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\ 23-Nov-2021, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , Published by:, , An ISO : 9001-2015 Company, , 9, Daryaganj, New Delhi-110002, Phone: 011- 40556600 (100 Lines), Website: www.fullmarks.org, E-mail: [email protected], © Publishers, All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means,, without permission. Any person who does any unauthorised act in relation to this publication may be liable to criminal, prosecution and civil claims for damages., , Branches:, • Chennai • Guwahati, Marketing Offices:, • Ahmedabad • Bengaluru • Bhopal • Dehradun • Hyderabad • Jaipur • Jalandhar • Kochi, • Kolkata • Lucknow • Mumbai • Patna • Ranchi, , NEW EDITION, , “This book is meant for educational and learning purposes. The author of the book has taken all reasonable care to ensure that the contents, of the book do not violate any existing copyright or other intellectual property rights of any person in any manner whatsoever. In the event, the author has been unable to track any source and if any copyright has been inadvertently infringed, please notify the publisher in writing, for corrective action.”, , Printed at:

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\ 23-Nov-2021, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , Note from the Publishers, Mathematics Standard-10 (Term-2) is based on the latest curriculum released by CBSE in July 2021, for 2021-22 (Term-2) Board Examination to be conducted between March - April 2022. It will certainly prove to, be a torch-bearer for those who toil hard to achieve their goal., This All-in-one Question Bank has been developed keeping in mind all the requirement of the students for Board, Examinations preparations like learning, practicing, revising and assessing., Salient Features of the Book:, • Each chapter is designed in ‘Topic wise’ manner where each topic is briefly explained with sufficient Examples, and Exercise., • In all chapters, Topic wise exercises cover Very Short Answers Type Questions (MCQs, Assertion - Reason, Type Questions and Stand Alone Questions), Short Answer Type Questions - I, Short Answer Type Questions - II,, Long Answer Type Questions and Case Study Based Questions as per the Special Scheme of Assessment, suggested by CBSE vide Circular No. Acad-51/2021. Answers and sufficient hints are also provided separately, at the end of each exercise., • Experts’ Opinion has been provided to suggest the students which type of questions are important for examination, point of view., • Chapterwise Important Formulae and Quick Revision Notes have been prepared for Quick Revision., • Common Errors by the students are provided to make students aware what errors are usually committed by them, unknowingly., • 3 Sample Papers (2 Solved & 1 Unsolved with Answers) are given for self assessment., • The book has been well prepared to build confidence among students., Suggestions for further improvement of the book, pointing out printing errors/mistakes which might have crept, in spite of all efforts, will be thankfully received and incorporated in the next edition., –Publishers, , (iii)

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\ 24-Nov-2021, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , CBSE CIRCULAR 2021-22, , osQUnzh; ekè;fed f'k{kk cksMZ, (f'k{kk ea=kky;] Hkkjr ljdkj osQ v/hu Lok;Ùk laxBu), , CENTRAL BOARD OF SECONDARY EDUCATION, (An Autonomous Organisation Under the Ministry of Education, Govt. of India), , CBSE/DIR (ACAD)/2021, , Date: July 05, 2021, Circular No: A cad-51/2021, , All the Heads of Schools affiliated to CBSE, Subject : Special Scheme of Assessment for Board Examination Classes X and XII for the Session 2021-22, COVID 19 pandemic caused almost all CBSE schools to function in a virtual mode for most part of the academic session of, 2020-21. Due to the extreme risk associated with the conduct of Board examinations during the second wave in April 2021,, CBSE had to cancel both its class X and XII Board examinations of the year 2021 and results are to be declared on the, basis of a credible, reliable, flexible and valid alternative assessment policy. This, in turn, also necessitated deliberations, over alternative ways to look at the learning objectives as well as the conduct of the Board Examinations for the academic, session 2021-22 in case the situation remains unfeasible., CBSE has also held stake holder consultations with Government schools as well as private independent schools from across, the country especially schools from the remote rural areas and a majority of them have requested for the rationalization, of the syllabus, similar to last year in view of reduced time permitted for organizing online classes. The Board has also, considered the concerns regarding differential availability of electronic gadgets, connectivity and effectiveness of online, teaching and other socio-economic issues specially with respect to students from economically weaker section and those, residing in far flung areas of the country. In a survey conducted by CBSE, it was revealed that the rationalized syllabus, notified for the session 2020-21 was effective for schools in covering the syllabus and helped learners in achieving learning, objectives in a less stressful manner., In the above backdrop and in line with the Board’s continued focus on assessing stipulated learning outcomes by making the, examinations competencies and core concepts based, student-centric, transparent, technology-driven, and having advance, provision of alternatives for different future scenarios, the following schemes are introduced for the Academic Session for, Class X and Class XII 2021-22., , 2. Special Scheme for 2021-22, A., , B., , C., , Academic session to be divided into 2 Terms with approximately 50% syllabus in each term: The syllabus, for the Academic session 2021-22 will be divided into 2 terms by following a systematic approach by looking into, the interconnectivity of concepts and topics by the Subject Experts and the Board will conduct examinations at, the end of each term on the basis of the bifurcated syllabus. This is done to increase the probability of having a, Board conducted classes X and XII examinations at the end of the academic session., The syllabus for the Board examination 2021-22 will be rationalized similar to that of the last academic, session to be notified in July 2021. For academic transactions, however, schools will follow the curriculum and, syllabus released by the Board vide Circular no. F.1001/CBSEAcad/ Curriculum/2021 dated 31 March 2021. Schools, will also use alternative academic calendar and inputs from the NCERT on transacting the curriculum., Efforts will be made to make Internal Assessment/ Practical/ Project work more credible and valid as, per the guidelines and Moderation Policy to be announced by the Board to ensure fair distribution of marks., , 3. Details of Curriculum Transaction, ●, ●, , ●, , ●, , ●, ●, , Schools will continue teaching in distance mode till the authorities permit inperson mode of teaching in schools., Classes IX-X: Internal Assessment (throughout the year-irrespective of Term I and II) would include the 3, periodic tests, student enrichment, portfolio and practical work/ speaking listening activities/ project., Classes XI-XII: Internal Assessment (throughout the year-irrespective of Term I and II) would include end of, topic or unit tests/ exploratory activities/ practicals/ projects., Schools would create a student profile for all assessment undertaken over the year and retain the evidences in, digital format., CBSE will facilitate schools to upload marks of Internal Assessment on the CBSE IT platform., Guidelines for Internal Assessment for all subjects will also be released along with the rationalized term, wise divided syllabus for the session 2021-22.The Board would also provide additional resources like sample, assessments, question banks, teacher training etc. for more reliable and valid internal assessments., , 4. Term II Examination/ Year-end Examination, ●, , At the end of the second term, the Board would organize Term II or Year-end Examination based on the, rationalized syllabus of Term II only (i.e. approximately 50% of the entire syllabus)., , (iv)

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\ 24-Nov-2021, , ●, ●, , ●, , ●, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , This examination would be held around March-April 2022 at the examination centres fixed by the Board., The paper will be of 2 hours duration and have questions of different formats (case-based/ situation based, open, ended- short answer/ long answer type)., In case the situation is not conducive for normal descriptive examination a 90 minute MCQ based exam will be, conducted at the end of the Term II also., Marks of the Term II Examination would contribute to the final overall score., , 5. Assessment / Examination as per different situations, A., , B., , C., , D., , , , In case the situation of the pandemic improves and students are able tocome to schools or centres, for taking the exams., Board would conduct Term I and Term II examinations at schools/centres and the theory marks will be distributed, equally between the two exams., In case the situation of the pandemic forces complete closure of schools during November-December, 2021, but Term II exams are held at schools or centres., Term I MCQ based examination would be done by students online/offline from home - in this case, the weightage, of this exam for the final score would be reduced, and weightage of Term II exams will be increased for declaration, of final result., In case the situation of the pandemic forces complete closure of schools during March-April 2022, but, Term I exams are held at schools or centres., Results would be based on the performance of students on Term I MCQ based examination and internal assessments., The weightage of marks of Term I examination conducted by the Board will be increased to provide year end, results of candidates., In case the situation of the pandemic forces complete closure of schools and Board conducted Term, I and II exams are taken by the candidates from home in the session 2021-22., Results would be computed on the basis of the Internal Assessment/Practical/Project Work and Theory marks, of Term-I and II exams taken by the candidate from home in Class X / XII subject to the moderation or other, measures to ensure validity and reliability of the assessment., In all the above cases, data analysis of marks of students will be undertaken to ensure the integrity of internal, assessments and home based exams., , CBSE CIRCULAR 2021-22, , osQUnzh; ekè;fed f'k{kk cksMZ, (f'k{kk ea=kky;] Hkkjr ljdkj osQ v/hu Lok;Ùk laxBu), , CENTRAL BOARD OF SECONDARY EDUCATION, (An Autonomous Organisation Under the Ministry of Education, Govt. of India), , NO.: F.1001/CBSE-Acad/Curriculum/2021, , July 22, 2021, Cir. No. A cad-53/2021, , All the Heads of Schools affiliated to CBSE, , Subject : Term wise syllabus for Board Examinations to be held in the academic session, 2021-22 for Secondary conduct of the Internal Assessment/Practicum/Project., This is in continuation to Board’s circular num regarding Special Scheme of Assessment for Board Examination, for Classes X and XII for the Session 2021- 22. The subjects for classes IX to XII are hereby notified vides, syllabus for term end board examinations, guidelines for the conduct of Internal Assessment/Practicum/Project, are also enclosed., Schools are requested to share the term wise syllabus and guidelines for the conduct of board examinations, and Internal Assessment / Practicum / Project available on CBSE Academic Website http://cbseacademic.nic., in at the link http:/chseacademic.nic.in/Term-wise-curriculum 2022.html with all their teachers and students., , (v)

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\ 24-Nov-2021, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , SYLLABUS, Units, I., II., III., IV., V., , SECOND TERM, Unit Name, , 90 Marks, 10, 09, 07, 06, 08, 40, 10, 50, , Algebra, Geometry, Trigonometry, Mensuration, Statistics & Probability, Total, INTERNAL ASSESSMENT, Total, , UNIT I: ALGEBRA, 1. Quadratic Equations, Standard form of a quadratic equation ax2 + 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., Situational problems based on quadratic equations related to day to day activities (problems on equations reducible, to quadratic equations are excluded), 2. Arithmetic Progressions, 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., (Applications based on sum to n terms of an A.P. are excluded), UNIT II: GEOMETRY, 3. Circles, 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., 4. Constructions, 1. Division of a line segment in a given ratio (internally)., 2. Tangents to a circle from a point outside it., , UNIT III: TRIGONOMETRY, 5. Some Applications of Trigonometry, 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°., UNIT IV: MENSURATION, , 6. Surface Areas and Volumes, 1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres, and right circular cylinders/cones., 2. Problems involving converting one type of metallic solid into another and other mixed problems. (Problems, with combination of not more than two different solids be taken)., UNIT VII: STATISTICS AND PROBABILITY, , 7. Statistics & Probability, Mean, median and mode of grouped data (bimodal situation to be avoided). Mean by Direct Method and Assumed, Mean Method only, INTERNAL ASSESSMENT, Periodic Tests, Multiple Assessments, Portfolio, Student Enrichment, Activities - practical work, , MARKS, 3, 2, 2, 3, , (vi), , TOTAL MARKS, 10 marks for the term

