = PQ – FD = (100 – 50) m = 50 m ½ In DPRD, ∠R = 90° PR ½ sin 45° = PD 1 50 = ⇒ 2 PD ⇒ PD = 50 2 = 50 × 1.414 = 70.7 m Thus, the distance of the bird from Deepak is 70.7 m. 1
COMMONLY MADE ERROR
45°
Many candidates express the answer in
3 significant figures or four significant figures which is not necessary if not asked in questions.
1
In DAPQ
PQ 1 = sin 30° = AP 2
1
ANSWERING TIP
1 PQ = (200) = 100 m 2
1
PQ = 100 – 50 = 50 m
1
In DPRD,
Students should practice mathematical calculations each day to reduce calculation errors.
1 PR = sin 45° = 2 PD
PD = (PR)( 2 ) = 50 2 m
[CBSE Marking Scheme, 2019]
1
Detailed Solution:
Q. 9. From a point P on the ground, the angle of elevation of the top of a tower is 30° and that of the top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find the height of the tower. (Use
3 = 1.732) A [CBSE OD, Set-III, 2019]
Sol. Let AB be a tower and BC be a flagstaff.
x 45°
x
1
y
1
[ 271
Heights and Distances
=
5( 3 + 1) 2
=
13.66 2
= 6.83
In ∆PAC, According to question, AC = tan 45° = 1 AP ⇒ x + 5 = y x 1 In ∆PAB, = tan 30° = y 3 1 x = 3 x + 5 5 ⇒ x = 3 -1
1
1½
Height of tower = 6.83 m
1
[CBSE Marking Scheme, 2019]
½
COMMONLY MADE ERROR Many candidates were not able to understand the language of question.
ANSWERING TIP Do sufficient practice of drawing correct
diagrams for problems based on Heights and Distances.
T
Q. 10. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given 3 = 1.732) [CBSE Delhi Term, 2019]
Sol.
opper Answer, 2019
272 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
5
T
Q. 11. As observed from the top of a 100 m high light house 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 light house, find the distance between the two ships. [Use 3 = 1·732] C + A [CBSE Delhi/OD, 2018]
opper Answer, 2018
Sol.
5
COMMONLY MADE ERROR Most candidates are unable to draw the diagram as per the given data and lose their marks. Some candidates do calculation errors while putting the values of write inaccurate answer.
3 = 1.73 instead of 1.732 and hence
ANSWERING TIP Students should do rounding off at the end while calculating the final answer.
[ 273
Heights and Distances
Q. 12. A man on the top of a vertical observation tower observes a car moving at uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point ? C + A [CBSE SQP, 2018] Sol. Let the speed of car by x m/minute In DABC, h y = tan 45° ⇒ h = y h In DABD, y + 12 x = tan 30°
⇒
=
h x+h
x = h
12 Time for remaining distance,
1
=
½
1 ½
h 3 = y + 12x
3
( 3 - 1) h ( 3 - 1) Speed = m/min
i.e.,
1
x + h = h 3
Thus,
⇒
⇒
= =
h
12 x
y =
⇒
y = 6x ( 3 + 1)
1
y 3 – y = 12x 3 −1
=
1 1
)
3 -1 12
(
3 +1
)
3 -1 12 ( 3 +1) 2
(
)
3 + 1 min
1
Q. 13. 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 tower is 50 meter high, find the height of the A [CBSE Comptt. Set-I, II, III, 2018] hill.
12 x( 3 + 1) 2
(
12
= 6
h
[CBSE Delhi Set-I, II, III, 2015]
Sol.
Hence, time taken from C to B = 6 ( 3 + 1) minutes [CBSE Marking Scheme, 2018] 1
Detailed Solution:
⇒ In DAPB,
Let AB be the tower of height h and x be the distance between two cars ∠AQB = 45° Now, in DABQ, AB tan 45° = 1 BQ h ⇒ 1 = BQ
Let AB = 50 m be the height of the tower and CD be the height of hill. 1
Now, in DABC,
∠ABC = 90°
tan 30° =
BQ = h
A
or,
or,
BC =
AB BC
1
50 50× 3 m = tan 30° 1
BC = 50 3 m
1
Again in DBCD, ∠BCD = 90° h
or,
P
30° x
45° Q
tan 30° =
AB PB
B
1
⇒
tan 60° =
DC BC
DC = BC tan 60°
1
= 50 3 × 3 m DC = 150 m
∴ The height of hill is 150 m.
1
274 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
COMMONLY MADE ERROR
Sol.
The concept of angle of depression is not clear to many students. That's why they are not able to draw the diagram correctly.
ANSWERING TIP The concept of angle of depression and
Let the speed of the boat be x m/min. \ Distance covered in 2 minutes = 2 x \ In DABC,
AB = tan 60° BC
Þ
150 = y
Þ
y =
Þ
y = 50 3 m.
In DABD,
Tower
1 In DDCA,
Þ
= tan 60° =
Þ Þ
3
½
x + y = 15 5 3 + y = 15 y = 15 − 5 3
(
1 ½
1
3 150 3 ...(i)
AB = tan 45° BD
Þ
150 =1 y + 2x
Þ
y + 2x = 150
½
...(ii) 1
50 3 + 2 x = 150
2x = 150 − 50 3
2x = 50 3 − 3
)
Hence, the distance between the points = 5 3 − 3 m.
)
[CBSE Marking Scheme, 2017] 1 Q. 15. A moving boat observed from the top of a 150 m high cliff, moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat. A [CBSE Delhi Set-I, 2017]
(
)
(
)
(
)
x = 25 3 − 3 m.
1
Speed of the boat = 25 3 − 3 m/min.
=
= 5 3 − 3 m
(
½
Substituting the value of y from (i) in (ii),
15 Þ x= 3 Þ x= 5 3 DC 15 = tan 45° = =1 In DDCB, CB x+y Þ
CD = 2x
Let BC be y m.
Q. 14. Two points A and B are on the same side of a tower and in the same straight line with its base. The angle of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points. C + A [CBSE Delhi Set-I, 2017] Sol.
DC CA 15 x
1
angle of elevation must be clear to the students.
=
(
)
25 3 − 3 × 60 1000 3 3 − 3 km/h 2
(
)
= 19.02 km/h 1 [CBSE Marking Scheme, 2017] Q. 16. 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. C + A [Foreign Set-II 2017]
[ 275
Heights and Distances
Sol. Let AB be the height of the aeroplane, then AB = 7500 m. Also let D and C be the positions of two ships on the same line. From the point A of an aeroplane, the angles of depression of two ships D and C are ∠OAD = 30° i.e., ∠BDA = 30° and ∠OAC = 45° i.e., ∠BCA = 45° Let distance between two ships DC = x m and BC = y m. (Aeroplane)
Let \ In ∆PQX,
PX = x m and PQ = h m QT = (h – 40) m h x h 3 = x
⇒ In ∆QTY,
Þ Þ In DABD,
Þ
= 7500
(
1 ½
1 ½
x + 7500 = 7500 3 x = 7500 3 − 7500
3 x ...(i) 1
h - 40 tan 45° = x h - 40 1 = x
( (
In DABC,
h =
½
⇒ x = h – 40 Solving (i) and (ii), x = 3 x – 40
7500 m
½
⇒
⇒
AB = tan 45° BC 7500 =1 y y = 7500 AB = tan 30° BD 7500 1 = x+y 3 7500 3 x + y =
tan 60° =
½
or
)
)
3 - 1 x = 40 x =
3 × 20
(
40 3 -1
)
(
= 20
(
)
3 +1 m
)
3 + 1 = 20 3 + 3 m
1
20(3 + 1.73) = 20 × 4.73 Hence, the height of tower is 94.6 m. Q. 18. The angle of elevation of the top B of a tower AB from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Find the height of the tower AB and A [CBSE SA-2, 2016] the distance XB. Sol. In ∆YCB, we have
3 −1
)
3x − x = 40
or ∴h=
...(ii) 1
tan 45° =
BC YC
= 7500(1.732 – 1) = 7500 × 0.732 = 5490 m Hence, the distance between two ships = 5490 m 1½ Q . 17. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use 3 = 1.73) Sol.
⇒ In ∆XAB,
U [CBSE OD Set I, III, 2016]
1
1 =
1
x YC
YC = x m XA = x m tan 60° =
AB XA
½
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
3 =
x + 40 x
½
In ∆XAB,
3x = x + 40
x 3 − x = 40 40
x =
=
= 20( 3 + 3) m
276 ]
3 -1
×
3 +1 3 +1
½
20( 3 + 1)
(
)
= 20 3 + 20 m ∴ Height of the tower, AB = x+ 40 = 20 3 + 20 + 40
½
= 20 3 + 60
½
20 × 4.732 = 94.64 m AB BX
½
3 AB = 2 BX
½
sin 60° =
BX =
20( 3 + 3)2 3
(
)
= 40 3 + 1 m = 40 × 2.732 m = 109.28 m ½ [CBSE Marking Scheme, 2016]
T
Q. 19. As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30° to 60°. Find the distance travelled by the ship during the period of C [CBSE OD, Set-II, 2016] observation. (Use 3 = 1.73)
opper Answer, 2017
Sol.
Q. 20. 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
5
finds the angles of elevation of their tops to be complementary, then find the height of the shorter post.
A [CBSE Term-2, 2015]
[ 277
Heights and Distances
2h = cot θ k 2
Sol. Let AB and CD be the two posts such that AB = 2 CD. Let M be the mid-point of CA. Let ∠CMD = θ and ∠AMB = 90° – θ
⇒
B
2h
D
h = tan θ k 2
⇒
⇒
A
M
tan θ =
1 Clearly,
CM = MA =
1 k 2
Let CD = h m, then AB = 2h m AB = tan (90° – θ) = cot θ Now, AM
1
2h = cot θ k 2
⇒
A
B
h2 =
⇒
h =
k 2 2
k 2 m 4
1
(ii) What should be the length of Ladder, when inclined at an angle of 60° to the horizontal ? (a) 4.28 m
3.7
(b)
3
m
(c) 3.7 m (d) 7.4 m Sol. Correct option: (a). Explanation: In DADC, BD sin 60° = BC
3 3.7 = 2 BC
Þ
BC =
3.7 × 2 3
BC = 4.28 m (approx)
(iii) How far from the foot of pole should she place the foot of the ladder ? (a) 3.7 (b) 2.14 1 (c) (d) None of these 3
=
4 marks each
Þ
k2 8
[CBSE Marking Scheme, 2015]
Þ
60° D C (i) What is the length of BD ? (a) 1.3 m (b) 5 m (c) 3.7 m (d) None of these Sol. Correct option: (c). Explanation: From figure, the electrician is required to reach at the point B on the pole AD. So, BD = AD – AB = (5 – 1.3) m = 3.7 m
4h 2h × =1 k k
∴
...(ii) 1
Q. 1. An electrician has to repaired an electric fault on the pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work (see figure)
2h k
Multiplying (i) and (ii),
Visual Case Based Questions Note: Attempt any four sub parts from each question. Each sub part carries 1 mark
...(i) 1
CD Also in ∆CMD, = tan θ CM
h C
4h cot θ = ⇒ k
Sol. Correct option: (b). Explanation: In DBDC, \
cot 60° =
DC BD
Þ
1 3
=
DC =
DC 3.7 3.7 3
Þ DC = 2.14 m (approx) (iv) If the horizontal angle is changed to 30°, then what should be the length of the ladder ? (a) 7.4 m (b) 3.7 m (c) 1.3 m (d) 5 m Sol. Correct option: (a). Explanation: In DBDC, BD \ sin 60° = BC Þ
1 3.7 = 2 BC
Þ BC = 3.7 × 2 = 7.4 m (v) What is the value of ÐB ? (a) 60° (b) 90° (c) 30° (d) 180° Sol. Correct option: (c). Explanation: In DADC, angle D is 90°. So, by angle sum property. ÐB + ÐD + ÐC = 180° or, ÐB = 180° – (90° + 60°) = 30° Q. 2. A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kings way), is about 138 feet (42 metres) in height.
Þ
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Now,
tan q =
AB BC
tan q =
42 42
tan q = 1 tan q = tan 45° q = 45° Hence, angle of elevation = 45° (ii) They want to see the tower at an angle of 60°. So, they want to know the distance where they should stand and hence find the distance. (a) 25.24 m (b) 20.12 m (c) 42 m (d) 24.64 m Sol. Correct option: (a). Explanation: Height of India gate = 42 cm Angle = 60° Let the distance between students and India gate = x m.
278 ]
Now,
tan q = tan 60° =
AB BC 42 x
42 3 = x 42 x = 3 x =
=
42 × 3 3× 3 42 3 3
= 14 3 m = 25.24 m
(i) What is the angle of elevation if they are standing at a distance of 42 m away from the monument ? (a) 30° (b) 45° (c) 60° (d) 0° Sol. Correct option: (b). Explanation: Height of Indian gate = 42 m Distance between students and Indian gate = 42 cm
(iii) If the altitude of the Sun is at 60° , then the height of the vertical tower that will cast a shadow of length 20 m is 20 m (a) 20 3 m (b) 3 15 m (d) 15 3 m (c) 3
[ 279
Sol. Correct option: (a). Explanation:
(i) The distance of the satellite from the top of Nanda Devi is
Heights and Distances
(a) 1118.36 km
(b) 577.52 km
(c) 1937 km
(d) 1025.36 km
Sol. Correct option: (a). Explanation:
Let, the height of tower = h AB Now, tan q = BC
tan 60° =
h 20
h 3 = 20 h = 20 3
(iv) The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is (a) 30° (b) 45° (c) 60° (d) 90° Sol. Correct option: (b). (v) The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is (a) corresponding angle (b) angle of elevation (c) angle of depression (d) complete angle Sol. Correct option: (a). Q. 3. A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka ,them being Nanda Devi(height 7,816 m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.
Now,
AG = cos q =
1937 km 2 AG AF
1937 cos 30° = 2 AF
3 1937 = 2 2AF
AF =
AF = 1118.36 km
1937 3
(ii) The distance of the satellite from the top of Mullayanagiri is
(a) 1139.4 km
(b) 577.52 km
(c) 1937 km
(d) 1025.36 km
Sol. Correct option: (c). Explanation: For DFPH,
cos q =
cos 60° =
PH FP 1937 2FP
1 1937 = 2 2FP FP = 1937 km
(iii) The distance of the satellite from the ground is
(a) 1139.4 km
(b) 577.52 km
(c) 1937 km
(d) 1025.36 km
Sol. Correct option: (b). (iv) What is the angle of elevation if a man is standing at a distance of 7816 m from Nanda Devi ? (a) 30° (c) 60° Sol. Correct option: (b).
(b) 45° (d) 0°
280 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Explanation: Height of Nanda Devi Mountain = 7816 m Distance between man and mountain = 7816 m.
tan q =
AB BC
tan q =
7816 7816
tan q = 1 tan q = tan 45° q = 45° (v) If a mile stone very far away from, makes 45° to the top of Mullanyangiri mountain. So, find the distance of this mile stone form the mountain.
(a) 1118.327 km
(b) 566.976 km
(c) 1937 km
(d) 1025.36 km
Sol. Correct option: (c).
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[ 281
HEIGHTS AND DISTANCES
Maximum Time: 1 hour
A
Q. 3. Evaluate: (sec A + tan A) (1 – sin A). 2 2 Q. 4. Find the value of 9 sec A – 9 tan A.
U
Q. 5. If the altitude of the Sun is 60°, what is the height of a tower which casts a shadow of length 30 m ? C + U [CBSE Term-2, Set (B1), 2011]
Q. 6. A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied at the ground. The height of the pole is 12 m and the angle made by the rope with ground level is 30°. A
(iii) If sin (A + B) =
(v) Which mathematical concept is used in this problem ? (a) Trigonometry (b) Triangle (c) Circle (d) Mensuration
Q. 7. If 4 cos θ = 11 sin θ, find the value of
11 cos θ − 7 sin θ . 11 cos θ + 7 sin θ
Q. 8. Prove that : –1 +
Rope
30°
sin A sin (90° - A) = – sin2 A. 2 cot (90° - A)
C
B
Give answer of the following questions: C + AE (i) The distance covered by the artist in climbing the top of the pole is : (a) 24 m (b) 36 m (c) 28 m (d) 22 m (ii) The length of BC is :
(a) 24 3 m
(b)
12 3 m
(c) 2 3 m
(d)
3m
3
U [CBSE Term-1, 2012]
Q. 9. If x sin q + y cos q = sin q cos q and x sin q = y cos q, prove that x2 + y2 = 1. A [CBSE Term-1, 2011] 3
U [CBSE Term-1, 2012] 2
12 m 3
3 , then the value of (A + B) is: 2
(a) 30° (b) 90° (c) 60° (d) 45° (iv) In DABC, given that ∠A = 60° and ∠C = 30°, then the value of sin A cos C + cos C sin A is: (a) 0 (b) ∞ 3 (c) 10 (d) 2
Q. 1. If DABC is right angled at C, then find the value of C cos(A + B). Q. 2. If cos 9a = sin a and 9a < 90°, then find the value U of tan 5a.
MM: 25
Q. 10. Simplify : sin θ sec (90° − θ ) tan θ tan (90° − θ) – · cos ec (90 ° − θ) cos θ cot (90°− θ) cot θ U [CBSE Term-1, 2011] 3
Q. 11. Two ships are approaching a light house from opposite directions. The angles of depression of two ships from top of the light house are 30° and 45°. If the distance between two ships is 100 m., find the height of light-house. A [Foreign set I, II, III, 2014] 5
Finished Solving the Paper ? Time to evaluate yourself !
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
UNIT 6: MENSURATION
c h a p te r
12
Areas related to circles
Syllabus ¾
¾ Motivate the area of a circle; area of the sectors and the segments of a circle. Problems based
on areas and perimeter/circumference of the plane figures of circles. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only. Plane figures involving triangles, simple quadrilaterals and circle.)
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Outside Delhi
A circle is a collection of all points in a plane which are at a constant distance from a fixed point in the same plane. A
.
.
O
.
Area related to circles
B
A line segment joining the centre of the circle to a point on the circumference of the circle is called its radius.
O
.
Radius r
A
284 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
A line segment joining any two points of a circle is called a chord. A chord passing through the centre of circle is called its diameter. A diameter is the largest chord of the circle. Here AB is a a diameter, which is a longest chord. A diameter of a circle divides a circle into two equal arcs, each known as a semi-circle.
A
.
O
.
Semi-circle
B
Diameter
O Chord A part of a circumference of circle is called an arc. An arc of a circle whose length is less than that of a semi-circle of the same circle is called a minor arc. An arc of a circle whose length is greater than that of a semi-circle of the same circle is called a major arc. The region bounded by an arc of a circle and two radii at its end points is called a sector. Major Sector
.
O
Minor Sector A chord divides the interior of a circle into two parts, each called a segment. Q
.
.
Major Segment O
Minor Segment Circles having the same centre but different radii are called concentric circles.
O
.
r
A
R
B
Two circles (or arcs) are said to be congruent if on placing one over the other cover each other completely.
The distance around the circle or the length of a circle is called its circumference or perimeter. The mid-point of the hypotenuse of a right triangle is equidistant from the vertices of the triangle. Angle subtended at the circumference by a diameter is always a right angle.
.
O
Angle described by minute hand in 60 minutes is 360°.
Angle described by hour hand in 12 hours is 360°.
[ 285
Areas related to circles
Know the Formulae
1. Circumference (perimeter) of a circle = πd or 2πr, where d is diameter and r is the radius of the circle. 2. Area of a circle = πr2. 1 3. Area of a semi-circle = πr2. 2 4. Perimeter of a semi-circle = πr + 2r = (π + 2)r 5. Area of a ring or an annulus = π(R + r)(R – r). where R is the outer radius and r is the inner radius. 2 πrθ prq 6. Length of arc, l = , where q is the angle subtended at centre by the arc. or 360° 180°
pr 2 q 360° 1 or area of sector = (l × r ), where l is the length of arc. 2
7. Area of a sector =
8. Area of minor segment =
πr 2 θ 1 2 - r sin θ. 360° 2
9. Area of major segment = Area of the circle – Area of minor segment = πr² – Area of minor segment. 10. If a chord subtends a right angle at the centre, then π 1 2 area of the corresponding segment = − r 4 2 11. If a chord subtends an angle of 60° at the centre, then π 3 2 area of the corresponding segment = r . 6 4
12. If a chord subtends an angle of 120° at the centre, then π 3 2 area of the corresponding segment = r . 4 3
13. Distance moved by a wheel in 1 revolution = Circumference of the wheel. Distance moved in 1 minute . 14. Number of revolutions in one minute = Circumference 15. Perimeter of a sector =
πrθ + 2r. 180°
Know the Facts An Indian mathematician Srinivas Ramanujan worked out the identity using the value of π correct to million places of decimals.
The Indian mathematician Aryabhatta gave the value of π as
62832 20000
“How I made a greater discovery” this mnemonic help us in getting the value of π = 3.14159 ............ .
No. of Letters
↓
Give it under separate reading with explanation how to use CAN
I
HAVE
A
SMALL
CONTAINER
OF
COFFEE
↓
↓
↓
↓
↓
↓
↓
↓
3
1
4
1
5
9
2
6
Archimedes calculated the area of a circle by approximating it to a square. Area of sector of a circle depends on two parameters-radius and central angle.
286 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
How is it done on the
GREENBOARD?
radius of circle = 7 cm Q.1. Find the area of the shaded re- or, gion in the given figure. Step II: Area of rectangle = 28 × 14 cm2 [Delhi - 2013] U = 392 cm2 22 Area of each circle = × (7)2 14 cm 7 = 154 cm2 Area of both circles = 2 × 154 cm2 = 308 cm2 Step III: Shaded area = Area of rectangle – Area of both circles = 392 – 308 = 84 cm2
28 cm
Solution: Step I: Length of rectangle = 28 cm Breadth of rectangle = 14 cm \ Diameter of circle = 14 cm
Very Short Answer Type Questions
Q. 1. In a circle of diameter 42 cm, if an arc subtends 22 an angle of 60° at the centre, where p = , then 7 what will be the length of arc. U [CBSE SQP, 2020-21]
θ ( 2 πr ) 360° 60° 22 = × 21 2 × 360° 7 Sol.
Length of arc =
½
1 mark each
22 60° ×2× × 21 360° 7 = 22 cm
=
½
COMMONLY MADE ERROR Some students used incorrect formula for
length of arc and some made mistakes in calculation.
= 22 cm ½ [CBSE Marking Scheme, 2020-21] Detailed Solution: Given, diameter of the circle = 42 cm 42 cm = 21 cm. Then, radius of the circle = 2 and angle subtended at the centre, q = 60°
ANSWERING TIP Remember the formula of length of arc
and the concept of angle subtended at the centre.
Q. 2. In the given figure, find the perimeter of the sector of a circle with radius 10.5 cm and of angle 22 60°. Take π = . 7 A
B
\ The length of arc, l =
l
θ × 2 πr 360°
½
U [CBSE OD Set-I, 2020]
[ 287
Areas related to circles
Sol. We have,
radius (r) = 10.5 cm
and
angle (q) = 60°
P B 10
O
O
Then, the length of arc APB
=
60° 22 ×2× × 10.5 360° 7
= 11 cm½ Now, the perimeter of the sector OAPBO = OA + length of an arc APB + BO = (10.5 + 11 + 10.5) cm ½
= 32 cm.
30°
A B Given, radius of a circle, OA = OB = 6 cm (Assuming in figure) ½ and central angle q = ∠AOB = 30° By using formula, area of the sector of a circle θ × πr 2 = 360° 30° × 3.14 × 6 × 6 = 360° = 9.42 cm2 ½
θ × 2 πr = 360°
6 cm
6 cm
cm
10.
.5
Sol.
5c m
A
Q. 3. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take p = 3.14) U [CBSE OD Set-III, 2020]
Short Answer Type Questions-I Q. 1. 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 A [CBSE Delhi Set-I, 2020] Sol. AB is an arc of a circle. Let radius of a circle be r cm.
2 marks each
Sol. Given, radius of circle (r) = 5.2 cm i.e., OA = OB = r = 5.2 cm and the perimeter of a sector = 16.4 cm As we know that perimeter of the sector 2 prq = 2r + 360° ⇒
16.4 = 2 × 5.2 +
2 p ´ 5.2 ´ q 360°
O O
\
cm
5.2
B
22 cm AB = 22 cm q = 60° 2 prq Length of arc = 360°
⇒
22 =
2 ´ 22 ´ r ´ 60° 7 ´ 360°
⇒
22 =
22 ´ r 21
θ
⇒ 1
⇒ 22 × r = 22 × 21 ⇒ r = 21 Hence, The radius of the circle (r) is 21 cm. 1 Q. 2. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector. A [CBSE Delhi Set-II, 2020]
A
cm
i.e., and
A
5.2
60°
B
2 p ´ 5.2 ´ q =6 360°
6 × 360° 1 2 π × 5.2 q × pr2 Now, area of sector = 360° 6 × 360° × p × (5.2)2 = 2 5 . 2 360 π × × ° = 15.6 sq. units. 1
⇒
q =
Q. 3. The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
A [CBSE Delhi Set-III, 2020]
288 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
\ Angle subtended in 35 minutes = 6° × 35 = 210° ½ Area of the face of the clock described by the minute hand in 35 minutes = Area of the sector (210°)
\ Total perimeter ofl the shaded region = AD + BC + lengths of the arcs (APB + DPC) = (21 + 21 + 66) cm = 108 cm. 1 Q. 5. Find the area of the square that can be inscribed in a circle of radius 8 cm. U [CBSE Board Term-2, 2015]
pr 2 q 360° 22 12 ´ 12 ´ 210° ´ = 7 360°
360° = 6° 60°
=
½
½
665280 = 2520 = 264 cm2. ½ Q. 4. 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 U [CBSE SQP, 2016]
D
Sol. Diameter of the circle = diagonal of square = 2 × 8 = 16 cm Let x be the side of square.
Sol. Angle subtended in 1 minute =
C
∴
x 2 = 16 or, x = 8 2
(
)
Area of square = x2 = 8 2 = 128 cm2 1 [CBSE Marking Scheme, 2015] Detailed Solution:
∴
Diameter of circle = 16 cm.
∴
Diagonal of square = 16 cm
Radius of the circle = 8 cm.
Let the side of square = x cm. x2 + x2 = (16)2
.
1
2
(Pythagoras theorem)
D
A
P
21 cm
O
12 B
A
Sol. Perimeter of the shaded region = AD + BC + lengths of the arcs of semi circles APB and CPD 1 æ 22 21 ö = 21 +21 + 2 ç ´ ÷ ½ è 7 2ø
x2 =
or,
16 × 16 = 128 2
Area of square = x2 = 128 cm2.
C [CBSE Board Term-2, 2015]
Sol. Side of square = 50 cm \ Area of square = 50 × 50 = 2500 cm2 D
and length of arc APB = length of arc DPC \ Length of arc (APB + DPC)
C
= ( 2πr )
= 2 ×
21 cm
22 21 × 7 2
= 22 × 3 = 66 cm
1
Q. 6. 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.
Given, ABCD is a square, then
\
2x2 = 16 × 16
Detailed Solution:
21 radius (r) = cm 2
1
C
or,
= 42 + 66 = 108 cm. ½ [CBSE Marking Scheme, 2016]
AB = BC = CD = DA \ Side of the square = 21 cm Since APB and DPC are are two semicircles.
B
cm
15 cm A
1
50 cm
B
Radius of quadrant = 15 cm.
½
[ 289
Areas related to circles
1 pr2 = pr2 4 = p × 15 × 15 22 × 225 = 7 Area of 4 quadrants = 4 ×
= 707.14 cm2 Area of circle = pr2 =
22 21 × 7 2
= 346.5 cm2 Area of remaining sheet = Area of square – 4(area of quadrant) – Area of circle ½ = 2500 – 707.14 – 346.5 = 1446.36 sq. cm 1 [CBSE Marking Scheme, 2015]
½
2
Short Answer Type Questions-II
Q. 1. In the figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.
A [CBSE SQP, 2020-21] [CBSE Delhi Set-I, II, III, 2016]
Sol. Area of 1 segment = area of sector – area of triangle ½
æ 90° ö 2 1 pr – ×7×7 360° ÷ø 2 1 22 1 ×7×7 ½ = ´ × 72 – 4 7 2 = ç è
3 marks each
Similarly, area of semicircle DOC = 77 cm2 Hence, the area of shaded region (Part W and Part Y) = Area of square – Area of two semicircles AOB and COD = 196 – 154 = 42 cm2 1 Therefore, area of four shaded parts (i.e., X, Y, W, Z) = 2 × 42 = 84 cm2 1 2 Q. 2. The area of a circular playground is 22176 cm . Find the cost of fencing this ground at the rate of A [CBSE OD Set-I, 2020] 50 per metre. Sol. Area of a circular play ground = 22176 cm2 i.e., pr2 = 22176 cm2 1 [where r is the radius of a play ground] 7 = 7056 ⇒ r2 = 22176 × 22 ⇒ r = 84 cm = 0.84 m 1 Cost of fencing this ground = ` 50 × 2pr 22 × 0.84 = ` 50 × 2 × 7 = ` 264. 1 Q. 3. Sides of a right triangular field are 25 m, 24 m and 7 m. At the three corners of the field, a cow, a buffalo and a horse are tied separately with ropes of 3.5 m each to graze in the field. Find the area of the field that cannot be grazed by these animals. A [CBSE SQP, 2020]
= 14 cm2 ½ Area of 8 segments = 8 × 14 = 112 cm2 ½ Area of the shaded region = 14 × 14 – 112 ½ = 196 – 112 = 84 cm2 ½ (each petal is divided into 2 segments) [CBSE Marking Scheme, 2020-21] Detailed Solution:
Sol.