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Praveen Kumar, , \ 24-Nov-2021, , Proof-1, , Reader’s Sign _______________________, , Date __________, , Mathematics 10, , GLIMPSE OF A CHAPTER, 2, , ., Arithmetic Progressions, , Topics covered, 1. Sequence / Progression, 3. Sum of First n Terms of an AP, , 2. Arithmetic Progressions and its nth term, , 1. Sequence/ProgreSSion, • Sequence/Progression: A sequence/progression is a succession of numbers or terms formed according to some pattern, or rule. Various numbers occurring in a sequence are called terms or elements., Consider the following arrangements of numbers:, 1 1 1 1, (i) 1, 8, 27, 64, 125, ..., (ii) 1, , , , , ..., (iii) 2, 4, 6, 8, 10, ..., 2 3 4 5, In each of the above arrangements, numbers are arranged in a definite order according to some rule. So, they are, sequences., A sequence is generally written as < an > : a1, a2, a3, ..., an where a1, a2, a3, ... are the first, second and third terms of, the sequence., • A sequence with finite number of terms or numbers is called a finite sequence., • A sequence with infinite number of terms or numbers is called an infinite sequence., Example 1. Write first four terms of each of the following sequence, whose general terms are:, (i) an = 3n – 7, (ii) an = (– 1)n+1 × 3n, Solution. (i), an = 3n – 7, \, a1 = 3 × 1 – 7 = 3 – 7 = – 4, a2 = 3 × 2 – 7 = 6 – 7 = – 1,, a3 = 3 × 3 – 7 = 9 – 7 = 2 and a4 = 3 × 4 – 7 = 12 – 7 = 5, (ii), an = (– 1)n+1 × 3n, 1+1, 1, fi, a1 = (– 1) × 3 = 3,, \, a2 = (– 1)2+1 × 32 = (–1)3 × 32 = –9,, a3 = (– 1)4 × 33 = 27 and a4 = (– 1)5 × 34 = – 81, Example 2. What is 18th term of the sequence defined by an =, Solution. We have,, Putting n = 18, we get, , an =, ,  xercise related to each topic dealt separately and, E, Questions included segregated into Very Short Answer, Type Questions, Short Answer Type Questions, Long, Answer Type Questions, and Case Study Based, Questions., Exercise 4.1, I. Very Short Answer Type Questions, , n(n − 3), ?, n+4, , n(n − 3), n+4, , 18 × (18 − 3), a18 =, 18 + 4, =, , 18 × 15, 135, =, 22, 11, , Exercise 2.1, I. Very Short Answer Type Questions, 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) If an = 5n – 4 is a sequence, then a12 is, (a) 48, (b) 52, (c) 56, (2) If an = 3n – 2, then the value of a7 + a8 is, (a) 39, (b) 41, (c) 47, , [1 Mark], , 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn such that ∠BAX is an acute angle and then at, equal distances points are marked on the ray such that the minimum number of these points is, (a) 8, (b) 10, (c) 11, (d) 12, (2) To divide line segment AB in the ratio m : n, draw a ray AX so that ∠BAX is an acute angle and then mark points, on ray AX at equal distance such that the minimum number of these points is, (a) m + n, (b) m – n, (c) m × n, (d) m ÷ n, (3) To divide a line segment AB in the ratio 5 : 6, we draw a ray such that ∠BAX is an acute angle, locate points A1,, A2, A3, ... at equal distances on the ray AX and join point B to, (a) A12, (b) A11, (c) A10, (d) A9, 2. Assertion-Reason Type Questions, In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): By geometrical construction, it is possible to divide a line segment in the ratio 3 : 1 ., 3, Reason (R): To divide a line segment in the ratio p : q, p and q must be positive integers., (2) Assertion (A): When a line segment is divided in the ratio 2 : 3, the number of parts in which the line is divided is, 5., Reason (R): To divide a line segment in the ratio p : q, we divide it into (p + q) parts., 3. Answer the following:, , [1 Mark], , (d) 62, (d) 53, , 33, , (1) Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4., [Delhi 2011], AP 3, = . [Foreign 2011], (2) Draw a line segment AB of length 6.5 cm. Find a point P on it such that, AB 5, , .E ach chapter is divided into topics and, explained separately., , (3) Geometrically divide a line segment of length 8.4 cm in the ratio 5 : 2., , [CBSE 2015], , (4) Draw a line segment of length 7.6 cm and divide it in the ratio 3 : 2., , [Foreign 2011], , (5) In the figure, if B1, B2, B3,…... and A1, A2, A3,….. have been marked at equal distances. In what ratio C divides AB?, [CBSE Standard SP 2020-21], , (6) To divide a line segment BC internally in the ratio 3 : 5, we draw a ray BX such that ∠CBX is an acute angle. What, will be the minimum number of points to be located at equal distances, on ray BX?, II. Short Answer Type Questions - I, [2 Marks], 4. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP : PB = 3 : 5., [NCERT Exemplar] [CBSE 2011], 5. Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5., [Delhi 2017], 6. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. [NCERT], , (3) The second term of the sequence defined by an = 3n + 2 is, (a) 2, (b) 4, (c) 6, (d) 8, 2. Assertion-Reason Type Questions, In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) B o t h as s er tio n ( A ) an d r e as o n ( R ) ar e tr u e a n d r eas o n ( R ) i s th e co r r ect ex p lan ati o n o f, assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of, assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): The arrangement of numbers, i.e., – 4, 16, – 64, 256, – 1024, 4096, ... form a sequence., Reason (R): An arrangement of numbers which are arranged in a definite order according to some rule, is called a, sequence., (2) Assertion (A): Sequence 1, 5, 9, 13, 17, 21, ... is a finite sequence., Reason (R): A sequence with finite number of terms or numbers is called a finite sequence., 3. Answer the following:, (1) Write down the first six terms of each of the following sequences, whose general terms are:, 2n + 1, (b) an = (– 1)n ⋅ 22n, (c) an =, (d) an = (–1)n – 1 ⋅ n2, (a) an = 5n – 3, n+2, , III. Short Answer Type Questions - II, E:\2020-2021\1. Delhi Mathematics-2021\14112021\1.Open_Files-Final\Ch-2 Folder\Ch-2, 7. Draw a line, segment ofAmitlength, 8 cm and divide it in the ratio 2 : 3., \ 17-Nov-2021, Proof-3, , 76, , Case Study Based Questions, , Answers, , I. Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day, practice it takes him 2 seconds less. He wants to do in 31 seconds., , 5 7 3 11 13, , , , ,, 4 5 2 7 8, (d) 1, –4, 9, –16, 25, –36, (2) –9765625, (1) (3) 10, (c) 1,, , (1), (1), , 2. Arithmetic Progression And its nth term, , . Topic wise concepts are presented to, remember them easily., , • An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed, number ‘d’ to the preceding term, except the first term ‘a’. This fixed number is known as, common difference of the AP. Common difference of an AP can be negative, positive or zero., The general form of an AP is a, a + d, a + 2d, a + 3d, ..., Examples:, (i) The sequence 1, 4, 7, 10, 13, ... is an AP whose first term is 1 and the common difference is equal to 3., (ii) The sequence 11, 7, 3, –1, ... is an AP whose first term is 11 and the common difference is equal to –4., • In the list of numbers a1, a2, a3, ... if the differences a2 – a1, a3 – a2, a4 – a3, ... give the same value, i.e., if ak +1 – ak is, the same for different values of k, then the given list of numbers is an AP., • The nth term an (or the general term) of an AP is an = a + (n – 1) d, where a is the first term, d is the common difference, and n is the number of terms. Also, d = an + 1 – an., • If three terms a, b and c are in AP, then b – a = c – b or 2b = a + c., • If l is the last term of an AP, then nth term from the end of the AP = l + (n – 1)(–d) = l – (n – 1)d., Example 1. In an AP, if d = – 4, n = 7, an = 4, then find the value of a., [CBSE Standard SP 2020-21, Delhi 2018], Solution. We have, an = 4 for n = 7, \, an = a + (n – 1) d fi 4 = a + 6(– 4) fi a = 28, Example 2. Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer., [NCERT Exemplar], Solution. Given AP is 31, 28, 25, ..., Here,, a = 31, d = 28 – 31 = – 3 = 25 – 28, For 0 be a term of this AP,, 0 = an for some ‘n’ ⇒ 0 = a + (n – 1)d, , 34, , Date __________, , IV. Long Answer Type Questions, [5 Marks], 30. 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 AP is zero., [AI2019], 31. The 19th term of an AP is equal to three times its sixth term. If its 9th term is 19, find the AP., [AI 2013], 32. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of, the AP., [Imp.], 33. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the, th, 15 term., [Imp.], 34. If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22nd term., , (2) Find the 10th term of the sequence defined by an = (–1)2n – 1 ⋅ 5n., (3) Find the difference between the 12th term and 10th term of the sequence whose general term is given by an = 5n – 1., , 1. (1) (c) 56, (1) (2) (b) 41, (1), (3) (d) 8, (1), 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A)., (1), (2) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) (a) 2, 7, 12, 17, 22, 27, (b) –4, 16, –64, 256, –1024, 4096, , [3 Marks], Reader’s Sign _______________________, , MatheMatics–10, , . Each topic is well explained, with relevant example for better, understanding., , 1. 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, …, 2. What is the minimum number of days he needs to practice till his goal is achieved?, (a) 10, (b) 12, (c) 11, (d) 9, 3. Which of the following term is not in the AP of the above given situation?, (a) 41, (b) 30, (c) 37, (d) 39, 4. 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, 5. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP is, (a) 6, (b) – 6, (c) 18, (d) –18, II. 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., , QUICK REVISION NOTES, , MatheMatics–10, , • Surface1.areas, and volumes, of various, The production, during first, year is solids are obtained by using these formulae., (a) 3000 TV sets, (b) 5000 TV sets, 2. The production during 8th year is, Name of(b)the, Solid, (a)S.No., 10500, 11900, , 38, , (c) 7000 TV sets, , Figure, (c), 12500, , (d) 10000 TV sets, , Lateral/Curved, (d) 20400, Surface, Area, , Total Surface, Area, , Volume, , MatheMatics–10, , (i), , 2, , Cube, , a, , 2, , 4a, , 6a, , a3, , 2 (l + b) h, , 2 (lb + bh + hl), , lbh, , 2prh, , 2pr (r + h), , pr2h, , prl, , pr (l + r), , 1 2, pr h, 3, , 4pr2, , 4pr2, , 4 3, pr, 3, , 2pr2, , 3pr2, , 2 3, pr, 3, , a, a, , (ii), , Cuboid, , h, b, , .Quite effective for a quick revision before exams., Have the complete essence of the chapter., , (iii), , Right circular cylinder, , l, , h, , r, , (iv), , Right circular cone, , l, , h, r, , r, , (v), , Sphere, , r, , IMPORTANT FORMULAE, , (vi), , • The nth term of an AP, an = a + (n – 1) d, • The nth term of an AP from end, an = l – (n – 1) d, • Sum of finite terms of an AP, n, n, Sn = [2a + (n – 1) d] or Sn = (a + an), 2, 2, , Hemisphere, r, , • When a solid or solids of one shape is (are) converted into solid or solids of another shape, then total volume of the solid(s), to be converted is (are) same to the total volume of the solid(s) into which the given solid(s) is (are) converted., • To find the number of solids of a given shape in which a given solid is to be converted, we divide the total volume of the, solid(s) to be converted by volume of one converted solid., • To find the surface area of a complex solid, i.e., a solid formed by combination of other solids, we add the curved surface, area of only visible portion of individual solid., • To find the volume of a complex solid, we add the volume of individual solid., , • If there are only n terms in an AP, then, an = l, the last term, n, Sn = (a + l), 2, Note:, an = Sn – Sn – 1, where, a = first term, n = number of terms, d = common difference, and an = nth term, l = last term., , qqq, , COMMON ERRORS, Errors, (i) Choosing not a suitable method to obtain the mean of, grouped data., , Corrections, Surface areaS and VolumeS 129, (i) The choice of method to be used depends on the numerical, values of xi and fi. If xi and fi are sufficiently small, the, direct method is an appropriate choice. If xi and fi are, numerically large numbers, choose the assumed mean, method., (ii) Interpreting incorrectly that mode of a given data is (ii) For a given data, mode may be equal, more than or less, always less than the mean of that data., than the mean of that data., (iii) Interpreting incorrectly that mode can be calculated for (iii) The mode can also be calculated for grouped data with, grouped data of equal size only., unequal class sizes., , ExpErts’ OpiniOn, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. Finding nth term of given AP., , 2. Finding nth term of given AP from the end., , 3. Finding n when nth term of an AP is given., , 4. Finding AP or nth term or both when its two terms are given., 5. Finding sum of first n terms of an AP., 6. Finding number of terms when sum of first n terms and AP are given., , .As these topics find place in the exam most of the, time it is suggested that student should put more, emphasis on these topics., , . Common errors have been tagged to clear confusions, with cautions and answers for productive learning., , Arithmetic Progressions 55, , (vii)

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\ 23-Nov-2021, , Praveen Kumar, , Proof-1, , Reader’s Sign _______________________, , Date __________, , CONTENTS, , 1. Quadratic Equations .............................................................................................................., , 9, , 2. Arithmetic Progressions ........................................................................................................ 33, 3. Circles .................................................................................................................................... 55, 4. Constructions ......................................................................................................................... 73, 5. Some Applications of Trigonometry ..................................................................................... 82, 6. Surface Areas and Volumes ................................................................................................... 103, 7. Statistics ................................................................................................................................. 126, • Sample Paper 1 (Solved) ...................................................................................................... 145, • Sample Paper 2 (Solved)....................................................................................................... 155, • Sample Paper 3 (Unsolved)................................................................................................... 165, , (viii)

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1, , Quadratic Equations, , Topics Covered, 1., 2., 3., 4., , Quadratic Equations, Solution of a Quadratic Equation by Factorisation, Solution of a Quadratic Equation by Quadratic Formula, Nature of Roots, , 1. Quadratic Equations, The second degree polynomial equations is commonly known as quadratic equation, i.e., if p(x) is a quadratic polynomial,, then p(x) = 0 is called a quadratic equation. The general form of quadratic equation in the variable x is ax2 + bx + c = 0,, where a, b, c are real numbers and a ≠ 0., For example, 2x2 + x – 150 = 0, 3x2 – 2x + 5 = 0, 4x – 3x2 + 2 = 0 are quadratic equations., Example 1. Check whether the following are quadratic equations:, (a) x(x + 2) – 3 = (x + 4) x, (b) (x + 2)3 = x3 – 4x2 + 2, Solution. (a) Since, x(x + 2) – 3 = x(x + 4), ⇒, x2 + 2x – 3 = x2 + 4x, ⇒, 2x + 3 = 0, This is linear equation not a quadratic equation., (b) (x + 2)3 = x3 – 4x2 + 2, ⇒, x3 + 6x2 + 12x + 8 = x3 – 4x2 + 2, ⇒, 10x2 + 12x + 6 = 0, ⇒, 5x2 + 6x + 3 = 0, This is a quadratic equation., Example 2. Represent the following situation in the form a quadratic equation:, (a) Abdul and Sneha together have 30 oranges. Both of them ate 3 oranges each and the product of the number of, oranges they have now is 120. We would like to find out how many oranges they had initially., (b) The area of a rectangular plot is 428 m2. The length of the plot (in metres) is two more than twice its breadth. We, need to find the length and breadth of the plot., Solution. (a) Let the number of number of left oranges with Abdul and Sneha be x and y respectively., Then,, x + y = 30 ⇒ y = 30 – x, The number of oranges left with both Abdul and Sneha are x – 3 and y – 3 respectively., The product of number of left oranges = 120, , ⇒, (x – 3) (y – 3) = 120, , ⇒, (x – 3) (30 – x – 3) = 120, (Q y = 30 – x), , ⇒, (x – 3) (27 – x) = 120, , ⇒, 27x – x2 – 81 + 3x = 120 ⇒ x2 – 30x + 201 = 0, (b) Let the breadth of the plot be x. Then the length of the plot = 2x + 2, Since,, area of the plot = 428 m2, (Given), \, x(2x + 2) = 428, ⇒, 2x2 + 2x – 428 = 0, , ⇒, x2 + x – 214 = 0, , Roots of a Quadratic Equation, A real number a is called a root of the quadratic equation ax2 + bx + c = 0, a π 0 if aa2 + ba + c = 0., In other words, x = a is a root or solution of the quadratic equation ax2 + bx + c = 0 as it satisfies it., Example 3. Determine whether 3 is a root of the equation, x2 − 4 x + 3 +, , x2 − 9 =, , 4 x 2 − 14 x + 6, , 9

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Solution., L.H.S. = x 2 − 4 x + 3 +, = 0 + 0 = 0, , x2 − 9 =, , (3)2 − 12 + 3 +, , Date __________, , 9−9, , R.H.S. = 4 x 2 − 14 x + 6 = 36 − 42 + 6 = 0, Since, L.H.S. = R.H.S., \ 3 is the root of the given quadratic equation., 1, 5, 2, Example 4. If, is a root of the equation x + kx − = 0 , then find the value of k., 2, 4, 1, Solution. It is given that, is a root of quadratic equation., 2, \ It must satisfy the quadratic equation:, 2, 5, 1, 1, 5, x 2 + kx − = 0 ⇒   + k   − = 0,  2,  2 4, 4, 1 k 5, 1 + 2k − 5, + − =0 ⇒, =0, 4 2 4, 4, ⇒, 2k – 4 = 0 ⇒ k = 2, Example 5. If a and b are the roots of the equation x2 + ax – b = 0, then find a and b., Solution. Here, A = 1, B = a, C = –b, B, \, Sum of the roots = a + b = −, = –a, A, C, Product of the roots = ab =, = –b, A, From (i) and (ii),, a + b = – a and ab = – b, fi, 2a = – b and a = – 1, ⇒, b = 2 and a = –1, ⇒, , ...(i), ...(ii), , Exercise 1.1, I. Very Short Answer Type Questions, 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) Which of the following is a quadratic equation?, (a) x2 + 2x + 1 = (4 – x)2 + 3, , [1 Mark], , 2, , (b) – 2x2 = (5 – x)  2 x − , , 5, , 3, (d) x3 – x2 = (x – 1)3, x = 7   (where k = – 1), 2, (2) Which of the following equations has 2 as a root?, (a) x­2 – 4x + 5 = 0, (b) x2 + 3x – 12 = 0, (c) 2x2 – 7x + 6 = 0, (d) 3x2 – 6x – 2 = 0, 2, (3) The roots of the quadratic equation x – 0.04 = 0 are, [CBSE Standard 2020], (a) ± 0.2, (b) ± 0.02, (c) 0.4, (d) 2, (4) The degree of quadratic equation is, (a) 0, (b) 1, (c) 2, (d) 5, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): The equation x2 + 3x + 1 = (x – 2)2 is a quadratic equation., , Reason (R): Any equation of the form ax2 + bx + c = 0 where a π 0, is a quadratic equation., (2) Assertion (A): (2x – 1)2 – 4x2 + 5 = 0 is not a quadratic equation., , Reason (R): x = 0, 3 are the roots of the equation 2x2 – 6x = 0., (c) (k + 1) x 2 +, , 10, , Mathematics–10

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3. Answer the following., 1, (1) If x = − is a solution of the quadratic equation 3x2 + 2kx – 3 = 0, find the value of k. , 2, (2) Find the value of k for which x = 3 is a solution of the equation kx 2 + 3 x − 4 = 0., , [Delhi 2015], , II. Short Answer Type Questions - I, [2 Marks], 2, 4. If x =, and x = –3 are roots of the quadratic equations ax2 + 7x + b = 0, find the values of a and b., 3, , [Delhi 2016], 5. Show that x = –2 is a solution of the equation 3x2 + 13x + 14 = 0., III. Short Answer Type Questions - II, Represent the following situations in the form of a quadratic equation (Q. 6 & Q.7):, , [3 Marks], , 6. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles, they have now is 124. We would like to find out how many marbles they had to start with., 7. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found, to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We, would like to find out the number of toys produced on that day.  , 2, 8. If one root of the quadratic equation 3x2 + px + 4 = 0 is , then find the value of p and the other root of the equation., 3, , Case Study Based Questions, I. 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., , 1. 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, , 2. Which of the following quadratic equations describes 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, , 3. The roots of the quadratic equation which describe the speed of Raj’s car are, (a) 15, – 20, , (b) 20, – 15, , (c) 20, – 25, , (d) 25, – 25, , 4. Which of the following quadratic equations has 2 as a root?, (a) x2 – 4x + 5 = 0, 5. The positive root of, (a) 5, , (b) x2 + 3x – 12 = 0, , (c) 2x2 – 7x + 6 = 0, , (d) 3x2 – 6x – 2 = 0, , (c) 3, , (d) –3, , 3 x 2 + 6 = 9 is, (b) –5, , Quadratic Equations, , 11

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , Date __________, , Answers and Hints, (d) x3 – x2 = (x – 1)3(1), (c) 2x2 – 7x + 6 = 0, (1), (a) ± 0.2, (1), (c) 2, (1), (d) Assertion (A) is false but reason (R) is true. (1), (b) Both assertion (A) and reason (R) are true but, reason (R) is not the correct explanation of, assertion (A)., (1), −1, 2, 3. (1) Q x =, is the solution of 3x + 2kx – 3 = 0, 2, , 1. (1), (2), (3), , (4), 2. (1), , (2), , 2, ,  −1,  −1, So, 3   + 2k   − 3 = 0,  2,  2, 3, − k − 3 = 0, 4, 3, k = −3, 4, , , fi, , fi, , fi, , k = −2, , (2) We have, , fi, , fi, , fi, , fi, , fi, , k, , 1, (1), 4, , kx 2 + 3 x − 4 = 0, , ( 3), , 2, , + 3, , [ x =, , b = 21 – 9a, , ...(iii)(1), , Putting the value of b from (iii) in (ii), we get, , , 4a + 9[21 – 9a] + 42 = 0, , , fi 4a + 189 – 81a + 42 = 0, , fi, a=3, Putting a = 3 in (iii), we have, , b = –6, So,, a = 3, b = – 6., 2, 6. x – 45x + 324 = 0, 7. x2 – 55x + 750 = 0, 8. , 3x2 + px + 4 = 0, , (1), (3), (3), (½), , 2, , 3k + 3 – 4 = 0, 3k – 1 = 0, 3k = 1, 1, k=, 3, , 4. Given quadratic equation is, , ax2 + 7x + b = 0, 2,  2,  2, a  + 7  + b = 0, ,  3,  3, , , fi, ,  2,  2, 3  + p   + 4 = 0,  3,  3, 4, 2p, , +, + 4 = 0, 3, 3, , p = –8, , 3x2 – 8x + 4 = 0, , 3x2 – 6x – 2x + 4 = 0, 2, , x=, or x = 2, 3, Hence,, x = 2, , , ( 3) − 4 = 0, , Hence, the required value of k is, , 4a + 42 + 9b, =0, 9, , fi, 4a + 9b + 42 = 0, ...(ii), 2, Also, a(–3) + 7(–3) + b = 0, [ x = –3 is the root of eq. (i)], , fi, 9a + b – 21 = 0, , fi, , 1, . (1), 3, ...(i), , (½), (½), (½), (½), (½), , Case Study Based Questions, I. 1. (a) 2(x + 5) km, 2. (c) x2 + 5x – 500 = 0, 3. (c) 20, – 25, 4. (c) 2x2 – 7x + 6 = 0, 5. (a) 5, , 2, is the root of eq. (i)], 3, , 2. Solution of a Quadratic Equation by Factorisation, To find the solution of a quadratic equation by factorisation method, we first express the given equation as product of two, linear factors by splitting the middle term. By equating each factor to zero, we get possible solutions/roots of the given, quadratic equation., Let the given quadratic equation be ax2 + bx + c = 0. Let the quadratic polynomial ax2 + bx + c be expressed as the product, of two linear factors say (px + q) and (rx + s) where, p, q, r, s are real numbers such that p ≠ 0, r ≠ 0., Then, ax2 + bx + c = 0 ⇒ (px + q)(rx + s) = 0, ⇒, Either (px + q) = 0 or (rx + s) = 0, −q, −s, ⇒, x=, or x =, p, r, Example 1. Solve the following quadratic equations by the factorisation method., (a) 7x2 = 8 – 10x, (b) x (x + 9) = 52, 2, (c) 3 (x – 4) = 5x, (d) x (x + 1) + (x + 2) (x + 3) = 42, (e) 3x 2 − 2 6 x + 2 = 0, , 12, , Mathematics–10, , [NCERT] [Imp.]

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7x2 = 8 – 10x ⇒ 7x2 + 10x – 8 = 0, 7x + 14x – 4x – 8 = 0 ⇒ 7x(x + 2) – 4 (x + 2) = 0, (7x – 4) (x + 2) = 0, 4, x = , x = –2, 7, x (x + 9) = 52 ⇒ x2 + 9x – 52 = 0, x2 + 13x – 4x – 52 = 0 ⇒ x (x + 13) – 4 (x + 13) = 0, (x + 13) (x – 4) = 0, x = –13, x = 4, 3(x2 – 4) = 5x ⇒ 3x2 – 5x – 12 = 0, 3x2 – 9x + 4x + 12 = 0 ⇒ 3x (x – 3) + 4 (x – 3) = 0, (3x + 4) (x – 3) = 0, −4, x=, , x=3, 3, x(x + 1) + (x + 2) (x + 3) = 42 ⇒ x2 + x + x2 + 3x + 2x + 6 – 42 = 0, 2x2 + 6x – 36 = 0 ⇒ x2 + 3x – 18 = 0, x2 + 6x – 3x – 18 = 0 ⇒ x(x + 6) – 3(x + 6) = 0 ⇒ (x + 6) (x – 3) = 0, x = –6, x = 3, , Solution. (a), , ⇒, , ⇒, , 2, , , \, (b) , , ⇒, , ⇒, , \, (c) , , ⇒, , ⇒, , \, (d) , , ⇒, , ⇒, , \, (e) , , ⇒ 3x, , (, , ), , 3x2 – 2 6x + 2 = 0, , ⇒, , 3x − 2 = 0, , ⇒, , 3x − 2 − 2, , (, , ), , , \, , x=, , 2, 3, , ,, , 3x 2 − 6 x − 6 x + 2 = 0, , (, , 3x − 2, , )(, , ), , 3x − 2 = 0, , 2, 3, , 5, 3, 4, ; x ≠ 0, − , for x., −3 =, x, +, 2, 3, 2, x, 4, − 3x, 5, 5, 4, −3 =, Solution., ⇒, =, x, 2x + 3, 2x + 3, x, (4 – 3x) (2x + 3) = 5x ⇒ 8x – 6x2 + 12 – 9x = 5x, 6x2 + 6x – 12 = 0 ⇒ x2 + x – 2 = 0, x2 + 2x – x – 2 = 0 ⇒ x (x + 2) – 1 (x + 2) = 0, ⇒, (x – 1) (x + 2) = 0 ⇒ x – 1 = 0 or x + 2 = 0, ⇒, x = 1 or x = –2, 16, 15, −1 =, ; x ≠ 0, − 1, Example 3. Solve for x:, x, x +1, 15, 16 − x, 15, 16, −1 =, Solution., ⇒, =, x +1, x +1, x, x, fi, (16 – x) (x + 1) = 15 x fi 16x – x2 + 16 – x = 15 x, fi, x2 + 15x – 15x – 16 = 0 fi x2 = 16, fi, x=±4, x − 4 x − 6 10, ; x ≠ 5, 7, +, Example 4. Solve for x :, =, 3, x−5 x−7, 10, x−4 x−6, +, Solution., = 3, x−5 x−7, 10, ( x − 4)( x − 7) + ( x − 6)( x − 5), x 2 − 7 x − 4 x + 28 + x 2 − 5 x − 6 x + 30 10, ⇒, = 3 ⇒, =, 3, ( x − 5)( x − 7), x 2 − 7 x − 5 x + 35, Example 2. Solve the equation, , ⇒, ⇒, ⇒, , 2 x 2 − 22 x + 58, , [Delhi 2014], , [AI 2014], , [AI 2014], , x 2 − 11x + 29, , 5, = 3, x − 12 x + 35, x − 12 x + 35, 2, 2, 3x – 33x + 87 = 5x – 60x + 175, 2x2 – 27x + 88 = 0 ⇒ 2x2 – 16x – 11x + 88 = 0, 2, , =, , 10, 3, , ⇒, , 2, , Quadratic Equations, , 13

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 2x (x – 8) – 11(x – 8) = 0 ⇒ (2x – 11) (x – 8) = 0, 2x – 11 = 0 or x – 8 = 0, 11, ⇒, x=, or x = 8, 2, Example 5. The sum of the squares of two consecutive odd numbers is 394. Find the numbers., [Foreign 2014], Solution. Let the two consecutive odd numbers be x and x + 2., \, x2 + (x + 2)2 = 394 ⇒ x2 + x2 + 4 + 4x = 394, ⇒, 2x2 + 4x + 4 = 394 ⇒ 2x2 + 4x – 390 = 0, ⇒, x2 + 2x – 195 = 0 ⇒ x2 + 15x – 13x – 195 = 0, ⇒, x (x + 15) – 13 (x + 15) = 0 ⇒ (x – 13) (x + 15) = 0, Either, x – 13 = 0 or x + 15 = 0 ⇒ x = 13 or x = – 15 (neglected), When first number x = 13, then second number x + 2 = 13 + 2 = 15., Example 6. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so, that the area of the grass surrounding the pond would be 1184 m2. Find the length and breadth of the pond. [NCERT, Exemplar], Solution. Let ABCD be a rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, same, around the pond., Now, length of lawn = 50 m, width of lawn = 40 m, \ Length of pond = (50 – 2x) m, width of pond = (40 – 2x) m, Since area of grass surrounds the pond = 1184 m2, fi, Area of lawn – Area of pond = 1184 m2, fi, 50 × 40 – (50 – 2x) (40 – 2x) = 1184, fi, 2000 – (2000 – 80x – 100x + 4x2) = 1184, fi, 4x2 – 180 x + 1184 = 0 or x2 – 45x + 296 = 0 fi x2 – 37x – 8x + 296 = 0, fi, (x – 37) (x – 8) = 0 fi x = 37, 8, Since x = 37 is not possible, as otherwise the length of pond will be negative., Hence, x = 8 is the required solution., \, Length of pond = 50 – 2 × 8 = 34 m, and, breadth of pond = 40 – 2 × 8 = 24 m, 1, Example 7. The difference of two natural numbers is 5 and the difference of their reciprocals is, . Find the numbers., 10, [Delhi 2014], Solution. Let the two natural numbers be x and y such that x > y., According to the question,, Difference of numbers,, x–y=5 ⇒ x=5+y, ...(i), Difference of their reciprocals,, 1, 1 1, − =, ...(ii), 10, y x, ⇒, ⇒, , Putting the value of (i) in (ii), 1, 1, 1, −, = 10, y 5+ y, , ⇒, , 1, 5+ y− y, = 10, y (5 + y ), , ⇒, 50 = 5y + y2 ⇒ y2 + 5y – 50 = 0, 2, ⇒, y + 10y – 5y – 50 = 0 ⇒ y (y + 10) – 5 (y + 10) = 0, ⇒, (y – 5) (y + 10) = 0, \, y = 5 or y = –10, Q y is a natural number., \, y=5, Putting the value of y in (i), we get, x = 5 + 5 = 10, Thus, the required numbers are 10 and 5., , 3, Example 8. The sum of two numbers is 15 and the sum of their reciprocals is, . Find the numbers., 10, Solution. Let the numbers be x and 15 – x., According to given condition,, , 14, , Mathematics–10

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3, 1, 1, x + 15 – x = 10, , ⇒, , 15 – x + x, 3, =, x (15 – x) 10, , ⇒, 150 = 3x (15 – x) ⇒ 50 = 15x – x2, ⇒, x2 – 15x + 50 = 0 ⇒ x2 – 5x – 10x + 50 = 0, ⇒, x (x – 5) – 10 (x – 5) = 0 ⇒ (x – 5) (x – 10) = 0, ⇒, x = 5 or 10., When x = 5, then 15 – x = 15 – 5 = 10, When x = 10, then 15 – x = 15 – 10 = 5, Hence, the two numbers are 5 and 10., Example 9. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been, 11 more than five times her actual age. What is her age now?, [NCERT Exemplar], Solution. Let the present age of Zeba be x years., Age before 5 years = (x – 5) years, According to given condition,, (x – 5)2 = 5x + 11, ⇒, x2 + 25 – 10x = 5x + 11 ⇒ x2 – 10x – 5x + 25 – 11 = 0, ⇒, x2 – 15x + 14 = 0    ⇒    x2 – 14x – x + 14 = 0, ⇒, x (x – 14) – 1 (x – 14) = 0 ⇒ (x – 1) (x – 14) = 0, ⇒, x – 1 = 0 or x – 14 = 0, ⇒, x = 1 or x = 14, But present age cannot be 1 year., \ Present age of Zeba is 14 years., Example 10. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4, hours 30 minutes. Find the speed of the stream. , [Delhi, 2017], Solution. Let the speed of stream be x km/hr., \ Speed of boat in upstream = (15 – x) km/hr., Speed of boat in downstream = (15 + x) km/hr., 30 (15 + x + 15 − x ) 9, 30, 30, 1 9, +, = 4 =, fi, =, 2, 15 − x 15 + x, 2 2, (15 − x )(15 + x ), , fi, 200 = 225 – x2 fi x = 5 (Rejecting –5), \, Speed of stream = 5 km/hr, Example 11. A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance, if its speed were 5 km/hour more. Find the original speed of the train., [CBSE Standard SP 2019-20], Solution. Let original speed of the train be x km/hr., 360, Time taken at original speed =, hours, x, 360, Time taken at increased speed =, hours, x+5, 48, 360 360, 1 , 4, 1, −, ATQ,, =, fi 360  −, =, fi x2 + 5x – 2250 = 0, , 60, x, x+5, x, x, +, 5, 5, , , fi, x = 45 or –50 (as speed cannot be negative), fi, x = 45 km/h, Example 12. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in, time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed., Solution. Let the usual speed of the plane be x km/hr., [CBSE 2018], 30, 1500, 1500, 2, \, =, fi x + 100x – 300000 = 0, −, 60, x, x + 100, fi, x2 + 600x – 500x – 300000 = 0 fi (x + 600)(x – 500) = 0, x ≠ – 600, \, x = 500, Speed of plane = 500 km/hr, , Quadratic Equations, , 15

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 1, Example 13. Two pipes running together can fill a cistern in 3, hours. If one pipe takes 3 hours more than the other, to fill it, find the time in which each pipe would fill the cistern. 13, [Delhi 2017] [HOTS], Solution. Let time taken by faster pipe to fill the cistern be x hrs., Therefore, time taken by slower pipe to fill the cistern = (x + 3) hrs, Since the faster pipe takes x minutes to fill the cistern., 1, \ Portion of the cistern filled by the faster pipe in one hour =, x, 1, Portion of the cistern filled by the slower pipe in one hour =, x+3, 1, 13, Portion of the cistern filled by the two pipes together in one hour = 40 =, 40, 13, 1, 1, 13, x + 3 + x 13, =, =, According to question, +, ⇒, x x + 3 40, x( x + 3), 40, ⇒, 40 (2x + 3) = 13x (x + 3) ⇒ 80x + 120 = 13x2 + 39x, ⇒, 13x2 – 41x – 120 = 0 ⇒ 13x2 – 65x + 24x – 120 = 0, ⇒, 13x (x – 5) + 24 (x – 5) = 0 ⇒ (x – 5) (13x + 24) = 0, Either, x – 5 = 0 or 13x + 24 = 0, −24, ⇒, x = 5 as x =, not possible., 13, \ The time taken by the two pipes is 5 hours and 8 hours respectively., 1, 3, 5, 1, +, =, , x ≠ − 1, − , − 4 ., Example 14. Solve for x:, x + 1 5x + 1 x + 4, 5, , Solution. Here,, ⇒, ⇒, , 1, 3, 5, +, =, x+4, x + 1 5x + 1, , ⇒, , [AI 2017], , 5 x + 1 + 3( x + 1), 5, =, x+4, ( x + 1)(5 x + 1), , [(5x + 1) + (x + 1)3](x + 4) = 5(x + 1)(5x + 1), (8x + 4)(x + 4) = 5(5x2 + 6x + 1), 17x2 – 6x – 11 = 0 ⇒ (17x + 11)(x – 1) = 0, x=, , ⇒, , − 11, , x =1, 17, , Exercise 1.2, I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, 4, 3, (1) The roots of the equation x 2 − 2 x + = 0 are, 3, 4, 2 3, 3 3, 1 1, ,, ,, ,−, (a), (b), (c), (d) None of these, 3 2, 4 4, 2 2, 2, (2) The required solution of 4x – 25x = 0 are, 12, 25, 5, 12, (a) x = 0, x =, (b) x = 0, x =, (c) x = 1, x =, (d) x = 1, x =, 7, 4, 9, 7, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., , 16, , Mathematics–10