Required Area = Area of triangle – Area of 3 sectors 1 × 24 × 7 = 84 m2 1 Area of Triangle = 2
Area of three sectors =
πr 2 360°
× (Sum of three angles of triangle) 22 × 7 × 7 × 180° = 7 × 2 × 2 × 360°
Side of a square = 14 cm \ Area of the square ABCD = 14 × 14 = 196 cm2 1 Area of semicircle AOB = πr 2 2 1 22 ×7×7 = × 2 7 = 77 cm2
22 21 21 × × 7 2 2
=
77 or 19.25 m2 1 4 \ Required Area = (84 – 19.25) m2 = 64.75 m2 1 [CBSE SQP Marking Scheme, 2020] =
1
290 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
18
17 m
Since DCBH is a right angled triangle, 1 then area of DCBH = × CB × BH 2 =
22 ö æ çè Use p = 7 ÷ø
m
A
21
cm
Area of sector ∠B =
Sol. Draw MO ⊥ AB
(r) = 3.5 m ∠C 2 πr Area of sector ∠C = 360°
=
½
3.5 m
A [CBSE Delhi Set-II, 2019]
H (Horse)
7m
=
Area of sector ∠H = =
cm
m
m
3.5 m
3.5 m
21
3.5
(Buffalo) B
3.5
m
B
120° O
5 3.
1 × 24 × 7 2
= 84 m2 ½ So, area of the field which cannot be grazed by three animals = Area of DCBH – Area of three sectors = (84 – 19.25) m2 = 64.75 m2. 1 Q. 4. Find the area of the segment shown in Figure, if radius of the circle is 21 cm and ∠AOB = 120°.
25 m
24 m
m 3.5
3.5 m
Detailed Solution: Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form. Given that radius of each sector C (Cow)
A
21
∠C π( 3.5)2 m 2 360°
M
cm
120° cm 21 O
B
∠B 2 πr 360° ∠B π( 3.5)2 m 2 360°
∠OAB = ∠OBA = 30°
∠H 2 πr 360°
1 OM 21 = ⇒ OM = sin 30° = 2 21 2
∠H π( 3.5)2 m 2 360°
cos 30° =
Sum of the areas of three sectors ½ ∠C ∠B 2 π( 3.5) + π( 3.5)2 = 360° 360° +
∠H π( 3.5)2 360°
∠C + ∠B + ∠H = 360° 22 × × 3.5 × 3.5 7 [Q ∠C + ∠B + ∠H = 180°] 180° × 38.5 = 360°
2
= 19.25 m
½
Area of DOAB = =
½
AM 3 21 = ⇒ AM = 3 21 2 2
21 1 1 × AB × OM = × 21 3 × 2 2 2 441 2 3 cm . 4
1
\ Area of shaded region
= Area (sector OAYB) – Area (DOAB) =
22 120 441 × 21 × 21 × − 3 1 7 360 4
3 2 = 462 − 441 4 cm = 271.3 cm2 (approx.)
½
[CBSE Marking Scheme, 2019]
[ 291
Areas related to circles
Detailed Solution: Given that, radius of a circle (r) = 21 cm and in DAOB, \
OA = OB ∠OAB = ∠OBA (Isosceles property)
So,
2∠OAB = 180° – 120°
(Angle sum property)
⇒
60° = 30° ∠OAB = 2
\
∠OAB = ∠OBA = 30°
In DOMB,
OM sin 30° = OB
A [CBSE OD Set-I, 2019]
Sol. Since OABC is a square.
B 15
C
1 MB = 2 21
⇒
OM =
21 2
and
cos 30° =
MV OB
cm
A cm
⇒
cm
½
O
∠OAB = 90°
Then, \ In DOAB,
Radius of quadrant = OB =
⇒
3 MB = 2 OB
⇒
MB =
\
AB = 2 × MB
21 3 2
½
⇒
152 + 152
(By Pythagoras theorem) 1
OB = 15 2 cm
Now, area of quadrant OQBP =
21 3 = 2 × 2 = 21 3 cm
=
θ × pr2 360° 90° × 3.14 × (OB)2 360°
Area of DOAB =
1 × AB × OM 2
1 × 3.14 × (15 2 )2 = 4
=
21 1 × 21 3 × 2 2
= 353.25 cm2
=
441 3 cm 2 4
½
θ × πr 2 Area of sector OACB = 360° =
= 225 cm2
½
Hence,
2
– area of square
= (353.25 – 225) cm2 = 128.25 cm2 1
= 462 cm Hence, area of shaded region = area of sector OACB
– area of DOAB
441 × 1.73 2 = 462 − cm 4 = 271.3 cm2.
Area of square = (OA) = (15)
½ 2
area of shaded region = area of quadrant
120° 22 × × 21 × 21 360° 7
2
½
Q. 5. In Figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use p = 3.14)
COMMONLY MADE ERROR Many candidates use incorrect formula for finding the area of shaded region.
ANSWERING TIP Remember all the formulae related to square and quadrant of the circle.
1
292 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
We know that, the length of the diagonal of a square is given by, d = a 2 ⇒ d = 2 2 ´ 2 = 4 cm ½ Since, the square is inscribed in a circle, hence the diagonal of square will be the diameter of the circle, d 4 = = 2 cm ½ i.e., radius, r = 2 2
Q. 6. In figure, ABCD is a square with side 2 2 cm and inscribed in a circle. Find the area of the shaded region. (Use p = 3.14) D
A
\ Area of the circle = pr2 = 3.14 × (2)2 = 12.56 cm2 ½ Therefore, area of shaded region = Area of circle – Area of the square = (12.56 – 8) = 4.56 cm2 1
C
B A [CBSE OD Set-I, 2019]
Sol.
BD =
( 2 2 )2 + ( 2 2 )2 =
16 = 4 cm 1
COMMONLY MADE ERROR
\ Radius of circle = 2 cm ½ \ Shaded area = Area of circle – Area of square ½ = 3.14 × 22 − (2 2 )2
Some candidates either use incorrect formula or made errors is calculation.
= 12.56 – 8 = 4.56 cm2 1 [CBSE Marking Scheme, 2019]
ANSWERING TIP
Detailed Solution:
Given, \
side of square, a = 2 2 cm. area of square = a2 = ( 2 2 )2 = 8 cm2 ½
Understand all the formula related to area of circle.
T
Q. 7. A wiper blade has length 21 cm, sweeps 120°. Calculate the area swept by two blades. [CBSE Delhi Region, 2019]
opper’s Answer, 2019
Sol.
3 Q. 8. Find the area of the shaded region in given figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. A [CBSE Delhi/OD, 2018] [Use p = 3.14] P A B
Q
S
D
R
C
[ 293
Areas related to circles
Sol.
Radius of each arc drawn = 6 cm
½
36 Area of one quadrant = (3.14) × 4
1 1
Area of four quadrants = 3.14 × 36 = 113.04 cm2 Area of square ABCD = 12 × 12 = 144 cm2 Hence, Area of shaded region = 144 – 113.04 = 30.96 cm2
½ [CBSE Marking Scheme, 2018]
T
Detailed Solution:
opper’s Answer, 2018
3
Q. 9. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 48 hours. C + A [CBSE Compt Set-I, II, III, 2018] [Foreign Set-I, II, III, 2015] Sol. Distance travelled by short hand in 48 hours = 4 × 2p × 4 cm = 32p cm.
1
Distance travelled by long hand in 48 hours = 48 × 2p × 6 cm = 576p cm
1
Total distance travelled = (32p + 576p) cm
= 608p cm
1
[CBSE Marking Scheme, 2018]
Detailed Solution: Short hand makes 4 rounds in 48 hours Long hand makes 48 rounds in 48 hours ½ Radius of the circle formed by short hand = 4 cm and radius of the circle formed by long hand = 6 cm Distance travelled by short hand in one round = circumference of the circle = 2 × 4 × p = 8p cm ½ Distance travelled by short hand in 4 rounds = 2 × 4 × 4p = 32p cm ½
Distance travelled by long hand in one round = 2 × p × 6 = 12p cm ½ Distance travelled by long hand in 48 rounds = 48 × 12p = 576p cm ½ Sum of the distances = 32p + 576p = 608p cm ½ Q. 10. The side of a square is 10 cm. Find the area between inscribed and circumscribed circles of the square. A [CBSE Comptt. Set-I, II, III, 2018] Sol. Radius of inner circle = 5 cm
½
1 Radius of outer circle = 5 2 cm Required area = Area of outer circle – Area of inner circle 1 A
B
D
C
= p [( 5 2 )2 - 5 2 ] = 25p cm2 ½ [CBSE Marking Scheme, 2018]
294 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Diameter of outer circle = Diagonal of square = 10 2 cm
10 2 \ Radius of outer circle = = 5 2 cm 1 2 Then, the required Area = Area of outer circle – Area of inner circle = p [( 5 2 )2 - ( 5)2 ] = 25p cm2 1 Q. 11. A wire, when bent in the form of an equilateral triangle, encloses an area of 121 3 cm2. If the wire is bent in the form of a circle, find the area 22 enclosed by the circle, Use π = 7
A [CBSE OD Set-I, II, III, 2017]
Sol. Let a be the side of triangle. 3 2 a 4
\ Area enclosed by the triangle = 3 2 a = 121 3 4 121 3 × 4 a2 = 3
Þ Þ
a = 22 cm ½ Perimeter of triangle = circumference of circle formed. \ 2pr = 22 × 3 1 22 × r = 22 × 3 2× Þ 7 22 × 3 × 7 21 = cm 22 × 2 2 Area enclosed by the circle = pr2 22 21 21 693 × × = = 7 2 2 2 Þ
r =
½
= 346.5 cm2 (Approx) 1 [CBSE Marking Scheme, 2017]
Q. 12. In adjoining fig, ABCD is a trapezium with AB || DC and ÐBCD = 30°. Fig. BGEC is a sector of a circle with centre C and AB = BC = 7 cm, DE = 4 cm and BF = 3.5 cm, then find the the area of the 22 ö æ shaded region. ç Use p = ÷ è 7ø 7 cm
A
B
G 3.5 cm
D
Sol. Given, AB = 7 cm, DE = 4 cm, and BF = 3.5 cm DC = DE + EC = 4 + 7 = 11 cm Area of trapezium ABCD 1 (Sum of || lines) × (distance between them) = 2 1 = (11 + 7 ) ´ 3.5 2 1 = ´ 18 ´ 3.5 2
= 31.5 cm2 1 Area of the sector BGEC 30° 22 1 × ×7×7 = = × 22 × 7 = 12.83 cm2 1 360° 7 12 (Approx.) Area of shaded region = Area of trapezium – Area of sector =31.5 – 12.83 1 = 18.67 cm2 (Approx.) [CBSE Marking Scheme, 2017] Q. 13. In the given figure ABCD is a trapezium with AB | | DC, AB = 18 cm and DC = 32 cm and the distance between AB and DC is 14 cm. If arcs of equal radii 7 cm taking A, B, C and D have been drawn, then find the area of the shaded region. 32 cm D C
A
B 18 cm A [Foreign Set-I, II, III, 2017]
Sol. Given, in trapezium ABCD, AB = 18 cm, CD = 32 cm, AB || CD and distance between || lines = 14 cm and the radius of each sector = 7 cm. 1 Area of trapezium ABCD = (18 + 32 ) × 14 2 1 = × 50 × 14 2 = 350 cm2 1 Let, ÐA = q1, ÐB = q2, ÐC = q3 and ÐD = q4 θ1 ar of sector A = πr 2 360 o θ 22 = 1o × ×7×7 7 360 θ = 1 o × 154 cm 2 360
Detailed Solution: Here, diameter of inner circle = side of the square = 10 cm 10 = 5 cm 1 \ Radius of inner circle = 2
ar of sector B =
θ2 × 154 cm 2 360 o
ar of sector C =
θ3 × 154 cm 2 360 o
ar of sector D =
θ4 × 154 360 o
ar of 4 sectors =
θ1 + θ 2 + θ 3 + θ 4 × 154 360 o
cm
C 4 cm E F A [CBSE OD Comptt. Set-I, II, III, 2017]
[ 295
Areas related to circles
= 2(28)2
360 o × 154 360 o
∴
BC = 28 2 cm
=
2
= 154 cm 1 \ Area of shaded region = 350 – 154 = 196 cm2 1 [CBSE Marking Scheme, 2017] Q. 14. ABDC is a quadrant of a circle of radius 28 cm and a semi-circle BEC is drawn with BC as diameter. 22 Find the area of the shaded region. Use π = 7
Radius of semi-circle drawn on BC = = 14 2 cm Area of semi-circle =
.
E
B
.
D
A
½
1 22 × (14 2 )2 × 2 7
= 616 cm2
1
1 × 28 × 28 Area of DABC = 2 = 392 cm2
½
1 22 × 28 × 28 × Area of of quadrant = 4 7 = 616 cm2 1 Area of the shaded region = Area of semi-circle + Area of D – Area of quadrant = 616 + 392 – 616 = 392 cm2. ½ [CBSE Marking Scheme, 2017]
C
A [CBSE SQP, 2017]
28 2 2
Sol. As ABC is a quadrant of the circle, ∠BAC will be measured 90°. In DABC, BC2 = AC2 + AB2 = (28)2 + (28)2
Q. 15. Three semicircles each of diameter are 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.
cm
cm
cm
Topper Answer, 2017 Sol.
A [CBSE OD Set II, 2017]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
296 ]
3
Q. 16. In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the 22 area of the shaded region. Use π = 7
O 60° C
A
B
T
D
A [CBSE OD Set III, 2017] [CBSE OD, Set-I, II, III, 2016]
opper Answer, 2017
Sol.
3
[ 297
Areas related to circles
Q. 17. Figure shows two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm, show that area of shaded region is 25 3 - π cm 2 . 6
O 60°
P
A
B A [CBSE S.A.2, 2016]
Sol. As OA = 17 cm, AP = 15 cm and ∆OPA is right triangle.
∴ Using Pythagoras theorem in ∆OPA,
=
A [CBSE Delhi Set I, II, III, 2016]
Sol. Given,
OP = OQ = PQ = 10 cm ∠POQ = 60°
and
Area of segment PAQM
½
æ 100 p 100 3 ö cm 2 . = ç 4 ÷ø è 6
1
25π cm 2 2
½
Area of semicircle =
25π 50 π Area of shaded region = − − 25 3 . 3 2
pö æ = 25 ç 3 - ÷ cm 2 . è 6ø
1
(17 )2 − (15)2
⇒
Area of the shaded region
1
OP = 8 cm = Area of the sector AOBA
– Area of ∆OPA 1
60° 1 ´ pr 2 - ´ b ´ h = 2 360°
1 60° 22 × × 17 × 17 - × 8 × 15 = 2 360° 7
= 151.38 – 60
= 91.38 cm2
1
Q . 19. In the given figure, O is the centre of circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. A [CBSE OD Set-I, II, III, 2016] ( = 3.14) A
[CBSE Marking Scheme, 2016] Q. 18. In the given figure, AOB is a sector of angle 60° of a circle with centre O and radius 17 cm. If AP ⊥ OB and AP = 15 cm, find the area of the shaded region.
Topper’s Answer, 2016
Sol.
OA 2 − AP 2
OP =
O
B
C
298 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 20. Find the area of minor segment of a circle of radius 14 cm, when its central angle is 60°. Also, find the area of corresponding major segment. 22
A [CBSE OD Set I, II, II, 2015]
(Use p = ) 7
Sol. Here, r = 14 cm, θ = 60° Then, the area of minor segment cm
O
14
θ 1 = πr 2 - r 2 sin θ 2 360
3 Sol. From the given figure
1 × 10 × 10 Area of right-angled DABC = 2 = 50 cm2 1 Area of quadrant APR of the circle of radius 7 cm 1 × p × (7)2 = 4
½
60°
[Q Area of quadrant =
B
A
o
½
308 2 2 = ( - 49 3 ) cm = 17.89 cm = 17.9 approx. 1 3 22 × 14 × 14 and the area of major segment = 7 – ( =
308 - 49 3 ) 3
1540 2 + 49 3 =598.10 cm 3
= 598 cm2 approx. 1 [CBSE Marking Scheme, 2015] Q. 21. A momento is made as shown in the figure. Its base PBCR is silver plated from the front side. Find the area which is silver plated. 22 (Use p = ) 7
P cm
B
cm
R cm
Q. 22. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle. 22
(Use p = ) 7
A [CBSE Term-2, 2015]
U [CBSE Term-2, 2015]
Sol. Let radius of the circle be r cm.
Diameter = 2r cm
Circumference = 2pr
½
Circumference = Diameter + 16.8
½
or,
2pr = 2r + 16.8
22 or, 2 r = 2r + 16.8 7
44 r = 2r + 16.8 7
or,
44r = 14r + 16.8 × 7
or,
30r = 117.6
or, \
C
1 22 × × 49 4 7
= 38.5 cm2 1 DABC \ Area of base PBCR = Area of – Area of quadrant APR = 50 – 38.5 = 11.5 cm2. 1 [CBSE Marking Scheme, 2015]
or,
A cm
=
60 3 22 1 × 14 × 14 × × 14 × 14 × = o 7 360 2 2
90° 2 πr ] 360°
r =
117.6 30
r = 3.92 cm
1 1
[CBSE Marking Scheme, 2015]
[ 299
Q. 23. In fig., APB and AQP are semi-circles, and AO = OB. If the perimeter of the figure is 47 cm, find the area
of the shaded region. Use π =
or,
22
or,
7
A [CBSE Delhi Set-I, II, III, 2015]
Areas related to circles
Q
or,
3πr + 2r = 47 cm 2
Q
3 22 r × + 2 = 47 cm 2 7
A
33 r + 2 = 47 cm 7
O
B
P
47 × 7 cm r= 47 or, r = 7 cm 1 Now, area of shaded region 1 A = area of circle – area of circle 4
or,
A
B
O
=
3 area of circle 4
=
3 × πr 2 4
P
Sol. Let ‘r‘ be the radius of given circle. Perimeter of given figure = 47 cm th
1 Perimeter of full circle – perimeter of circle 4 = 47 – 2r ½ 1 2 πr − ( 2 πr ) = 47 – 2r ½ or, 4
3 22 = × × 7 × 7 cm2 4 7 3 = × 77 cm2 2
= 115.5 cm2 1 [CBSE Marking Scheme, 2015]
Long Answer Type Questions Q. 1. Find the area of the shaded region in the given figure if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Q
O
R
P
A [CBSE OD Set-I, 2020] Sol. Given, PQ = 24 cm, PR = 7 cm We know that the angle in the semicircle is right angle. Here, ∠RPQ = 90° 1 In DRPQ, RQ2 = PR2 + PQ2 (By Pythagoras theorem) ⇒ RQ2 = (7)2 + (24)2 = 49 + 576 = 625 \ RQ = 25 cm 1 1 × RP × PQ \ Area of DRPQ = 2 1 × 7 × 24 = 84 cm2 1 = 2
and area of semi-circle = =
1 × pr2 2
1 22 25 × × 2 7 2
2
=
5 marks each 11 × 625 6875 = cm 7×4 28
1
Now, area of shaded region = area of semi-circle – area of DRPQ 6875 6875 − 2352 – 84 = = 28 28 4523 = 28 = 161.54 cm2. 1 Q. 2. 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. C + A [CBSE Delhi Set-I, II, 2017]
O'
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
= 308 cm2 1 Area the shaded region = Area of square + Area of two circles – area of two quadrants = 784 + 1232 – 308 = 1708 cm2 Hence, the area of shaded region = 1708 cm2 1 [CBSE Marking Scheme, 2017]
Sol. Given, the side of the square = 28 cm. Area of the square = 28 × 28 = 784 cm 28 = 14 cm Radius of each circle = 2 22 × 14 × 14 \ Area of two circles = 2 × 7 = 1232 cm2 o Area of the 2 quadrants =
90 o ´ p ´ 14 ´ 14 ´ 2 360
300 ]
1
1
1
Q. 3. In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠ BOD = 90°. Find the area of the shaded region. A B O C
D
C + A [CBSE OD Set-III, 2017]
Topper Answer, 2017
Sol.
5
Q. 4. In the given figure, DABC is a right angled triangle in which ÐA = 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.
C + A [CBSE OD Set-II, 2017]
3c
m
A
B
4c
m C
[ 301
Areas related to circles
Sol. In DABC, ÐA = 90°, AB = 3 cm, and AC = 4 cm \ By using Pythagoras theorem, BC = AB 2 + AC 2 = 3 2 + 4 2 = 5 cm. 1 3 Area of semicircle with radius cm + Area of semi 2 2
Q. 6. Four equal circles are described at the four corners of a square so that each touches two others. The shaded area enclosed between the circles is 24 cm 2 . Find the radius of each circle. 7 A [CBSE SQP, 2016]
Sol.
2
p3 p4 4 -circle with radius cm = + ½ 2 2 2 2 2 5 cm – Area of Area of semicircle with radius 2
D
C
1
2
π 5 1 DABC = − × 3 × 4 2 2 2 ...(i) 1
Area of shaded region 2
=
π 3 π 2 25 2 + ( 2 ) − π − 6 cm 2 2 2 8
1
π 9 25 = + 4 − + 6 2 4 4 =
π 9 16 − 25 + +6 2 4 4
=
π 9 9 − +6 2 4 4
1
= 6 cm2 [CBSE Marking Scheme, 2017] ½ Q. 5. A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width of 0.7 m. Find the expenditure of cementing the path. If its cost is ` 110 per sq. m. C + A [Foreign Set-I, II, III, 2017]
24 90° or, (2r)2 – 4 × πr 2 = 360° 7 22 24 4r 2 - r 2 = or, 7 7
5c D
θ
O
10 cm P
m
The width of path = 0.7 m \Radius of park with path = 3.5 + 0.7 = 4.2 m Area of the path = p(4.2)2 – p(3.5)2 22 = (17.64 − 12.25) 7
1
Q. 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
½ 1
1
5c
Given, the diameter of park = 7 m 7 \ Radius = = 3.5 m 2
1
24 28r 2 - 22r 2 = or, 7 7 2 6r = 24 or, r2 = 4 1 or, r =± 2 or, Radius of each circle is 2 cm (r cannot be negative) [CBSE Marking Scheme, 2016]
Sol.
Let r cm be the radius of each circle. 24 cm2 Area of square – Area of 4 sectors = 7
m
25 2 = π − 6 cm 8
B
A
1 ½
22 × 5.39 = 22 × 0.77 = 7
= 16.94 m2 1 Cost of the cementing the path = 16.94 × 110 = ` 1863.40 1 [CBSE Marking Scheme, 2017]
B C + A [CBSE Delhi Set I, II, III, 2016]
Sol. In right angled DOAP,
5 1 cos θ = = or, θ = 60° 10 2 So, ∠AOB = 60° + 60° = 120° Reflex ∠AOB = 240° ∴ Length of arc 2 × 3.14 × 5 × 240 o ADB = 360 o = 20.93 cm
1 1
1
302 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
θ ∵ l = 360° × 2 πr
Hence length of elastic in contact = 20.93 cm
AP =
OP 2 − OA 2
Area (∆OAP + ∆OBP) = Area of DOAP + Area of DOBP 1 = 2 × × 5 × 5 3 2 = 25 3 = 43.25 cm2
[By using Pythagoras theorem]
½
25 × 3.14 × 120 Area of sector OACB = 360 o = (10 ) − ( 5) = 75 = 26.16 cm2 ½ Now, AP = 5 3 cm Shaded Area = 43.25 – 26.16 = 17.09 cm2 1 Q. 8. The given fig. shows a sector OAP of a circle with centre O, containing Ðq. AB is perpendicular to the radius πθ − 1 OA and meets OP produced at B. Prove that the perimeter of shaded region is r tan θ + sec θ + 180 o 2
o
2
B
P
θ O
r
A
Topper Answer, 2016
Sol.
C + A [CBSE OD Set-I,II, III, 2015, 16]
[ 303
Areas related to circles
Q. 9. 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). S
R
Now, area of sector OQR = =
90° × πr 2 360°
1 22 × × 882 m 2 4 7
= 693 m2
1
And area of right angled DQOR O
=
=
P
Q
C + A [CBSE OD Set-I, II, III, 2015]
Sol. Radius of circle with centre O is OR. Let OR = x
1 × 21 2 × 21 2 2
= 441 m2
½
Area of flower bed (shaded part) with QR
= area of sector OQR – area of DQOR
x2 + x2 = (42)2 or, x = 21 2 m 1 (Using pythagoras theorem) Area of the flower bed = Area of segment of circle with centre angle 90° 22 90 o × 21 2 × 21 2 × = 7 360 o
∴
– 1 × 21 2 × 21 2 1 2
= 693 – 441 = 252 m2 2 2 \ Area of flower beds = 2 × 252 = 504 m . 1 [CBSE Marking Scheme, 2015] Detailed Solution: S
1 × OQ × OR 2
= (693 – 441) m2 = 252 m2 ½ Similarly, area of flower bed with PS = 252 m2 Hence Total area of both flower beds with QR and PS = (252 + 252) m2 = 504 m2.
1
Q. 10. 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.
R 42 m
O
42 m
42 m
xm
1
xm
C + A [CBSE Term-2, 2015] Sol. Let r be the radius of incircle.
P
Q
In DBAC,
Let OR be x cm, then OQ = x cm
Since, diagonals of a square bisect each other at right angles,
then
∠QOR = 90°
In DQOR,
x2 + x2 = (42)2
[Using Pythagoras theorem]
⇒ 2x2 = 1764 ⇒ ⇒
i.e., radius (r) of each semi-circle = 21 2 m
BC 2 − AB2
(By using Pythagoras theorem)
=
10 2 − 6 2
=
100 − 36
=
64
= 8 cm
x2 = 882 x = 21 2 m
AC =
1
Here, ∠A = 90°, ∠P = 90° and ∠Q = 90° Then
∠O = 360° – (90° + 90° + 90°)
= 90°
1
304 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
\ OPAQ is a square.
Q. 11. Two circular beads of different sizes are joined together such that the distance between their centres is 14 cm. The sum of their areas is 130 p cm2. Find the radius each bead. C + A [CBSE Term-2, 2015]
A
rc
m rc
m
P
90°
Q
C
8 r
Then,
m r c
m rc
8
-r
m
O
rc
O
B
r
10 cm
Sol. Let the radii of the circles are r1 cm and r2 cm.
BC = 10 = 8 – r + 6 – r
1
(By using the tangent properties)
or,
2r = 8 + 6 – 10
or,
2r = 4 or, r = 2 cm
\
Area of circle = pr2 =
Now,
=
area of DABC =
22 ×2×2 7
88 = 12.57 cm2 7
1
1 ×8×6 2
= 24 cm2
1
\Area of shaded region = Area of DABC – Area of circle = 24 – 12.57 = 11.43 cm2
1
r1 cm
Q. 1. A horse is tied to a peg at one corner of a square shaped grass field of sides 15 m by means of a 5 m long rope (see the given figure). C + AE
O'
\ r1 + r2 = 14 (Given) ...(i) and sum of their areas = pr12 + pr22 130p = p(r12 + r22) (Given) 1 or, 130p = p(r12 + r22) 2 \ r1 + r22 = 130 ...(ii) 1 (r1 + r2)2 = r12 + r22 + 2r1r2 or, (14)2 = 130 + 2r1r2 or, 2r1r2 = 196 – 130 = 66 1 (r1 – r2)2 = r12 + r22 – 2r1r2 = 130 – 66 = 64 or, r1 – r2 = 8 ...(iii) 1 From (i) and (iii), 2r1 = 22 or, r1 = 11 cm and r2 = 14 – 11 = 3 cm. 1 [CBSE Marking Scheme, 2015]
Visual Case Based Questions ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark
r2 cm
4 marks each Explanation: Area of square = (side)2 = 15 × 15 = 225 m2
1
(ii) The area of that part of the field in which the horse can graze. (a) 19.625 m2
(b) 19.265 m2
(c) 19 m2
(d) 78.5 m2
Sol. Correct option: (a) Explanation: From the figure, it can be observe that the horse can graze a sector of 90° in a circle of 5 m radius. Area that can be grazed by horse = Area of sector
(i) What is the area of the grass field? (a) 225 m2 (b) 225 m (c) 255 m2 (d) 15 m Sol. Correct option: (a)
=
90° × πr 2 360°
=
1 × 3.14 × 5 × 5 4
= 19.625 m2
1
[ 305
(iii) The grazing area if the rope were 10 m long instead of 5 m. (d)
78.5 m2
Explanation: Area that can be grazed by the horse when length of rope is 10 m long 90° = × πr 2 360°
1
(b) 110 mm
(c) 50 mm
(d) 10 mm
= 110 mm
1
(iii) What is the total length of silver wire required? (b) 825 mm
(d) 78.5 m2
(c) 285 mm
(d) 852 mm
Sol. Correct option: (c)
= 58.875 m
Explanation: Length of wire required = 110 + 5 × 35
2
2
1
(v) The given problem is based on which concept?
= 110 + 175 = 285 mm.
(a) Coordinate geometry (b) Area related to circles (c) Circle (d) None of these Sol. Correct option: (b)1 Q. 2. In a workshop, 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 C + AE
(i) What is the radius of the circle? (b)
5 mm 2
(d) 10 mm
1
(iv) What is the area of the each sector of the brooch?
385 (a) mm2 2
(b)
358 mm2 4
585 (c) mm2 4
(d)
385 mm2 4
Sol. Correct option: (d) Explanation: It can be observed from the figure that an angle of each 10 sectors of the circle is subtending at the centre of the circle. \ Area of each sector =
36 × πr 2 360°
=
1 22 35 35 × × × 10 7 2 2
=
385 mm 2 4
the given figure.
22 35 × 7 2
(a) 528 mm
= (78.5 – 19.625) m
(a) 100 mm
(b) 58.875 m2
Explanation: Increase in grazing area
Sol. Correct option: (a)
(ii) What is the circumference of the brooch?
= 2 ×
Sol. Correct option: (b)
35 mm (a) 2 (c) 35 mm
1
Explanation: Circumference of brooch = 2pr
1 × 3.14 × 10 × 10 4
(iv) The increase in the grazing area if the rope were 10 m long instead of 5 m.
(c) 58 m2
35 mm 2
Sol. Correct option: (b)
= 78.5 m2
(a) 58.758 m2
=
=
Diameter 2
Sol. Correct option: (d)
Explanation: Radius of circle =
(c) 225 m2
(b) 785 m2
(a) 7.85 m2
Areas related to circles
1
(v) The given problem is based on which mathematical concept ?
(a) Areas Related to circles (b) Circles (c) Construction (d) none of these Sol. Correct option: (a).
1
306 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 3. AREAS RELATED TO CIRCLES Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu.
During the festival of Onam , your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below. Observe these carefully.
Explanation:
radius = 7 cm
diameter = 2 × 7 cm
= 14 cm
side of square = 14 cm + 14 cm + 14 cm
= 42 cm.
Area of square = side2
= (42 cm)2 = 1764 cm2 (iv) Area of each circular design is
(a) 124 cm2
(b) 132 cm2
2
(d) 154 cm2
(c) 144 cm
Sol. Correct option: (d). Explanation:
radius = 7 cm
Area of each circular design = pr2 =
22 ×7×7 7
= 154 cm2
(v) Area of the remaining portion of the square ABCD is (a) 378 cm2
(b) 260 cm2
2
(d) 278 cm2
(c) 340 cm
Sol. Correct option: (a). Design I: This design is made with a circle of radius 32 cm leaving equilateral triangle ABC in the middle as shown in the given figure.
Design II: This Pookalam is made with 9 circular design each of radius 7 cm.
Area of 9 circular design = 9 × pr2 = 9 ×
Refer Design I: (i) The side of equilateral triangle is
(a) 12 3 cm (b) 32 3 cm
(c) 48 cm
(d) 64 cm
Sol. Correct option: (b).
Explanation:
(ii) The altitude of the equilateral triangle is
22 ×7×7 7
= 1386 cm2
Area of square = 1764 cm2
Area of remaining portion of square ABCD = Area of square – Area of 9 circular design 2
(a) 8 cm
(b) 12 cm
= 1764 cm – 1386 cm2
(c) 48 cm
(d) 52 cm
= 378 cm2
Sol. Correct option: (c).
Q. 4.
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 brooch are shown below. Observe them carefully.
Refer Design II:
(iii) The area of square is
(a) 1264 cm2
(b) 1764 cm2
(c) 1830 cm2
(d) 1944 cm2
Sol. Correct option: (b).
A Brooch
[ 307
Areas related to circles
= 112 + 88 = 200 mm
(ii) The area of each sector of the brooch is
(a) 44 mm2 (b) 52 mm2
(c) 77 mm2 (d) 68 mm2
Sol. Correct option: (c). Explanation: Area of each sector of Broof =
1 × Area of Brooch 8
=
1 × πr 2 8
=
1 22 × × 14 × 14 8 7
= 77 mm2
Refer to Design B
(iii) The circumference of outer part (golden) is
(a) 48.49 mm
(b) 82.2 mm
(c) 72.50 mm
(d) 62.86 mm
Sol. Correct option: (d). (iv) The difference of areas of golden and silver parts is
(a) 18p (b) 44p
(c) 51p (d) 64p
Sol. Correct option: (c). Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts.
Design B: Brooch b is made two colours i.e. 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. Refer to Design A (i) The total length of silver wire required is
(a) 180 mm
(b) 200 mm
(c) 250 mm
(d) 280 mm
(v) A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm ?
(a) 2
(b) 3
(c) 4
(d) 5
Sol. Correct option: (c). Explanation: Circumference of silver part of Brooch = 44 cm
2pr = 44 mm 2×
22 × r = 44 7
Sol. Correct option: (b).
Explanation:
Diameter = 28 mm
radius of whole Brooch
radius = 14 mm
r = 7 mm.
= 7mm + 8 mm
Total length of wire = length of 4 diameter
= 10 mm.
Circumference of outer edge
+ circumference of circle.
= 4 × 28 + 2pr2 22 × 14 = 112 + 2 × 7
= 2pr = 2 ×
22 × 10 7
308 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
=
440 mm 7
let the number of revolutions = n Now, According to question,
O
n.2pr = 80p
n.
440 = 80p 7
n.
440 22 = 80 × 7 7 n = 4
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sURFACE aREAS AND VOLUMES
[ 309
C H A P TE R
13
SURFACE AREAS AND VOLUMES
Syllabus Surface areas and volumes of combinations of any two of the following : cubes, cuboids, spheres, hemispheres and right circular cylinders, cones. Frustum of a cone. ¾¾ Problems involving converting one type of metallic solid into another and other mixed problems. Problems with combination of not more than two different solids. ¾
¾
Trend Analysis 2018 List of Concepts Surface Areas and volumes
Delhi
Outside Delhi
2 Q (3 M) 1 Q (4 M)
2019 Delhi
Outside Delhi
1 Q (4 M) 1 Q (3 M) 1 Q (4 M)
Delhi
Outside Delhi
1 Q (1 M)
3 Q (2 M) 1 Q (4 M)
1 Q (1 M)
Conversion of one type of metallic solid into another Frustum of a cone
2020
1 Q (4 M)
1 Q (4 M)
2 Q (4 M)
1 Q (1 M)
TOPIC - 1
Surface Areas and Volumes TOPIC - 1
Revision Notes A sphere is a perfectly round geometrical object in three-dimensional space .
Surface Areas and Volumes Page No. 310
TOPIC - 2 Scan to know more about this topic
Conversion of One type of Metallic Solid into Another Page No. 327
Surface area and volumes
Frustum of a Cone Page No. 332
[ 311
sURFACE aREAS AND VOLUMES
A Cone is a three dimensional geometric shape tapers smoothly from a flat base to a point called the apex or vertex. Scan to know more about this topic
O
A cylinder is a solid or a hollow object that has a circular base and a circular top of the same size. A hemisphere is half of a sphere.
Combined and Conversion of Solids
Know the Formulae Cuboid:
Lateral surface area or area of four walls = 2(l + b)h
Total surface area = 2(lb + bh + hl)
Volume = l × b × h Diagonal =
2
2
l +b +h
2
Here, l is length, b is breadth and h is height of the cuboid Cube:
Lateral surface area or area of four walls = 4 × (edge)2
Total surface area = 6 × (edge)2
Volume = (edge)3
Diagonal of a cube =
3 × edge.
Right circular cylinder:
Area of base or top face = πr²
Area of curved surface or curved surface area = perimeter of the base × height
= 2πrh
Total surface area (including both ends) = 2πrh + 2πr² = 2πr(h + r)
Volume = (Area of the base × height)= πr²h
Here, r is the radius of base and h is the height of the right circular cylinder. Right circular hollow cylinder: Total surface area = (External surface area + internal surface area) + (Area of brim)
= (2πRh + 2πrh) + 2(πR² – πr²)
= [2πh(R+ r) + 2π(R² – r²)] = [2π(R + r)(h + R – r)]
Volume of the material used = (External volume) – (Internal volume)
R
Curved surface area = (2πRh + 2πrh)= 2πh(R + r)
= πR²h – πr²h = πh(R² – r²) Here, R and r are the external and internal radii and h is the right circular height of the right circular hollow cylinder.
312 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Right circular cone: h2 + r 2
Slant height, l =
l
Area of curved surface = πrl = πr h 2 + r 2
Total surface area = Area of curved surface + Area of base
= πrl + πr² = πr(l + r) Volume =
1 3
πr²h
Here, r, h and l are the radius, vertical height and slant height respectively of the right circular cone. Sphere:
Surface area = 4πr² Volume =
4 3
πr3
Here, r is the radius of the sphere. Spherical shell: Surface area (outer) = 4πR2
Volume of material =
=
4 3 4 3
πR3 –
4 3
πr3
π(R3 – r3)
R r
Here, R and r are the external and internal radii of the spherical shell. Hemisphere:
Area of curved surface = 2πr²
Total surface area = Area of curved surface + Area of base
= 2πr² + πr² = 3πr²
Volume =
2 3
πr3
Here, r is the radius of the hemisphere.
Know the Terms The platonic solids also called the regular solids or regular polyhedra. 5 such solids are : dodecahedron, icosahedron, octahedron and tetrahedron.
Greek mathematician Plato equated tetrahedron with the ‘element’ fire, the cube with earth, the icosahedron with water, the octahedron with air and dodecahedron with the stuff of which the constellations and heavens were made. The stone of platonic solids are kept in Ashmolean Museum in Oxford. The fonds of Archimedes carried a sculpture consisting of a sphere and cylinder circumscribing it.
[ 313
sURFACE aREAS AND VOLUMES
How is it done on the
GREENBOARD?
Q.1. The circumference of the base of a conical tent is 44 m. If the height of tent is 24 m, find the length of the canvas used in making the tent, if the width of the canvas is 2 m. 22 Use π = 7 U Solution: Step I: Diagrammatic representation: Given that h = 24 m Circumference = 44 m height, h = 24 m
Circumference = 44 m
Step II: Calculation of radius: Given that circumference = 44 m
or,
2pr = 44 44 44 × 7 or, r = = =7m 2π 2 × 22 Step III: Calculation of slant height: l2 = r2 + h2 l2 = (7)2 + (24)2 = 49 + 576 l2 = 625 or, l = 25 m Step IV: Calculation for curved surface area: C.S.A. = prl 22 = × 7 × 25 m2 7 = 550 m2 Step V: Calculation of length of canvas: We know that area of rectangle = length × breadth or, 550 = l × 2 or, l = 225 m
Very Short Answer Type Questions
Given,
h1 r1 1 3 = and = h2 r 3 1 2 2
2
æ 3ö æ 1ö ær ö æh ö 3 = ç 1 ÷ ç 1 ÷ = ç ÷ ç ÷ = è 1ø è 3ø 1 è r2 ø è h2 ø
Hence, ratio of their volumes is 3 : 1. ½ Q. 2. Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas. R [CBSE OD Comptt. Set I, II, III, 2018]
Sol. Let the sides of two cubes be a and A, then a3 1 = (given) ½ 3 27 A
Q. 1. Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their A [CBSE Delhi Set-I, 2020] volumes ? Sol. Let h1 and h2 be height and r1, r2 be radii of two cones, then ratio of their volumes 1 2 pr1 h1 = 3 ½ 1 2 pr2 h2 3
1 mark each
a 1 = A 3 We know that the surface area of a cube is 6(side)2.
⇒
6a2
2
æ 1ö 1 = ç ÷ = . ½ 2 è ø 3 6A 9 [CBSE Marking Scheme, 2018]
\ Ratio of sufrace areas =
Q. 3. If the volume of a cube is 8 cm3, then what is the A surface area of a cube. Sol. Given, volume of a cube = 8 cm3 Let the side of a cube be a cm, then a3 = 8 a =
3
8
=
3
2 × 2 × 2 = 2 cm
⇒
314 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Then, Surface area of a cube = 6a2
= 6 × 2 × 2 = 24 cm2.
Short Answer Type Questions-I Q. 1. The volume of a right circular cylinder with 1 its height equal to the radius is 25 cm3. Find the 7 22 height of the cylinder. Use p = 7 A [CBSE OD Set-I, 2020]
Sol. Given,
1 Volume of a right circular cylinder = 25 cm 7 176 7 where h is height and r is radius then 22 176 × h2 × h = 7 7
i.e.,
⇒
pr2h =
h3 =
1
on squaring both of the sides, we get 4r2 = h2 + r2 ⇒ 4r2 – r2 = h2 ⇒ 3r2 = h2 ⇒
½
Slant height of a cone,
r 2 + h2 So, curved surface area of a cone = prl l =
2
2
½
i.e., 2pr = pr h + r (Given) ⇒ 2r =
h2 + r 2
1 3
= 1 : 3 .
½
Sol. Given, height (h) = 14 cm and Base radius (r) = 6 cm Volume of the remaining solid = Volume of a right circular cylinder – Volume of a right circular cone 1 = pr2h –
1 2 pr h 3
=
2 2 pr h 3
=
2 22 × × 6 × 6 × 14 3 7
=1056 cm3.
1
Q. 4. Isha is 10 years old girl. On the result day, Isha and her father Suresh were very happy as she got first position in the class. While coming back to their home, Isha asked for a treat from her father as a reward for her success. They went to a juice shop and asked for two glasses of juice.
2
=
1 r = ⇒ 3 h Hence, the ratio of the radius and the height
1
= pr h 2 + r 2
h2
A [CBSE OD Set-III, 2020]
Hence, height of the cylinder = 2 cm. 1 Q. 2. A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, then find the ratio of the radius and the height of the conical part. A [CBSE OD Set-II, 2020] Sol. Let ABC be a cone, which is mounted on a hemisphere. Given: OC = OD = r cm ½ Curved surface area of the hemispherical part 1 = (4pr2) 2
r2
Q. 3. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base removed. Find the volume of the remaining solid.
176 = 8 = 2 3. 22
= 2pr2
2 marks each
Aisha, a juice seller, was serving juice to her customers in two type of glasses. Both the glasses had inner radius 3 cm. The height of both the glasses was 10 cm.
[ 315
sURFACE aREAS AND VOLUMES
First type: A Glass with hemispherical raised bottom.
\ Capacity of conical part 1 2 = πr h 3 1 × π × ( 3)2 × 1.5 3
=
= 4.5p cm3
Now, capacity of Juice in second glass Second type: A glass with conical raised bottom of height 1.5 cm.
Isha insisted to have the juice in first type of glass and her father decided to have the juice in second type of glass. Out of the two, Isha or her father Suresh, who got more quantity of juice to drink
and by how much ?
U [CBSE SQP, 2020]
Sol. Capacity of first glass = pr2H –
2 pr3 3
Capacity of second glass = pr2H –
1 2 pr h 3
= p(3)2 × 10 – 4.5p = 85.5 × 3.14 = 268.47 cm3
i.e., Suresh got 268.47 – 226.08 = 42.39 cm3 more juice than Isha. ½
1 1
= p × 3 × 3(10 – 0.5) = 85.5p cm3 ∵ Suresh got 42.39 cm3 more quantity of juice. [CBSE SQP Marking Scheme, 2020]
3 cm
\
= pr2H
2 3 πr 3
πr r + h
⇒
⇒
2 3 ½ = π × ( 3) 3 \ Capacity of Juice in first glass
3 cm
2
=
8 5
½
2
=
4 5
=
16 25
2
r +h h2 2
2
r +h
⇒
r
2 2
h
=
½
9 16
r 3 ½ = h 4 [CBSE Marking Scheme, 2018]
Detailed Solution:
1.5 cm
2 2 3 = π × ( 3) × 10 − π × ( 3) 3
l
= 72p cm3 = 72 × 3.14 cm
2
⇒ 9h2 = 16r2
Capacity of hemispherical part
= 9p × 8
½
⇒ 25h2 = 16r2 + 16h2
= p × (3)3 × 10
2 × 3] 3
2 πrh
10 cm
Capacity of Juice in cylindrical part
2 = π × ( 3) [10 −
r 2 + h2
Slant height of cone =
h
Height (H) = 10 cm
=
U [CBSE O.D. Comptt. Set-I, II, III, 2018]
Sol. Let r be the radii of bases of cylinder and cone and h be the height
Detailed Solution: In case of first type glass, we have radius (r) = 3 cm
½
So, Suresh used second glass for drinking juice, so he got more quantity of juice.
Q. 5. A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the ratio between radius of their bases to their height is 3 : 4.
= p × 9(10 – 2) = 72p cm3
1 = πr 2 H − πr 2 h 3
3
½
h
3 cm
= 226.08 cm3 In case of second glass, we have. height of bottom part (h) = 1.5 cm
½ r
316 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Let r be the radii of bases of cylinder and cone and h be the height, then Curved surface area of cylinder = 2prh.
In such type of problems, mostly students
and curved surface area of a cone = prl = πr × h 2 + r 2 where,
½
h2 + r 2
l =
COMMONLY MADE ERRORS write incorrect formulas of surface area of cylinder and cone and also they do errors in calculation. Students write the formula of cylinder in place of cones and vice-versa.
Given, 2 πrh 2
2
=
8 5
=
4 5
πr × h + r ⇒
⇒
h h2 + r 2
ANSWERING TIP
1
Adequate practice and remembering of formulae is necessary.
Q. 6. A cylindrical glass tube with radius 10 cm has water upto a height of 9 cm. A metal cube of 8 cm edge is immersed completely. By how much the water level will rise in the glass tube ?
h2
16 = r 2 + h 2 25
⇒ 25h2 = 16r2 + 16h2 2
2
2
2
⇒ 25h – 16h = 16r ⇒ 9h = 16r
A [CBSE Term-2, 2015]
Sol. L et the height of water raised measured be h cm. ½ ∴ Volume of water displaced in cylinder = π(10)2h ½ Volume of cube = 8 × 8 × 8 cm3 ½ ∴ π(10)2h = 8 × 8 × 8 8×8×8×7 ½ ∴ h = 22×10×10 = 1.629 cm. [CBSE Marking Scheme, 2015]
2
r2
9 = h 2 16 r 3 ⇒ = ½ h 4 Hence, the ratio between radius of their bases to their heights is 3 : 4. Proved. ⇒
Short Answer Type Questions-II Q. 1. A solid is in the form of a cylinder with hemispherical end. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. 22 Find the total volume of the solid. Use π = 7
Detailed Solution:
A [CBSE OD Set-1, 2019]
Sol.
3 marks each
Height of cylinder = 20 – 7 = 13 cm.
1
Height of the cylinder (h) = (20 – 7) cm = 13 cm 1 7 cm Radius of circular part (r) = 2 Volume of solid = Volume of cylinder + 2 × Volume of hemisphere ½ æ 2p 3 ö 2 r 1 V = pr h + 2 × ç è 3 ÷ø 4 ö æ = pr2 ç h + r ÷ è 3 ø
2
3
4 7 7 Total volume = p × 13 + π cm3 1 2 3 2 22 49 4 7 × 13 + . cm3 = 7 4 3 2 77 × 53 6 = 680.17 cm3 (Approx) 1 [CBSE Marking Scheme, 2019] =
[ 317
sURFACE aREAS AND VOLUMES
22
7
7é
4
Q. 2. A Cylindrical tank of radius 40 cm is filled upto height 3.15 m by an other cylindrical pipe with the 1 hour. Calculate the diameter rate of 2.52 km/h in 2 of cylindrical pipe ?
7ù
´ ´ 13 + ´ ú = 7 2 2 êë 3 2û
77 53 3 = cm 2 3
T
= 680.17 cm3 (Approx)
Sol.
½
opper's Answer, 2019
C + A [CBSE Delhi Region, 2019]
[CBSE Delhi Set-I, II, III, 2015]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
3
318 ]
Q. 3. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in fig. 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. C+ A [CBSE Delhi/OD, 2018]
COMMONLY MADE ERROR Mostly students are unable to find the radius of hemisphere also they use the 22 and they value of p, 3.14 in place of 7
Sol. Total surface Area of article = CSA of cylinder + CSA of 2 hemispheres CSA of cylinder = 2prh 22 × 3.5 × 10 = 2 × 7 = 220 cm2 1 Surface area of two hemispherical scoops 22 × 3.5 × 3.5 = 4 × 7 = 154 cm2 1 \ Total surface area of article = 220 + 154 = 374 cm2 1 [CBSE Marking Scheme, 2018]
do errors in calculation, they subtract the area of hemisphere from T.S.A. of cylinder in place of adding these.
ANSWERING TIP They should read the question clearly and use right formula and correct calculation for which good practice is necessary.
Q. 4. 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 A [CBSE Delhi/O.D. 2018] much canvas cloth is required to just cover the heap ? Sol.
½
Radius of conical heap = 12 m 1 22 × 12 × 12 × 3.5 m3 Volume of rice = × 3 7 = 528 m3 Area of canvas cloth required = prl l = 12 2 + ( 3.5)2 = 12.5 m
1 ½
22 × 12 × 12.5 7 = 471.4 m2 [CBSE Marking Scheme, 2018] 1
\
Area of canvas required =
T
Detailed Solution:
opper's Answer, 2018
[ 319
sURFACE aREAS AND VOLUMES
3
COMMONLY MADE ERROR Sometimes the students find TSA of the canvas in place of CSA
ANSWERING TIP They should have clear idea about C.S.A and T.S.A and volume.
T
Q. 5. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The A [CBSE Delhi/OD, 2017] total height of the toy is 15.5 cm. Find the total surface area of the toy.
Sol.
opper's Answer, 2017
320 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
3
3 Q. 6. The 4 th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical A [CBSE Delhi Set-I, 2017] vessel.
Q. 8. The given figure is a decorative block, made up of two solids – a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the
total surface area of the block. Use π =
[CBSE OD Set-II, 2016]
22
. 7
Sol. Radius of conical vessel = 5 cm and its height = 24 cm 1 Volume of this vessel = πr 2 h 3 1 × π × 5 × 5 × 24 3 = 200p cm3. Internal radius of cylindrical vessel = 10 cm Let the height of emptied water be h. \ Volume of water in cylinder 3 = × Volume of cone 4 =
1
3 1 × Volume of cone 4 Þ p × 10 × 10 × h = 150p Þ h = 1.5 cm Hence the height of water = 1.5 cm 1 [CBSE Marking Scheme, 2017]
Þ pr2h =
Q. 7. Rampal decided to donate canvas for 10 tents conical in shape with base diameter 14 m and height 24 m to a centre for handicapped person's welfare. If the cost of 2 m wide canvas is ` 40 per meter, find the amount by which Rampal helped the centre. C + A [CBSE OD Comptt. Set-I, II, III, 2017] [CBSE OD Set-I, II, III, 2015, 2016] Sol. Diameter of tent = 14 m and height = 24 m \ radius of tent = 7 m
Slant height =
=
A [CBSE Delhi Set I, II, III, 2016]
Sol. Surface area of block 22 3.5 3.5 22 3.5 3.5 × × +2× × × 1+½+½ = 216 − 7 2 2 7 2 2 = 225.625 cm2. [CBSE Marking Scheme, 2016] 1 Detailed Solution: Given, side of cube = 6 cm diameter of hemisphere = 3.5 cm 3.5 ½ radius of hemisphere = 2
Total surface area of cube = 6a2 = 6 × (6)2 = 216 cm2 ½ Total surface area of solid = TSA of cube – Area of circle + CSA of hemisphere. 2
h 2 + r 2 = 24 2 + 7 2 576 + 49 = 25 m
1
Surface area of the tent = prl 22 × 7 × 25 = 7 2
= 550 m 1 Surface area of 10 tents = 550 × 10 = 5500 m2 40 = ` 110000 Total cost = 5500 × 2 Hence, the amount by which Rampal helped the centre = ` 110000 1 [CBSE Marking Scheme, 2017]
2
æ 3.5 ö æ 3.5 ö = 216 cm2 – p. ç ÷ cm2 + 2.p. ç ÷ cm2 1 è 2 ø è 2 ø 22 3.5 3.5 22 3.5 3.5 ö æ ´ ´ +2´ ´ ´ = çè 216 cm2 7 2 2 7 2 2 ÷ø 77 77 ö æ + ÷ cm2 = ç 216 è 8 4ø
æ 1728 - 77 + 154 ö 2 = ç ÷ø cm è 8
1805 = 225.625 cm2 8
1 Q. 9. 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 water level into the cylindrical vessel rises by 5 3 cm. Find the diameter of the cylindrical vessel. 9 =
A [CBSE OD Set–II, 2016]
[ 321 T
sURFACE aREAS AND VOLUMES
opper's Answer, 2016
Sol.
Diameter of earth dug out = 4 m Radius of earth dug out, r = 2 m Depth of the earth, h = 21 m, Volume of earth dug out = pr2h 22 × 2 × 2 × 21 = 7 = 264 m3 1 Width of embankment = 3 m Outer radius of ring = 2 + 3 = 5 m Let the height of embankment be h \ Volume of embankment = Volume of earth dug out p(R2 – r2)h = 264 22 × ( 25 − 4 ) × h = 264 1 7 264 × 7 =4 h= 22 × 21 ∴ Height of embankment = 4 m. 1 [CBSE Marking Scheme, 2016]
Sol.
Q. 11. The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of 22 the cylinder. Use π = 7 U [CBSE Delhi Set I, 2016]
Sol. Here r + h = 37 and 2pr(r + h) = 1628 or, 2pr × 37 = 1628 1628 or, 2pr = 37 r = 7 cm h = 30 cm.
or, and
Hence, volume of cylinder = pr2h =
½ ½
½ ½ 22 × 7 × 7 × 30 7
1 = 4620 cm3. [CBSE Marking Scheme, 2016] Q. 12. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in A [CBSE OD Set I, II, III, 2015] this transfer.
Q. 10. A well of diameter 4 m dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankU [CBSE Delhi Set I, II, III, 2016, 2015] ment.
3
Sol.
Volume of bowl =
2 3 πR 3
2 3 3 Volume of liquid in bowl = p ´ (18 ) cm ½ 3 Volume of liquid after wastage 90 2 = π × (18 )3 × cm 3 ½ 3 100 Volume of one bottle = pr2h Volume of liquid in 72 bottles = p × (3)2 × h × 72 cm3 ½ Volume of bottles = volume of liquid after wastage
322 ]
2 90 π × (18 )3 × 3 100
=
1 π × ( 30 )2 × 60 3
× (3)2 × h × 72 =
p
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
= 18000p cm3 1 Volume of water left in cylinder = Volume of cylinder – volume of cone = 648000p – 18000p = 630000p cm3 630000 × 22 3 m = 1000000 ×7 1 = 1.98 m3. [CBSE Marking Scheme, 2015]
2 90 π × (18 )3 × 3 100 h = π × ( 3)2 × 72
or, Hence, the height of bottle = 5.4 cm. ½+1 [CBSE Marking Scheme, 2015] Q. 13. A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weights, a conical hole is drilled in the cylinder. The conical hole has a radius of 3 8 cm and its depth cm. Calculate the ratio 2 9
Q. 15. The rain water from 22 m × 20 m roof drains into cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills of the volume of metal left in the cylinder to the 4 volume of metal taken out in conical shape. th of cylindrical vessel then find the rainfall 5 A [Foreign Set I, II, III, 2015] A [Foreign Set I, II, III, 2015] in cm. Sol. Volume of cylinder = pr2h = p(3)2 × 5 Sol. Volume of water collected in cylindrical vessel = 45p cm3 ½ 4 7 3 2 2 = × π × (1) × m 1 1 1 3 8 2 5 Volume of conical hole = πr 2 h = π × 3 3 2 9 44 3 m 1 = 2 5 1 = p cm 3 3 Let the rainfall is h m. Volume of rain water from roof = 22 × 20 × h m3 133 p 2 cm 3 Metal left in cylinder = 45p - π = 44 3 3 or, 22 × 20 × h = 5 1 44 1 1 or, h = × = m Again, the required ratio 5 22 × 20 50 Volume of metal left = 1 × 100 = 2 cm 1 = Volume of metal taken out 50 133 [CBSE Marking Scheme, 2015] π = 3 = 133 : 2. ½ Q . 16. A hollow cylindrical pipe is made up of copper. 2 It is 21 dm long. The outer and inner diameters of π 3 the pipe are 10 cm and 6 cm respectively. Find the volume of copper used in making the pipe.
A [CBSE Board Term-2, 2015]
Sol. Height of cylindrical pipe, h = 21 dm = 210 cm 10 = 5 cm External radius, R = 2
Hence, Volume of metal left : Volume of metal taken out = 133 : 2. [CBSE Marking Scheme, 2015] Q. 14. A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right-circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder in 22 . cubic metre. Use π = 7
A [Foreign Set I, II, III, 2015]
Sol. Volume of water in cylinder = Volume of cylinder = pr2h = p × (60)2 × 180 = 648000p cm3 1 Water displaced on dropping cone = Volume of solid cone 1 2 = πr h 3
Internal radius, r =
6 = 3 cm 2
1
Volume of copper used in making the pipe = (Volume of external cylinder) – (Volume of internal cylinder) = pR2h – pr2h 1 = ph (R2 – h2) 22 ´ 210 ´ ( 52 - 32 ) = 7 =
22 ´ 16 ´ 210 7
= 10560 cm3. 1 [CBSE Marking Scheme, 2015]
[ 323
sURFACE aREAS AND VOLUMES
Q. 17. A glass is in the shape of a cylinder of radius 7 cm and height 10 cm. Find the volume of juice in litre 22 required to fill 6 such glasses. Use π = 7
5 cm3 . Find the height 6 of the toy. Also find the cost of painting the hemisphere part of the toy at the rate of ` 10 per 22 cm2. Use π = 7 in the making of toy is 166
A [Board Term-2, 2015]
Sol.
Radius of the glass = 7 cm Height of the glass = 10 cm Volume of 1 glass = pr2h 22 ×7 ×7 ×10 = 7
A [CBSE Delhi, Set-I, II, III, 2015]
Sol. Given, radius of cone = radius of hemisphere = r r = 3.5 cm. 1001 5 = cm 3 ½ Total volume, V = 166 cm 3 Mch-9-107 6 6
= 1540 cm3 1 ∴ Volume of juice to fill 6 glasses = 6 × 1540 = 9240 cm3 1 9240 = 9.240 litre. 1 ∴ Volume in litre = 1000
Let the height of cone be h. Total volume = Volume of cone + Volume of hemisphere
[CBSE Marking Scheme, 2015]
Q. 18. In fig., 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 22 area of the remaining block. Use π = 7
h 3.5 cm
A [CBSE Delhi Set-I, II, III, 2015]
1001 1 2 = πr 2 h + πr 3 6 3 3
or,
1001 1 2 = π( 3.5)2 h + π( 3.5)3 6 3 3
or,
1001 1 = π[12.25 h + 2 × 42.875] 6 3
7 cm
5 cm
or, 10 cm
or, or, 12.25 h = 159.25 – 85.75 73.5 = 6 cm or, h = 12.25 Height of the toy = 6 + 3.5 = 9.5 cm.