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(d) Assertion (A) is false but reason (R) is true., 2, 1, and − ., 3, 2, 2 1, 2, , Reason (R): 6x – x – 2 = 0 ⇒ 2x(3x – 2) + (3x – 2) = 0 ⇒ (3x – 2) (2x + 1) = 0 ⇒ x = , −, 3 2, 2, 2, (2) Assertion (A): If x − ( 3 + 1) x + 3 = 0, then x − 3 x − x + 3 = 0, (1) Assertion (A): When the quadratic equation 6x2 – x – 2 = 0 is factorised, we get its roots as, , , ⇒ x( x − 3 ) − 1( x − 3 ) = 0 ⇒, , ( x − 3 ) ( x − 1) = 0, , ⇒ x=, , 3,1, , Reason (R): If we can factorise ax + bx + c, a π 0 into a product of two linear factors, then the roots of the quadratic, equation ax2 + bx + c = 0 can be found by equating each factor to zero., 3. Answer the following., , Find the roots of the following quadratic equations by factorisation [(1) to (4)]:, (1) 3 x 2 + 10 x + 7 3 = 0, [Imp.] (2) (x – 3)(2x + 3) = 0, 2, , (3) 3x2 – 2ax – a2 = 0, (4) 3a2x2 + 8abx + 4b2 = 0, 2, (5) Find the roots of the equation x + 7x + 10 = 0., II. Short Answer Type Questions - I,   Find the roots of the following quadratic equations by factorisation (Q4 to Q10)., 4. Solve for x: 4 3 x 2 + 5 x − 2 3 = 0, 5. Solve for x: x 2 − ( 2 + 1) x + 2 = 0, , [CBSE Standard SP 2020-21], [2 Marks], [Delhi 2013], [Foreign 2013], , 6. Solve for x:, , 2 x + 9 + x = 13., , [AI 2016], , 7. Solve for x:, , 6 x + 7 − (2 x − 7) = 0., , [AI 2016], , 8. Solve for x:, , 3 x 2 − 2 2 x − 2 3 = 0., , 9. Solve for x:, 10. Solve for x:, , 1, 1, 1, −, = , x π 3, – 5., x−3 x+5 6, , [Foreign 2016], , 3 x 2 + 14 x − 5 3 = 0, , III. Short Answer Type Questions - II, 11. Solve for x:, , [Foreign 2016], , x +1 x − 2, 2x + 3, +, =4−, ; x ≠ 1, − 2, 2 ., x −1 x + 2, x−2, , [3 Marks], [Delhi 2016], , 3, . Find the numbers.[Delhi 2014], 28, 5, 13. The difference of two natural numbers is 5 and the difference of their reciprocals is, . Find the numbers.[Delhi 2014], 14, 12. The difference of two natural numbers is 3 and the difference of their reciprocals is, , IV. Long Answer Type Questions, , [5 Marks], , 14. Solve the equation for x:, , 3x − 4, 7, 5, 4, +, = ,x≠ ., 7, 3x − 4 2, 3, , [Foreign 2010], , 15. Solve the equation for x:, , 1, 2, 5, +, =, , x ≠ −1, − 2, − 4 ., x +1 x + 2 x + 4, , [Foreign 2012], , 16. Some students planned a picnic. The total budget for food was `2,000. But 5 students failed to attend the picnic and, thus the cost of food for each member increased by `20. How many students attended the picnic and how much did each, student pay for the food?, [Foreign 2010], 17. A two-digit number is such that the product of its digits is 14. When 45 is added to the number, the digits interchange, their places. Find the number., [Foreign 2011], 18. Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to, fill the tank separately. Find the time in which each tap can separately fill the tank., [Foreign 2012], 1, minutes. If one pipe takes 5 minutes more than the other to fill the tank separately,, 9, find the time in which each pipe would fill the tank separately., [AI 2016], , 19. Two pipes running together can fill a tank in 11, , Quadratic Equations, , 17

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 20. 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 2016], 21. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream, to the same spot. Find the speed of the stream., [CBSE Standard 2020, CBSE 2018, AI 2013], 22. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s, present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha, and Nisha., [NCERT Exemplar], 23. 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 total journey, what is the original average, speed?, [CBSE 2018], 24. Solve the following equation:, 1, 1, −, , = 3, x ≠ 0, 2, [CBSE Standard SP 2019-20], x x−2, 25. Find two consecutive positive integers sum of whose squares is 365., [CBSE Standard SP 2019-20], 26. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more, than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the, rectangular park and of altitude 12 m. Find the length and breadth of the park., [CBSE 2016], 27. In a flight of 600 km, an aircraft was slowed down due to bad weather. The average speed of the trip was reduced by 200, km/hr and the time of flight increased by 30 minutes. Find the duration of flight., [CBSE Standard 2020], , Case Study Based Questions, I. 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., , 1. 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, 2. What is the relation between speed, distance and time?, (distance), (speed ), (a) speed =, , (b) distance =, time, time, (c) time = speed × distance , (d) speed = distance × time, 3. Which is the correct quadratic equation for the speed of the stream?, (a) x2 + 30x − 200 = 0 (b) x2 + 20x − 400 = 0 (c) x2 + 30x − 400 = 0 (d) x2 – 20x − 400 = 0, 4. What is the speed of stream?, (a) 20 km/hour, (b) 10 km/hour, (c) 15 km/hour, (d) 25 km/hour, 5. How much time boat took in downstream?, (a) 90 minutes, (b) 15 minutes, (c) 30 minutes, (d) 45 minutes, , Answers and Hints, 1. (1) (b), , 18, , 3 3, ,, 4 4, , (1) (2) (b) x = 0, x =, , Mathematics–10, , 25, 4, , (1), , 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1)

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(2) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), 7, −3, 3. (1) − 3 , −, (1) (2) 3,, (1), 2, 3, −a, −2b −2b, (3) a,, (1) (4), ,, (1), 3, a 3a, (5), x2 + 7x + 10 = 0, 2, x + 5x + 2x + 10 = 0, (½), , (x + 5)(x + 2) = 0, , x = –5, x = –2, (½), 2, 4. Consider, 4 3x + 5 x − 2 3 = 0, , fi, , 4 3 x 2 + 8 x − 3 x − 2 3 = 0, , (1), , , fi 4 x( 3 x + 2) − 3 ( 3 x + 2) = 0, 3, 2, fi x=, (1), and −, 4, 3, 5. Consider:, x 2 − ( 2 + 1) x + 2 = 0, , x 2 − 2 x − x + 2 = 0, , , fi, , (1), , ( x − 2 )( x − 1) = 0, , , fi, , fi, , x=, , 2 or x = 1, , (1), , 2 x + 9 = (13 – x), 6. 2 x + 9 + x = 13 fi, Squaring both sides, we get, , fi, 2x + 9 = 169 – 26x + x2(1), 2, , fi, x – 28x + 160 = 0, , fi, (x – 20)(x – 8) = 0, , fi, x = 20 or x = 8, , fi, x = 8, [as x = 20 does not satisfy the equation] (1), 7. , , 6 x + 7 − ( 2 x − 7) = 0, , , fi, 6 x + 7 = 2x – 7, Squaring both sides, we get, , fi, 6x + 7 = 4x2 – 28x + 49, (1), 2, , fi, 4x – 34x + 42 = 0, , fi, 2x2 – 17x + 21 = 0, 2, , fi 2x – 14x – 3x + 21 = 0, , fi, (2x – 3)(x – 7) = 0, 3, , fi, x = 7 or x =, fi x=7, 2, 3, [as x = does not satisfy the equation] (1), 2, 8. , 3 x 2 − 2 2 x − 2 3 = 0, , fi, , 3 x 2 + 2 x − 3 2 x − 2 3 = 0, , , fi, , − 2, , − 6, ×, =, and x =, 3, 3, 3, 3, , 1, 1, 1, −, =, 6, x−3 x+5, , fi, , 6 (1), , x + 5− x+3, , =, , x + 2 x − 15, , fi x2 + 2x – 15 = 48 fi x2 + 2x – 63 = 0, , fi (x + 9)(x – 7) = 0 fi x = 7 or x = –9, 2, , , fi, , 3 x 2 + 15 x − x − 5 3 = 0, , , fi, , 3x( x + 5 3 ) − ( x + 5 3 ) = 0, , , fi, , ( x + 5 3 ) ( 3 x − 1) = 0, , Either, , 1, (1), 6, (1), , x + 5 3 = 0, , or, , (1), , 3x − 1 = 0, , , fi, , x = −5 3, 1, x=, (1), 3, 2x + 3, x +1 x − 2, +, = 4−, x−2, x −1 x + 2, x + 1 x − 2 2x + 3, +, +, = 4  , (1), x −1 x + 2 x − 2, , or, 11. , , fi, , , ( x + 1)( x + 2)( x − 2) + ( x − 2) 22 ( x − 1), + (2 x + 3)( x − 1)( x + 2), , fi, =4, ( x − 1)( x + 2)( x − 2), , fi (x + 1)(x2 – 4) + (x – 1)(x2 + 4 – 4x), , + (2x + 3)(x2 + x – 2) = 4(x – 1)(x2 – 4) (1), , fi, 5x2 + 19x – 30 = 0, , fi, (x + 5)(5x – 6) = 0  fi  x = –5, 6, or, x = (1), 5, 12. 4, 7, (3), 13. 7, 2, (3), 5, 4, 3x − 4, 7, 14. Given that:, = ,x≠, +, 2, 3, 7, 3x − 4, 3x − 4, =y, 7, 5, 1, , fi The given equation becomes y + = (1), 2, y, Let us consider:, , 2y 2 – 4y – y + 2 = 0, 1, , fi (2y – 1)(y – 2) = 0 fi y =, or 2, (1), 2, 3x − 4, 1, 3x − 4, , =, or, = 2, (1), 7, 2, 7, , fi, 6x – 8 = 7, fi 3x = 18, 15, , fi, x=, (1) fi x = 6, (1), 6, 15. Given that:, 1, 2, 5, +, , =, ; x π –1, –2, –4, x +1 x + 2, x+4, , fi, , , fi, , ( 3 x + 2 )( x − 6 ) = 0, , , \   x =, 9. , , (1), , 3 x 2 + 14 x − 5 3 = 0, , 10. , , , fi, , fi, , fi, , fi, , fi, , 2y2 – 5y + 2 = 0, , x + 2 + 2x + 2, x + 3x + 2, (3x + 4)(x + 4), 3x2 + 16x + 16, 2x2 – x – 6, (2x + 3)(x – 2), 2, , =, , fi, , 5, (1), x+4, , 5(x2 + 3x + 2), (1), 5x2 + 15x + 10, (1), 0, 0, (1), −3, x = 2 or x =, (1), 2, =, =, =, =, , Quadratic Equations, , 19

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\ 23-Nov-2021, , Amit, , Proof-3, , 16. Case I. Let number of students = x, and cost of food for each member = ` y, , Then, x × y = 2,000, ...(i)(1), , Case II. New number of students = x – 5, New cost of food for each member = `(y + 20), Then, (x – 5)(y + 20) = 2,000, , fi, xy + 20x – 5y – 100 = 2,000, ...(ii)(1), Solving (i) and (ii), we get, , \, x = –20, 25, (1), , x = –20 is rejected because number of students can’t be, negative., So,, x = 25, , \, y = 80, (1), , Number of students = 25, Cost of food for each student = `80.(1), 17. 27, (5), 18. 9 hrs., (5), 19. 20 minutes, 25 minutes., , Hints: Solve same as Example 13., (5), 20. Let P be the position of the pole., , ∠APB = 90° (angle in a semicircle), , , By Pythagoras Theorem,, , AB2 = AP2 + PB2, , fi, 172 = AP2 + PB2 ...(i)(1), Now,, AP – PB = 7, ...(ii), , fi, (AP – PB)2 = 49, , fi AP2 + PB2 – 2AP · PB = 49, ...(iii)(1), From (i) and (iii), we have, , 172 – 2AP · PB = 49, , fi, AP · PB = 120, ...(iv), From (ii) and (iv), we have, , 120 = PB(7 + PB), (1), Let, PB = x, , 120 = x(7 + x), , fi, x2 + 7x – 120 = 0, , fi, (x – 8)(x + 15) = 0, , fi, x=8, or, x = –15, (Rejected), , \, PB = x = 8 m,, , AP = 15 m, (2), 21. 6 km/h., (5), 22. Nisha’s age = 5 years, Asha’s age = 27 years  , (5), 23. Let the original average speed of train be x km/hr., 63, 72, +, Therefore, = 3, (1), x, x+6, , fi, x2 – 39x – 126 = 0, (1), , fi, (x – 42)(x + 3) = 0, (1), , x ≠ –3, , \, x = 42, (1), Original speed of train is 42 km/hr., (1), , 20, , Mathematics–10, , Reader’s Sign _______________________, , Date __________, , 24. , , 1, 1, −, =3, x x−2, , , , x−2−x, 3, = (1), x ( x − 2), 1, , , , , , 3x2 – 6x = –2, (1), 3x – 6x + 2 = 0, (1), 6 ± 12, x=, (1), 6, 2, , 3+ 3 3− 3, ,, (1), 3, 3, 25. Let two consecutive positive integers be x and x + 1, , ∴, x2 + (x + 1)2 = 365, (1), 2, , ⇒, x + x − 182 = 0, (1), , (x + 14)(x − 13) = 0, (1), , ∴, x = 13, (1), Hence, two consecutive positive integers are 13 and 14., (1), 26. Let ABCD is a rectangular park and CDE is a triangular, park (isosceles triangle)., =, , , , Rectangle:, , Let breadth = x, , Then its length = x + 3, So,, area = x(x + 3), (1), , Triangle:, 1, , Area of triangle = × Base × Altitude, 2, 1, = × CD × Altitude (1), 2, 1, = × x × 12, 2, = 6x, ATQ,, x(x + 3) = 4 + 6x(1), , x2 + 3x = 4 + 6x, 2, , x – 3x – 4 = 0, , x2 – 4x + x – 4 = 0, x(x – 4) + 1(x – 4) = 0, , (x – 4)(x + 1) = 0, Either, x – 4 = 0 or x + 1 = 0, , x = 4 or x = –1, (1), Since x cannot be negative., So, x = 4 is the solution, Thus, Breadth = x = 4 m, and  length = x + 3 = 4 + 3 = 7 m, (1), 27. 1 hr, (5)