15 cm
Sol. Total surface area = 2(lb + bh + hl) + 2prh – 2pr2 7 Here, l = 15 cm, b = 10 cm, h = 5 cm, r = cm 2
TSA of cuboidal block = 2(15 × 10 + 10 × 5 +5 × 15) = 550 cm2 1 CSA of cylinder = 2prh 22 7 × ×5 = 2 × 7 2 2 = 110 cm 1 22 7 7 Area of two circular bases = 2 × × × 7 2 2 = 77 cm2 ½ ½ Required area = 550 + 110 – 77 = 583 cm2. [CBSE Marking Scheme, 2015] Q. 19. A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used
1001 × 3 × 7 = 12.25h + 85.75 6 × 22 21021 = 12.25h + 85.75 132
½
½ ½
Curved surface area of hemisphere = 2pr2 22 = 2 × × 3.5 × 3.5 7
= 77 cm2 ½ Cost of painting = ` 10 × 77 = ` 770 ½ [CBSE Marking Scheme, 2015] Q. 20. A solid is consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm. It 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 cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. [Use p = 3.14]
A [Board Term-2, 2015]
324 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Given, height of cone, h = 120 cm,
Volume of hemisphere=
radius of cone r = 60 cm.
2 3 πr 3
Radius of hemisphere = 60 cm.
=
1 Volume of cone = πr 2 h 3
=
= 452160 cm3 ½ Total volume = Volume of cone + Volume of hemisphere = 452160 + 452160 = 904320 cm3 ½ Height of cylinder = 180 cm, radius = 60 cm. Volume of water in the cylinder = Volume of cylinder = pr2h = 3.14 × 60 × 60 × 180 = 2034720 cm3 ½ Water left in the cylinder = Volume of water in cylinder – Volume of (cone + sphere) = 2034720 – 904320 = 1130400 cm3 ½ [CBSE Marking Scheme, 2015]
1 × 3.14 × 60 × 60 × 120 3
= 3.14 × 60 × 60 × 40 = 452160 cm3
1
120 cm
O
180 cm
60 cm B
60 cm
A
y
Long Answer Type Questions
A [CBSE OD Set-III, 2020]
Sol. Here, radius (r) = 7 cm and height of a cone (h) = 3.5 cm \ Volume of the soild = Volume of hemisphere + volume of a cone 1 2 1 = pr3 + pr2h 1 3 3
1
2 22 1 22 × × (7)3 + × × (7)2 × 3.5 1 3 7 3 7
=
1 [2156 + 539] 3
=
1 × 2695 3
= 898.33 cm3.
1
5 marks each Q. 2. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/hour. How much area will it irrigate in 30 minutes; if 8 cm standing water is needed ?
A [CBSE Delhi Set-I, II, III, 2019, 2018]
[CBSE OD Set-III, 2020, 2017]
Sol. Length of canal covered in 30 min = 5000 m. 1½ \ Volume of water flown in 30 min = 6 × 1.5 × 5000 m3 2 If 8 cm standing water is needed 6 ´ 1.5 ´ 5000 = 562500 m2. 1½ then area irrigated = 0.08 [CBSE Marking Scheme, 2019]
Q. 1. A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 22 3.5 cm, find the volume of the solid. Take π = 7
=
2 × 3.14 × 60 × 60 × 60 3
Detailed Solution: Canal is of the shape of cuboid, where Breadth = 6 m Depth = 1.5 m and speed of water = 10 km/hr Length of water moved in 60 minutes = 10 km 1 × 10 km Length of water moved in 1 minute = 60 1
Length of water moved in 30 minutes 30 = × 10 = 5 km = 5000 m 60
1
Now, volume of water in canal = Length × Breadth × Depth 1 = 5000 × 6 × 1.5 m3
[ 325
sURFACE aREAS AND VOLUMES
Now, volume of water in canal
Q . 4. 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)
= volume of water in area irrigated 5000 × 6 × 1.5 m3 = Area irrigated × 8 cm 1 8 5000 × 6 × 1.5 m3 = Area irrigated × m 100 5000 × 6 × 1.5 × 100 2 \ Area irrigated = m 8 = 5.625 × 105 m2
A [CBSE OD Set-I, 2019] [Board Term-2, 2012]
Sol.
1
8 cm 60 cm
COMMONLY MADE ERROR 1
220 cm
Some Candidates made mistakes for calculating the length of water.
ANSWERING TIP
= 99475.2 + 12057.6 = 111532.8 cm3
Q. 3. Water is flowing at the rate of 15 km/h through a cylindrical pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time the level of water in pond rise by 21 cm ? A [CBSE SQP, 2020, 2018]
Sol. Quantity of water flowing through pipe in 1 hour 7 7 × 15000 m3 × 100 100
2½
Mass =
111532.8 × 8 kg 1000
Detailed Solution: To find mass of pole, we need to find volume of pole. Volume of pole = Volume of small cylinder + Volume of large cylinder
Now, for small cylinder, 21 7 7 Radius, r = 8 cm × × 15000 ÷ π × Required time = 50 × 44 × Height, h = 60 cm 100 100 100 1 Volume of small cylinder
7 7 π × 100 × 100 × 15000
= p r2 h1
= 2 hours 2½ [CBSE Marking Scheme, 2020]
Detailed Solution: Speed of water flowing through the pipe = 15 km/hr = 15000 m/hr
1
2
Volume of water flowing in 1hr = pR H
=
1
22 7 7 × × × 15000 m3 7 100 100
= 231 m3
1
[CBSE Marking Scheme, 2019]
1
1
= 892.262 kg
Total volume = 3.14 (12) (220) + 3.14(8)2(60) cm3 1
be encouraged to find length of water.
=p×
12 cm 2
To avoid calculation errors students must
1
= 3.14 × (8)2 × 60 = 3.14 × 64 × 60 = 12057.6 cm3 Volume of large cylinder, Height = h2 = 220 cm
Radius = R =
1
diameter 2
24 = 12 cm = 2
½
r=8 cm
Value of water in the tank when the depth is 21 cm = lbh
½
h1 =60 cm
21 m3 = 50 × 44 × 100 = 462 m3
1
462 = 2 hrs. Time taken to fill 462 m3 = 231
½
24 cm
h2 =220 cm
1
326 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Volume of large cylinder = pR2 h2 = 3.14×(12)2 × 220 = 3.14 × 144 × 220 = 99475.2 cm3 1 Now, Volume of pole = Volume of small cylinder + Volume of large cylinder = 12057.6 + 99475.2 = 111532.8 cm3 1 Since, 1 cm3 of iron has 8 gm mass So, given volume of iron has mass = 111532.8 × 8 gm = 892262.4 gm 892262.4 kg = 1000
= 892.2624 kg = 892.262 kg
½
COMMONLY MADE ERROR Many candidates use formula for finding volume of cylinder, 2pr2h instead of pr2h.
ANSWERING TIP Remember the correct formula for finding volume of cylinder.
Q. 5. In Figure, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4·2 cm. Find 4.2 cm
6 cm 6 cm 6 cm
(a) the total surface area of the block. 22 (b) the volume of the block formed. Take π = 7
T
Sol.
opper's Answer, 2019
A [CBSE Delhi Region, 2019]
[ 327
sURFACE aREAS AND VOLUMES
5 Q. 6. Water is flowing at the rate of 5 km/hour through a pipe of diameter 14 cm into a rectangular tank of dimensions 50 m × 44 m. Find the time in which the level of water in the tank will rise by 7 cm. A [CBSE Delhi Comptt. Set-I, II, III, 2017]
2
22 7 7 × × 5000 t = 50 × 44 × 7 100 100 22 7 7 7 × × × 5000t = 50 × 44 × Þ 7 100 100 100
Þ
1 1
50 × 44 t= =2 22 × 50 Hence, time taken to fill the tank = 2 hours. 1 [CBSE Marking Scheme, 2017] Þ
T
Q. 7. A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top,
1 2 πr h 3
½
1 π × ( 5 )2 × 8 3
½
Sol. Volume of water in cone =
Sol. Speed of water in pipe = 5 km/hour In an hour length of water = 5000 m Let time taken to fill the tank be t hour. \ Total length of water = t × 5000 m 1 Volume of water flown = Volume of water in tank 1 Þ pr2h = l × b × h
which is open, is 5 cm. 100 spherical lead balls are dropped into vessel. One-fourth of the water flows out of the vessel. Find the radius of a spherical ball. [Foreign Set I, II, III, 2015]
=
200 ½ π cm 3 3 Volume of water flown out 1 200 50 p= p cm 3 1 = ´ 4 3 3 Let the radius of one spherical ball be r cm 1½ 4 3 50 πr × 100 = π ∴ 3 3 =
or,
r3 =
50 1 = 4 × 100 8
1 = 0.5 cm 1 2 [CBSE Marking Scheme, 2015] r =
OPIC - 2
Conversion of One Type of Metallic Solid into Another Revision Notes
While converting one metallic object into another, the volume will remain same by assuming no wastage of metal. Total surface area always be different from the original. Total surface area of the solid formed by the combination of solids is the sum of the curved surface areas of each individual solid. The solids having the same curved surface do not necessarily have the same volume.
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Conversion of solid from one shape to another
328 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Know the Formulae Name of solid
Volume
Cube
V = a3
Cuboid
V=l×b×h
Total surface Area
2
V = pr h
Cylinder
2
2
Lateral surface Area
TSA = 6a2
LSA = 4a2
TSA = 2(lb + bh + hl)
LSA = 2h(l + b)
TSA = 2pr(h + r)
CSA = 2prh
Hollow cylinder (R > r)
V = p(R – r )h
Cone
V=
1 2 pr h 3
TSA = pr(l + r)
CSA = prl
Sphere
V=
4 3 pr 3
TSA = 4pr2
CSA = 4pr2
Hemisphere
V=
2 3 pr 3
TSA = 3pr2
CSA = 2pr2
How is it done on the
TSA = 2p(R + r)(h + R – r)
GREENBOARD?
Q.1. A spherical ball of radius 3 cm Step II: Volume of spherical ball, is melted and recast into a cone of 4 V = π(3)3 cm3 same radius. Calculate the height of 3 A cone ? 1 Volume of cone, V' = π(3)2 h 3 Solution Step I: According to question Step I: Diagram: r=3 cm
h
4 π π × 27 = × 9 × h 3 3 4 × 27 =h 9
h = 12 cm \ Height of cone = 12 cm.
r=3 cm
Very Short Answer Type Questions Q. 1. Find the number of solid spheres of diameter 6 cm can be made by melting a solid metallic cylinder of height 45 cm and diameter 4 cm. Sol. Let the number of spheres be n. Radius of sphere, r1 = 3 cm radius of cylinder, r2 = 2 cm
Volume of spheres = Volume of cylinder
A
1 mark each 4 3 π r = π r22 h 3 1 4 22 22 n× × × (3)3 = × (2)2 × 45 3 7 7 n×
or, or, 36n = 180 180 or, n = =5 36
Thus, the number of solid spheres = 5.
1
[ 329
sURFACE aREAS AND VOLUMES
Sol. Let the radius of spherical ball be R. Volume of spherical ball = Volume of three balls 4 4 π R3 = π [(3)3 + (4)3 + (5)3] 3 3 3 or, R = 27 + 64 + 125 or, R3 = 216 or, R = 6 cm. 1 Q. 3. 12 solid spheres of the same size are made by melting a solid metallic cone of base radius 1 cm and height of 48 cm. Find the radius of each A sphere.
Sol. No. of spheres = 12 Radius of cone, r = 1 cm Height of the cone = 48 cm ∴ Volume of 12 spheres = Volume of cone Let the radius of sphere be R cm 4 1 12 × π R3 = π r2h 3 3
Q. 2. Three solid metallic spherical balls of radii 3 cm, 4 cm and 5 cm are melted into a single spherical A ball, find its radius.
4 1 or, 12 × π R3 = π × (1)2 × 48 3 3 16R3 = 16 or, R3 = 1 or, R = 1 cm 1 [CBSE Marking Scheme, 2014]
Short Answer Type Questions-I Q. 1. A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed. A [Foreign Set-I, II, 2017]
Sol. Volume of cuboid = 9 × 8 × 2 cm3 Volume of cube = 2 × 2 × 2 cm3 Let number of recast cubes be n. \ Volume of n cubes = Volume of cuboid n × 2 × 2 × 2 = 9 × 8 × 2 9×8×2 n = 2×2×2
½ ½
1
= 18 Hence, number of cubes recast = 18. [CBSE Marking Scheme, 2017] Q. 2. A solid metallic cylinder of radius 3.5 cm and height 14 cm is melted and recast into a number of small 7 solid metallic balls, each of radius cm. Find the 12 number of balls so formed. A [CBSE S.A.2, 2016] Sol. Let the number of recast balls be n radius of cylinder, R = 3.5 cm height of cylinder, h = 14 cm 7 cm radius of recast balls, r = 12
\
Þ
Þ
Volume of n balls = Volume of cylinder 4 2 1 n πr 3 = pR h 3 n×
4 7 7 7 = 3.5 × 3.5 × 14 × × × 3 12 12 12
3.5 × 3.5 × 14 × 3 × 12 × 12 × 12 n= 4×7×7×7 = 0.5 × 0.5 × 2 × 3 × 3 × 12 × 12 = 648 Hence, number of recast balls = 648 1 [CBSE Marking Scheme, 2016]
2 marks each
Q. 3. A 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 will the level of water rise in the A [CBSE SQP, 2016] cylindrical vessel ? Sol. Diameter of sphere = 6 cm Diameter of cylindrical vessel = 12 cm 4 3 Volume of sphere = πr 3 4 = ×π×3×3×3 3
= 36p cm3 1 \ Volume of sphere = Increase in volume of cylinder 36p = p(6)2 × h h = 1 cm \ Level of water rise in vessel = 1 cm. 1 [CBSE Marking Scheme, 2016]
Q. 4. Find the number of coins of 1.5 cm diameter and 0.2 cm thickness to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm. A [CBSE SQP, 2016] Sol.
Þ
Volume of coin = πr2h 22 × (0.75)2 × 0.2 cm3 ½ = 7 22 Volume of cylinder = × (2.25)2 × 10 cm3 ½ 7 Volume of cylinder No. of coins = ½ Volume of coin æ 22 ö 2 çè 7 ´ ( 2.25) ´ 10÷ø = æ 22 ö 2 çè 7 ´ ( 0.75) ´ 0.2÷ø ½ = 450 [CBSE Marking Scheme, 2016]
330 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Short Answer Type Questions-II
A [CBSE OD Set-III, 2019]
Sol. Radius of first sphere =3 cm 4 p (3)3. d = 1 Mass 3
{d= density}
4 p (r)3. d = 7 3
Mass
r3 = 7(3)3
⇒
½
4 4 4 ⇒ p (3)3 + p. (3)3.7 = pR3 3 3 3
1
R3 = (3)3(1+7) R = 3(2) = 6 ½ Diameter = 12 cm 1 [CBSE Marking Scheme, 2019]
⇒ ⇒ Detailed Solution:
Weight of smaller sphere = 1 kg
Weight of larger sphere = 7 kg
Radius of smaller sphere, r = 3 cm
\
Volume of smaller sphere =
1
4 π × 10.5 × 10.5 × 10.5 3
= 4p × 3.5 × 10.5 × 10.5 cm3 1 Radius of one recast cone = 3.5 cm and height = 3 cm 1 \ Volume = π × 3.5 × 3.5 × 3 3 = p × 3.5 × 3.5 cm3 1 Let the number of recast cones be n. \ n × p × 3.5 × 3.5 cm3 = 4 × p × 3.5 × 10.5 × 10.5 cm3 4 × 3.5 × 10.5 × 10.5 Þ n = 3.5 × 3.5 = 126 Hence, number of recast cones = 126. 1 [CBSE Marking Scheme, 2017] Q. 3. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of same height and same base radius is hollowed out. Find the total surface area of the remaining solid. (Take p = 3.14) U [CBSE OD Comptt. Set-I, II, III, 2017]
4 p(3)3 3
Sol. Height of cylinder = height of cone = 8 cm radius of cylinder = radius of cone = 6 cm
=
4 3 pr 3
Sol. Radius of given sphere = 10.5 cm 4 \ Volume of sphere = πr 3 3 =
let radius of 2nd sphere be r cm
Q. 2. A metallic solid sphere of radius 10.5 cm is melted and recasted into smaller solid cones each of radius 3.5 cm and height 3 cm. How many cones A [CBSE Delhi Set-II, 2017] will be made ?
Q. 1. Two spheres of same metal weighs 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
3 marks each
4 = p(27) 3 = 36p cm3 ½ Since, 1 kg metal sphere occupies 36p cm3 space. Now, weight of recasted metal sphere = (1 + 7) kg = 8 kg \ 8 kg sphere occupies
4 pR3 3
(R is the radius of new sphere.) 4 pR3 3
⇒
8 × 36p =
⇒
R3 =
⇒
R3 = 216
⇒
R3 = 6 × 6 × 6
⇒
1
8 ´ 36 ´ 3 4
8 2 + 6 2 = 64 + 36
8 2 + 6 2 == 64 + 36 = 10 cm ½ Total surface area of remaining solid = Curved surface area of cylinder + Surface area of cone + Area of top cylinder = 2prh + prl + pr2 = pr(2h + l + r) 1 22 × 6 ( 2 × 8 + 10 + 6 ) = 7 =
22 × 6 × 32 7
= 603.43 1 Hence total surface area of remaining solid
R = 6 cm
Hence, diameter of new sphere, D = 2R = 2 × 6 = 12 cm.
\ Slant height of cone =
½
= 603.43 cm2
[CBSE Marking Scheme, 2017] ½
[ 331
sURFACE aREAS AND VOLUMES
4 22 3 1 22 35 35 × × r = 504 × × × × ×3 3 7 3 7 20 20
U [CBSE OD Set-I, II, III, 2015]
1 2 πr h 3
or,
or,
∴
21 r3 = 2
½
3
r = 10.5 cm ½
Diameter = 21 cm
Sol. Volume of cone =
1 Since, Volume of sphere = Volume of 504 cones
Q. 4. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its 22 surface area. Use π = 7
2
and surface area = 4pr
Volume of metal in 504 cones 1 22 3.5 3.5 × × ×3 ½ = 504 × × 3 7 2 2
= 4 ×
= 1386 cm2
4 4 22 Volume of Sphere = πr 3 = × × r 3 3 3 7
Sol. Volume of cuboidal block = l × b × h = 15 × 10 × 3.5 = 525 cm3 1 Volume of one cone 1 2 1 = πr h 3 =
1 22 × × 0.5 × 0.5 × 2.1 cm 3 3 7
= 0.55 cm3 1 Volume of 4 cones = 0.55 × 4 = 2.2 cm3 1 Volume of wood remaining in pen stand = 525 – 2.2 = 522.80 cm3. 1 [CBSE Marking Scheme, 2017] Q. 2. A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.
A [CBSE OD Comptt. Set-II, III, 2017]
Sol.
5 marks each
Let the number of toys recast be n. 1 \ Volume of n conical toys = Volume of cylinder 1 1 n × π × 3 × 3 × 9 = p × 6 × 6 × 15 3 6 × 6 × 15 1 3×9 n = 20 Hence the number of toys = 20. 1 [CBSE Marking Scheme, 2017]
Sol. Volume of cylinder = pr2h
\
its diameter = 12 cm radius of cone = 3 cm
and height = 9 cm
22 42 42 ´ ´ ´ 10 7 10 10
1 ½
Volume of metal scooped out = 2 × volume of hemisphere = 2 ×
2 × πr 3 3
=
4 3 πr 3
=
4 22 42 × × 3 7 10
3
radius = 6 cm
=
= 554.40 cm3
Given, height of cylinder = 15 cm
and
n =
Q. 3. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. 22 A [CBSE OD Set I, II, III, 2015] Use π = 7
A [CBSE Delhi Comptt. Set-I, II, III, 2017]
½
[CBSE Marking Scheme, 2015]
Long Answer Type Questions Q. 1. From a rectangular block of wood, having dimensions 15 cm × 10 cm × 3.5 cm, a pen stand is made by making four conical depressions. The radius of each one of the depression is 0.5 cm and the depth 2.1 cm. Find the volume of wood left in the pen stand.
22 × 10.5 × 10.5 7
1
= 310.46 cm3
1
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Volume of rest of cylinder = 554.40 – 310.46
10 cm
Let the length of wire be l cm \ Volume of wire = pr2l and pr2l = 243.94 cm3 22 7 7 ´ ´ ´ l = 243.94 cm3 7 10 10
332 ]
½
= 243.94 m3 ½
T
l =
4.2 cm
½
243.94 × 10 × 10 22 × 7 l = 158. 4 cm ½ [CBSE Marking Scheme, 2015]
or,
½
OPIC - 3
Frustum of a Cone
Revision Notes When the smaller conical portion of a given right circular cone, which is sliced through by a plane parallel to its base, is removed, the resulting solid is called a Frustum of Right Circular Cone.
Volume of a frustum of a cone =
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1 ph(r12 + r22 + r1r2) 3
Curved surface area of a frustum = pl(r1 + r2)
Frustum of a cone
Total surface area = pl(r1 + r2) + p(r21 + r22) where r1 and r2 are the radii of two ends and h is the height. Upper part containing the vertex
frustum of a right cone
Very Short Answer Type Questions
=
22 × 45 × 35 7
⇒ Curved surface area of bucket = 22 × 45 × 5 cm2 = 4,950 cm2 1 Q. 2. What is the slant height of the frustum of a right circular cone of height 16 cm with radii of its A circular ends as 8 cm and 20 cm ?
Sol.
8 cm
16 cm
Q. 1. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm respectively. Find the curved surface area of the bucket. A Sol. Here, r1 = 7 cm, r2 = 28 cm, l = 45 cm Curved surface area of bucket = πl(r1 + r2) 22 × 45(7 + 28) = 7
1 mark each
l
h
d 12 cm 20 cm
[ 333
sURFACE aREAS AND VOLUMES
Q. 3. The slant height of a bucket is 26 cm. The diameter of upper and lower circular ends are 36 cm and A 16 cm. Find the height of the bucket. Sol. Here, l = 26 cm, upper radius = 18 cm, lower radius = 8 cm d = difference in radius = 18 – 8 = 10 cm. Let h be the height of bucket
20 cm
h
l
8 cm
=
=
=
∴
Slant height of the frustum l = h 2 + d 2
l2 - d2
h =
=
26 2 - (18 - 8 )2
16 2 + ( 20 - 8 )2
=
( 26 )2 - (10 )2
(16 )2 + (12 )2
=
676 - 100
256 + 144
=
576
= 400 = 20 cm.
1
Short Answer Type Questions-I
5 = 2.236
12 cm
6 cm
A [CBSE Delhi Set I, II, III, 2015]
Sol. Let r be the radius of the top., h = 12 – 4 = 8 cm 4 12 = r 6 ∴
½
r = 2 cm l =
=
2
h + (R - r )
2
( 8 )2 + ( 6 - 2 )2
=
64 + 16
=
80
22 [36 + 4 + 71.552] 7
=
22 × 111.552 7
= 350.59 cm2. (Approx) 1 [CBSE Marking Scheme, 2015]
4 cm r
=
= 4 5 = 4 × 2.236 = 8.944 cm (Approx) ½ Total surface area of frustum = π[R2 + r2 + l(R + r)] 22 [(6)2 + (2)2 + 8.944 (6 + 2)] = 7
Q. 2. Milk in a container, which is in the form of frustum of a cone of height 30 cm and the radii of whose lower and upper circular ends are 20 cm and 40 cm respectively, is to be distributed in a camp for flood victims. If this milk is available at the rate of ` 35 per litre and 880 litre of milk is needed daily for a camp, find how many such containers of milk are needed for a camp and what cost will it put on the donor agency for this. AE [Foreign Set I, II, III, 2015]
Sol. Volume of the milk container =Volume of frustum 1 = πh[ R 2 + r 2 + Rr ] 3
22 and Use π = 7
2 marks each
Q. 1. In fig from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid.
1
= 24 cm.
=
1 π × 30( 40 2 + 20 2 + 40 × 20 ) 3
= 10p(1600 + 400 + 800) 22 = 10 × × 2800 7 = 88000 cm3 = 88 litre
1
880 = 10 Number of containers needed = 88 Cost of milk = ` 880 × 35 = ` 30800 1 [CBSE Marking Scheme, 2015]
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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2 1 h Sol. Volume of upper cone = π r × 3 2 2
=
Q. 3. If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, find the ratio of the volume of the upper part and A [CBSE Term-2, 2011] the cone.
1 r 2h π 8 3 1 Volume of cone = π r2 h 3 =
h/2
h
½
h/2
½
1 r 2h π× Volume of upper part of cone 8 = 3 1 2 Volume of cone πr h 3 1 = 8
= 1 : 8 1 [CBSE Marking Scheme, 2011]
r
1 r2 h π × 3 4 2
Short Answer Type Questions-II Q. 1. A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts. Sol. \
A [CBSE Delhi Set-I, 2020]
DABC ∼ DAPQ (By AA Similarity) r h = 1 2h 4
⇒
½
r1 = 2 cm A
h r2 C
B
2h
1
h
P
r2 = 4 cm
Volume of smaller cone =
Q
1 2 pr h 3 1
1 = p(2)2h 3 1 × 4ph = 3 Volume of frustum =
½
1 ph(r12 + r22 + r1r2) 3
1 = p × h(22 + 42 + 2 × 4) 3 1 ½ = ph × 28 3
3 marks each
\
1 ´ 4 ph Required ratio = 3 1 ´ 28 ph 3
1 4 = = ½ 7 28 Q. 2. The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum be 8 cm, find the whole surface area of the A [CBSE SQP, 2016] frustum. (Use π = 3.14) Sol. Let R and r be the radii of the circular ends of the frustum. (R > r) 2πR = 207.24 R = 207.24/(2 × 3.14) R = 33 cm ½ 2πr =169.56 cm r = 169.56/(2 × 3.14) ½ r = 27 cm l2 = h2 + (R – r)2 ½ = 82 + (33 – 27)2 ½ l = 10 cm Whole surface area of the frustum = π(R2 + r2 + (R + r)l) = 3.14 [(33)2 + (27)2 + (33 + 27)10] = 3.14 (1089 + 729 + 600) = 3.14 × 2418 cm2 = 7592.52 cm2. 1 [CBSE Marking Scheme, 2016] Q. 3. A cone is cut by a plane parallel to the base and upper part is removed. If the curved surface area 1 times the curved surface area of upper cone is 9 of original cone. Find the ratio of line segment to which the cone’s height is divided by the plane. A
[ 335
sURFACE aREAS AND VOLUMES
Now
A
Sol.
l r
O
L
h B
∴ Substituting (ii) in (i), h h 1 × = or, H H 9
H
R
D
or,
C
Curved surface of upper cone 1 = Curved surface of original conee 9 πrl 1 or, = π RL 9
or,
rl 1 = RL 9
∆AOB ∼ ∆ACD(by AA similarity) r h l = = ...(ii) 1 R H L
or,
...(i) 1
h2
1 = H2 9 h 1 = H 3
Height of upper cone 1 1 Hence, = = Height of lower frustum 3 1 2 ∴ Ratio of the line segments OA : OC = 1 : 2.
Long Answer Type Questions Q. 1. A bucket in the from of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity of the bucket. Also find the cost of milk which can completely fill the bucket at the rate of ` 40 per 22 litre. Take π = 7 C + A [CBSE Delhi Set- I, 2020]
Sol. Height of frustum, h = 30 cm, R1 = 20 cm and R2 = 10 cm
5 marks each the cost of milk which can completely fill the bucket, at the rate of ` 40 per litre. (Use p = 3.14)
C + A [CBSE Delhi Set- II, 2020] Sol. Try yourself similar to Q.1. of LATQ. ` 418. Q. 3. Find the curved surface area of the frustum of a cone, the diameters of whose circular ends are 20 m and 6 m and its height is 24 m. A [CBSE OD Set- I, 2020] Sol. Here, h = 24 m 20 r1 = = 10 m ½ 2
Capacity of the bucket 1 = ph[R12 + R22 + R1R2] 3
1
=
220 [400 + 100 + 200] 7
=
220 × 700 7
= 22 litre.
6 = 3 m 2
½
((r - r ) + h ) = ((10 - 3) + ( 24 ) ) 1
2
2
2
2
= 625 = 25 m curved surface area = π(r1 + r2)l 22 (10 + 3)25 = 7
= 22000 cm3 1 = 22000 × litre 1000
r2 =
slant height (l) =
1 22 × 30[(20)2 + (10)2 + 20 × 10] 1 = × 3 7
1
[ 1 litre = 1000 cm3] 1
Total cost of milk which can completely fill the bucket = ` 40 × 22 = ` 880. 1 Q. 2. A bucket is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find
1
2
1 ½ ½ 1
22 × 13 × 25 = 7 7150 = 7
= 1021.42 m2
1
Q. 4. A petrol tank is in the form of a frustum of a cone of height 20 m with diameters of its lower and upper ends as 20 m and 50 m respectively. Find the cost of petrol which can fill the tank completely at the rate of ` 70 per litre. Also find the surface area A [CBSE SQP, 2020] of the tank.
336 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Capacity of tank =
1 p × 20 × (102 + 252 + 10 3 × 25) m3
= p × 20 × 325 m3 = p × 20 × 325 × 1000 litre 2 Cost of petrol = 3.14 × 20 × 325 × 70 × 1000 = ` 1428700 1
Slant height =
2
20 + ( 25 − 10 )
2
= 25 m 1 Surface area of tank = p × 25(10 + 25) m2 = 2747.5 m2 1 [CBSE Marking Scheme, 2020] Detailed Solution: Volume of container = Volume of frustum 1 = ph(r12 + r22 + r1r2) 3
ere, r1 = 20 cm, r2 = 8 cm and l = ? H We know that,
l = 25 cm. Surface area of frustum = p(r1 + r2)l = 3.14 × (25 + 10) × 25
1
r1 = 25 m
h = 20 m
½
r2 = 10 m
ere, H
h = height = 20 m r1 = radius of upper end = 25 m r2 = radius of lower end = 10 m 1 Volume of container = ph(r12 + r22 + r1r2) 3
1 × 3.14 × 20(252 + 102 = 3
+ 25 × 10) =
3.14 × 20 × 975 3
1
= 3.14 × 20 × 325 = 20410 m3 = 20410 × 1000 litre ½ [ 1 m3 = 1000 litre] = 2,04,10,000 litre Now, Cost of 1 litre petrol = ` 70 Cost of 2,04,10,000 litre petrol = ` 70 × 20410000 = ` 1428700000 ½ Now, Surface area of tank = surface area of frustum = p(r1 + r2) l
l =
h 2 + ( r1 − r2 )2
l =
20 2 + ( 25 − 10 )2
l =
20 2 + (15)2
l =
400 + 225
l =
625
l =
252
½
= 3.14 × 35 × 25 = 2747.5 cm2.
1 Q. 5. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use π = 3.14) C + A [CBSE Delhi Set- I, II, III, 2019 CBSE Delhi Set I, II, III, 2016]
Sol. Given, Volume of the bucket = 12308.8 cm3 r1 = 20 cm and r2 = 12 cm Let h be the height of bucket. We have,
ph 2 2 ( r1 + r2 + r1r2 ) 3
1
Volume of bucket =
⇒ 12308.8 = 3.14 × ⇒ h = 12 308.8 ×
h (202+122+ 20 × 12) ½ 3
3 = 15 cm 3.14 ´ 784
½
\ Height of the bucket, h = 15 cm. Now, l = h 2 + ( r1 - r2 )2 =
15 2 + 8 2 = 225 + 64 = 17 cm 1
\ Surface area of the metal sheet used 1
= p r22 + p ( r1 + r2 )l 2
= 3.14 × (12) +3.14(20+12)17 = 3.14(144+544) = 2160.32 sq. cm.
½
Hence, height of the bucket is 15 cm and area of the metal sheet used is 2160.32 sq. cm. ½ [CBSE Marking Scheme, 2019]
Q. 6. The diameters of the lower and upper ends of a bucket in the form of a frustum of the cone are 10 cm and 30 cm respectively. If its height is 24 cm, find :
(i) The area of the metal sheet used to make the bucket.
(ii) Why we should avoid the bucket made by ordinary plastic ? [Use p = 3.14]
A [CBSE Delhi/OD, 2018] [CBSE SQP, 2018]
[ 337
sURFACE aREAS AND VOLUMES
Sol. Here r1 = 15 cm, r2 = 5 cm and h = 24 cm (i) Area of metal sheet = CSA of the bucket + Area of lower circle = pl(r1 + r2) + pr22 2
24 + (15 − 5)
1
2
= 26 cm 1 1 \ Surface area of metal sheet = 3.14 (26 × 20 + 25) cm2 2 1 = 1711.3 cm (ii) We should avoid use of plastic because it is non-degradable or similar value. 1 [CBSE Marking Scheme, 2018]
where
l =
opper's Answer, 2018
T
Detailed Solution:
5
338 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Let the radius of cone be r2 and cut of cone be r1 Height of the cone = 10 cm 1 And the height the cone cut off = 5 cm DAOC ~ DAO'D r 10 AO = 2 = \ r 5 AO ' 1 Þ r2 = 2r1 1 1 2 Volume of cut off cone = πr1 × 5 3
COMMONLY MADE ERROR In problems related to frustum, students
write incorrect formula and do wrong calculation.
ANSWERING TIP Students should learn the formula clearly
by adequate practice and do the correct calculation.
Q. 7. A man donates 10 aluminium buckets to a orphanage. A bucket made of aluminium is of height 20 cm and has its upper and lower ends of radius 36 cm and 21 cm respectively. Find the cost of preparing 10 buckets if the cost of aluminium sheet is ` 42 per 100 cm2. Write your comments on the act of the man. A [CBSE OD Comptt. Set-I, II, III, 2018] Sol. Surface area of bucket = p(r1 + r2) l + pr12 l =
=
2
h + ( r2 − r1 )
5 2 pr1 sq. units 3
Volume of original cone =
1 2 π ( 2r1 ) × 10 3
=
40 2 πr1 sq. units 3
Volume of frustum = Volume of original cone – Volume of cut of cone 40 2 5 2 = pr1 - pr1 3 3 =
35 2 πr1 sq. units 3
35πr12 7 = 1 5πr12
2
20 2 + ( 36 − 21)2
= 625 = 25 cm 1 \ Surface area of 1 bucket 22 = [(36 + 21) × 25 + 212] 7 =
=
22 × 1866 cm2 7
1
22 × 18660 cm2 7
1
Hence the ratio of two parts = 7 : 1 1 [CBSE Marking Scheme, 2017] Q. 9. The height of a cone is 30 cm. From its topside a small cone is cut by a plane parallel to its base. If 1 of the cone then at volume of smaller cone is 27
22 18660 × 42 × 1 Cost of aluminium sheet = ` 7 100 = ` 24631.20 Any relevant comment 1 [CBSE Marking Scheme, 2018]
Q. 8. The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of volume of the two parts. A [CBSE Delhi Set-I, II, III, 2017] Sol.
Surface are of 10 buckets =
Ratio of two parts =
1
what height it is cut from the base ? A [CBSE Delhi Set-II, 2017] Sol. Let the radii of smaller cone and original cone be r1 and r2 respectively and the height of smaller cone be h.
DABC ~ DAPQ (By AA Similarity) ½
r1 h = ...(i) 1 r 30 2 Volume of smaller cone 1 = × Volume of original cone 27
Þ
A
A
h r1 O'
B
O
1
D
C
P
C
B
r2
30 cm
Q
½
[ 339
sURFACE aREAS AND VOLUMES
Þ Þ
2
r1 h 1 r × 30 = 27 2 2 1 h h = × 30 30 27 h r1 Using 30 = r From (i) 2
3 h = 1 30 27 30 × 30 × 30 1 Þ h3 = 27 h = 10 cm ½ Hence, required height = (30 – 10) = 20 cm. ½ [CBSE Marking Scheme, 2017]
Þ
Q. 10. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base if the frustum so obtained be drawn into 1 a wire of uniform diameter cm, find the length 16 of the wire. [Foreign Set-I, 2017] Sol. Total height of cone = 20 cm and vertex angle = 30° Let the radius of cone be r2
1 r2 = tan 30° Þ 3 20 20 r2 = cm 3
\
1
The height of the cone cut off = 10 cm Let its radius be r1 r1 10 Þ = tan 30° Þ r1 = cm. 10 3 1 30°
2
1 = π × l 32
1
1 1 2 1 2 × πr2 × 30 πr1 × h = 27 3 3
1 Þ 1 ´ 10 é 100 + 400 + 200 ù = ×l ê ú 3 3 3 û 32 × 32 ë 3
1
1 700 1 1 ´ 10 ´ = × ×l 3 3 32 32 32 ´ 32 ´ 700 ´ 10 l= 3´3
Þ Þ
= 796444.44 cm. Hence, the length of wire = 7964.44 m. 1 [CBSE Marking Scheme, 2017] Q. 11. A right circular cone is divided into three parts trisecting its height by two planes drawn parallel to the base. Show that volumes of the three portions starting from the top are in the ratio A [Foreign Set-III, 2017] 1 : 7 : 19.
Þ
Sol. Let the radii of three cones from top be r1, r2 and r3 respectively.
Let the height of given cone be 3h. So, the height of cone ADE = 2h. and height of cone ABC = h r1 h = r2 2 h
\ DABC ~ DADE,
Þ
2r1 = r2
DABC ~ DAFG
r1 h = r3 3 h
3r1 = r3
Volume of cone ABC =
1
1
1 2 πr1 h 3
A
B
h
r1
C
2h
10 cm 3h
r2
r1
h 20 cm 1 32
E
D
1
r3 G
F
Let the length of wire be l 1 cm Its radius = 32
r2
Volume of frustum = Volume of wire 1 Þ π × h (r1 )2 + (r2 )2 + (r1r2 ) = pr2l 3 éæ 10 ö 1 10 20 ù æ 20 ö ú ´ Þ ´ 10 ´ p êç ÷ + ç ÷ + è ø è ø 3 3 3 3 úû êë 3 2
2
Volume of cone ADE =
1 2 π (r2 ) 2 h 3
=
1 2 p ( 2r1 ) 2 h 3
½
1 2 1 2 Volume of frustum BCED = p 4 r1 2 h - pr1 h 3 3 =
7 2 πr1 h 3
1
340 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Volume of frustum DEGF
=
1 2 1 2 = πr3 .3h − r2 .2 h 3 3
19 2 πr1 h 3
1 2 7 2 19 2 πr1 h : πr1 h : πr1 h 1 3 3 3
=
1 1 2 2 π (3r1 ) 3h − ( 2r1 ) .2 h 3 3
=
1 2 πr1 h ( 27 − 8 ) 3
Hence, required ratio = 1 : 7 : 19.
Visual Case Based Questions
Ratio =
[CBSE Marking Scheme, 2017]
4 marks each
ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark Q. 1. Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below. After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551 m2. Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1 m2, the students put the tents. The radius of the tent is 7 m.
(i) The volume of cylindrical cup is (a) 295.75 cm3 (b) 7415.5 cm3 3 (c) 384.88 cm (d) 404.25 cm3 Sol. Correct option: (d). Explanation: diameter = 7 cm radius = 3.5 cm height = 10.5 cm
½
[ 341
sURFACE aREAS AND VOLUMES
Volume of cylindrical cup = pr2h 22 × 3.5 × 3.5 × 10.5 = 7 = 404.25 cm2 (ii) The volume of hemispherical cup is (a) 179.67 cm3 (b) 89.83 cm3 3 (c) 172.25 cm (d) 210.60 cm3 Sol. Correct option: (b). (iii) Which container had more juice and by how much? (a) Hemispherical cup, 195 cm3 (b) Cylindrical glass, 207 cm3 (c) Hemispherical cup, 280.85 cm3 (d) Cylindrical glass, 314.42 cm3 Sol. Correct option: (d). (iv) The height of the conical tent prepared to accommodate four students is (a) 18 m (b) 10 m (c) 24 m (d) 14 m Sol. Correct option: (c). Explanation: Radius = 7 m Area of conical tent = 551 m2 – 1 m2 = 550 m2 prl = 551 22 × 7 r 2 + h 2 = 550 7
student =
1 × 154 m 2 4
= 38.5 m2 Q. 2. A The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure mounted on it. 22 Take π = 7
22 × 7 7 2 + h 2 = 550 7 550 7 2 + h2 = 22 50 7 2 + h2 = 2 7 2 + h 2 = 25 72 + h2 = (25)2 h2 = 625 – 49 h2 = 576 h =
576
= 24 m (v) How much space on the ground is occupied by each student in the conical tent (a) 54 m2 (b) 38.5 m2 2 (c) 86 m (d) 24 m2 Sol. Correct option: (b). Explanation: Area of Base of conical tent = pr2 22 = ×7×7 7 = 154 m2
(i) Calculate the volume of the hemispherical dome if the height of the dome is 21 m: (a) 19404 cu. m (b) 2000 cu. m (c) 15000 cu. m (d) 19000 cu. m Sol. Correct option: (a). Explanation: height of hemispherical dome = Radius of hemispherical dome = 21 m. 2 3 Volume of dome = πr 3 2 22 × 21 × 21 × 21 = × 3 7
Area of occupied by each
= 19,404 m3 (ii) The formula to find the Volume of Sphere is: 2 4 3 (b) πr (a) πr 3 3 3 (c) 4pr2 (d) 2pr2 Sol. Correct option: (b).
(iii) The cloth require to cover the hemispherical dome if the radius of its base is 14m is: (a) 1222 sq.m (b) 1232 sq.m (c) 1200 sq.m (d) 1400 sq.m Sol. Correct option: (b). (iv) The total surface area of the combined figure i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8 m × 6 m × 4 m is (a) 1200 sq. m (b) 1232 sq. m (c) 1392 sq.m (d) 1932 sq. m Explanation: Total surface Area of Combined figure = 2pr2 + 2(lb + bh + hl) – lb 22 × 14 × 4 + 2( 8 × 6 + 6 × 4 + 4 × 8 ) − 8 × 6]m 2 = 2× 7 = [1232 + 208 – 48] m2 = 1392 m2 Sol. Correct option: (c). (v) The volume of the cuboidal shaped top is with dimensions mentioned in question 4. (a) 182.45 m3 (b) 282.45 m3 3 (c) 292 m (d) 192 m3 Sol. Correct option: (d). Explanation: Volume of the cuboidal shaped top = l × b × h = 8 m × 6 m × 4 m = 192 m3. Q. 3. On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’s cube and strawberry icecream for you. Observe the figures and answer the questions:
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(i) The length of the diagonal if each edge measures 6 cm is (a) 3 3 (b) 3 6
(d) 6 3 (c) 12 Sol. Correct option: (d). (ii) Volume of the solid figure if the length of the edge is 7 cm is: (a) 256 cm3 (b) 196 cm3 3 (c) 343 cm (d) 434 cm3 Sol. Correct option: (c). (iii) What is the curved surface area of hemisphere (ice cream) if the base radius is 7 cm ? (a) 309 cm2 (b) 308 cm2 2 (c) 803 cm (d) 903 cm2 Sol. Correct option: (b). (iv) Slant height of a cone if the radius is 7 cm and the height is 24 cm___ (a) 26 cm (b) 25 cm (c) 52 cm (d) 62 cm Sol. Correct option: (b). (v) The total surface area of cone with hemispherical ice cream is (a) 858 cm2 (b) 885 cm2 2 (c) 588 cm (d) 855 cm2 Sol. Correct option: (a). Q. 4. A carpenter made a wooden pen stand. It is in the shape of cuboid with four conical depressions to hold pens. The dimensions of the cuboid 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. (See Figure).
C + AE
(i) What is the volume of cuboid? (a) 525 cm3 (b) 225 cm3 (c) 552 cm3 (c) 255 cm3 Sol. Correct option: (a). Explanation: For cuboid l = 15 cm, b = 10 cm and h = 3.5 cm Volume of the cuboid = l × b × h = 15 × 10 × 3.5 = 525 cm3 (ii) What is the volume of a conical depression ? 11 11 cm3 (b) cm3 (a) 3 30
342 ]
3 (c) cm3 11
(d)
30 cm3 11
1
[ 343
sURFACE aREAS AND VOLUMES
Sol. Correct option: (b). Explanation: For conical depression: r = 0.5 cm, h = 1.4 cm Volume of conical depression 1 22 × 0.5 × 0.5 × 1.4 = × 3 7 11 1 cm 3 = 30 (iii) What is the total volume of conical depressions? (a) 1.74 cm3 (b) 1.44 cm3 3 (c) 1.47 cm (d) 1.77 cm3 Sol. Correct option: (c). Explanation: Volume of four conical depressions 11 1 = 4 × = 1.47 cm 3 30
(iv) What is the volume of wood in the entire stand? (b) 532.53 cm3 (a) 522.35 cm3 (c) 523.35 cm3 (d) 523.53 cm3 Sol. Correct option: (d). Explanation: Volume of the wood in the entire stand = Volume of cuboid – Volume of 4 conical depressions = 525 – 1.47 1 = 523.53 cm3 (v) The given problem is based on which mathematical concept? (a) Triangle (b) Surface Area and Volumes (c) Height and Distances (d) None of these Sol. Correct option: (b). 1
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344 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, MATHEMATICS (STANDARD), Class – X
Maximum Time: 1 hour Q. 1. If the circumferences of two concentric circles forming a ring are 88 cm and 66 cm respectively. Find the width of the ring. U [CBSE Delhi, 2013] 1 Q. 2. Find the diameter of a circle whose area is equal to the sum of areas of two circles of diameter 16 cm A [CBSE Term-2, 2012] 1 and 12 cm. Q. 3. The curved surface area of a cylinder is 264 m2 and 3 its volume is 924 m . Find the ratio of its height to A [CBSE Term-2, 2014] 1 its diameter. Q. 4. A thin wire is in the shape of a circle of radius 77 cm. It is bent into a square. Find the side of the 22 square. Taking, π = U1 7
Q. 5. What is the diameter of a circle whose area is equal to the sum of the areas of two circles of radii U [CBSE Term-2, 2012] 1 40 cm and 9 cm ? Q. 6. CASE STUDY BASED QUESTIONS In a class activity Sheena done a block painting on a square handkerchief as shown in figure. She made nine designer circles each of radius 7 cm. C + AE A
B
D
C
(i) What is the area of nine circles? (a) 154 cm2 (b) 145 cm2 2 (c) 1386 cm (d) 1836 cm21 (ii) What is the side of the square? (a) 22 cm (b) 44 cm (c) 24 cm (d) 42 cm 1
MM: 25
(iii) What is the area of square? (a) 1764 cm2 (b) 1674 cm2 2 (c) 1476 cm (d) 1746 cm21 (iv) What is the area of remaining portion of square? (a) 837 cm2 (b) 378 cm2 2 (c) 738 cm (d) 383 cm21 (v) The given problem is based on which mathematical concept (a) Areas Related to circles (b) Circles (c) Construction (d) none of these 1 Q. 7. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand from 9 U [CBSE Term-2, 2012] 2 a.m. to 9.35 a.m. Q. 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 volume of the remaining solid to the nearest cm3. 22 U [CBSE Term-2, 2012]2 Use π = 7 Q. 9. The radii of two right circular cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 4. Calculate the ratio of their curved surface areas and ratio of their volumes. A [CBSE Term-2, 2012] 3 Q. 10. In the figure, DACB is in the semi-circle. Find the area of shaded region given that AB = 42 cm. A [CBSE Term-2, 2014] 3 Q. 11. A cone is cut by a plane parallel to the base and upper part is removed. If the CSA of the remainder 15 of the CSA of whole cone, find the ratio is 16 of the line segments to which the cone’s height is divided by the plane. [CBSE Term-2, 2014]
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Statistics
[ 345
UNIT 7: Statistics and Probability
c h a p te r
14
Statistics
Syllabus ¾
¾
Mean, Median and Mode of grouped data (bimodal situation to be avoided). Cumulative frequency graph.
Trend Analysis 2018 List of Concepts
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1 Q (3 M) 1 Q (4 M)
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Cumulative Frequency Graph
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T
Mean, Median and Mode
OPIC - 1
Mean, Median and Mode Revision Notes TOPIC - 1
Statistics deals with the collection, presentation and analysis of numerical data. Three measures of central tendency are : (i) Mean, (ii) Median and (iii) Mode Mean: In statistics mean means the arithmetic mean of the given items. Sum of given items i.e., Mean = No. of Items
Mean, Median and Mode Page No. 346
TOPIC - 2 Cumulative Frequency
where, the Greek letter S(sigma) means ‘Summation of ’. Graph Page No. 368 Median: It is defined as the middle most or the central value of the variable in a set of observations, when the observations are arranged either in ascending or descending order of their magnitudes. It divides the arranged series in two equal parts i.e., 50% of the observations lie below the median and the remaining are above the median.
[ 347
Statistics
Mode: Mode is the observation which occurred maximum times. In ungrouped data, mode is the observation having maximum frequency. In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. To find the mode of grouped data, locate the class with the maximum frequency. This class is known as the modal class. The mode of the data is a value inside the modal class. Note: (a) If the series has only one mode, then it is known as Unimodal. (b) If the series has two modes, then it is known as Bimodal. (c) If the series has three modes, then it is known as Trimodal. (d) Mode may or may not be defined for a given series.
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Mean Median and Mode
Know the Formulae Mean: (a) For Raw Data: If n observations x1, x2, ..., xn are given, then their arithmetic mean is given by : n x = x1 + x 2 +...+ xn = 1 x ∑ i n n i =1
(b) For Ungrouped Data: If there are n distinct observations x1, x2,..., xn of variable x with frequencies f1, f2,..., fn respectively, then the arithmetic mean is given by: n
x =
f1x1 + f2 x 2 + ... + fn xn = f1 + f2 + f3 + ... + fn
∑ fx i =1 n
i i
fi ∑ i =1 (c) For Grouped Data: (i) To find the mean of grouped data, it is assumed that the frequency of each class-interval is centred around its mid-point. (ii) Direct Method:
Σfi xi Mean ( x ) = Σfi where the xi (class mark) is the mid-point of the ith class interval and fi is the corresponding frequency. (iii) Assumed Mean Method or Short-cut Method: Mean ( x ) = a +
Σfi di Σfi
where a is the assumed mean and di = xi – a are the deviations of xi from a for each i. (iv) Step-Deviation Method:
Σfi ui Mean ( x ) = a + h Σfi
where a is the assumed mean, h is the class-size and ui =
Median: (a) For Ungrouped Data:
xi - a · h
th
If n is odd,
n + 1 Median = term 2 th
If n is even,
th
æ nö æn ö çè 2 ÷ø term + çè 2 + 1÷ø term Median = 2
348 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(b) For Grouped Data: n Let n = f1 + f2 + f3 + ... + fn. First of all find n and then the class in which lies. This class is known as the 2 2 median class. Median of the given distribution lies in this class. Median of the grouped data can be calculated using the formula: n - c. f . 2 Median (Me) = l + ×h f where l = lower limit of median class, f = frequency of median class, n = number of observations, c.f. = cumulative frequency of the class preceding the median class, h = class-size or width of the class-interval. Mode of Grouped Data: Mode of the grouped data can be calculated by using the formula: f1 - f0 ×h Mode (M) = l + 2 f1 - f0 - f2 where l = lower limit of the modal class, h = width or size of the class-interval, f1 = frequency of the modal class, f0 = frequency of the class preceding the modal class, f2 = frequency of the class succeeding the modal class. Empirical Relation Between Mean, Median and Mode: (i) Mode = 3 Median – 2 Mean 1 2 Mode + Mean Median = 3 3 (ii) Mean =
3 1 Median – Mode 2 2
(iii) Note: For calculating the mode and median for grouped data, it should be ensured that the class-intervals are continuous before applying the formula. Same condition also apply for construction of an ogive. Further, in case of ogives, the scale may not be the same on both the axes.
How is it done on the
GREENBOARD?
Q.1. Find mean of the following data : x f
1 6
2 8
3 4
4 5
5 2
Solution Step I: x 1 2 3
4 5 Total
Step II: f 6 8 4
fx 6 16 12
Hence,
Very Short Answer Type Questions Q. 1. Find the class-marks of the classes 10 – 25 and 35 – 55. A [CBSE OD Set-I, 2020] 10 + 25 Sol. Class mark of 10 – 25 = 2 35 = = 17.5 ½ 2
5 2 25
20 10 64
Mean =
Σfx Σf
64 x = 25 mean x = 2·56
1 mark each
and Class mark of 35 – 55 =
35 + 55 2
=
90 = 45. 2
½
[ 349
Statistics
Q. 2. Find the class marks of the classes 15 – 35 and 45 – 60. A [CBSE OD Set-II, 2020] 15 + 35 Sol. Class mark of 15 – 35 = 2 50 = = 25 ½ 2 45 + 60 and Class mark of 45 – 60 = 2 105 = = 52.5. ½ 2 Q. 3. Find the class marks of the classes 20 – 50 and 35 – 60. A [CBSE OD Set-III, 2020] 20 + 50 Sol. Class mark of 20 – 50 = 2 70 = = 35 ½ 2 35 + 60 and Class mark of 35 – 60 = 2 95 = = 47.5. ½ 2 Q. 4. Consider the following frequency distribution of the heights of 60 students of a class. Heights (in cm)
No. of students
150-155
15
155-160
13
160-165
10
165-170
8
170-175
9
175-180 5 Find the upper limit of the median class in the given data. A [CBSE SQP 2020-21] Sol.
Heights (in cm)
No. of students
Cumulative frequency
150-155
15
15
155-160
13
15 + 13 = 28
160-165
10
28 + 10 = 38
165-170
8
38 + 8 = 46
170-175
9
46 + 9 = 55
175-180
5
55 + 5 = 60
Since total frequency is 60. N = 30 2 And cumulative frequency greater than or equal to 30 lies in class 160-165. So, median class is 160-165. \ Upper limit of median class is 165. 1
Q. 5. Following distribution gives cumulative frequencies of 'more than type' : Marks obtained
More than or equal to 5
More than or equal to 10
More than or equal to 15
More than or equal to 20
30
23
8
2
Number of students (cumulative frequency)
U [CBSE Term- 1, 2015]
Change the above data to a continuous grouped frequency distribution. Sol.
C.I.
5 – 10
10–15
15–20
More than 20
f
7
15
6
2
[CBSE Marking Scheme, 2015] 1
Q. 6. In the following frequency distribution, find the median class. Height (in cm)
140 – 145
145 – 150
150 – 155
155 – 160
160 – 165
165 – 170
Frequency
5
15
25
30
15
10 U [CBSE Term- 1, 2015]
Sol. Height
Frequency
c.f.
140 – 145
5
5
145 – 150
15
20
150 – 155
25
45
155 – 160
30
75
160 – 165
15
90
165 – 170
10
100
N = Sf = 100
½
350 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
N = 100 N 100 = = 50 2 2
⇒ The cumulative frequency just greater than 50 is 75 and the corresponding class is 155 – 160. Hence, median class is 155 – 160.
½
Q. 7. Find the median of the data, using an empirical relation when it is given that Mode = 12.4 and Mean = 10.5. U [CBSE Term- 1, 2015] 1 2 Sol. Median = Mode + Mean 3 3 =
1 2 (10.5) (12.4) + 3 3
=
12.4 21 + 3 3
=
12.4 + 21 33.4 = 3 3
=
33.4 = 11.13 (Approx) 3
1
Short Answer Type Questions-I
2 marks each
Q. 1. Find the mean of the following distribution : Class
3–5
5–7
7–9
9 – 11
Frequency
5
10
10
7
11 – 13 8 A [CBSE Delhi Set-I, 2020]
Sol. Class
Frequency (f)
Mid-Value (x)
f×x
3–5
5
4
20
5–7
10
6
60
7–9
10
8
80
9 – 11
7
10
70
8
12
11 – 13
∑f= 40 \
96 ∑fx = 326
mean =
=
Sfx Sf
1
326 = 8.15. 40
1
Q. 2. Find the mode of the following data: Class
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
Frequency
6
8
10
12
6
120 – 140
5
3
A [CBSE Delhi Set-I, 2020]
Sol. Since the modal class is the class having the maximum frequency. \ Modal class = 60 – 80 f1 - f0 \ Mode = l + × h 2 f1 - f0 - f2 Hence, l = 60, f1 = 12, f0 = 10, f2 = 6 and h = 20
100 – 120
12 - 10 × 20 Mode = 60 + 2 ´ 12 - 10 - 6
½ ½ ½
[ 351
Statistics
= 60 +
2 ´ 20 24 - 16
40 = 60 + 8 = 60 + 5 = 65. Q. 3. Compute the mode for the following frequency distribution :
½
Size of items (in cm)
0–4
4–8
8 – 12
12 – 16
16 – 20
20 – 24
24 – 28
Frequency
5
7
9
17
12
10
6
A [CBSE OD Set-I, 2020]
Sol. Here,
Modal class = 12 – 16
\ l = 12, f1 = 17, f0 = 9, f2 = 12 and h = 4
1
f1 − f0 Mode = l + ×h 2 f 1 − f0 − f2
17 − 9 ×4 = 12 + 2 × 17 − 9 − 12 = 12 +
8´4 13
1
= 12 + 2.46 = 14.46. (Approx) Q. 4. Find the mode of the following frequency distribution : Class
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
40 – 45
Frequency
3
8
9
10
3
2
Sol. Here,
modal class = 30 – 35
\ l = 30, f0 = 9, f1 = 10, f2 = 3 and h = 5 f1 − f0 Mode = l + × h 2 f1 − f0 − f2
1
10 − 9 ×5 = 30 + 2 × 10 − 9 − 3 = 30 +
5 = 30 + 0.625 8 1
= 30.625. Q. 5. Find the mode of the following distribution : Class
25 – 30
30 – 35
35 – 40
40 – 45
45 – 50
50 – 55
Frequency
25
34
50
42
38
14
U [CBSE Outside Delhi Set-I, 2019]
Sol.
Maximum frequency = 50, class (modal) = 35 – 40. ½
f1 − f0 ×h Mode = l + 2 f 1 − f0 − f2
= 35 +
50 − 34 × 5 100 − 34 − 42
= 35 +
16 × 5 = 38.33 (Approx) 24
1 [CBSE Marking Scheme, 2019] ½
352 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 6. Find the unknown values in the following table : Class Interval
Frequency
Cumulative Frequency
0 – 10
5
5
10 – 20
7
x1
20 – 30
x2
18
30 – 40
5
x3
40 – 50
x4
30 U [Board Term- 1, 2016]
Sol. and
x1 = 5 + 7 = 12 x2 = 18 – x1 = 18 – 12 = 6 x3 = 18 + 5 = 23 x4 = 30 – x3 = 30 – 23 = 7
[CBSE Marking Scheme, 2016] ½ × 4 = 2
Q. 7. The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is A [CBSE Term- 1, 2016] 100. It was later found that it is 110 not 100. Find the true mean and median. Sol.
Mean =
Σfx Σf
50 =
Σfx 100
⇒ ⇒
Σfx = 5000 Correct, Σfx' = 5000 – 100 + 110 = 5010 5010 Correct Mean = 100
\
1
1 [CBSE Marking Scheme, 2016]
= 50.1 Median will remain same i.e. median = 52.
Q. 8. The data regarding marks obtained by 48 students of a class in a class test is given below: Marks obtained
0–5
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
40 – 45
45 – 50
1
0
2
0
0
10
25
7
2
1
Number of students
A [CBSE Term- 1, 2015]
Calculate the modal marks of students. Sol. Modal class is 30 – 35, l = 30, f1 = 25, f0 = 10, f2 = 7 and h = 5 25 - 10 f1 - f0 Mode = l + 2 f - f - f × h ⇒ Mode = 30 + 50 - 10 - 7 × 5 1 0 2 = 30 + 2.27 or 32.27 approx.
[CBSE Marking Scheme, 2015] 2
Q . 9. Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.