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Case Study Based Questions, , (distance), I. 1. (c) (20 – x)km/hr, 2. (a) speed =, time, 2, 3. (c) x + 30x − 400 = 0 4. (b) 10 km/hour, , 5. (c) 30 minutes, , 3. Solution of a Quadratic Equation by Quadratic Formula, b 2 − 4ac, , if b2 – 4ac ≥ 0., 2a, This formula for finding the roots of a quadratic equation is often referred to as the quadratic formula., The expression b2 – 4ac is called the discriminant of the quadratic equation and generally denoted by D. If b2 – 4ac, < 0, then the equation will have no real roots., As this formula was given by an ancient Indian mathematician Sridharacharya around AD 1025, it is known as, Sridharacharya’s formula for determining the roots of the quadratic equations ax2 + bx + c = 0., The roots of a quadratic equation ax2 + bx + c = 0 are given by, , −b ±, , Example 1. Find the solution of the quadratic equations by quadratic formula., 1 2, (i), x − 11 x + 1 = 0 (ii) – x2 + 7x – 10 = 0, 2, Solution. (i), , [Imp.], , 1 2, x − 11 x + 1 = 0 ⇒ x2 – 2 11 x + 2 = 0, 2, , Here, a = 1, b = −2 11 , c = 2, , Using the formula, x =, , , x=, , , ⇒, , x=, , −b ±, , b 2 − 4ac, , we get, 2a, , 2 11 ± 6, + 2 11 ± 44 − 8, =, 2, 2, 11 + 3, 11 − 3, , Hence, the roots of the given quadratic equation are 11 + 3 and 11 − 3, (ii) – x2 + 7x – 10 = 0, Here, a = –1, b = 7, c = –10, Using the formula, x =, , , x=, , −b ±, , b 2 − 4ac, , we get, 2a, , −7 ±, , 49 − 40, −7 ± 9, =, −2, −2, , −7 + 3 −4, −7 − 3 −10, =, =2,or, =, =5, −2, −2, −2, −2, Roots are 2 and 5., , ⇒, , x=, , Example 2. Using the quadratic formula, solve the equation, a2b2x2 – (4b4 – 3a4) x – 12a2b2 = 0, , [CBSE 2006], , Solution. Comparing given equation with Ax2 + Bx + C = 0, we get, A = a2b2, B = –(4b4 – 3a4) and C = –12a2b2., \ B2 – 4AC = (4b4 – 3a4)2 – 4 × a2b2 × (–12a2b2), = 16b8 + 9a8 – 24a4b4 + 48a4b4, = 16b8 + 9a8 + 24a4b4 = (4b4 + 3a4)2, ⇒ B2 − 4AC = 4b4 + 3a4, , Quadratic Equations, , 21

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Exercise 1.3, I. Very Short Answer Type Questions, , [1 Mark], , 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) The discriminant of the equation 9x2 + 6x + 1 = 0 is, (a) 0, (b) 1, (c) 2, , (d) 3, , (2) If D is the discriminant of the equation x2 + 2x – 4, then 2D is:, (a) 20, (b) 40, (c) 60, , (d) 80, , (3) The discriminant of the quadratic equation 4x2 – 6x + 3 = 0 is:, (a) 12, , (b) 84, , (c) 2 3, , (4) The roots of the quadratic equation ax2 + bx + c = 0 are given by, (a) < 0, , (b) ≤ 0, , (d) –12, −b ±, , (c) > 0, , b 2 − 4ac, if b2 – 4ac, 2ac, (d) ≥ 0, , (5) The quadratic formula was given by an ancient Indian mathematician., (a) Sridharacharya, , (b) Aryabhata, , (c) Brahmagupta, , (d) None of these, , 2. Assertion-Reason Type Question, , In the following question, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., a, (1) Assertion (A): The values of x are – , a for a quadratic equation 2x2 + ax – a2 = 0., 2, , Reason (R): For quadratic equation ax2 + bx + c = 0, x =, , − b ± b 2 − 4ac, 2a, , 3. Answer the following:, (1) Write the discriminant of the quadratic equation (x + 5)2 = 2(5x –3)., 2, 1, (2) Find the discriminant of the quadratic equation: 4 x 2 − x −, = 0., 3, 16, II. Short Answer Type Questions - I, 4. Find the roots of the equation ax2 + a = a2x + x., 5. Solve the following quadratic equation for x: 4x2 – 4a2x + (a4 – b4) = 0., 6. Solve the following quadratic equation for x: 9x2 – 6b2x – (a4 – b4) = 0., 7. Solve the following quadratic equation for x: 4x2 + 4bx – (a2 – b2) = 0., 8. Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0., III. Short Answer Type Questions - II, Solve the following using quadratic formula (Q. 9 to 11):, 9. 2 3 x 2 − 5 x + 3 = 0, 10. 3 x + 2 5 x − 5 = 0, 1, 1, 11, 11., −, =, , x ≠ – 4, 7, x + 4 x − 7 30, 12. Find the roots of quadratic equation: x 2 − 3 5 x + 10 = 0, 2, , [2 Marks], [CBSE 2012], [Delhi 2015], [Delhi 2015], [AI 2015], [AI 2015], [3 Marks], [Imp.], [Foreign 2011] [Imp.], [CBSE Standard 2020] [Imp.], [All India 2017], , 13. Find the roots of quadratic equation: 5 5 x 2 + 30 x + 8 5, 14. Solve for x: 4x2 – 4ax + (a2 – b2) = 0. , [Delhi 2012], 15. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of large diameter takes 8 hours less than the smaller, one to fill the tank separately. Find the time in which each tap can separately fill the tank., [Foreign 2016], , Quadratic Equations, , 23

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , IV. Long Answer Type Questions, [5 Marks], 16. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it, having an area of 111 sq m., Find the width of the path. , [CBSE 2013, 2012] [Imp.], 17. At ‘t’ minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less, t2, minutes. Find t., [NCERT Exemplar], 4, 18. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number., [NCERT Exemplar], x − 3 x − 5 10, +, 19. Solve for x:, = ; x ≠ 4, 6 [AI 2014], x−4 x−6 6, than, , 20. Solve for x:, , x − 2 x − 4 10, +, = ; x ≠ 3, 5 [AI 2014], x−3 x−5 3, ,  3x − 1 ,  2 x + 3, 1 3, 21. Solve for x: 3 , − 2, = 5; x ≠ , −, 3 2,  2 x + 3 ,  3 x − 1 , , [Foreign 2014], , 22. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find, the two numbers., [Delhi 2010], 23. 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., , Case Study Based Questions, I. Water Distribution System: Delhi Jal Board (DJB) is the main body of the Delhi Government which supplies drinking, water in the National Capital Territory of Delhi. Distribution system is well knit and properly planned. Maintenance of, underground pipe and hose system is also performed at regular interval of time. Many rivers and canals are inter-connected, in order to ensure un-interrupted water supply. It has been meeting the needs of potable water for more than 16 million, people. It ensures availability of 50 gallons per capita per day of pure and filtered water with the help of efficient network, of water treatment plants and pumping stations., In our locality, DJB constructed two big reservoir labelled as Reservoir–A and Reservoir–B., Reservoir–A: In order to fill it, department uses two pipes of different diameter., , Reservoir–B: Department uses two taps to store water in this reservoir., , Refer to Reservoir-A, 1, 1. Two pipes running together can fill the reservoir in 11 minutes. If one pipe takes 5 minutes more than the other to fill, 9, the reservoir, the time in which each pipe alone would fill the reservoir is, (a) 10 min, 12 min, (b) 25 min, 20 min, (c) 15 min, 18 min, (d) 22 min, 28 min, 2. Two pipes running together can fill a reservoir in 6 minutes. If one pipe takes 5 minutes more than the other to fill the, reservoir, the time in which each pipe would fill the reservoir separately is, (a) 8 min, 6 min, (b) 10 min, 15 min, (c) 12 min, 16 min, (d) 16 min, 18 min, Refer to Reservoir-B, 3, 3. Two water taps together can fill a reservoir in 9 hours. The tap of larger diameter takes 10 hours less than the smaller, 8, one to fill the reservoir separately. The time in which each tap can separately fill the reservoir will be, (a) 15 hrs, 25 hrs, (b) 20 hrs, 22 hrs, (c) 14 hrs, 18 hrs, (d) 18 hrs, 16 hrs, 1, 4. Two taps running together can fill the reservoir in 3 minutes. If one tap takes 3 minutes more than the other to fill, 13, it, how many minutes each tap would take to fill the reservoir?, (a) 12 min, 15 min, (b) 6 min, 9 min, (c) 18 min, 14 min, (d) 5 min, 8 min, 5. If two tapes function simultaneously, reservoir will be filled in 12 hours. One tap fills the reservoir 10 hours faster than, the other. The time that the second tap takes to fill the reservoir is given by, (a) 25 hrs, (b) 28 hrs, (c) 30 hrs, (d) 32 hrs, II. A Hill Station: In the last summer, I enjoyed a tour to a hill station at Shimla. I was accompanied by my five friends and, enjoyed the natural beauties of mountains, rivers, streams, forests etc. The beginning of the tour was the most adventurous, , 24, , Mathematics–10

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itself! How amazingly my group win the bet! Actually, the story is that my two friends along with me prefered train to go, to Shimla, but other three were forcing for a car or a bus. At last the consensus was reached and we were divided ourselves, in two groups of 3 each and started for Shimla at the same time. It was decided that the group who reach the destination, first, would be declared as the winner, and runner up the group have to bear the expanses of the tour. I named my group,, ‘Group A’ while the second group was named as ‘Group B’. Luckily we reached Shimla 1 hour before the Group-B and, enjoyed the trip for absolutely FREE!! How thrilling it was the tour!, Refer to Group-A, 1. An express train takes 1 hour less than a passenger train to travel 132 km between Delhi and Shimla (without taking into, consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than, that of the passenger train, the average speeds of the two trains will be, (a) 33 km/h, 44 km/hr , (b) 40 km/h, 45 km/h, (c) 30 km/h, 38 km/h , (d) 42 km/h, 62 km/h, 2. An express train makes a run of 240 km at a certain speed. Another train whose speed is 12 km/hr less takes an hour longer, to make the same trip. The speed of the express train will be, (a) 60 km/h, (b) 50 km/h, (c) 65 km/h, (d) 48 km/h, 3. A journey of 192 km from Delhi to Shimla takes 2 hours less by a super fast train than that by an ordinary passenger train., If the average speed of the slower train is 16 km/hr less than that of the faster train, average speed of super fast train is, (a) 50 km/h, (b) 48 km/h, (c) 55 km/h, (d) 60 km/h, Refer to Group-B, 4. A deluxe bus takes 3 hours less than a ordinary bus for a journey of 600 km. If the speed of the ordinary bus is 10 km/hr, less than that of the deluxe bus, the speeds of the two buses will be, (a) 35 km/h, 42 km/h , (b) 42 km/h, 52 km/h, (c) 40 km/h, 50 km/h , (d) 30 km/h, 58 km/h, 5. A bus travels a distance of 300 km at a uniform speed. If the speed of the bus is increased by 5 km an hour, the journey, would have taken two hours less. The original speed of the bus will be, (a) 20 km/h, (b) 15 km/h, (c) 22 km/h, (d) 25 km/h, , Answers and Hints, 1. (1) (a) 0, (1) (2) (b) 40, (1), (3) (d) –12, (1) (4) (d) ≥(1), (5) (a) Sridharacharya, (1), 2. (1) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) , (x + 5)2 = 2(5x – 3), , ⇒ x2 + 25 + 10x = 10x – 6, , ⇒, x2 + 31 = 0, , ⇒ x2 + 0x + 31 = 0, ∴ , D = (0)2 – 4 × 1 × 31, = 0 – 124 = –124, (1), (2) 3328(1), 1, 4. a, (2), a, 5. 4x2 – 4a2x + (a4 – b4) = 0, , fi x=, , , 4a 2 ± 16a 4 − 4 × 4 × (a 4 − b 4 ), 2×4, −B ±, Q x=, , , fi x=, , 4a ± 4b, a ±b, =, 2×4, 2, 2, , 2, , 2, , a 2 + b2, a 2 − b2, or, (1), 2, 2, , 6. 9x2 – 6b2x – (a4 – b4) = 0, , fi x=, , 6b 2 ±, , 36b 4 + 4 × 9 × (a 4 − b 4 ), 2×9, , , , Q x=, , , fi x=, , 6b 2 ±, , −B ±, , B2 − 4AC, 2A, , 36b 4 + 36a 4 − 36b 4, (1), 2×9, , , fi x=, , 6b 2 ± 36a 4, 2×9, , , fi x=, , b2 ± a 2, 3, , fi, , fi x=, , x=, , 6b 2 ± 6a 2, 2×3×3, , b2 + a 2, b2 − a 2, or, (1), 3, 3, , 7. 4x2 + 4bx – (a2 – b2) = 0, B2 − 4AC, 2A, , 4a 2 ± 16b 4, , fi x=, (1), 2×4, 2, , , fi x=, , x=, , −4b ± 16a 2 −4b ± 4a −b ± a, =, =, (1), 2×4, 8, 2, , , fi x=, , −b + a −b − a, ,, (1), 2, 2, , , , 8. x2 – 2ax – (4b2 – a2) = 0, , \, , x=, , − ( −2a ) ± 16b 2 2a ± 4b, =, = a ± 2b (1), 2 ×1, 2, , Quadratic Equations, , 25

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , , fi x = a + 2b or a – 2b.(1), 3 1, ,, 2, 3, 11. 2, 1, , 9., , 5, , − 5 (3), 3, , (3) 10., , (3), , x=, , 3 5± 5, −b ± D, =, (1), 2 ×1, 2a, , , fi x=, , 3 5+ 5, 3 5− 5, or, (1), 2, 2, , , fi x=, , 4 5, 2 5, or, 2, 2, , 12. , , , fi x = 2 5 or x = 5 (1), 13. , , fi, , 5 5 x 2 + 30x + 8 5 = 0, 5 5 x 2 + 20x + 10x + 8 5 = 0, , , fi, , 5x ( 5 x + 4) + 2 5 ( 5 x + 4) = 0, , , fi, , ( 5 x + 4) (5x + 2 5) = 0, , , , x=, , (1), (1), , –4 5, –2 5, (1), or, 5, 5, , − B ± B2 − 4AC, 14. Roots are =, 2A, =, , −( −4a) ± ( −4a) 2 − 4 × 4(a 2 − b 2 ), (1), 2×4, , 4a ± 16a 2 − 16a 2 + 16b 2, (1), 8, 4a ± 4b a ± b, =, =, (1), 8, 2, =, , 15. Let x be the time taken by larger diameter tap., , \ x + 8 be the time taken by smaller diameter tap., 1, 1, 10, ATQ,, =, +, x x+8, 96, 96 , , Q 9 hrs 36 min = hrs (1), 10, , , , fi, , fi, , 10x2 – 112x – 768 = 0, 5x2 – 56x – 384 = 0, , , fi x=, , 56 ± (56) 2 − 4 × 5 × ( −384), (1), 2×5, , , fi x=, , 56 + 104, 56 − 104, or, 10, 10, , 4. Nature of Roots, , Date __________, , , fi x = 16 or x = – 4.8 (Rejected), Hence, time taken by larger and smaller taps are 16 hrs, and 24 hrs respectively., (1), 16. 1.5 m, (5) 17. 14 minutes, (5), 18. 12, (5) 19. 2 ± 10 (5), 7, 20., (5) 21. 0, –7, (5), ,6, 2, 22. 9 and 13, (5), 23. Let the sides of two squares in metres be x and y, respectively (where x > y)., , Given:, Sum of areas of two squares = 544 m2, , ⇒, x2 + y2 = 544, ...(i) (1), Also, difference of their perimeters, = 32 m, , ⇒, 4x – 4y = 32, , ⇒, x–y=8, , ⇒, y = x – 8, ...(ii) (1), Substituting the value of y for equation (ii) in equation (i),, we get, , x2 + (x – 8)2 = 544, 2, 2, , ⇒ x + x – 16x + 64 – 544 = 0, , ⇒, 2x2 – 16x – 480 = 0, , ⇒, x2 – 8x – 240 = 0, (1), , ⇒, , x=, , =, , −( −8) ±, 8±, , ( −8) 2 − 4 × 1 × ( −240), 2 ×1, , 64 + 960, 2, , 8 ± 1024, (1), 2, 8 ± 32, =, = 4 ± 16, 2, , ⇒, x = 4 + 16 = 20, or, x = 4 – 16 = –12 (rejected), From (ii), y = 20 – 8 = 12, Thus, the sides of two squares are 20 m and 12 m. (1), =, , Case Study Based Questions, I. 1. (b) 25 min, 20 min 2. (b) 10 min, 15 min, 3. (a) 15 hrs, 25 hrs 4. (d) 5 min, 8 min, 5. (c) 30 hrs, II. 1. (a) 33 km/h, 44 km/hr, 2. (a) 60 km/h, 3. (b) 48 km/h, 4. (c) 40 km/h, 50 km/h, 5. (d) 25 km/h, , We know that the roots of the equation ax2 + bx + c = 0, a ≠ 0 are given by, x =, The quadratic equation ax2 + bx + c = 0 has, , 26, , Mathematics–10, , −b ±, , b 2 − 4ac, , where, b2 – 4ac = D, 2a