Pocket money (in `)
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
100 – 120
120 – 140
Number of students
2
2
3
12
18
5
2 A [CBSE Term- 1, 2015]
Sol. Class Interval
Frequency
0 – 20
2
20 – 40
2
40 – 60
3
60 – 80
12
[ 353
Statistics
80 – 100
18
100 – 120
5
120 – 140
2
Total
44
Here,
Modal Class = 80 – 100 l = 80, f1 = 18, f2 = 5, f0 = 12 and h = 20
f1 - f0 Mode = l + ×h 2 f1 - f0 - f2
\
18 - 12 = 80 + × 20 36 - 12 - 5 = 80 +
6 × 20 19
= 80 + 6.31 = 86.31 (approx.) Hence, mode = 86.31. Q. 10. The mean of the following frequency distribution is 25. Find the value of p.
2
Class interval
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
Frequency
4
6
10
6
p U [CBSE Term- 1, 2015]
Sol. Class-Interval
Mid-Point xi
fi
fixi
0 – 10
5
4
20
10 – 20
15
6
90
20 – 30
25
10
250
30 – 40
35
6
210
40 – 50
45
p
45p
∑fi = 26 + p
∑fixi = 570 + 45p
Mean, x =
∑ fi xi ∑ fi
570 + 45 p 25 = 26 + p 650 + 25p = 570 + 45p 650 – 570 = 45p – 25p
⇒ ⇒ ⇒ \
[CBSE Marking Scheme, 2015] 2
p = 4 (Approx)
Short Answer Type Questions-II
3 marks each
Q. 1. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.
Class
0–5
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
Frequency
12
a
12
15
b
6
6
4 A [CBSE SQP, 2020-21]
354 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Class
Frequency (f)
Cumulative frequency (c.f.)
0–5
12
12
5 – 10
a
12 + a
10 – 15
12
24 + a
15 – 20
15
39 + a
20 – 25
b
39 + a + b
25 – 30
6
45 + a + b
30 – 35
6
51 + a + b
35 – 40
4
55 + a + b
Total
70
According to question,
1
55 + a + b = 70 a + b = 15
...(i) ½
N - cf ´h Median = l + 2 f
Q
½
35 - 24 - a ´5 16 = 15 + 15
\
1 =
11 − a 3 ½
a = 8 Substituting the value of a in equation (i), we get 8 + b = 15 ⇒ b = 15 – 8 ⇒ b = 7. Q. 2. The mode of the following data is 67. Find the missing frequency x. Class
½
40 – 50
50 – 60
60 – 70
70 – 80
80 – 90
5
x
15
12
7
Frequency
Sol. From the table of given question, the modal class is the class having the maximum frequency, i.e.,
modal class = 60 – 70
Then, l = 60, f1 = 15, f0 = x, f2 = 12 and h = 10 Mode = l +
\
f1 - f0 ´h 2 f1 - f2 - f0
67 = 60 +
7 =
½
15 - x ´ 10 30 - 12 - x
½
15 - x ´ 10 18 - x
½
7 × (18 – x) = 10(15 – x)½
126 – 7x = 150 – 10x
3x = 150 – 126½
3x = 24
x = 8.½ Q. 3. Find the mode of the following frequency distribution. Class Frequency
0 – 10
10 – 20
20 – 30
30 – 40
8
10
10
16
40 – 50
50 – 60
60 – 70
12
6
7
A [CBSE Delhi Set- I, 2019] [Board Term-I, 2016]
[ 355
Statistics
Sol. Modal class is 30 – 40
½
f1 − f0 Mode = l + ×h 2 f1 − f0 − f2
\
16 − 10 = 30 + × 10 32 − 10 − 12 = 36.
2 [CBSE Marking Scheme, 2019] ½
Detailed Solution:
⇒
Class
Frequency
0 – 10
8
10 – 20
10
20 – 30 30 – 40 40 – 50
10 16 12
50 – 60
6
60 – 70
7
Modal-class = 30 – 40 l = 30, f0 = 10, f1 = 16, f2 = 12, h = 10 æ f1 - f0 ö h Mode = l + ç è 2 f1 - f0 - f2 ÷ø
16 - 10 æ ö ´ 10 = 30 + ç è 2 ´ 16 - 10 - 12 ÷ø æ
6
ö
´ 10 = 30 + ç è 32 - 22 ÷ø
æ 6ö
= 30 + ç ÷ ´ 10 è 10 ø = 30 + 6 Q. 4. The mean of the following distribution is 53. Find the missing frequency k ? Class
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
Frequency
12
15
32
k
13 U [CBSE Delhi Set-II, 2019]
Sol.
Class Interval
Frequency (fi)
Class Marks (xi)
fixi
0 – 20
12
10
120
20 – 40
15
30
450
40 – 60
32
50
1600
60 – 80
k
70
70k
80 – 100
13
90
1170
Sfi = 72 +k
Given, \ ⇒
⇒ =
1½
Sfi = 3340 + 70k median = 53 S fx Median = i i S fi
3340 + 70 k 72 + k 53(72 + k) = 3340 + 70k 3816 + 53k = 3340 + 70k 53 =
1
356 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
⇒ 70k – 53k = 3816 – 3340 ⇒ 17k = 476 ⇒ k = 28 Hence, value of k is 28.
½ [CBSE Marking Scheme, 2019]
Q. 5. The marks obtained by 100 students in an examination are given below : Marks
30 – 35
35 – 40
40 – 45
45 – 50
50 – 55
55 – 60
60 – 65
14
16
28
23
18
8
3
Number of Students
Find the mean marks of the students.
A [CBSE OD, Set-I, 2019]
Sol. xi
32.5
37.5
42.5
47.5
52.5
57.5
62.5
fi
14
16
28
23
18
8
3
ui
–3
–2
–1
0
1
2
3
fiui
–42
–32
–28
0
18
16
9,
Mean = 47.5 –
½ Sfi = 110
½
Sfiui = –59
1
59 × 5 = 47.5 – 2.68 = 44.82 110
1
Note: If N is taken as 100, Ans. 44.55 If some one write, data is wrong, give full 3 marks.
Accept. [CBSE Marking Scheme, 2019]
Detailed Solution: xi - A h
fiui
32.5
–3
–42
16
37.5
–2
–32
40 – 45
28
42.5
–1
–28
45 – 50
23
47.5
0
0
50 – 55
18
52.5
1
18
55 – 60
8
57.5
2
16
60 – 65
3
62.5
3
9
Marks
Number of students (fi)
Mid Values (xi)
30 – 35
14
35 – 40
ui =
Sfiui = –59
N = Sfi = 110
2
Here,
Assumed mean, A = 47.5
Mean = A +
Sfi ui ´h N
= 47.5 +
( -59 ) ×5 110
= 47.5 – 2.682 = 44.818
1
Q. 6. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean A [CBSE Delhi, 2019] number of days a student was absent. Number of days Number of students
0–6
6 – 12
12 – 18
18 – 24
24 – 30
30 – 36
36 – 42
10
11
7
4
4
3
1
[ 357 T
Statistics
opper Answer, 2019
Sol.
3
Q. 7. The table below show the salaries of 280 persons :
Salary (in thousand `)
5 – 10
No. of Persons
49
10 – 15 15 – 20 20 – 25 133
63
15
25 – 30
30 – 35
35 – 40
40 – 45
45 – 50
6
7
4
2
1
U [CBSE Delhi/OD 2018]
Calculate the median salary of the data. Sol. Salary (in thousand `)
No. of Persons
c.f.
5 – 10
49
49
10 – 15
133 = f
182
15 – 20
63
245
20 – 25
15
260
358 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
25 – 30
6
266
30 – 35
7
273
35 – 40
4
277
40 – 45
2
279
45 – 50
1
280
N 280 = = 140. 2 2
1
Median class = 10 – 15. hæN
ö
Median = l + çè - c. f .÷ø f 2
= 10 +
5 (140 – 49) 133
= 10 +
5 × 91 133
1
= 13.42 Hence, median salary is ` 13.42 thousand or ` 13420 (approx).
[CBSE Marking Scheme, 2018] 1
opper Answer, 2018
3
T
Detailed Solution:
Q. 8. If the mean of the following data is 14.7, find the values of p and q. Class
0–6
6 – 12
12 – 18
18 – 24
24 – 30
30 – 36
Frequency
10
p
4
7
q
4
36 – 42
Total
1
40
U [CBSE Term-I, 2016]
Sol. Class
xi
fi
xi f i
0–6
3
10
30
6 – 12
9
p
9p
[ 359
Statistics
12 – 18
15
4
60
18 – 24
21
7
147
24 – 30
27
q
27q
30 – 36
33
4
132
36 – 42
39
1
39
Sfi = 26 + p + q = 40 Sfi = 40 26 + p + q = 40 p + q = 14 Σxi fi Mean, x = Σfi 408 + 9 p + 27 q 14.7 = 40
Total Given, ⇒ ⇒ ∴
Σxifi = 408 + 9p + 27q
...(i) ½ ½
⇒ ⇒ 588 = 408 + 9p + 27q ⇒ 180 = 9p + 27q ⇒ p + 3q = 20 Subtracting eq. (i) from eq. (ii), 2q = 6 ⇒ q = 3 Putting the value of q in eq. (i), p = 14 – q = 14 – 3 = 11 Hence, p = 11, q = 3 Q. 9. Find the mean of the following distribution :
1 ...(ii)
½ ½
Height (in cm)
Less than 75
Less than 100
Less than 125
Less than 150
Less than 175
Less than 200
No. of students
5
11
14
18
21
28
Height (in cm)
Less than 225
Less than 250
Less than 275
Less than 300
No. of students
33
37
45
50 U [CBSE Term- 1, 2016]
Sol. ui = xi - a
Class-Interval Height (in cm)
Frequency fi
xi
50 – 75
5
62.5
–5
– 25
h
75 – 100
6
87.5
–4
– 24
100 – 125
3
112.5
–3
–9
125 – 150
4
137.5
–2
–8
150 – 175
3
162.5
–1
–3
175 – 200
7
187.5 = a
0
0
200 – 225
5
212.5
1
5
225 – 250
4
237.5
2
8
250 – 275
8
262.5
3
24
5
287.5
4
275 – 300
∑ fi = 50 Here,
fiui
20 ∑ fiui = –12
2
∑ fiui = – 12; N = 50 and h = 25 ∑ fi ui ×h Mean = a + N Mean = 187.5 +
-12 × 25 = 187.5 – 6 = 181.5. 50
1
360 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 10. The mean of the following distribution is 48 and sum of all the frequencies is 50. Find the missing frequencies x and y. Class
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
Frequency
8
6
x
11
y
Sol.
U [CBSE Term- 1, 2016, 2015]
ui =
xi - a h
C.I.
fi
xi
fiui
20 – 30
8
25
–2
– 16
30 – 40
6
35
–1
–6
40 – 50
x
45 = a
0
0
50 – 60
11
55
1
11
60 – 70
y
65
2
2y
Total
Σfi = 25 + x + y
Σfiui = 2y – 11
Sf u Mean = a + i i × h Sfi
48 = 45 +
⇒ Þ
48 – 45 =
⇒ ⇒ Also ⇒ ⇒ \ x = 12 and y = 13.
2 y − 11 × 10 50
2 y − 11 5
1
3 × 5 = 2y – 11 15 = 2y – 11 y = 13 Σfi = 25 + x + y = 50 x + y = 25 x = 25 – 13 = 12
1
[CBSE Marking Scheme, 2016] 1
Q. 11. Find the median of the following data : Height (in cm)
Less than 120
Number of students
12
Less than 140 Less than 160 26
34
Less than 180
Less than 200
40
50 U [CBSE Term- 1, 2015]
Sol. Height (in cm)
Frequency
c.f.
less than 120
12
12
120 – 140
14
26
140 – 160
8
34
160 – 180
6
40
180 – 200
10
50
Total
N = 50 N = 50 N ⇒ Median = 2 50 = 25 = 2 So, Median Class = 120 – 140. N - c. f . 2 Median = l + ×h f Here,
[ 361
Statistics
25 - 12 = 120 + × 20 14 = 120 +
260 14
= 120 + 18.57 \ Median = 138.57.
[CBSE Marking Scheme, 2015] 3
Long Answer Type Questions
5 marks each
Q. 1. The median of the following data is 525. Find the values of x and y, if total frequency is 100. Class Frequency 0 – 100
2
100 – 200
5
200 – 300
x
300 – 400
12
400 – 500
17
500 – 600
20
600 – 700
y
700 – 800
9
800 – 900
7
900 – 1000
4 A [CBSE Delhi Set-I, 2020]
Sol. Class Interval
Frequency
Cumulative frequency
0 – 100
2
2
100 – 200
5
7
200 – 300
x
7+x
300 – 400
12
19 + x
400 – 500
17
36 + x
500 – 600
20
56 + x
600 – 700
y
56 + x + y
700 – 800
9
65 + x + y
800 – 900
7
72 + x + y
900 – 1000
4
76 + x + y
Total Also, ⇒ Given, ⇒ Now,
⇒
N = 100 76 + x + y = 100 x + y = 100 – 76 = 24 Median = 525, which lies between class 500 – 600. Median class = 500 – 600 N − c. f . Median = l + 2 × h f 100 − ( 36 + x ) × 100 525 = 500 + 2 20
1 ...(i) 1
1
362 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
⇒
25 = (50 – 36 – x) 5
⇒
14 – x =
⇒
25 =5 5 1
x = 14 – 5 = 9
Putting the value of x in eq. (i), we get Hence, x = 9 and y = 15.
1
y = 24 – 9 = 15
Q. 2. 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
3
6
9
13
f
5
4
U [CBSE OD Set-I, 2020] [CBSE Delhi/O.D. 2018]
Sol. Class
Class mark (x)
Frequency (f)
fx
11 – 13
12
3
36
13 – 15
14
6
84
15 – 17
16
9
144
17 – 19
18
13
234
19 – 21
20
f
20 f
21 – 23
22
5
110
23 – 25
24
4
96
Σf = 40 + f
Σfx = 704 + 20f
Σf = 40 + f Σfx = 704 + 20f 704 + 20 f Mean = 18 = 40 + f ⇒ ⇒
720 + 18 f = 704 + 20 f f = 8
Detailed Solution:
Topper Answer, 2018
1 1 1 1 1 [CBSE Marking Scheme, 2018]
[ 363
Statistics
5
Q. 3. Daily wages of 110 workers, obtained in a survey, are tabulated below : Daily Wages (in `) Number of Workers
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
200 – 220
220 – 240
10
15
20
22
18
12
13
A [CBSE SQP, 2020-21]
Compute the mean daily wages and modal daily wages of these workers. Sol. Daily Wages (in `)
Number of Workers (f1)
xi
ui
fiui
100 – 120
10
110
–3
– 30
120 – 140
15
130
–2
– 30
140 – 160
20
150
–1
– 20
160 – 180
22
170 = A
0
0
180 – 200
18
190
1
18
200 – 220
12
210
2
24
220 – 240
13
230
3
39
Total
110
1 1
1 × 20 = ` 170.19 (approx.) Mean daily wages = 170 + 110
Mode = 160 +
2
22 - 20 × 20 = ` 166.67 (approx.) [CBSE Marking Scheme, 2020] 2 44 - 20 - 18
Detailed Solution: 1. Calculation of mean: Daily Wages (Class interval)
Class mark (xi)
No. of Workers (fi)
fi . xi
100 – 120
110
10
1100
120 – 140
130
15
1950
140 – 160
150
20
3000
160 – 180
170
22
3740
180 – 200
190
18
3420
200 – 220
210
12
2520
220 – 240
230
13
2990
Total
∑fi = 110
å fi xi 18720 = 170.182 = Mean, x = å fi 110 Hence, mean daily: wages are ` 170.182 (Approx)
∑fi xi = 18720
1 1 ½
364 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2. Calculation of mode Daily Wages (in `)
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
200 – 220
220 – 240
10
15
20 → f1
22 → fm
18 → f2
12
13
Number of Workers
Here, maximum frequency, fm = 22. So, corresponding class 160 – 180 is modal class. l = lower boundary of modal class = 160 fm = maximum frequency = 22. f1 = frequency of pre-modal class = 20. f2 = frequency of past modal class = 18. h = width of modal class = 180 – 160 = 20
½
fm - f1 Mode = l + h fm - f1 - f2
\
1
22 - 20 é ù = 160 + ê ú ´ 20 ë 2(22) - 20 - 18 û
= 160 +
2 × 20 6
= 160 + 6.67 \ Mode = 166.67 Therefore, modal wages of workers is ` 166.67. (Approx) Q. 4. If the median of the following frequency distribution is 32.5. Find the values of f1 and f2. Class
1
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
Total
f1
5
9
12
f2
3
2
40
Frequency
A [CBSE Delhi Set-I, 2019]
Sol. Class
Frequency
Cumulative frequency
0 – 10
f1
f1
10 – 20
5
f1 + 5
20 – 30
9
f1 + 14
30 – 40
12
f1 + 26
40 – 50
f2
f1 + f2 + 26
50 – 60
3
f1 + f2 + 29
2
f1 + f2 + 31
60 – 70
1
40 Median = 32.5 ⇒ median class is 30 – 40. 10 32.5 = 30 + ( 20 − 14 − f1 ) 12
Now ⇒ Also ⇒
1 1 1
f1 = 3 f1 + f2 + 31 = 40 f2 = 6
[CBSE Marking Scheme, 2019] 1
Detailed Solution: Class
Frequency (f)
Cumulative Frequency (c.f.)
0 – 10
f1
f1
10 – 20
5
f1+5
20 – 30
9
f1+14
30 – 40
12
f1+26
[ 365
Statistics
40 – 50
f2
f1+f2+26
50 – 60
3
f1+f2+29
60 – 70
2
f1+f2+31
N = Sf = 40
1
Now, ⇒ ⇒
f1 + f2 + 31 = 40 f1+f2 = 9 f2 = 9 – f1 ...(i) ½
Given that median is 32.5, which lies in 30 – 40 Hence, median class = 30 – 40. N 40 = = 20, f = 12 and c.f. = 14 + f1 2 2 Now, median = 32.5 Here; l = 30,
N −c f . . 2 l+ × h = 32.5 f
⇒
⇒
½
1
é 20 - (14 + f1 ) ù 30 + ê ú × 10 = 32.5 12 ë û æ 6 - f1 ö çè 12 ÷ø ×10 = 2.5
⇒
⇒
60 - 10 f1 = 2.5 12 60 – 10 f1 = 30
⇒
10 f1 = 30
⇒
f1 = 3
⇒
1
From eq (i), we get f2 = 9 – 3 = 6 Hence, f1 = 3 and f2 = 6
COMMONLY MADE ERROR Some candidates use incorrect formula for median. Some get confused that which formulae has to be applied to find median.
ANSWERING TIP Students should read the question carefully and keep in mind the known and unknown data. Q. 5. Find the mode of the following distribution of marks obtained by the students in an examination : Marks obtained Number of students
0 – 20
20 – 40
40 – 60
60 – 80
80 – 100
15
18
21
29
17
Given the mean of the above distribution is 53, using empirical relationship estimate the value of its median. U [CBSE SQP, 2017-18]
Sol.
\
Modal class = 60 – 80 f1 − f0 ×h Mode = l + 2 f1 − f0 − f2
1
366 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
½
Here, l = 60, f1 = 29, f0 = 21, f2 = 17 and h = 20
Mode = 60 +
= 60 +
Empirical relationship, \
29 − 21 × 20 2 × 29 − 21 − 17
1
8 × 20 58 − 38 ½ 1
Mode Mode 3 median 3 median
= 60 + 8 = 68 = 3 median – 2 mean = 68 and mean = 53 (given) = mode + 2 mean = 68 + 2 × 53 174 = 58 Median = 3
1
Hence, median = 58. Q. 6. Literacy rates of 40 cities are given in the following table. If it is given that mean literacy rate is 63.5, then find the missing frequencies x and y. Literacy rate (in %) 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80 80 – 85 85 – 90 Number of cities
1
2
3
x
y
6
8
4
3
2
U [CBSE Term- 1, 2016]
2
Sol.
xi - 62.5 5
fi
fiui
37.5
–5
1
–5
40 – 45
42.5
–4
2
–8
45 – 50
47.5
–3
3
–9
50 – 55
52.5
–2
x
– 2x
55 – 60
57.5
–1
y
–y
60 – 65
62.5 = a
0
6
0
65 – 70
67.5
1
8
8
70 – 75
72.5
2
4
8
75 – 80
77.5
3
2
6
80 – 85
82.5
4
3
12
85 – 90
87.5
5
2
10
C.I.
xi
35 – 40
ui =
Sfiui = 22 – 2x – y
Sfi =31 + x + y
Total
2
Σfi = 31 + x + y = 40 x + y = 9 Σfiui = 22 – 2x – y Σf u Mean = a + i i × h Σfi ( 22 − 2 x − y ) 63.5 = 62.5 + ×5 40
Here, ⇒ \ ⇒
⇒ 2x + y = 14 Solving eqns, (i) and (ii), we get x = 5 and y = 4.
...(i) ½ 1
...(ii) 1 [CBSE Marking Scheme, 2016] ½
Q. 7. Monthly expenditures on milk in 100 families of a housing society are given in the following frequency distribution : Monthly expenditure (in `) Number of families
0 – 175
175 – 350
350 – 525
525 – 700
700 – 875
875 – 1050
1050 – 1225
10
14
15
21
28
7
5
Find the mode and median for this distribution.
A [CBSE Term-1 2016]
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Statistics
Sol. C.I. 0 – 175 175 – 350 350 – 525 525 – 700 700 – 875 875 – 1050 1050 – 1225
f 10 14 15 21 28 7 5 N = 100 N Median = 2 100 = 50 = 2 \ Median class = 525 – 700 N − c. f . 2 ×h Median = l + f
c.f. 10 24 39 60 88 95 100 1
50 − 39 1 × 175 21 11 × 175 = 525 + 21 175 × [50 – 39] \ Median = 525 + 21 = 525 + 91.6 = 616.6 (Approx) and Modal class = 700 – 875. f1 − f0 1 Mode = l + 2 f1 − f0 − f2 l = 700, f0 = 21, f1 = 28, f2 = 7 and h = 175 28 − 21 × 175 Mode = 700 + 1 2 × 28 − 21 − 7 7 × 175 = 700 + 28 = 700 + 43.75 = 743.75. 1 Q. 8. 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)
05 – 07
07 – 09
Number of students 70 120 Find mean and median of the above data.
= 525 +
09 – 11
11 – 13
13 – 15
15 – 17
32
100
45
28
17 – 19 5 A [CBSE Term- 1, 2015]
Sol. C. I.
fi
c.f.
xi
05 – 07 07 – 09 09 – 11 11 – 13 13 – 15 15 – 17 17 – 19
70 120 32 100 45 28 5
70 190 222 322 367 395 400
6 8 10 12 = a 14 16 18
Σf = 400
ui =
xi − a h –3 –2 –1 0 1 2 3
fiui –210 –240 –32 0 45 56 15 Σfiui = – 366
368 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2
Let a = Assumed mean = 12 Sfi ui Mean, x = a + Sf ´ h i
- 366 183 × 2 = 12 − = 12 − 1.83 = 10.17 400 100
Mean = 12 +
1
N 2 400 = 200 = 09 – 11 = 2 Median class =
N
- c. f .
Median = l + 2 f
\
Median = 9 +
×h
200 - 190 10 ×2= 9+ × 2 = 9 + 0.625 = 9.625 32 32
2
T
[CBSE Marking Scheme, 2015]
OPIC - 2
Cumulative Frequency Graph
Revision Notes Scan to know more about this topic
Cumulative Frequency Distribution:
(i) Cumulative frequency of a particular value of the variable (or class) is the sum (total) of all the frequencies up to that value (or class).
(ii) There are two types of cumulative frequency distributions:
(a) Cumulative frequency distribution of less than type.
(b) Cumulative frequency distribution of more than type.
For example: Class
Frequency Cumulative
Cumulative frequency graph
Less than type
More than type
interval
(No. of
frequency
Marks Out
(marks)
Students)
(c.f.)
of 50
0 – 10
2
2
Less than 10
10 – 20
10
12
Less than 20
2 + 10 = 12
More than 10
20 – 30
25
37
Less than 30
12 + 25 = 37
More than 20
58 – 10 = 48
30 – 40
20
57
Less than 40
37 + 20 = 57
More than 30
48 – 25 = 23
40 – 50
3
60
Less than 50
57 + 3 = 60
More than 40
23 – 20 = 3
c.f.
Marks Out
c.f.
of 50 2=2
0 or More than 0
60 – 0 = 60 60 – 2 = 58
Cumulative frequency curve or an Ogive curve : The graphical representation of a cumulative frequency distribution is called the cumulative frequency curve or ogive. There are two methods to construct ogives: (i) Less than ogive: In this method, an ogive is cumulated upward. Scale the cumulative frequencies along the Y-axis and exact upper limits along the X-axis. The scale along the Y-axis should be such as may accommodate the total frequency. Step I. Form the cumulative frequency table. Step II. Mark the actual upper class limits along the X-axis.
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Step III. Mark the cumulative frequency of the respective classes along the Y-axis. Step IV. Plot the points (upper limits, corresponding cumulative frequency.) By joining these points on the graph by a free hand curve, we get an ogive of 'less than' type. More than ogive: In this method, an ogive is cumulated downward. Scale the cumulative frequencies along the Y-axis and the exact lower limits along the X-axis. Step I. Scale the cumulative frequencies along the Y-axis and the actual lower limits along the X–axis. Step II. Plot the ordered pairs (lower limit, corresponding cumulative frequency). To complete an ogive, we also plot the ordered pair (upper limit of the highest class). Step III. Join these plotted points by a smooth curve. The curve so obtained is the required 'more than' type ogive. Median: (ii)
Ogive can be used to estimate the median of the data. There are two methods to get the median:
N , where N is the total frequency on cumulative frequency axis (i.e., Y-axis). 2 Draw a line parallel to X-axis to cut the ogive at a point. From this point draw a line perpendicular to the X-axis to get another point. The abscissa of this point gives median.
(i) Mark a point corresponding to
(ii) Draw both the ogives (less than and more than ogive) on the same graph paper which cut each other at a point. From this point draw a line perpendicular to the X-axis, to get another point. The point at which it cuts X-axis, gives the median.
How is it done on the
GREENBOARD?
Q.1. Form a more than frequency table of the following data : x f
1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 3 4 1 2 6 2 4 3
Solution Step I: x 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9
f 3 4 1 2 6 2 4 3
c.f. (More than) 22 + 3 18 + 4 17 + 1 15 + 2 9+6 7+2 3+4 3
Very Short Answer Type Questions Q. 1. Which central tendency is obtained by the abscissa of point of intersection of less than type and more than type ogives ? R [CBSE Term- 1, 2015] Sol. Median. 1
1 mark each
Q. 2. What is abscissa of the point of intersection of the “Less than type“ and of the “More than type“ cumulative frequency curve of a grouped data ? R
370 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. The abscissa of the point of intersection of the “Less than type“ and “More than type“ cumulative frequency curve of a grouped data is median. 1
Q. 3. What is the other name of cumulative frequency curve ? R Sol. Ogive.
1
Short Answer Type Questions-I
2 marks each
Q. 1. Given below is a frequency distribution table showing daily income of 100 workers of a factory : Daily income of workers (in `)
200 – 300
300 – 400
400 – 500
500 – 600
600 – 700
Number of workers 12 18 35 20 Convert this table to a cumulative frequency distribution table of 'more than type'.
15
C + U [CBSE Term- 1, 2016]
Sol. Cumulative frequency distribution table (more than type) Daily income of workers (in `)
Number of workers
More than 200
100
More than 300
88
More than 400
70
More than 500
35
More than 600
15
More than 700
0
[CBSE Marking Scheme, 2016] 2
Q. 2. The given distribution shows the number of runs scored by the batsmen in inter-school cricket matches :
Runs scored
0–50
50–100
100–150
150–200
200–250
Number of batsmen
4
6
9
7
5 C + A [CBSE Term- 1, 2015]
Draw a ' more than type' ogive for the above data.
Sol. Units on x-axis : 1 cm = 50, y-axis : 1 cm = 5 Runs scored
c.f.
More than 0 More than 50 More than 100 More than 150 More than 200 More than 250
31 27 21 12 5 0
1
On On
250
Runs Scored
X
1
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Statistics
Q. 3. The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :
Age (in years)
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
Number of patients
60
42
55
70
53
20 U
Form the “less than type” cumulative frequency distribution table.
Sol. Age (in years)
Number of Patients
Less than 20
60
Less than 30
102
Less than 40
157
Less than 50
227
Less than 60
280
Less than 70
300
2
Short Answer Type Questions-II
3 marks each
Q. 1. A TV reporter was given a task to prepare a report on the rainfall of the city Dispur of India in a particular year. After collecting the data, he analyzed the data and prepared a report on the rainfall of the city. Using this report, he drew the following graphs for a particular time period of 66 days. 70 1
CumulativeFrequency
60 50 40 30 20 11
10 0
0
10
20
30 40 50 Rainfallincm curve 2 curve 1
607
0
Based on the above graph, answer the following questions: (i) Identify less than type ogive and more than type ogive from the given graph. (ii) Find the median rainfall of Dispur. (iii) Obtain the Mode of the data if mean rainfall is 23.4 cm.
C [CBSE SQP 2020-21]
Sol. (i) Curve 1 – Less than ogive, Curve 2 – More than ogive 1 (ii) Median Rainfall = 21 cm 1 (iii) 3 Median = Mode + 2 mean \ Mode = 16.2 cm [CBSE SQP Marking Scheme, 2020] 1 Detailed Solution: \ median = 21 (iii) We have, mean = 23.4 cm (Given) N 70 \ mode = 3 median – 2 mean = = 35 From the given graph, (By Empirical formula) 2 2 = 3 × 21 – 2 × 23.4 Now, locate the point on the ogive where ordinate is 35. = 63 – 46.8 The X-coordinate corresponding to this ordinate is 21 = 16.2. 1
372 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 2. By changing the following frequency distribution 'to less than type' distribution, draw its ogive. Classes Frequency
0 – 15 6
15 – 30 8
30 – 45 10
45 – 60 6
60 – 75 4
C + A [CBSE OD Comptt. Set-I, II, III, 2018]
Sol. Classes
Frequency
Classes
Cumulative frequency
0 – 15
6
Less than 15
6
15 – 30
8
Less than 30
14
30 – 45
10
Less than 45
24
45 – 60
6
Less than 60
30
60 – 75
4
Less than 75
34
1
Y
Cumulative frequency
35
25
(45, 24)
20 15
(30, 14)
10 (15, 6)
5 0
(75, 34) (60, 30)
30
15
30 45 60 Upper limits
75
X
[CBSE Marking Scheme, 2018] 2 C+ A
Q. 3. From the frequency distribution table from the following data : Marks (out of 90)
Number of students (c.f.)