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−b + b 2 − 4ac, −b − b 2 − 4ac, and b =, 2a, 2a, −b, (ii) two equal real roots (i.e., coincident roots), if b2 – 4ac = 0. Roots are given by, ., 2a, 2, (iii) no real roots, if b – 4ac < 0., (i) two distinct real roots a and b, if b2 – 4ac > 0 where a =, , Formation of Quadratic Equation with given Roots, If a and b are the two roots of a quadratic equation, then the formula to construct the quadratic equation is x2 – (a + b), x + ab = 0., That is x2 – (sum of roots)x + product of roots = 0, Note: Let a and b be the two roots of quadratic equation ax2 + bx + c = 0. Then the formula to get sum and product of, the roots of a quadratic equation are:, Coefficient of x, b, a+b = − = −, a, Coefficient of x 2, c, Constant term, ab =, =, a, Coefficient of x 2, Example 1. What will be the nature of roots of quadratic equation 2x2 + 4x – 7 = 0?, Solution. Q, 2x2 + 4x – 7 = 0, Here, a = 2, b = 4, c = –7, D = b2 – 4ac = 16 – 4 × 2 × (–7) = 16 + 56 = 72 > 0, Hence, roots of quadratic equation are real and unequal., Example 2. If ax2 + bx + c = 0 has equal roots, find the value of c., Solution. For equal roots, D=0, i.e.,, , 2, b2 – 4ac = 0 ⇒ b = 4ac ⇒ c =, , b2, 4a, , Example 3. State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer., Solution. We have (x + 1) (x – 2) + x = 0 ⇒ x2 – x – 2 + x = 0 ⇒ x2 – 2 = 0, , D = b2 – 4ac = 0 – 4(1) (–2) = 8 > 0, \ Given equation has two distinct real roots., Example 4. Write the set of values of k for which the quadratic equation 2x2 + kx + 8 has real roots., [Imp.], Solution. For real roots, D ≥ 0, ⇒, b2 – 4ac ≥ 0 ⇒ k2 – 4(2) (8) ≥ 0, ⇒, k2 – 64 ≥ 0 ⇒ k2 ≥ 64 ⇒ k < –8 and k > 8, Example 5. Find the value of k for the quadratic equation kx (x – 2) + 6 = 0, so that it has two equal roots., Solution. We have, kx (x – 2) + 6 = 0, [NCERT] [Imp.], ⇒, kx2 – 2kx + 6 = 0, Here, a = k, b = –2k, c = 6, For equal roots,, D=0, 2, i.e.,, b – 4ac = 0 ⇒ (–2k)2 – 4 × k × 6 = 0, ⇒, 4k2 – 24k = 0 ⇒ 4k (k – 6) = 0, Either, 4k = 0 or k – 6 = 0 ⇒ k = 0 or k = 6, But k ≠ 0 (because if k = 0, then given equation will not be a quadratic equation)., So,, k = 6., Example 6. If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has, equal roots, then find the value of k. , [Foreign 2014], Solution. Q –5 is a root of the equation 2x2 + px – 15 = 0, \, 2 (–5)2 + p (–5) – 15 = 0, ⇒, 50 – 5p – 15 = 0 or 5p = 35 or p = 7, Again p (x2 + x) + k = 0 or 7x2 + 7x + k = 0 has equal roots, \, D=0, 49 7, 2, =, i.e.,, b – 4ac = 0 or 49 – 4 × 7k = 0 ⇒ k =, 28 4, , Quadratic Equations, , 27

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:, (a) 3x2 – 4 3x + 4 = 0, (b) 2x2 – 6x + 3 = 0, [NCERT] [Imp.], 2, Solution. (a) We have, 3x − 4 3x + 4 = 0, , Here, a = 3, b = −4 3 and c = 4, 2, Therefore,, D = b2 – 4ac = (−4 3 ) − 4 × 3 × 4 = 48 − 48 = 0, Hence, the given quadratic equation has real and equal roots., , Thus,, , x=, , −b 4 3 2 3, =, =, 2a, 6, 3, , Hence, equal roots of the given equation are, , 2 3, 2 3, and, 3, 3, , (b) We have,, 2x2 – 6x + 3 = 0, Here a = 2, b = –6, c = 3, Therefore,, D = b2 – 4ac = (–6)2 – 4 × 2 × 3 = 36 – 24 = 12 > 0, Hence, given quadratic equation has real and distinct roots., Thus,, , a=, , −b + b 2 − 4ac − ( −6) + 12 6 + 2 3 3 + 3, =, =, =, 2, 2a, 2×2, 4, , and, , b=, , −b − b 2 − 4ac −( −6) − 12 6 − 2 3 3 − 3, =, =, =, 2a, 2×2, 4, 2, , Hence, roots of given equation are, , 3+ 3, 3− 3, and, ., 2, 2, , Example 8. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other., [AI 2017], Solution. Let the roots of the given equation be a and 6a., Thus the quadratic equation is (x – a)(x – 6a) = 0, fi, x2 – 7ax + 6a2 = 0, ...(i), Given equation can be written as, 14, 8, x2 − x +, =0, ...(ii), p, p, Comparing the coefficients in (i) and (ii) 7a =, , 14, 8, and 6a2 =, p, p, , Solving to get p = 3., Example 9. Find the value(s) of k for which the equation x2 + 5kx + 16 = 0 has real and equal roots., Solution. For roots to be real and equal, b2 – 4ac = 0, [CBSE SP 2018-19], 8, ⇒, (5k)2 – 4 × 1 × 16 = 0 ⇒ k = ±, 5, Example 10. If the roots of the equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c., Solution. Since the equation (a – b) x2 + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant, D = (b – c)2 – 4 (a – b) (c – a) = 0, 2, ⇒, b + c2 – 2bc – 4 (ac – a2 – bc + ab) = 0, 2, ⇒, b + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0, ⇒, 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0, 2, 2, 2, ⇒ (2a) + (–b) + (–c) + 2 (2a) (–b) + 2 (–b) (–c) + 2 (–c) 2a = 0, ⇒, (2a – b – c)2 = 0, ⇒, 2a – b – c = 0, ⇒, 2a = b + c. Hence Proved., Example 11. If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots, show that c2 = a2 (1 + m2). [Delhi 2017], Solution. The given equation is (1 + m2) x2 + (2mc) x + (c2 – a2) = 0, Here, A = 1 + m2, B = 2mc and C = c2 – a2, Since the given equation has equal roots, therefore, D = 0 ⇒ B2 – 4AC = 0, , 28, , Mathematics–10

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⇒, (2mc)2 – 4 (1 + m2) (c2 – a2) = 0, 2 2, ⇒, 4m c – 4(c2 – a2 + m2c2 – m2a2) = 0, ⇒, m2c2 – c2 + a2 – m2c2 + m2a2 = 0, [Dividing throughout by 4], ⇒, –c2 + a2 (1 + m2) = 0 ⇒ c2 = a2 (1 + m2), Hence Proved., Example 12. Form a quadratic equation whose roots are 2 and 3., Solution., Sum of roots = 2 + 3 = 5, Product of roots = 2 × 3 = 6, So, quadratic equation can be formed as x2 – (sum of roots)x + product of roots = 0, x2 – 5x + 6 = 0, , Exercise 1.4, , I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) For what value of k, the equation 9x2 – 24x + k = 0 has equal roots?, (a) 12, (b) 16, (c) 18, (d) 20, (2) The values of k for which the quadratic equation (k + 1)x2 + 2(k – 1)x + (k – 2) = 0 has equal roots, is:, (a) k = 2, (b) k = 3, (c) k = 0, (d) None of these, (3) The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is, (a) 4, (b) ± 4, (c) – 4, (d) 0, (4) The value(s) of k, for which the roots of the equation 3x2 + 2k + 27 = 0 are real and equal are, (a) k = 9, (b) k = ± 9, (c) k = –9, (d) k = 0, (5) The value of k, for which the equation 2x2 – 10x + k = 0 has real roots is, 25, 25, 25, 25, (a) k ≤, (b) k ≥, (c) k =, (d) k >, 2, 2, 2, 2, (6) If one root of the equation (k − 1)x2 – 10x + 3 = 0 is the reciprocal of the other, then the value of k is, (a) 1, (b) 2, (c) 3, (d) 4, (7) If the quadratic equation x2 – 2x + k = 0 has equal roots, then value of k is, (a) 1, (b) 2, (c) 3, (d) 0, 2, (8) If quadratic equation 3x – 4x + k = 0 has equal roots, then the value of k, 4, 2, 1, (a), (b), (c) 3, (d), 3, 3, 3, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., 8, (1) Assertion (A): The equation 8x2 + 3kx + 2 = 0 has equal roots, then the value of k is ± ., 2, 2, 3, , Reason (R): The equation ax + bx + c = 0 has equal roots if D = b – 4ac = 0., , (2) Assertion (A): The roots of the quadratic equation x2 + 2x + 2 = 0 are imaginary., , Reason (R): If discriminant D = b2 – 4ac < 0, then the roots of quadratic equation ax2 + bx + c = 0 are imaginary., 3. Answer the following., (1) Find the value of p, so that the quadratic equation px(x – 3) + 9 = 0 has equal roots.  , [CBSE 2014], (2) For what values of k, the roots of the equation x2 + 4x + k = 0 are real?, [Delhi 2019], (3) Find the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other. [Delhi 2019], (4) Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, hence find the nature of its roots., [NCERT], (5) If x = 3 is one root of the quadratic equation x2 – 2kx– 6 = 0, then find the value of k.  , [CBSE 2018], (6) For what values of k, the equation 9x2 + 6kx + 4 = 0 has equal roots?, [CBSE Standard SP 2020-21], (7) For what value(s) of ‘a’ quadratic equation 3ax2 – 6x + 1 = 0 has no real roots?, [CBSE Standard SP 2020-21], II. Short Answer Type Questions - I, , [2 Marks], 25, = 0 has two distinct real roots or not. Justify your answer., 4. State whether the quadratic equation 4x – 5x +, 16, [NCERT Exemplar], 2, , Quadratic Equations, , 29

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\ 23-Nov-2021, , 5., 6., 7., 8., 9., 10., , Amit, , Proof-3, , Reader’s Sign _______________________, , Find the value of k so that the quadratic equation kx(3x – 10) + 25 = 0, has two equal roots.  , For what value of k does the quadratic equation (k – 5)x2 + 2(k – 5)x + 2 = 0 have equal roots?, Find the value(s) of k so that the quadratic equation 2x2 + kx + 3 = 0 has equal roots., Find the value(s) of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots., Find the value(s) of k so that the quadratic equation 3x2 – 2kx + 12 = 0 has equal roots.  , Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.  , , Date __________, , [Delhi 2011], [Foreign 2011], [Delhi 2012], [Delhi 2012], [Delhi 2012], [AI 2014], , III. Short Answer Type Questions - II, [3 Marks], 11. Find the nature of the roots of the following quadratic equations. If the real roots exist, then also find them., (a) 4x2 + 12x + 9 = 0, (b) 3x2 + 5x – 7 = 0, (c) 7y2 – 4y + 5 = 0, [Imp.], 2, 12. 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, the value of k., [CBSE (F) 2014], 2, 13. Find the value of p for which the quadratic equation (p + 1)x – 6(p + 1)x + 3(p + 9) = 0, p π –1 has equal roots. Hence,, find the roots of the equation., [Delhi 2015], 14. Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k – 1)x + x2 = 0 has equal roots. Hence, find, the roots of the equation., [Delhi 2015], 2, 15. The roots a and b of the quadratic equation x – 5x + 3(k – 1) = 0 are such that a – b = 1. Find the value k., [CBSE Standard SP 2020-21], 16. Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0 has equal roots. Also find these roots., , [Delhi 2014], IV. Long Answer Type Questions, [5 Marks], 1, 1, 3, +, = 1, x ≠ , 5 has real roots. If real roots exist, find them.[NCERT Exemplar], 17. Find whether the equation, 2x − 3 x − 5, 2, 18. Check whether the equation 5x2 – 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., [CBSE SP 2018], , Answers and Hints, 1. (1) (b) 16, (3) (b) ± 4, (5) (a) k ≤, , 25, 2, , (1) (2) (b) k = 3, (1) (4) (b) k = ± 9, , (1), (1), , (1) (6) (d) 4, , (1), , 4, (1), 3, 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A). , (1), , (7) (a) 1, , (1) (8) (d), , , (2) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A). , (1), 3. (1) , px(x – 3) + 9 = 0, , fi, px2 – 3px + 9 = 0, When roots are equal,, , D = b2 – 4ac = 0, , 9p2 – 36p = 0, , fi, 9p(p – 4) = 0, , fi, p = 0, p = 4, But, pπ0, [ In quadratic equation, a π 0], , \, p = 4, (1), (2) For real roots, D ≥ 0, , fi, b2 – 4ac ≥ 0, , 30, , Mathematics–10, , , fi (4)2 – 4 × 1 × k ≥ 0, , fi, 16 – 4k ≥ 0, fi, 16 ≥ 4k, k ≤ 4, , (1), 1, (3) Let the roots of the given equation be a and, ., α, 1, c k, = = & k = 3 (1), ., a, a a 3, (4) –8, no real roots, (1), (5) x = 3 is one root of the equation, , ∴, 9–6k–6=0, , 1, , , ⇒, k=, (1), 2, 2, (6) , 9x + 6kx + 4 = 0, , (6k)2 – 4 × 9 × 4 = 0, (½), 2, , 36k = 144, , ⇒, k2 = 4, , k = ±2, (½), (7) , 3ax2 – 6x + 1 = 0, (½), 2, , (–6) – 4(3a)(1) < 0, , 12a > 36, , ⇒, a > 3, (½), 4. No,, D = 0, (2), 5., kx(3x – 10) + 25 = 0, , fi 3kx2 – 10kx + 25 = 0