More than or equal to 80
4
More than or equal to 70
6
More than or equal to 60
11
More than or equal to 50
17
More than or equal to 40
23
More than or equal to 30
27
More than or equal to 20
30
More than or equal to 10
32
More than or equal to 0
34
Sol. Marks (out of 90)
Number of students (c.f.)
C.I.
Number of students (fi)
More than or equal to 0
34
0 – 10
34 – 32 = 2
More than or equal to 10
32
10 – 20
32 – 30 = 2
More than or equal to 20
30
20 – 30
30 – 27 = 3
More than or equal to 30
27
30 – 40
27 – 23 = 4
More than or equal to 40
23
40 – 50
23 – 17 = 6
More than or equal to 50
17
50 – 60
17 – 11 = 6
More than or equal to 60
11
60 – 70
11 – 6 =5
More than or equal to 70
6
70 – 80
6–4=2
More than or equal to 80
4
80 – 90
4–0=4
3
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Statistics
Long Answer Type Questions
5 marks each
Q. 1. The following table gives production yield per hectare (in quintals) of wheat of 100 farms of a village :
Production yield/hect. 40 – 45 45 – 50 50 – 55 55 – 60 No. of farms 4 6 16 20 Change the distribution to 'a more than' type distribution and draw its ogive.
60 – 65 30
65 – 70 24
C + A [CBSE Delhi & OD Set-I, 2020]
Sol. Production yield/hectare More than 40 More than 45 More than 50 More than 55 More than 60 More than 65 More than 70
c.f. 100 96 90 74 54 24 0 Y 100
2½
Units on x-axis 1 cm = 5 and y-axis 1 cm = 10
(40, 100) (45, 96) (50, 90)
90 80
(55, 74)
70
c.f.
60 (60, 54)
50
'more than' 0give
40 30
(65, 24)
20 10 0
40 45
50 55 60 65 70
Production (lower limit)
2½
X
Q. 2. Change the following data into ‘less than type’ distribution and draw its ogive : Class Interval 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 Frequency 7 5 8 10 6 6 8 [CBSE Delhi Set-I, 2019] A Sol. Less than 40
Less than 50
Less than 60 Less than 70 Less than 80 Less than 90 Less than 100
c.f. 7 12 20 30 36 42 50 1½ 2 Plotting of points (40, 7), (50, 12), (60, 20), (70, 30), (80, 36), (90, 42) and (100, 50) 1½ Joining the points to get the curve [CBSE Marking Scheme, 2019] 2 Detailed Solution: Classes
Cumulative frequency
Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100
7 7 + 5 = 12 12 + 8 = 20 20 + 10 = 30 30 + 6 = 36 36 + 6 = 42 42 + 8 = 50
3
374 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2
Scale: at X-axis, 1 small division = 10 units at Y-axis, 1 small division = 5 units Q. 3. The marks obtained by 100 students of a class in an examination are given below : Marks No. of Students
0–5
5 – 10
10 – 15
15 – 20
20 – 25
25 – 30
30 – 35
35 – 40
40 – 45
45 – 50
2
5
6
8
10
25
20
18
4
2
A [CBSE Delhi Set-I, 2019]
Draw ‘a less than’ type cumulative frequency curve (ogive). Hence find median. Sol. Less than type distribution is as follows: Marks Less than 5
No. of students 2
Less than 10
7
Less than 15
13
Less than 20
21
Less than 25
31
Less than 30
56
Less than 35
76
Less than 40
94
Less than 45
98
Less than 50
100
1½
Plotting of points (5, 2), (10, 7) (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98), (50, 100) 1½ Joining to get the curve 1 Getting median from graph (approx. 29) [CBSE Marking Scheme, 2019] 1 Detailed Solution: Marks Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50
No of students 2 5 6 8 10 25 20 18 4 2
c.f. 2 7 13 21 31 56 76 94 98 100
2
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We plot the points A(5, 2), B(10, 7), C(15, 13), D(20, 21), E(25, 31), F(30, 56), G(35, 76), H(40,94), I(45,98) and
1
J(50,100)
Join AB, BC, CD, DE, EE, FG, GH, HI, IJ and JA with a free hand to get the same representing the “less than type“ frequency curve (ogive)
Scale : On x-axis 1 cm = 5 marks On y-axis 1 cm = 10 studetns J
2
N 100 = = 50 2 2 Now, locate the point on the ogive where ordinate is 50. The X-coordinate corresponding to this ordinate is 28.8.
From graph,
Therefore, the required median on the graph is 28.8. Q. 4. The following distribution gives the daily income of 50 workers of a factory : Daily income (in `)
200 – 220
220 – 240
240 – 260
260 – 280
280 – 300
Number of workers
12
14
8
6
10
Convert the distribution above to a ‘less than type’ cumulative frequency distribution and draw its ogive. A [CBSE Delhi Set-III, 2019]
Sol. Less than type distribution is as follows Daily income
Number of workers
Less than 220
12
Less than 240
26
Less than 260
34
Less than 280
40
Less than 300
50
Plotting of points (220,12), (240, 26), (260, 34)(280, 40) and (300, 50)
3 [CBSE Marking Scheme, 2019] 2
Joining to get curve Detailed Solution: Daily Income (in `)
Cumulative Frequency
Less than 220
12
Less than 240
26
Less than 260
34
Less than 280
40
Less than 300
50
3
376 ]
2 Q. 5. Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than type’ ogive for this distribution.
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Class interval
20 – 30
30 – 40
40 – 50
50 – 60
60 – 70
70 – 80
80 – 90
10
8
12
24
6
25
15
T
Frequency
Sol.
opper Answer, 2017
[ 377 T
Statistics
opper Answer, 2017
5 Q. 6. The following distribution gives the daily income of 50 workers of a factory :
Daily Income (in `)
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
Number of workers
12
14
8
6
10
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. A [CBSE Delhi/OD, 2018]
Sol. Cumulative frequency distribution table less than type is Daily Income (in `)
Cumulative Frequency (c.f.)
Less than 100
0
Less than 120
12
Less than 140
26
Less than 160
34
Less than 180
40
Less than 200
50
Daily Income (in `)
Number of Workers (f)
Cumulative Frequency (c.f.)
100 – 120
12
12
120 – 140
14
26
140 – 160
8
34
160 – 180
6
40
180 – 200
10
50
378 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Less than Daily Income in (`)
Number of Workers (c.f.)
100
0
120
12
140
26
160
34
180
40
200
50
3
[CBSE Marking Scheme, 2018] 2
COMMONLY MADE ERROR Mostly candidates are unable to find frequency from c.f. They commit errors to understand the difference between 'less than Type' or 'more than Type'. Also they commit errors to draw its ogive.
ANSWERING TIP Adequate practice of these types of problems is necessary. Also, students should do proper practice to draw the ogive.
Q. 7. The following data indicates the marks of 53 students in Mathematics. Marks
Number of Students
0 – 10
5
10 – 20
3
20 – 30
4
30 – 40
3
40 – 50
3
50 – 60
4
60 – 70
7
70 – 80
9
80 – 90 7 90 – 100 8 Draw less than type ogive for the data above and hence find the median. Sol. Marks Number of Students c.f. 0 – 10 5 5 10 – 20
3
8
20 – 30
4
12
A [CBSE SQP, 2018]
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Statistics
30 – 40
3
15
40 – 50
3
18
50 – 60
4
22
60 – 70
7
29
70 – 80
9
38
80 – 90
7
45
90 – 100 Correct table Drawing correct Ogive Median = 64
8
53
2 2 [CBSE Marking Scheme, 2018] 1
Detailed Solution: Marks obtained
Number of Students (c.f.)
Less than 10
5
Less than 20
8
Less than 30
12
Less than 40
15
Less than 50
18
Less than 60
22
Less than 70
29
Less than 80
38
Less than 90
45
Less than 100
53
2
Y X Y
26.5
66.4
2
Here,
So,
So, median class will be 60 – 70.
X
N = 53 N 53 = = 26.5 2 2 N 2 − c. f . Median = l + × h f
= 60 +
26.5 − 22 × 10 7
= 60 +
45 7
Median = 66.4. (Approx)
1
380 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Q. 8. On the annual day of school, age-wise participation of students is given in the following frequency distribution table : Age (in years)
Number of students
Less than 6
2
Less than 8
6
Less than 10
12
Less than 12
22
Less than 14
42
Less than 16
67
Less than 18
76 A [CBSE Term-1 2016]
Find the median of the students and get the median graphically. Sol. Age of student
C.I.
c.f.
f
Less than 6
4–6
2
2
Less than 8
6–8
6
4
Less than 10
8 – 10
12
6
Less than 12
10 – 12
22
10
Less than 14
12 – 14
42
20
Less than 16
14 – 16
67
25
Less than 18
16 – 18
76
9
N = 76
Now,
N 76 = = 38 2 2
unit unit
Hence, median class will be (12 – 14)
38
Here, l = 12, c.f. = 22, h = 2, f = 20
M
æN ö çè 2 - c. f .÷ø ´h Median = l + f
So,
= 12 +
(38 - 22) ´2 20
2
N Age (in years)
= 13.6
1
From the graph, median = 13.6. Q. 9. In annual day of a school, age-wise participation of students is shown in the following frequency distribution : Age of student (in years)
5–7
7–9
9 – 11
11 – 13
13–15
15–17
17–19
Number of students
20
18
22
25
20
15
10
Draw a 'less than type' ogive for the above data and from it find the median age. Sol. Students
c.f.
Less than 7
20
Less than 9
38
Less than 11
60
A [CBSE Term-1, 2015]
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Statistics
Less than 13
85
Less than 15
105
Less than 17
120
Less than 19
130 Y
X Y
unit unit
X
This curve is the required cumulative frequency curve or an ogive of the less than type. Here, N = 130, N 130 So, = = 65 2 2
2
1
Now, we locate the point on the ogive whose ordinate is 65. The X-co-ordinate corresponding to this ordinate is 11.4. Hence, the required median on the graph is 11.4. 1 Q. 10. In an orchard, the number of apples on trees are given below : Number of apples Number of trees
More than or equal to 50
More than or equal to 60
More than or equal to 70
More than or equal to 80
More than or equal to 90
More than or equal to 100
More than or equal to 110
60
55
39
29
10
6
2
Draw a 'more than type' ogive and hence obtain median from the curve.
C [CBSE Term-1, 2015]
Sol. Number of Apples
c.f.
More than 50
60
More than 60
55
More than 70
39
More than 80
29
More than 90
10
More than 100
6
More than 110
2 1
382 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
2
Number of Apples This curve shows cumulative frequency on an ogive of the 'more than type'. Here N = 60, N 60 = 30 = So, 2 2
1
Now, we locate the point on the ogive whose ordinate is 30. The x-co-ordinate corresponding to this ordinate is 79. Hence, the required median on the graph is 79.
Visual Case Based Questions
1
4 marks each
ote: Attempt any four sub parts from each question. Each sub part carries 1 mark N Q. 1. COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.
The following tables shows the age distribution of case admitted during a day in two different hospitals Table 1 Age (in years)
5 – 15
15 – 25
25 – 35
35 – 45
45 – 55
55 – 65
No. of cases
6
11
21 Table 1
23
14
5
Age (in years)
5 – 15
15 – 25
25 – 35
35 – 45
45 – 55
55 – 65
42
24
12
No. of cases 8 16 10 Refer to table 1 (i) The average age for which maximum cases occurred is (a) 32.24 (b) 34.36 (c) 36.82 Sol. Correct option: (c). Explanation: Since, highest frequency is 23. So, modal class is 35 – 45. f1 − f0 ×h Now, Mode = l + 2 f1 − f0 − f2 Here, l = 35, h = 10, f1 = 23, f0 = 21, f2 = 14,
(d) 42.24
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Mode = 35 +
⇒
23 − 21 × 10 46 − 21 − 14
= 35 +
2 × 10 11
= 35 +
20 11
= 35 + 1.81 = 36.818 ≈ 36.82 (ii) The upper limit of modal class is (a) 15 (b) 25 (c) 35 Sol. Correct option: (d). (iii) The mean of the given data is (a) 26.2 (b) 32.4 (c) 33.5 Sol. Correct option: (d). Explanation: Age (in years)
Class marks (xi)
(d) 45
(d) 35.4
frequency (fi)
Deviation di = (xi – a)
fidi
5 – 15
10
6
–3
–15
15 – 25
20
11
6
66
25 – 35
30
21
16
336
35 – 45
40
23
26
598
45 – 55
50
1
18
55 – 65
60
46
230
14 → a 5
Sfidi = 1,716
Sfi = n = 80 Σfi di Mean (x–) = a + Σfi
Now,
1716 80 = 14 + 21.45 = 35.45 Refer to table 2 (iv) The mode of the given data is (a) 41.4 (b) 48.2 (c) 55.3 Sol. Correct option: (a). (v) The median of the given data is (a) 32.7 (b) 40.2 (c) 42.3 Sol. Correct option: (b). Explanation:
= 14 +
Age (in years)
frequency (fi) (No. of cases)
5 – 15
8
8
15 – 25
16
24
25 – 35
10
34
35 – 45
42 (frequency)
45 – 55
24
100
55 – 65
12
112
Sfi = n = 112
(d) 64.6 (d) 48.6
C.f.
76 (Nearest to
n ) 2
384 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
n = 112 = 56. 2 2
Now, l = 35 (lower limit of median class) Cf = 34 (Preceding to median class)
n − cf 2 ×h Median = l + f
Here,
56 − 34 × 10 = 35 + 42 22 = 35 + × 10 42 11 = 35 + × 10 21 = 35 +
110 21
= 40.25 Q. 2.
Electricity Energy Consumption
Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption). Tariff
: LT - Residential
Bill Number
: 384756
Type of Supply
: Single Passes
Connected lead
: 3 kW
Mater Reading Date
: 31-11-13
Mater Reading
: 65700
Previous Mater Reading
: 65500
Previous Reading : 31-10-13 Date
Units consumed : 289
A survey is conducted for 56 families of a Colony A. The following tables gives the weekly consumption of electricity of these families. Weekly consumption (in units)
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
No. of families
16
12
18
6
4
0
Table 1
Weekly consumption (in units)
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
No. of families
0
5
10
20
40
5
Refer to data received from Colony A (i) The median weekly consumption is (a) 12 units
Sol. Correct option: (c).
(b) 16 units
(c) 20 units
(d) None of these
[ 385
Statistics
Explanation: Weekly consumption (in units)
frequency (fi) (No. of families)
0-10
8
C.f. 16 28
10-20 (Median class)
12 (frequency)
(Nearest to
20-30
18
46
30-40
6
52
40-50
4
56
50-60
0
56
n ) 2
Sfi = n = 56 n = 56 = 28 2 2
Now, l = 10, Cf = 16, f = 12, h = 10 Here,
n − cf 2 ×h Median = l + f
28 − 16 × 10 = 10 + 12 12 = 10 + × 10 12 = 10 + 10 = 20 Hence, median weekly consumption = 20 units. (ii) The mean weekly consumption is (a) 19.64 units (b) 22.5 units (c) 26 units (d) Sol. Correct option: (a). (iii) The modal class of the above data is I (a) 0-10 (b) 10-20 (c) 20-30 (d) Sol. Correct option: (c). Refer to data received from Colony B (iv) The modal weekly consumption is (a) 38.2 units (b) 43.6 units (c) 26 units (d) Sol. Correct option: (b). (v) The mean weekly consumption is (a) 15.65 units (b) 32.8 units (c) 38.75 units (d) Sol. Correct option: (c). Q. 3. The weights (in kg) of 50 wrestlers are recorded in the following table : Weight (in kg) No. of Wrestless
None of these 30-40
32 units 48 units
100 – 110
110 – 120
120 – 130
130 – 140
140 – 150
4
14
21
8
3
386 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(i) What is the upper limit of modal class. (a) 120 (b) 130 (c) 100 (d) 150 Sol. Correct option: (b). Explanation: Modal Class = 120 – 130 Upper limit = 130 (ii) What is the mode class frequency of the given data (a) 21 (b) 50 (c) 25 (d) 80 Sol. Correct option: (a). Explanation: Mode class frequency of the given data is 21. (iii) How many wrestlers weights have more than 120 kg weight ? (a) 32 (b) 50 (c) 16 (d) 21 Sol. Correct option: (a). Explanation: No. of wrestlers with more than 120 kg weight= 21 + 8 + 3 = 32 (iv) What is the class mark for class 130 – 140 ? (a) 120 (b) 130 (c) 135 (d) 150 Sol. Correct option: (c). Explanation: For class mark of 130 – 140, 130 + 140 = 2 =
1
1
1
270 = 135 2
1
(v) Which method is more suitable to find the mean of the above data ? (a) Direct method (b) Assumed mean method (c) Step-Deviation method (d) None of these Sol. Correct option: (c). 1 Q. 4. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows. Speed (in km/h) Number of players
85 – 100
100 – 115
115 – 130
130 – 145
11
9
8
5
(i) What is the modal class of the given data? (a) 85-100 (b) 100-115 (c) 115-130 (d) 130-145 Sol. Correct option: (a). Explanation: Modal class is the class with highest frequency i.e., 85 – 100 (ii) What is the value of class interval for the given data set? (a) 10 (b) 15 (c) 5 (d) 20 Sol. Correct option: (b). Explanation: The value of class interval = 100 – 85 = 15 again = 115 – 100 = 15 and = 130 – 115 = 145 – 130 (iii) What is the median class of the given data? (a) 85-100 (b) 100-115 (c) 115-130 (d) 130-145 Sol. Correct option: (b). Explanation: N = Number of observations = 33 Median of 33 observations = 17.5th observation, which lies in class 100 – 115 (iv) What is the median of bowling speed? (a) 109.17 km/hr (Approx) (b) 109.71 km/hr (Approx) (c) 107.17 km/hr (Approx) (d) 109.19 km/hr (Approx)
1
1
1
[ 387
Statistics
Sol. Correct option: (a). n − c. f . 2 Median = l + ×h f
Explanation:
l = 100, f = 9, c.f. = 11, h = 100 – 85 = 15
n − c. f . 2 Median = l + ×h f
33 − 11 2 × 15 = 100 + 9 100 + (16.5 − 11) = 9 × 15 5.5 × 15 = 100 + 9 82.5 = 100 + 9 = 100 + 9.166 = 109.17 km/h (Approx) Hence, the median bowling speed is 109.17 km/h (Approx) (v) What is the sum of lower limit of modal class and upper limit of median class? (a) 100 (b) 200 (c) 300 (d) 400 Sol. Correct option: (b). Explanation: Lower limit of modal class = 85 and upper limit of median class =115 sum = 85 + 115 = 200 5. 100 m RACE A stopwatch was used to find the time that it took a group of students to run 100 m.
Time (in sec) No. of students
0-20
20-40
40-60
60-80
80-100
8
10
13
6
3
1
1
C + AE [CBSE SQP, 2020-21]
(i) Estimate the mean time taken by a student to finish the race. (a) 54 (b) 63 (c) 43
(d) 50
Sol. Correct option: (c).
[CBSE Marking Scheme, 2020-21] 1
Explanation: Time (in sec)
x
f
fx
0-20
10
8
8
80
20-40
30
10
18
300
40-60
50
13
31
650
60-80
70
6
37
420
80-100
90
3
40
270
Total
cf
40 Mean =
1720 = 43 40
1720
½
388 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
(ii) What will be the upper limit of the modal class ? (a) 20 (b) 40 (c) 60
(d) 80
Sol. Correct option: (c). [CBSE Marking Scheme, 2020-21] 1 Explanation: Modal class = 40-60 Upper limit = 60 (iii) The construction of cumulative frequency table is useful in determining the (a) Mean (b) Median (c) Mode (d) All of the above Sol. Correct option: (b). Explanation: The construction of c.f. table is useful in determining the median. 1 (iv) The sum of lower limits of median class and modal class is (a) 60 (b) 100 (c) 80 (d) 140 Sol. Correct option: (c). [CBSE Marking Scheme, 2020-21] 1 Explanation: Median class = 40-60 ½ Modal class = 40-60 Therefore, the sum of the lower limits of median and modal class = 40 + 40 = 80 ½ (v) How many students finished the race within 1 minute? (a) 18 (b) 37 (c) 31 (d) 8 Sol. Correct option: (c). [CBSE Marking Scheme, 2020-21] 1 Explanation: Number of students who finished the race within 1 minute = 8 + 10 + 13 = 31
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Probability
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c h a pte r
15
Probability
Syllabus Classical definition of probability. Simple problems on single events (not using set notation).
Trend Analysis L
2018 ist of Concepts
Delhi
Outside Delhi
2019 Delhi
Outside Delhi
1 Q (2 M)
Problems based on tossing a coin Problems Based on Throwing a Die
1 Q (2 M)
Problems Based on Playing cards
1 Q (2 M)
Problems Based on selection of an 1 Q (2 M) object from a Bag etc.
2020 Delhi
Outside Delhi
1 Q (1 M) 1 Q (2 M) 2 Q (1 M)
1 Q (2 M)
1 Q (1 M) 2 Q (2 M)
1 Q (1 M) 2 Q (1 M) 1 Q (2 M)
Revision Notes Probability is a branch of mathematics that deals with calculating the likelihood of a given event’s occurrence. A random experiment is an experiment or a process for which the outcome cannot be predicted with certainty. e.g.,
Scan to know more about this topic
(i) tossing a coin, (ii) throwing a dice, (iii) selecting a card and (iv) selecting an object etc. Outcome associated with an experiment is called an event. e.g., (i) Getting a head on tossing a coin, (ii) getting a face card when a card is drawn from a pack of 52 cards. The events whose probability is one are called sure/certain events.
Basic Concepts
The events whose probability is zero are called impossible events. An event with only one possible outcome is called an elementary event. In a given experiment, if two or more events are equally likely to occur or have equal probabilities, then they are called equally likely events. Probability of an event always lies between 0 and 1. Probability can never be negative and more than one.
[ 391
Probability
A pack of playing cards consists of 52 cards which are divided into 4 suits of 13 cards each. Each suit consists of an ace, one king, one queen, one jack and 9 other cards numbered from 2 to 10. Four suits are spades, hearts, diamonds and clubs. King, queen and jack are face cards. The sum of the probabilities of all elementary events of an experiment is 1. Two events A and B are said to be complementary to of each other if the sum of their probabilities is 1. Probability of an event E, denoted as P(E), is given by: Number of outcomes favourable to E P(E) = Total possible number of outcomes
For an event E, P( E ) = 1 – P(E), where the event E representing ‘not E’ is the complement of the event E. For A and B two possible outcomes of an event, (i) If P(A) > P(B), then event A is more likely to occur than event B. (ii) If P(A) = P(B), then events A and B are equally likely to occur.
Know the Facts
The experimental or empirical probability of an event is based on what has actually happened while the theoretical probability of the event attempts to predict what will happen on the basis of certain assumptions. As the number of trials in an experiment go on increasing, we may expect the experimental and theoretical probabilities to be nearly the same. When we speak of a coin, we assume it to be ‘fair ‘ i.e., it is symmetrical so that there is no reason for it to come down more often on one side than the other. We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’, we mean that the coin is allowed to fall freely without any bias or interference. In the case of experiment we assume that the experiments have equally likely outcomes. A deck of playing cards consists of 4 suits : spades (♠), hearts (♥), diamonds (♦) and clubs (♣). Clubs and spades are of black colour, while hearts and diamonds are of red colour. The first book on probability ‘The Book on Games of Chance’ was written by Italian mathematician J. Cardan. The classical definition of probability was given by Pierre Simon Laplace.
How is it done on the
GREENBOARD?
Q.1. One dice and one coin are tossed simultaneously. Write the sample space. Find the probability of getting : (i) Prime number on dice (ii) head (iii) head and even number Solution: Step I: Writing the sample space: {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)} Step II: Finding prime number on dice : prime numbers are 2, 3 and 5 \ favourable cases are : A = {(H, 2), (H, 3), (H, 5), (T, 2), (T, 3), (T, 5)} n(A) = 6 \ Required probability
6 1 n(A) = = n(S ) 12 2 Step III: Finding cases having head: B = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)} n(B) = 6 n(B ) \ Required probability = n(S ) 6 1 = = 12 2 Step IV: Finding head and even number: C = {(H, 2), (H, 4), (H, 6)} n(C) = 3 n(C ) \ Required probability = n(S ) 3 1 = = 12 4 =
392 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Very Short Answer Type Questions
Q. 1. Find the probability of getting a doublet in a U [CBSE SQP, 2020-21] throw of a pair of dice. [CBSE Delhi Set-II, 2020]
1 1 6 Doublet means same number on both dice. [CBSE Marking Scheme, 2020] Detailed Solution: The outcomes when two dice are thrown together 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) ½ Total number of outcomes = n(S) = 36 Favourable outcomes are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) i.e., n(E) = 6 n( E) 6 1 = ½ Required Probability = = n(S) 36 6 Sol. Probability of getting a doublet =
Q. 2. Find the probability of getting a black queen when a card is drawn at random from a wellshuffled pack of 52 cards. U [CBSE SQP, 2020-21] 2 1 Sol. Probability of getting a black queen = = 1 52 26
Q. 5. If the probability of winning a game is 0.07, what is the probability of losing it ? Sol. P(winning the game) = 0.07 P(losing the game) = 1 – 0.07 = 0.93 1 Q. 6. A die thrown once. What is the probability of getting an even prime number ? U [CBSE Delhi Set-II, 2020] Total possible outcome n(S) = 6 Favourable outcome = {2} n(E) = 1 n( E) 1 = . 1 P(getting an even prime number) = 6 n(S)
Sol. i.e.,
Q. 7. If a number x is chosen at random from the numbers – 3, – 2, – 1. 0, 1, 2, 3, then find the U [CBSE OD Set-I, 2020] probability of x2 < 4. Sol. x 2
x
Detailed Solution: Total no. of cards = 52 No. of black queens = 2
2 1 = 1 52 26 Q. 3. A letter of english alphabet is chosen at random. What is the probability that the chosen letter is a U [CBSE Delhi Set-I, 2020] consonant. Sol. In English language, there are 26 alphabets. Consonants are 21. So, total no. of outcomes, n(S) = 26 and favourable outcomes, n(E) = 21 ½ \ The probability of choosing a consonant 21 n( E) = ½ = 26 n(S) So, Probability of black queen =
Q. 4. A die is thrown once. What is the probability of getting a number less than 3 ? A [CBSE Delhi Set-I, 2020] Sol. Total possible outcomes = n(S) = 6 Favourable outcomes = 1, 2 i.e., n(E) = 2 n( E) \ P(number less than 3) = n(S) 2 1 = . = 6 3
1
1 mark each
–3
–2
–1
0
1
2
3
9
4
1
0
1
4
9
Total possible outcomes = n(S) = 7 Favourable outcomes = x2 < 4 i.e., x = – 1, 0, 1 n(E) = 3 P(x2 < 4) =
n( E) 3 = . n(S) 7
1
Q. 8. What is the probability that a randomly taken leap year has not 53 Sundays ?
A [CBSE OD Set-I, 2020]
Sol. Number of days in a leap year = 366
Number of weeks =
366 = 52.28 7
So, there will be 52 weeks and 2 days So, every leap year has 52 Sundays Now, the probability depends on remaining 2 days The possible pairing of days are Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday. ½ There are total 7 pairs and out of 7 pairs, only 2 pairs have Sunday. The remaining 5 pairs does not include Sunday. Therefore, the probability of has not 53 Sundays in 5 ½ a Leap year is 7 Q. 9. A die is thrown once. What is the probability of getting a prime number ?
U [CBSE OD Set-I, 2020]
[ 393
Sol. Total possible outcomes = n(S) = 6 Favourable outcomes = {2, 3, 5} i.e., n(E) = 3 n( E) Probability = n(S)
=
3 1 = . 6 2
Probability
½
½
Q. 10. 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 U [CBSE SQP, 2020-21] less than or equal to 1.
Sol.
[Delhi Set-I, II, III, 2017]
3 11
[CBSE SQP Marking Scheme, 2020] 1
Detailed Solution: Given numbers, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5 Total outcomes = 11 ½ Squaring all the numbers we get 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25 Favourable outcomes Probability, P(E) = Total number of outcomes 3 P(E) = 11 3 Hence, Probability = . ½ 11
T
Q. 11. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap ? R [CBSE OD 2017]
opper Answer, 2017
Sol.
n( A ) 2 1 = = 1 n(S) 6 3 Q. 13. Out of 200 bulbs in a box, 12 bulbs are defective. One bulb is taken out at random from the box. What is the probability that the drawn bulb is not U [CBSE SQP, 2016] defective ? Sol. Total no. of cases = 200 Favourable cases = 200 – 12 = 188 188 ∴ Required probability = 200
∴
P(A) =
=
47 1 50
Q. 14. What is the probability that a non-leap year has 53 U [CBSE Term-2, 2015] Mondays ?
Sol. There are 365 days in a non-leap year. Q 365 days = 52 weeks + 1 day \ One day can be M, T, W, Th, F, S, S =7 ½ 1 ½ \ P(53 Mondays in non-leap year) = 7 [CBSE Marking Scheme, 2015] Q. 15. Two different dice are tossed together. Find the probability that the product of the number on the top A [CBSE OD Set-I, II, III, 2015] of the dice is 6. Sol. Product of 6 are (1, 6); (2, 3); (6, 1); (3, 2) No. of possible outcomes = 4 Total number of chances = 6 × 6 = 36 4 1 = P(Product = 6) = 36 9
Q. 12. A die is thrown once. Find the probability of U [CBSE S.A.-2 2016] getting ‘’at most 2.’’ Sol. S = {1, 2, 3, 4, 5, 6} n(S) = 6 A = {1, 2} n(A) = 2
1
[CBSE Marking Scheme, 2015] 1
394 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Sol. Total number of points = 8
Since, the factors of 8 are, 1, 2, 4 and 8.
= (1 × 8), (2 × 4), (8 × 1), (4 × 2)
Q. 16. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8 ? U [Foreign Set I, II, III, 2015]
No. of favourable outcomes = 4 \ P(Factors of 8) =
No. of favourable outcomes ½ Total no. of possible outcomes
=
4 1 = ½ 8 2
[CBSE Marking Scheme, 2015]
Short Answer Type Questions-I
U [CBSE Delhi Set-I, 2020] Sol. Try yourself, similar to Q. 7, of very short answer type questions.
Q. 2. ‘A child has a die whose six faces show the letters as shown below : A
B
C
D
E
A
The die is thrown once. What is the probability of getting (i) A, (ii) D ? C + A [CBSE OD Set-I, 2020] Sol. Total possible outcomes, n (S) = 6 (i) Let E1 = getting event letter A, then n(E1) = 2 n( E1 ) 2 = \ Probability = = 1 1 n(S) 6 3 (ii) Let E2 = getting event letter D, then n(E2) = 1 n( E2 ) 1 1 = . \ Probability = n(S) 6
Sol. Total number of outcomes = 8 ½ Favourable number of outcomes (HHH, TTT) = 2 ½ 2 1 ½ Prob. (getting success) = or 8 4 1 3 ½ \ Prob. (losing the game) = 1 − = 4 4 [CBSE Marking Scheme, 2019]
Q. 1. If a number x is chosen at random from the number – 3, – 2, – 1, 0, 1, 2, 3. What is probability that x2 ≤ 4 ?
Q. 3. A child has a die whose six faces show the letters as shown below : A
B
C
C
C
COMMONLY MADE ERROR Most of the students commit errors in writing correct total outcomes. Sometimes, favourable outcomes are also incorrectly written by students.
ANSWERING TIP Remember to list the favourable outcomes of tossing a coin 3 times.
Q. 5. A die is thrown twice. Find the probability that (i) 5 will come up at least once. (ii) 5 will not come up either time. U [CBSE OD Set-III, 2019]
Sol. E1 {(1, 5),(2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)} 11 1 P(5 will come at least once ) =P(E1) = 36 11 25 P(5 will not come either time) = 1– = 1 36 36 [CBSE Marking Scheme, 2019]
A
2 marks each
The die is thrown once. What is the probability of getting (i) A, (ii) C ? C + A [CBSE OD Set-II, 2020] Sol. Total possible outcomes n(S) = 6 (i) Let E1 = getting event letter A, then n(E1) = 2 n( E1 ) 2 1 = = 1 \ Probability = n(S) 6 3 (ii) Let E2 = Getting event letter C, then n (E2) = 3 n( E2 ) 3 1 = . \ Probability = = n(S) 6 2
1
Q. 4. 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.
U [CBSE Delhi Set-I, 2019]
COMMONLY MADE ERROR In die based problems, students often get confused in favorable outcomes.
ANSWERING TIP List all the events of throwing a die and not assume in mind.
[ 395
Probability
Q. 6. The probability of selecting a blue marble at random from a jar that contains only blue, black 1 and green marbles is . The probability of select5 ing a black marble at random from the same jar is 1 . If the jar contains 11 green marbles, find the 4
P(black marbles) =
or, Again,
total number of marbles in the jar.
U [CBSE OD, Set-I, 2019]
[CBSE OD, Set-I, II, 2015]
Sol.
P(blue marble) =
1 1 , P(black marble) = 5 4
11 1 1 \ P(green marble) = 1 – + = 5 4 20
1
Let total number of marbles be x 11 × x = 11 ⇒ x = 20 then 20
1
[CBSE Marking Scheme, 2019]
Detailed Solution: Let x and y be the number of Blue and Black marbles. No. of green marbles = 11 Total number of marbles = x + y + 11 ½ According to the problem,
or,
1 4
y 1 = x + y + 11 4
x = 3y – 11 1 P (blue marble) = 5 x = 1 x + y + 11 5
x =
y + 11
4 Now, from equation (i) and (ii), we have
...(i) ½
3y - 11 =
...(ii) ½
y + 11 4
or, 12y - 44 = y + 11 or, y = 5 From equation (i), x = 3 × 5 - 11 x = 4 Hence, total number of marbles in the jar = x + y + 11 = 4 + 5 + 11 = 20.
½
T
Q. 7. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number. U [CBSE Delhi Region, 2019]
opper Answer, 2019
Sol.
2
396 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
T
Q. 8. 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 ? [CBSE Delhi, 2019]
opper Answer, 2019
Sol.
2
Sol. Probability of either a red card or a queen 26 + 2 28 = = 52 52 28 P(neither red card nor a queen) = 1 52
1
24 6 52 - 28 = 1 = 52 52 13 [CBSE Marking Scheme, 2018] Q. 11. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime =
number.
U [CBSE SQP, 2018]
Sol. Total number of outcomes = 36 1 Favourable outcomes are (1, 2), (2, 1), (1, 3), (3, 1), (1, 5), (5, 1) i.e., 6 6 1 or 1 Required probability = 36 6 [CBSE Marking Scheme, 2018]
Sol. No. of all possible outcomes = 62 = 36 No. of favourable outcomes = 26 ½ (4, 2)(4, 3)(4, 4)(4, 5)(5, 1)(5, 2)(5, 3)(6, 1)(6, 2)(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)(4, 1) 26 13 ∴ P(Product appears is less than 18) = = 1½ 36 18 [CBSE Marking Scheme, 2017]
Sol. Total number of outcomes = 98 (i) Favourable outcomes are 8, 16, 24, ...., 96, i.e., 12 ½ 12 6 1 Probability (integer is divisible by 8) = or 98 49 6 (ii) Probability (integer is not divisible by 8) = 1 49 43 = 49 [CBSE Marking Scheme, 2018] ½ Q . 10. 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. U [CBSE SQP, 2018] [CBSE O.D., Set-I, II, III, 2016]
Q. 12. Two different dice are thrown together. Find the probability that the product of the number appeared is less than 18. U [Foreign Set-I, II, III, 2017]
Q. 13. A letter of English alphabet is chosen at random, find the probability that the letter so chosen is : (i) a vowel, (ii) a consonant. A [CBSE Delhi Term-2, Set-I, II, III, 2015] Sol. Since, total number in english alphabet is 26. in which 5 vowels and 21 consonants. 5 1 (i) P(a vowel) = 26
Q. 9. An integer is chosen at random between 1 and 100. Find the probability that it is : (i) divisible by 8. (ii) not divisible by 8. U [CBSE Delhi/OD, 2018]
21 (ii) P(a consonant) = 26 [CBSE Marking Scheme, 2015] 1
Q. 14. A bag contains cards with numbers written on it from 1 - 80. A card is pulled out at random. Find the probability that the card shows a perfect square. U [CBSE S.A.-2, 2016] [CBSE Delhi Set-I, II, III, 2016] Sol. S = {1, 2 ............... 80} n(S) = 80 ½ A = {1, 4, 9, 16, 25, 36, 49, 64} n(A) = 8 ½ n( A ) 8 P(A) = n(S) = 80 1 = 1 10
[ 397
Probability
Short Answer Type Questions-II
\ No. of favourable outcomes = 9 P(Selected card bears perfect square numbers) 9 1 = = 90 10
Q. 1. An integer is chosen between 70 and 100. Find the probability that it is. (i) a prime number U [CBSE SQP, 2020-21] (ii) divisible by 7.
3 marks each
1
[CBSE Marking Scheme, 2017] Q. 4. Two different dice are thrown together. Find the probability that the number obtained. (i) have a sum less than 7. (ii) have a product less than 16. (iii) is a doublet of odd numbers. U [CBSE Delhi Set-I, II, III, 2017]
6 29 (ii) Numbers, which are divisible by 7 = 77, 84, 91, 98 i.e., favourable outcomes = n(E1) = 4 n( E1 ) \ P(number divisible by 7) = n(S)
(ii) Product less than 16 = (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), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2) No. of favourable out outcomes = 25 \ P(have a product less than 16) 25 1 = 36
1
=
=
½
4 . 29
1
Q. 2. From all the two digit numbers a number is chosen at random. Find the probability that the chosen number is a multiple of 7.
U [CBSE OD Comptt. Set-III, 2017]
Sol. All possible outcomes are 10, 11, 12 ....... 98, 99. No. of all possible outcomes = 90. 1 All favourable outcomes are 14, 21, 28 ..... 98 No. of favourable outcomes = 13 1 13 1 \ P(getting a number multiple of 7) = 90 [CBSE Marking Scheme, 2017] Q. 3. A box contains cards, number 1 to 90. A card is drawn at random from the box. Find the probability that the selected card bears a : (i) Two digit number. (ii) Perfect square number U [Delhi Comptt. Set-I, 2017]
Sol. No. of all possible outcomes = 90 (i) No. of cards having 2 digit number = 90 – 9 = 81 \ No. of favourable outcomes = 81 P(selected card bears two digit number) No. of favourable outcomes 81 9 = 1 = = No. of all possible outcomes 900 10 (ii) Perfect square numbers between 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81 1
Sol. Total number of all possible outcomes = 62 = 36 (i) The sum less than 7 = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1) No. of favourable outcomes = 15 15 5 = 1 P(have sum less than 7) = 36 12
(iii) Doublet of odd numbers = (1, 1), (3, 3), (5, 5) No. of favourable outcomes = 3 \ P(a doublet of odd number) 3 1 = 1 = 36 12 [CBSE Marking Scheme, 2017] Q. 5. 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. [CBSE Delhi Set-I, II, III, 2017] Sol. Peter throws two dice together \ Total number of possible outcomes = 62 = 36 ½ He get 25 only when he gets (5, 5) \ No. of favourable outcomes = 1 ½ 1 1 P(getting the numbers of product 25) = 36 Rina throws one dice \ Total number of all possible outcomes = 6 The number where square is 25 is 5 \ No. of favourable outcomes = 1 ½ 1 P(getting a number whose square is 25) = 6 1 1 ½ > 6 36 Hence, Rina has better chances to get the number 25. [CBSE Marking Scheme, 2017]
Total number of integers = 29 6 (i) Prob. (prime number) = 29 4 (ii) Prob. (number divisible by 7) = 29 [CBSE SQP Marking Scheme, 2020] 3 Detailed Solution: Total number of outcomes = n(S) = 29 (i) Total prime numbers between 70 and 100 = 71, 73, 79, 83, 89, 97 i.e., favourable outcomes = n(E) = 6 ½ n( E) \ P(prime number) = n(S)
Sol.
398 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
T
Q. 6. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. C [CBSE OD Set-III, 2017]
opper Answer, 2017
Sol.
U [CBSE Delhi Set I, 2016]
Sol. (i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e., 9 outcomes. 1 9 1 ½ P(a prime number on each die) = or 36 4
(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes 1 6 1 or ½ P(a total of 9 or 11) = 36 6 [CBSE Marking Scheme, 2016] Q. 8. Two different dice are thrown together. Find the probability of :
(i) getting a number greater than 3 on each dice.
(ii) getting a total of 6 or 7 of the numbers on two dice.
C + A [CBSE Delhi Set II, 2016]
Sol. (i) Favourable outcomes are (4, 4) (4, 5) (4, 6) (5, 4) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6) 1 \ No. of favourable outcomes = 9 9 1 or ½ P(a number > 3 on each dice) = 36 4
(ii) Favourable outcomes are (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) 1 \ No. of favourable outcomes = 11 11 . ½ P(a total of 6 or 7) = 36
[CBSE Marking Scheme, 2016]
Q. 9. A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but ‘Kewal’ another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that : (i) Ramesh will buy the selected shirt ? (ii) ‘Kewal ‘ will buy the selected shirt ? C + A [CBSE Delhi Set III, 2016]
Sol. (i) Number of good shirts = 88 1 88 22 ½ P(Ramesh buys the shirt) = or 100 25 (ii) Number of shirts without major defect = 96 1 96 24 ½ P(Kewal buys a shirt) = or 100 25 [CBSE Marking Scheme, 2016] Q. 10. Three different coins are tossed together. Find the probability of getting (i) exactly two heads. (ii) at least two heads U [CBSE OD Set I, 2016] (iii) at least two tails.
Q. 7. In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice ? (ii) a total of 9 or 11 ?
3
Sol. Possible outcomes are {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒
(i)
∴
n(S) = 8
Exactly two heads = {HHT, HTH, THH} n(P1) = 3 P1 =
n( P1 ) 3 = n(S) 8
1
(ii) At least two heads {HHT, HTH, THH, HHH}
n(P2) = 4
P2 =
n( P2 ) 4 1 = = n(S) 8 2
1
(iii) At least two tails {TTH, THT, HTT, TTT}
n(P3) = 4
P3 =
[CBSE Marking Scheme, 2016]
n( P3 ) 4 1 = = 1 n(S) 8 2
Q. 11. A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh will win the game if all the tosses show the same result, (i.e. either all three heads or all three tails) and loses the game otherwise. Find the probability that Ramesh will lose the game.
C + A [Foreign Set-I, 2016]
[ 399
Probability
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
No. of total outcomes = 8
Same result on all the tosses (A) = HHH, TTT,
\ P (Ramesh will lose the game) =
1 1
No. of favourable outcomes = 2
Sol. (a) Total number of cards = 52 Number of non-face cards = 52 – 12 = 40 40 10 = P(non-face cards) = 52 13 (b) Number of black kings = 2 Number of red queens = 2 4 1 = P(a black King or a red queen) = 52 13 (c) Number of spade cards = 13 13 1 = P(Spade cards) = 52 4
Sol. Total outcomes are
6 3 8-2 = = 8 4 8
1
[CBSE Marking Scheme, 2016]
Q. 12. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting (a) Non face card, (b) Black king or a Red queen, (c) Spade card. C + A [CBSE SQP, 2016]
1
1
1
[CBSE Marking Scheme, 2016]
T
Q. 13. 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 drawn card is : (i) a black king, (ii) a card of red colour, (iii) a card of black colour. C [CBSE Term-II, 2015, 2016]
opper Answer, 2016
Sol.
Q. 14. A bag contains 18 balls out of which x balls are red.
(ii) Total number of balls = 18 + 2 = 20
(i) If one ball is drawn at random from the bag, what is the probability that it is not red ?
(ii) If 2 more red balls are put in the bag, the probability 9 of drawing a red ball will be times the probability 8
of drawing a red ball in the first case. Find the value
of x.
Sol. (i)
C + A [Foreign Set I, 2015]
P(red ball) =
x 18
x 18 - x = P(no red ball) = 1 18 18
1
3
red balls are = x + 2 P(red balls) =
x+2 20
½
Now, According to the question,
⇒
⇒
⇒
x+2 9 x = × 20 8 18
180x = 144x + 288 36x = 288 x =
288 =8 36
1½
[CBSE Marking Scheme, 2015]
400 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Long Answer Type Questions
Sol. Total number of cards = 52 – 3 = 49
1
13 P(spade) = 49
1
(i)
(ii)
P(black king) =
1 49
1
(iii)
P(club) =
10 49
1
(iv)
P(Jack) =
3 49
1
[CBSE Marking Scheme, 2018] Detailed Solution : Total number of cards = 52 – 3 = 49 1 (i) Number of spade cards = 13 13 1 \ P(getting a spade) = 49
(ii) \
(iii)
\
(iv)
\
Number of black king = 2 – 1 = 1 1 1 P(getting a black king) = 49 Number of club card = 13 – 3 = 10 10 P(getting a club card) = 1 49 Number of jacks = 4 – 1 = 3 3 P(getting a jack) = 1 49
Q. 2. A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is (i) a prime number (ii) a composite number (iii) a number divisible by 3 A [CBSE Comptt. I, II, III,2018] [CBSE Term-2, Set I, II, 2015]
Sol. (i) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 i.e. 8. 8 2 or P(prime number) = 20 5 (ii) Composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, i.e., 11 11 P(Composite number) = 20
1½
2
[CBSE Marking Scheme, 2018]
COMMONLY MADE ERROR Many candidates commit the following
errors. (i) Total outcomes of event are incorrect. (ii) Favourable outcomes are incorrect. (iii) The results are not given in the simplest form. e.g.
8 2 = 20 5
ANSWERING TIP All necessary outcomes must be listed before finding probability and all answers must be in the simplest from.
Q. 3. Two different dice are rolled together once. Find the probability of numbers coming on the tops whose product is a perfect square.
A [CBSE OD Comptt. Set-I, 2017]
Sol. No. of all possible outcomes = 62 = 36
1
All favourable outcomes are (2, 2), (3, 3), (4, 4) (5, 5), (6, 6), (1, 1), (4, 1), (1, 4) \ No. of favourable outcomes = 8
1½
\ P(getting the numbers, whose product is a 8 2 1½ = . perfect square) = 36 9
[CBSE Marking Scheme, 2017]
Q. 4. A box contains 125 shirts of which 110 are good 12 have minor defects and 3 have major defects Ram Lal will buy only those shirts which are good while Naveen will reject only those which have major defects. A shirt is taken out at random from the box. Find the probability that :
(i) Ram Lal will buy it.
(ii) Naveen will buy it.
A [CBSE OD Set-III, 2017]
Sol. No. of all possible outcomes = 125 (i) Ramlal will buy a good shirt \ No. of favourable outcomes = 110
1½
(iii) Numbers divisible by 3 from 1 to 20 are 3, 6, 9, 12, 15, 18 i.e., 6 6 3 or P(number divisible by 3) = 20 10
Q. 1. The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well shuffled. A card is selected from the remaining cards. Find the probability of getting a card. (i) of spade (ii) of black king (iii) of club (iv) of jacks U [CBSE Comptt. Set I, II, III, 2018]
5 marks each
½ ½
[ 401
Probability
=
\ P(Ramlal will buy a shirt) No. of favourable outcomes No. of all possible outcomes
No. of discs with two digits numbers = 90 – 9 = 81 \ No. of favourable outcomes = 81 P(a disc with two digit number) 1½
(ii) Naveen will reject the shirt which have major defects. \ No. of favourable outcomes = 125 – 3 = 122
1½
[CBSE Marking Scheme, 2017]
Required probability =
1
9 49 13 49 10 49
1
1 49
1½
[CBSE Marking Scheme, 2017]
Q. 7. From a deck of 52 playing cards, Jacks and kings of red colour and Queen and Aces of black colour are removed. The remaining cards are mixed and a card is drawn at random. Find the probability that the drawn card is :
(i) A black Queen. (ii) A card of red colour.
(iii) A Jack of black colour. (iv) A face card.
A [CBSE OD Comptt. Set-I, 2017]
Sol. No. of all possible outcomes = 52 – (2 + 2 + 2 + 2)
\ P(getting a black Queen) =
1
1
[CBSE Marking Scheme, 2017]
Q. 6. A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number, (ii) number divisible by 5.
18 1 = 90 5
1 0 =0 44
Hence it is an impossible event
(iv) There is only one queen of diamond \ Required probability =
½
(i) No. of black Queens in the deck = 0
(iii) No. of cards of clubs = 13 – 3 = 10 \ Required probability =
½
= 44
(ii) No. of card of heart in the deck = 13 \ Required probability =
=
Sol. No. of all possible outcomes = 52 – 3 = 49 No. of face cards = 12 – 3 = 9
½
\ P(a disc with a number divisible by 5)
A [CBSE Delhi Comptt. Set-II, 2017]
(i)
= 81 = 9 90 10
\ No. of favourable outcomes = 18
Q. 5. The king, queen and jack of clubs are removed from a deck of 52 cards. The remaining cards are mixed together and then a card is drawn at random from it. Find the probability of getting (i) a face card, (ii) a card of heart, (iii) a card of clubs (iv) a queen of diamond.
No. of favourable outcomes No. of all possible outcomes
5, 10, 15 .... 85, 90
No. of favourable outcomes = No. of all possible outcomes
=
(ii) The numbers divisible by 5 from 1 to 90 are
\ P(Naveen will buy the shirt)
½
No. of favourable outcomes = No. of all possible outcomes
122 = 125
½
(i) Discs with two digit number are 10, 11, ....... 90
\ P(Ramlal will buy a shirt)
110 22 = = 125 25
Sol. Total number of discs in the box = 90
\ No. of all possible outcomes = 90
A [Foreign Set-I, II, III, 2017]
(ii)
1
No. of red cards = 26 – 4 = 22 22 1 = 44 2
\
P(getting a red card) =
(iii)
No. of Jacks (black) = 2
1
\ P(getting a black coloured Jack) =
2 1 = 44 22
1
(iv) No. of face cards in the deck = 12 – 6 = 6 \ P(getting a face card) =
6 3 = 44 22
1
[CBSE Marking Scheme, 2017]
Q. 8. Two different dice are thrown together. Find the probability that the numbers obtained have (i) even sum, and (ii) even product. A [CBSE OD Set-III, 2017]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
T
402 ]
opper Answer, 2017
Sol.
5
[CBSE Marking Scheme, 2017]
Q. 10. In fig. a disc on which a player spins an arrow a twice. The fraction is formed, where ‘a‘ is the b number of sector on which arrow stops on the first spin and ‘b‘ is the number of the sector in which the arrow stops on second spin, On each spin, each sector has equal chance of selection by the arrow. a > 1. Find the probability that the fraction b
Sol. Total number of possible outcomes = 100 (i) Number of even numbers from 1 to 100 = 50 \ P(card taken out has an even number) 50 1 = = 100 2 (ii) Multiple of 13 from 1 to 100 are 13, 26, 39, 52, 65, 78 and 91. No. of favourable outcomes = 7 7 \ P(card taken out has multiple of 13) = 100
(iv) Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19. No. of all favourable outcomes = 8 \ P(card taken out has a prime number less than 20) 8 2 = 1 = 100 25
Q. 9. Cards on which numbers 1, 2, 3 ...... 100 are written (one number on one card and no number is repeated) and put in a bag and are mixed thoroughly. A card is drawn at random from the bag. Find the probability that card taken out has. (i) An even number (ii) A number which is a multiple of 13. (iii) A perfect square number (iv) A prime number less than 20. A [CBSE Delhi Comptt. Set-I, 2017] [CBSE SQP, 2016]
1½
2
1½
(iii) Perfect square numbers from 1 to 100 are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. No. of all favourable outcomes = 10 \ P(card taken out has a perfect square number) 10 1 = = 100 10
3
1
4
6 5
A [Foreign Set I, II, III, 2016]
[ 403
a > 1, when a = 1, b can not take any value, b
a = 2, b can take 1 value, a = 3, b can take 2 values, a = 4, b can take 3 values
When a = 5, b can take 4 values,
a = 6, b can take 5 values.
2½
Total possible outcomes = 36 ½ a + + + + 1 2 3 4 5 1 \ P > 1 = 36 b 15 5 or 1 = 36 12 [CBSE Marking Scheme, 2016]
Sol. For
Probability
Q. 11. 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. A [CBSE OD Set-II, 2016] Sol.
[Topper’s Answer, 2016] 5 . 12. A card is drawn at random from a well-shuffled Q deck of playing cards. Find the probability that the card drawn is : (i) a card of spade or an ace. (ii) a black king. (iii) neither a jack nor a king. (iv) either a king or a queen. A [CBSE OD Set-I, II, III, 2015]
A [CBSE Term-2, 2015]
Sol. (i) Cards of spade or an ace = 13 + 3 = 16 Total no. of cards = 52 16 4 = P(spade or an ace) = 52 13
Q. 13. A bag contains 15 balls of which x are blue and the remaining are red. If the number of red balls are increased by 5, the probability of drawing the red balls doubles. Find : (i) P(red ball) (ii) P(blue ball) (iii) P(blue ball if of 5 extra red balls are actually added)
1½
(ii)
Black kings = 2 2 1 = 1 P(a black king) = 52 26 (iii) Jack or king = 4 + 4 = 8 52 - 8 44 11 = = 1 P(neither jack nor a king) = 52 52 13 (iv) King or queen = 4 + 4 = 8 8 2 = 1½ P(either a king or a queen) = 52 13
Sol. According to the question,
[CBSE Marking Scheme, 2015]
⇒ ⇒
15 − x 20 − x =2 15 20 x 2x 1= 215 20
1
2x x − = 2 –1 15 20 8 x - 3x =1 60
⇒ ⇒ 5x = 60 \ x = 12 \ Blue balls = 12 and red balls = 3
1
404 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
P(red ball) =
(ii)
1 1
(iii) P(blue ball if 5 red balls are added) 12 3 = = 20 5
3 1 = 15 5 12 4 = P(blue ball) = 15 5
(i)
[CBSE Marking Scheme, 2015]
Visual Case Based Questions ote: Attempt any four sub parts from each N question. Each sub part carries 1 mark Q. 1. On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card.
4 marks each Sol. Correct option: (a). Explanation: No. of face card = 13 Total no of cards = 52 Probability of getting a face card No. of face cards = Total no. of cards =
(i) Find the probability of getting a king of red colour. 1 1 (b) (a) 26 13 1 1 (c) (d) 52 4 Sol. Correct option: (a). Explanation: No. of cards of a king of red colour = 2 Total no. of cards = 52 Probability of getting a king of red colour No. of king of red colour = Total number of cards 2 1 = = 52 26 (ii) Find the probability of getting a face card. 1 1 (b) (a) 26 13 2 3 (d) (c) 13 13 Sol. Correct option: (d). (iii) Find the probability of getting a jack of hearts. 1 1 (b) (a) 26 52 3 3 (c) (d) 52 26 Sol. Correct option: (b). (iv) Find the probability of getting a red face card. 3 1 (b) (a) 13 13
(c)
1 52
(d)
1 4
1
12 3 = 52 13
(v) Find the probability of getting a spade. 1 1 (b) (a) 26 13 (c)
1 52
(d)
1 4
Sol. Correct option: (d). No. of face card = 13 Total no of cards = 52 Probability of getting a face card No. of face cards = Total no. of cards =
13 1 = 52 4
Q. 2. Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.
[ 405
Probability
(i) Ravi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8? 1 5 (b) (a) 26 36
1 (d) 0 18 Sol. Correct option: (b). Explanation: The outcomes when two dice are thrown together 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) Total outcomes = 36 No. of outcomes when the sum is 8 = 5 5 Probability = 36
(c)
(ii) Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13? 5 (a) 1 (b) 36 (c)
(iii) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12 ? 5 (a) 1 (b) 36
O
1 0 18 (d)
Sol. Correct option: (a). Explanation: No. of outcomes when the sum is less than or equal to 12 = 36
Total outcomes = 36
probability =
36 =1 36
(iv) Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7 ? 5 5 (b) (a) 9 36
(c)
1 6
(d) 0
Sol. Correct option: (c).
1 0 18 (d)
Sol. Correct option: (d). Explanation: No. of outcomes when the sum is 13 = 0 Total outcomes = 36 0 Probability = =0 35
(c)
(v) Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8 ? 5 (a) 1 (b) 36 (c)
1
18 Sol. Correct option: (d).
(d)
5 18
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406 ]
Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, mathematics (Standard), Class – X
Maximum Time: 1 hour
MM: 25
Q. 1. For the following distribution:
Class
0–5
5 – 10
10 – 15
15 – 20
20 – 25
Frequency
10
15
12
20
9
Find the sum of lower limits of median class and modal class.
A
Q. 2. Consider the following distribution :
Marks Obtained
0 or more
10 or more
20 or more
30 or more
40 or more
50 or more
Number of students
63
58
55
51
48
42
(i) Calculate the frequency of the class 30 – 40. Q. 3. From the number 3, 5, 5, 7, 7, 7, 9, 9, 9, 9, one number is selected at random, what is the probability that the selected number is mean ?
R [CBSE Term-2, 2012]
Q. 5. A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that number is divisible by both 2 and 3.
U [Foreign Set I, II, III, 2014]
Q. 6. Visually Case Based Questions
U [CBSE Term-2, 2012]
Q. 4. A girl calculates the probability of her winning the game in a match and find it 0.08. What is the probability of her losing the game ?
A [Board Term- I, 2014]
(ii) Calculate the class mark of the class 10 – 20.
(1 mark each)
A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random.
(ii) Find the probability that lost piece is triangle. (a)
4 (b) 9
(c)
1 3
(d)
5 9 5 18
(iii) Find the probability that lost piece is square.
4 (b) 9
(a)
(c)
1 3
(d)
5 9 5 18
(iv) Find the probability that lost piece is square of blue colour.
4 (b) 9
(a)
5 9
1 5 (d) 3 18 (v) Find the probability that lost piece is triangle of red colour. (c)
(a)
4 (b) 9
(c)
1 3
(d)
5 9 5 18
Q. 7. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of the red ball, find the number of blue balls in the bag. C + A [Board Term-2, 2012] [Board Term-2, 2012]
C + AE (i) How many triangles are of red colour and how many squares are of red colour?
(a) 5, 4
(b) 4, 5
(c) 5, 5
(d) 8, 6
Q. 8. Two different dice are tossed together. Find the probability. (i) that the number on each dice is even. (ii) that the sum of numbers appearing on the two dice is 5. C + A [CBSE OD Set-I, 2014] [CBSE Term-2, 2012]
[ 407
Probability
Q. 9. If the median for the following frequency distribution is 28.5, find the value of x and y : Class
Frequencies
0 – 10
5
10 – 20
x
20 – 30
20
30 – 40
15
40 – 50
y
50 – 60
5
Total
60 U [CBSE Term-1, 2013]
Q. 10. Find the mode of the following data : Marks
Below 10
Below 20
Below 30
Below 40
Below 50
Number of students
8
20
45
58
70
Q. 11. The following table gives the weight of 120 articles :
Weight (in kg)
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
Number of students
14
17
22
26
23
18
Change the distribution to a ‘more than type’ distribution and draw its ogive. A [CBSE Term-1, Set-48, 2012]
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0704
HYDERABAD
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