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D = (–10k)2 – 4 × 3k × 25, = 100k2 – 300k, , For equal roots,, , D = 0, , For equal roots,, (1), , 2, , , fi, , 100k – 300k = 0, , , fi, , 100k(k – 3) = 0, , , fi, , k = 0 or k = 3, , But, , k π 0,, , So,, , k = 0 (Rejected), k = 3, , 6. k = 7, 7. , 2x + kx + 3, For equal roots, D, , fi, b2 – 4ac, , fi, k2 – 24, , (1), , 0, 0, 0, 0, , (1), , x2 – 4kx + k = 0, , Since given equation has equal roots,, , \, , D = 0, , , , 16k2 – 4k = 0, , , fi, , 4k(4k – 1) = 0, , 9. , , (1), , 3x2 – 2kx + 12 = 0, , , fi, , fi, , 4(k – 36) = 0, k = 6 and k = –6, , 10. For equal roots,, , fi, , 9k(k – 4) = 0, k = 0 or k = 4, , kx + 1 – 2(k – 1)x + x = 0, (k + 1)x2 – 2(k – 1)x + 1 = 0, , So, discriminant,, , D=0, , 2, , , fi {–2(k – 1)} – 4 × (k + 1) × 1 = 0, , fi, , 4(k2 – 2k + 1) – 4(k + 1) = 0, , , fi, , 4k2 – 12k = 0, , , fi, , 4k(k – 3) = 0, , , fi, , k = 3 (as k π 0), , (1), , (1), (3), , 16. For equal roots,, , , (1), , D = 0, , 2, , {2(k + 1)} – 4(3k + 1) · 1 = 0, , (1), , 2, , , fi, , 4(k + 2k + 1) – 12k – 4 = 0, , , fi, , 4k2 + 8k + 4 – 12k – 4 = 0, , , fi, , 4k2 – 4k = 0, , , fi, , 4k(k – 1) = 0, k = 0, 1, , 17. Yes,, , (1), , (1), , 8!3 2, (5), 2, Discriminant = b2 – 4ac, , = 76 > 0, 5x2 – 6x – 2 = 0, , (1), , Multiplying both sides by 5, we get, (1), , D=0, , 9k – 36k = 0, , 14. , , , , , (5x)2 – 2 × (5x) × 3 = 10, , , fi (5x)2 – 2 × (5x) × 3 + 32 = 10 + 32(1), , 2, , , fi, , 2, , So, the given equation has two distinct real roots, , 2, , , fi, , , fi, , (1), , 4k – 144 = 0, k = ±6, , (3), 2, , = 36 – 4 × 5 × (–2), , 2, , , fi, , (1), , , fi, , 18., , Since given equation has equal roots, so, D = 0, , k = 1, , , fi, , 1, k = 0 and k = (1), 4, , , fi, , , fi, , 15. k = 3, , k = ± 2 6 (1), , , fi, 8. , , =, =, =, =, , 16 = 16k, , , Q Above equation has equal roots,, , (2), 2, , , fi, 13. 3, 3., , [Q In quadratic equation, a π 0], Hence,, , D=0, , (4)2 – 4 × 4 × k = 0, , , fi, , (1), (1), , –3 –3, (1), ,, 2 2, (b) Real and distinct roots:, 11. (a) Real and equal roots:, , –5 + 109 –5 – 109, , (1), ,, 6, 6, (c) No real roots, (1), 12. , 3(2)2 + p(2) – 8 = 0, , fi, 12 + 2p – 8 = 0, , fi, p = –2, ...(i)(1), So, equation becomes, , 4x2 + 4x + k = 0, [using (i)](1), , , fi, , fi, , (5x – 3)2 = 19, 5x – 3 = ± 19, , , fi, , x=, , 3 ± 19, (1), 5, , 2, ,  3 + 19 ,  3 + 19 , Verification: 5 , − 6, ,  −2,  5 ,  5 , =, , 9 + 6 19 + 19 18 + 6 19 10, − = 0 (1), −, 5, 5, 5, 2, ,  3 − 19 ,  3 − 19 , Similarly, 5 , − 6, ,  −2 =0, 5 , ,  5 , (1), , Quadratic Equations, , 31

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , Date __________, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. To find the roots of a quadratic equation by factorisation., 2. To find the roots of a quadratic equation by quadratic formula., 3. To find the nature of roots of quadratic equation., 4. To find the value of unknown when nature of roots is given., , IMPORTANT FORMULAE, For quadratic equation ax2 + bx + c = 0,, , –b ! D, , where D = b2 – 4ac, 2a, • If D > 0, there are two distinct real roots a and b which are given by, –b + D, –b – D, , a=, ,b=, 2a, 2a, –b, • If D = 0, there are two equal roots and each equal root is given by x =, 2a, • If D < 0, then there is no real roots., • Roots are given by x =, , QUICK REVISION NOTES, •• A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a π 0., •• A real number a is said to be a root of the quadratic equation ax2 + bx + c = 0, if aa2 + ba + c = 0. The zeroes of the, quadratic polynomial ax2 + bx + c = 0 and the roots of the quadratic equation ax2 + bx + c = 0, are the same., •• If we can factorise ax2 + bx + c, a π 0 into a product of two linear factors, then the roots of the quadratic equation, ax2 + bx +c = 0 can be found by equating each factor to zero., , –b ! b 2 – 4ac, •• Quadratic formula: The roots of a quadratic equation ax + bx + c = 0 are given by, , provided, 2a, 2, b – 4ac  0., •• A quadratic equation ax2 + bx + c = 0 has, (i) two distinct real roots, if b2 – 4ac > 0, (ii) two equal roots (coincident roots) if b2 – 4ac = 0., (iii) no real roots, if b2 – 4ac < 0., 2, , COMMON ERRORS, Errors, , Corrections, , (i) While finding the discriminant, taking incorrectly the, value of a and b with variables., , (i) Remember that coefficient of x2 is a and coefficient of x is b., , (ii) Incorrectly considering two consecutive even integers, as x and x + 1., , (ii) Remember that difference between two consecutive even, integers is 2. So, we should take x and x + 2., , (iii) Incorrectly writing difference between two positive, 1 1, integers and difference between their reciprocals as (iii) Attention that when x > y, x < y and will be negative., , 1, , 1, , x – y = ___ and x – y = ___., (iv) In factorisation method, equating factors incorrectly., For example: (x – 4) (x – 2) = 4, fi x – 4 = 4, x – 2 = 4., , (iv) It is not correct. We should equating factors to zero only., For example, if (x – 4) (x – 2) = 0, then x – 4 = 0, x – 2 = 0., , qqq, , 32, , Mathematics–10

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2, , Arithmetic Progressions, , Topics Covered, 1. Sequence / Progression, 3. Sum of First n Terms of an AP, , 2. Arithmetic Progressions and its nth term, , 1. Sequence/Progression, • Sequence/Progression: A sequence/progression is a succession of numbers or terms formed according to some pattern, or rule. Various numbers occurring in a sequence are called terms or elements., Consider the following arrangements of numbers:, 1 1 1 1, (i) 1, 8, 27, 64, 125, ..., (ii) 1, , , , , ..., (iii) 2, 4, 6, 8, 10, ..., 2 3 4 5, In each of the above arrangements, numbers are arranged in a definite order according to some rule. So, they are, sequences., A sequence is generally written as < an > : a1, a2, a3, ..., an where a1, a2, a3, ... are the first, second and third terms of, the sequence., • A sequence with finite number of terms or numbers is called a finite sequence., • A sequence with infinite number of terms or numbers is called an infinite sequence., Example 1. Write first four terms of each of the following sequence, whose general terms are:, (i) an = 3n – 7, (ii) an = (– 1)n+1 × 3n, Solution. (i), an = 3n – 7, , \, a1 = 3 × 1 – 7 = 3 – 7 = – 4, a2 = 3 × 2 – 7 = 6 – 7 = – 1,, , a3 = 3 × 3 – 7 = 9 – 7 = 2 and a4 = 3 × 4 – 7 = 12 – 7 = 5, (ii) , an = (– 1)n+1 × 3n, , fi, a1 = (– 1)1+1 × 31 = 3,, , \, a2 = (– 1)2+1 × 32 = (–1)3 × 32 = –9,, , a3 = (– 1)4 × 33 = 27 and a4 = (– 1)5 × 34 = – 81, Example 2. What is 18th term of the sequence defined by an =, Solution. We have,, Putting n = 18, we get, , an =, a18 =, , =, , n(n − 3), ?, n+4, , n(n − 3), n+4, 18 × (18 − 3), 18 + 4, 18 × 15 135, =, 22, 11, , Exercise 2.1, I. Very Short Answer Type Questions, 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) If an = 5n – 4 is a sequence, then a12 is, (a) 48, (b) 52, (c) 56, (2) If an = 3n – 2, then the value of a7 + a8 is, (a) 39, (b) 41, (c) 47, , 33, , [1 Mark], , (d) 62, (d) 53

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , (3) The second term of the sequence defined by an = 3n + 2 is, (a) 2, (b) 4, (c) 6, (d) 8, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): The arrangement of numbers, i.e., – 4, 16, – 64, 256, – 1024, 4096, ... form a sequence., , Reason (R): An arrangement of numbers which are arranged in a definite order according to some rule, is called a, sequence., (2) Assertion (A): Sequence 1, 5, 9, 13, 17, 21, ... is a finite sequence., , Reason (R): A sequence with finite number of terms or numbers is called a finite sequence., 3. Answer the following:, (1) Write down the first six terms of each of the following sequences, whose general terms are:, 2n + 1, (a) an = 5n – 3, (b) an = (– 1)n ⋅ 22n, (c) an =, (d) an = (–1)n – 1 ⋅ n2, n+2, (2) Find the 10th term of the sequence defined by an = (–1)2n – 1 ⋅ 5n., (3) Find the difference between the 12th term and 10th term of the sequence whose general term is given by an = 5n – 1., , Answers, 1. (1) (c) 56, (1) (2) (b) 41, (1), (3) (d) 8, (1), 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (d) Assertion (A) is false but reason (R) is true. (1), , 3. (1) (a) 2, 7, 12, 17, 22, 27, (b) –4, 16, –64, 256, –1024, 4096, 5 7 3 11 13, (c) 1, , , , ,, 4 5 2 7 8, (d) 1, –4, 9, –16, 25, –36, (2) –9765625 , (3) 10, , (1), (1), (1), , 2. Arithmetic Progression and its nth Term, • An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed number ‘d’ to, the preceding term, except the first term ‘a’. This fixed number is known as common difference of the AP. Common, difference of an AP can be negative, positive or zero., The general form of an AP is a, a + d, a + 2d, a + 3d, ..., Examples:, (i) The sequence 1, 4, 7, 10, 13, ... is an AP whose first term is 1 and the common difference is equal to 3., (ii) The sequence 11, 7, 3, –1, ... is an AP whose first term is 11 and the common difference is equal to –4., • In the list of numbers a1, a2, a3, ... if the differences a2 – a1, a3 – a2, a4 – a3, ... give the same value, i.e., if ak +1 – ak is, the same for different values of k, then the given list of numbers is an AP., • The nth term an (or the general term) of an AP is an = a + (n – 1) d, where a is the first term, d is the common difference, and n is the number of terms. Also, d = an + 1 – an., • If three terms a, b and c are in AP, then b – a = c – b or 2b = a + c., • If l is the last term of an AP, then nth term from the end of the AP = l + (n – 1)(–d) = l – (n – 1)d., Example 1. In an AP, if d = – 4, n = 7, an = 4, then find the value of a., [CBSE Standard SP 2020-21, Delhi 2018], Solution. We have, an = 4 for n = 7, \, an = a + (n – 1) d fi 4 = a + 6(– 4) fi a = 28, Example 2. Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer., [NCERT Exemplar], Solution. Given AP is 31, 28, 25, ..., Here,, a = 31, d = 28 – 31 = – 3 = 25 – 28, For 0 be a term of this AP,, 0 = an for some ‘n’ ⇒ 0 = a + (n – 1)d, ⇒, 0 = 31 + (n – 1) (– 3) ⇒ 31 – 3n + 3 = 0, ⇒, , 34, , – 3n = – 34, , Mathematics–10, , fi, , n=, , 34, 1, = 11, 3, 3

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which is not possible as n cannot be a fraction., Therefore, 0 cannot be a term of this AP., Example 3. Find the 12th term from the end of the AP: –2, – 4, – 6, ..., – 100., Solution. Let a be the first term, d the common difference and l the last term of AP., Here, a = –2, d = (–4 + 2) = –2, l = –100 and n = 12, \, nth term from end = l – (n – 1) d, fi, 12th term from end = –100 – (12 – 1) (–2) = –100 + 24 – 2 = –78, Example 4. For what value of x: 2x, x + 10 and 3x + 2 are in AP?, Solution. Since, given numbers are in AP., So,, (x + 10) – 2x = (3x + 2) – (x + 10), ⇒, –x + 10 = 2x – 8 or 3x = 18 or x = 6, 5, −5, th, Example 5. Find the 25 term of the AP: −5,, , 0, , ..., 2, 2, 5, −5, − ( −5) =, Solution. We have,, a = –5, d =, 2, 2, Q, an = a + (n – 1) d, 5, 5, \, a25 = (– 5) + (25 – 1) = (−5) + 24   = − 5 + 60 = 55, 2, 2, , [Imp], , Example 6. Find the 20th term from the last term of the AP: 3, 8, 13,..., 253., [NCERT] [Imp.], Solution. Given, last term = l = 253, And, common difference = d = 8 – 3 = 5 = 13 – 8, \, 20th term from end = l – (n – 1) × d = 253 – 19 × 5 = 253 – 95 = 158, Example 7. Which term of the AP: 3, 8, 13, 18, ..., is 78?, [NCERT] [Imp.], Solution. Let an be the required term of the AP: 3, 8, 13, 18,..., Here, a = 3, d = 8 – 3 = 5 and an = 78, Now,, an = a + (n – 1)d, ⇒, 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n – 1) × 5, 75, ⇒, 75 = (n – 1) × 5 ⇒, =n–1, 5, ⇒, 15 = n – 1 ⇒ n = 15 + 1 = 16, Hence, 16th term of given AP is 78., Example 8. The sum of the 5th and 7th terms of an AP is 52 and the 10th term is 46. Find the AP., , [NCERT Exemplar] [Imp.], Solution. Let the first term and the common difference of an AP be ‘a’ and ‘d’., \, a5 = a + 4d and a7 = a + 6d, So,, a5 + a7 = 2a + 10d = 52 ⇒ 2a + 10d = 52, ...(i), Also,, a10 = a + 9d = 46 ⇒ a + 9d = 46, ...(ii), From (i) and (ii),, d = 5 and a = 1, So, the AP is as follows 1, 6, 11, 16, 21, ..., Example 9. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term., , [CBSE SP 2018-19] [NCERT] [Imp.], Solution. Let a be the first term and d be the common difference., Since, given AP has 50 terms, so n = 50, , a3 = 12 ⇒ a + (3 – 1) d = 12, ⇒, a + 2d = 12, ...(i), Also,, a50 = 106 ⇒ a + (50 – 1) d = 106, ⇒, a + 49d = 106, ...(ii), Subtracting (i) from (ii), we get, 94, =2, 47d = 94 ⇒ d =, 47, Putting the value of d in equation (i), we get, a + 2 × 2 = 12 ⇒ a = 12 – 4 = 8, , Arithmetic Progressions, , 35

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Here,, a = 8, d = 2, So, 29th term of the AP is given by, a29 = a + (29 – 1)d = 8 + 28 × 2 ⇒ a29 = 8 + 56 ⇒ a29 = 64, st, Example 10. Find the 31 term of an AP whose 11th term is 38 and the 16th term is 73., Solution. Let the first term be a and common difference be d., Now, given, a11 = 38 ⇒ a + (11 – 1)d = 38, ⇒, a + 10 d = 38, Also,, a16 = 73 fi a + (16 – 1) d = 73, ⇒, a + 15d = 73, Now, subtracting (ii) from (i), we get, a + 10 d = 38, a + 15 d = 73, – –    –, –5 d = –35 ⇒ 5 d = 35, 35, =7, ⇒, d=, 5, Putting the value of d in equation (i), we get, a + 10 × 7 = 38 ⇒ a + 70 = 38, ⇒, a = 38 – 70 ⇒ a = –32, We have, a = –32 and d = 7, Therefore,, a31 = a + (31 – 1)d = a + 30 d, ⇒, a31 = (–32) + 30 × 7 = –32 + 210 ⇒ a31 = 178, Example 11. The first term of an AP is x and its common difference is y. Find its 12th term., Solution., a12 = a + 11d = x + 11y., , [NCERT], , ...(i), ...(ii), , Exercise 2.2, , I. Very Short Answer Type Questions, 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, , [1 Mark], , (1) In an AP, if d = – 4, n = 7, an = 4, then a is, (a) 6, (b) 7, , (c) 20, , (d) 28, , (2) The nth term of the AP: a, 3a, 5a, ... is, (a) na, (b) (2n – 1)a, , (c) (2n + 1)a, , (d) 2na, , (3) The first term of an AP is p and the common difference is q, then its 10th term is, (a) q + 9p, (b) p – 9q, (c) p + 9q, (d) 2p + 9p, 4, (4) If , a, 2 are three consecutive terms of an AP, then the value of a is, 5, 2, 7, 5, 5, (a), (b), (c), (d), 7, 5, 7, 2, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): Common difference of the AP: –5, –1, 3, 7, ... is 4., , Reason (R): Common difference of the AP : a, a + d, a + 2d, ... is given by d = 2nd term – 1st term., (2) Assertion (A): If nth term of an AP is 7 – 4n, then its common difference is – 4., , Reason (R): Common difference of an AP is given by d = an+1 – an., (3) Assertion (A): Common difference of an AP in which a21 – a7 = 84 is 14., , Reason (R): nth term of an AP is given by an = a + (n – 1) d., , 36, , Mathematics–10

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3. Answer the following:, (1) Write first four terms of the AP, whose first term and the common difference are given as follows:, , a = 10, d = 10, (2) Find the 10th term of the AP: 2, 7, 12, ..., [NCERT] [Imp.], (3) In the given AP, find the missing terms: ......., 13, ......., 3., [NCERT], (4) Find the 6th term from the end of the AP: 17, 14, 11, ..., –40., [Imp.], (5) Which term of the AP: 21, 18, 15, ... is zero?, [Delhi 2008 (C)] [Imp.], (6), (7), (8), (9), (10), (11), (12), (13), (14), (15), (16), , Write the next term of the AP: 8 , 18 , 32 , ..........., [AI 2008], Find a, b, and c such that the numbers a, 7, b, 23, c are in AP., [NCERT Exemplar], Find the 9th term from the end (towards the first term) of the AP: 5, 9, 13, ..., 185.  , [Delhi 2016], For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an AP?, [Delhi 2016], For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form an AP?, [Foreign 2016], Find the eleventh term from the last term of the AP: 27, 23, 19, ..., –65., [CBSE Sample Paper 2018], If the first three terms of an AP are b, c and 2b, then find the ratio of b and c., [CBSE Standard SP 2019-20], Find the value of x so that –6, x, 8 are in AP., Find the 11th term of the AP: –27, –22, –17, –12, ... ., The nth term of an AP is (7 – 4n), then what is its common difference?, Find the common difference of the AP whose first term is 12 and fifth term is 0., , II. Short Answer Type Questions - I, [2 Marks], 4. Find how many integers between 200 and 500 are divisible by 8., [AI 2017], 1, 1, 3, 5. Which term of the progression 20, 19 , 18 , 17 , ... is the first negative term?, [AI 2017], 4, 2, 4, 6. Is –150 a term of the AP: 17, 12, 7, 2, ...?, [Delhi 2011], 7. Find the number of two-digit numbers which are divisible by 6., [AI 2011], th, 8. Which term of the AP: 3, 14, 25, 36, ... will be 99 more than its 25 term?, [AI 2011], 9. Which term of the AP: 3, 15, 27, 39, ... will be 120 more than its 21st term?, [Delhi 2019], 10. How many natural numbers are there between 200 and 500, which are divisible by 7?, [AI 2011], 11. How many two-digit numbers are divisible by 7?, [Foreign 2011], 12. How many two digits numbers are divisible by 3?, [Delhi 2019], 1, 1, 1, ,, and, 13. If, are in AP, find the value of x., [Foreign 2011], x+2 x+3, x+5, 14. How many three digit numbers are divisible by 11?, [AI 2012], 15. In an AP, the first term is 12 and the common difference is 6. If the last term of the AP is 252, find its middle term., , [Foreign 2017], 16. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5., [AI 2014], 17. The 4th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term., [AI 2016], 18. Find the middle term of the AP: 6, 13, 20, ..., 216., [Delhi 2015], 19. The nth term of an AP is 6n + 2. Find its common difference., [Delhi 2008], 20. Find the 10th term from end of the AP: 4, 9, 14, ..., 254., [Imp.], 2, 2, 2, 21. Determine k so that k + 4k + 8, 2k + 3k + 6, 3k + 4k + 4 are three consecutive terms of an AP. [NCERT Exemplar], 22. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both., , [CBSE Standard SP 2019-20], III. Short Answer Type Questions - II, [3 Marks], 23. Which term of the AP: 115, 110, 105, ..... is its first negative term?, 24. If the 9th term of an AP is zero, prove that its 29th term is double of its 19th term.  , [NCERT Exemplar], 25. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle., [NCERT Exemplar], 26. For what value of n, the nth term of two APs: 63, 65, 67, ... and 3, 10, 17, ... are equal., [NCERT], 27. The 8th term of an AP is 37 and its 12th term is 57. Find the AP., [Imp.], 28. The pth, qth and rth terms of an AP are a, b and c respectively. Show that a(q – r) + b(r – p) + c(p – q) = 0., , [Foreign 2016], th, 29. If the n terms of two APs: 23, 25, 27, ... and 5, 8, 11, 14, ... are equal, then find the value of n., , Arithmetic Progressions, , 37

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________ Date __________, , IV. Long Answer Type Questions, [5 Marks], 30. 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 AP is zero. , [AI2019], 31. The 19th term of an AP is equal to three times its sixth term. If its 9th term is 19, find the AP., [AI 2013], 32. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of, the AP., [Imp.], 33. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the, 15th term., [Imp.], 34. If 4 times the 4th term of an AP is equal to 18 times the 18th term, then find the 22nd term., , Case Study Based Questions, I. Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day, practice it takes him 2 seconds less. He wants to do in 31 seconds., , 1. 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, …, 2. What is the minimum number of days he needs to practice till his goal is achieved?, (a) 10, (b) 12, (c) 11, (d) 9, 3. Which of the following term is not in the AP of the above given situation?, (a) 41, (b) 30, (c) 37, (d) 39, 4. 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, 5. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP is, (a) 6, (b) – 6, (c) 18, (d) –18, II. 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., , 1. The production during first year is, (a) 3000 TV sets, (b) 5000 TV sets, 2. The production during 8th year is, (a) 10500, (b) 11900, , 38, , Mathematics–10, , (c) 7000 TV sets, , (d) 10000 TV sets, , (c) 12500, , (d) 20400

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3. The production during first 3 years is, (a) 12800, (b) 19300, (c) 21600, 4. In which year, the production is 29,200?, (a) 10th year, (b) 12th year, (c) 15th year, 5. The difference of the production during 7th year and 4th year is, (a) 6600, (b) 6800, (c) 5400, , (d) 25200, (d) 18th year, (d) 7200, , Answers and Hints, , 1. (1) (d) 28    (1) (2) (b) (2n – 1)a(1), 7, (3) (c) p + 9q  (1) (4) (a) = (1), 5, 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (3) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) 10, 20, 30, 40 (1) (2) 47(1), (3) 18, 8, (1) (4) –25(1), (5) 8, (1) (6) 50 or 5 2 (1), (7) a = –1, b = 15, c = 31, (1), (8) Reversing the given AP, we get, 185, 181, 174, ..., 9, 5, , Ninth term a9 = a + (9 – 1)d, = 185 + 8 × (– 4), = 185 – 32, = 153, (1), (9) Given that k + 9, 2k – 1 and 2k + 7 are in AP, Then,, , (2k – 1) – (k + 9) = (2k + 7) – (2k – 1), , fi, k – 10 = 8, , fi, k = 18, (1), (10) Given that 2k + 1, 3k + 3 and 5k – 1 are in AP., So, (3k + 3) – (2k + 1) = (5k – 1) – (3k + 3), , fi, k + 2 = 2k – 4, , fi, 2k – k = 2 + 4, , fi, k = 6, (1), (11) a11 = –25, (1), (12) b, c and 2b are in AP, 3b, , ⇒, c=, 2, , ∴, b : c = 2 : 3, (1), (13) 1, (1), (14) 23, (1), (15) , an = 7 – 4n, , ⇒, a1 = 7 – 4 × 1 = 3, , ⇒, a2 = 7 – 4 × 2 = 7 – 8 = –1, , a3 = 7 – 4 × 3 = 7 – 12 = –5, Now,, a2 – a1 = –1 – 3 = –4, , a3 – a2 = –5 – (–1), = –5 + 1 = –4, So, the common difference of AP is –4.  , (1), (16) , A5 = a1 + 4d = 0, , 12 + 4d = 0, , d = –3, (1), , 4. AP formed is 208, 216, 224, ...., 496, , an = 496, (1), , fi 208 + (n – 1) × 8 = 496, , fi, n = 37, (1), −3, 5. Here d =, (½), 4, Let the nth term be first negative term.,  −3 , , \ 20 + (n − 1)   < 0 fi 3n > 83, (1),  4, 2, , fi, n > 27, 3, Hence, 28th term is first negative term., (½), 6. Let, an = –150, , a + (n – 1)d = –150, , fi 17 + (n – 1)(–5) = –150, (1), , fi, (n – 1)(–5) = – 167, , fi, , n=, , 167 + 5 172, 2, =, = 34, 5, 5, 5, , Here, n is not a natural number., , \ –150 is not a term of the given AP., (1), 7. Two-digit numbers which are divisible by 6 are 12, 18, 24,, ..., 96, , Q Last term,, , an = 96, , fi, 12 + (n – 1)6 = 96, (1), , fi, (n – 1)6 = 96 – 12 = 84, , fi, n = 15, , \ There are 15 two-digit numbers divisible by 6. (1), 8. Let an be the term which is 99 more than 25th term of given, AP., ATQ,, an = a25 + 99, , fi, a + (n – 1)d = a + 24d + 99, (1), , fi, 11(n – 1) = 24 × 11 + 99, , fi, n = 34, Hence, 34th is the required term.(1), 9. AP: 3, 15, 27, 39, ..., , a = 3, d = 15 – 3 = 12, , a21 = a + 20d = 3 + 20 × 12, = 3 + 240 = 243, 120 more than a21 = 243 + 120 = 363(1), Let 363 be nth term., So,, 363 = 3 + (n – 1) 12, , ⇒, 360 = 12(n – 1), , 30 = n – 1 ⇒ n = 31, Thus, 31st term of the given AP is 120 more than its 21st, term.(1), , Arithmetic Progressions, , 39

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\ 23-Nov-2021, , Amit, , Proof-3, , 10. Natural numbers between 200 and 500 which are divisible, by 7 are as 203, 210, 217, ..., 497, Let above are n numbers and an = 497, , a + (n – 1)d = 497, (1), , fi 203 + 7(n – 1) = 497, , fi, n = 43, , \ There are 43 natural numbers between 200 and 500, divisible by 7., (1), 11. Two-digit numbers which are divisible by 7 are 14, 21, 28,, ..., 98., Let, an = 98, , fi, a + (n – 1)d = 98, (1), , fi, 14 + 7(n – 1) = 98, , n = 13, Hence, there are 13 two-digit numbers which are divisible, by 7.(1), 12. 2-digit numbers divisible by 3 are 12, 15, 18, ..., 99 which, is in AP., So,, an = 99, d = 15 – 12 = 3, Now,, an = a + (n – 1) d(1), , fi, 99 = 12 + (n – 1) 3, , fi, 87 = 3 (n – 1), , fi, 29 = n – 1, , fi, n = 30, Thus, 30, 2-digit numbers are divisible by 3.(1), 13. Given term are in AP, So,, , fi, , 2, 1, 1, +, =, x+3, x+2 x+5, 2, ( x + 5) + ( x + 2), =, (1), x+3, ( x + 2)( x + 5), , , fi 2x2 + 14x + 20 = 2x2 + 13x + 21, , \, x = 1(1), 14. Three-digit numbers which are divisible by 11 are 110,, 121, 132, ..., 990, Let, an = 990(1), , fi, a + (n – 1)d = 990, , fi 110 + 11(n – 1) = 990, , \, n = 81, Hence, there are 81 three-digit numbers which are divisible, by 11.(1), 15. Let, an = 252 = last term, , fi, a + (n – 1)d = 252, , fi, 12 + (n – 1)6 = 252, , fi, n = 41(1), , \ Since number of terms is odd, so only one middle, term., ,  41 + 1, Now,middle term = ,  2 , =, , \ 21st term, a21 =, =, =, =, , 40, , 21st term, a + 20d, 12 + 20 × 6, 132, middle term value.(1), , Mathematics–10, , Reader’s Sign _______________________, , Date __________, , 16. Numbers between 101 and 999 which are divisible by both, 2 and 5 (i.e., by 10) are 110, 120, 130, ... 990., Now,, an = a + (n – 1)d(1), , fi, 990 = 110 + (n – 1)10, , fi, n = 89, , \ Natural numbers which are divisible by 2 and 5 both, are 89.(1), 17. , a4 = a + (4 – 1)d, , 0 = a + 3d, , fi, a = –3d[Q Given, a4 = 0] (1), Now, a25 = a + (25 – 1)d = a + 24d, = –3d + 24d = 21d = 3 × 7d, , Hence,, a25 = 3 × a11, [Q Since a11 = a + (11 – 1)d = –3d + 10d = 7d] (1), 18. Given AP is 6, 13, 20, ..., 216, , nth term, an = 216, , fi, a + (n – 1)d = 216, , fi, 6 + 7(n – 1) = 216, , fi, 7n = 217, , fi, n = 31, (1), Since, the number of terms in AP are 31, so, the middle, most term is 16th term., (31 + 1), , , = 16 th term , Q middle term =, 2, , , , \ 16th term, a16 = a + 15d = 6 + 15 × 7 = 111., (1), 19. 6, (2) 20. 209, (2) 21. k = 0, (2), 22. 110, 120, 130, … , 990, , an = 990, , ⇒ 110 + (n − 1) × 10 = 990, , ∴, n = 89, (2), 23. 25th term, (3) 25. 40°, 60°, 80°, (3), 26. 13, (3) 27. 2, 7, 12, 17, 22, ...  , (3), 28. Let A and d be the first term and common difference of, the given AP, then, , ap = A + (p – 1)d = a, ...(i), , aq = A + (q – 1)d = b, ...(ii), , ar = A + (r – 1)d = c, ...(iii), Now, subtracting (i) and (ii), we get, , (p – q)d = a – b, a b, , p – q = − (1), d d, Multiplying by ‘c’ on both sides,, ca cb, −, , c(p – q) =, d, d, Now, (ii) - (iii), we get, , (q – r)d = b – c, b c, , q–r= −, d d, Multiplying by ‘a’ on both sides,, ab ac, , a(q – r) =, −, d, d, , ...(iv), , ...(v)(1)

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Now, (iii) - (i), we get, , (r – p)d = c – a, c a, , (r – p) = −, d d, Multiplying by ‘b’ on both sides,, bc ba, −, , (r – p)b =, d, d, , ...(vi), , Adding (iv), (v) and (vi), we get, , a(q – r) + b(r – p) + c(p – q), ab ac bc ba ca cb, − + −, + −, = 0 (1), =, d, d, d, d, d, d, 29. , AP1 = 23, 25, 27, ..., Here,, a1 = 23, , d1 = 25 – 23 = 27 – 25 = 2, th, , \, n term = a1 + (n – 1)d1, = 23 + (n – 1)2, , AP2 = 5, 8, 11, 14, ..., (1), Here,, a2 = 5, , d2 = 8 – 5 = 11 – 8 = 3, , \, nth term = a2 + (n – 1)d2, = 5 + (n – 1)3, (1), Now, 23 + (n – 1)2 = 5 + (n – 1)3, , ⇒, 23 + 2n – 2 = 5 + 3n – 3, , ⇒, 3n – 2n = 23 – 2 – 5 + 3, , ⇒, n = 26 – 7 = 19, (1), 30. We know that, an = a + (n – 1)d, From the given conditions,, m[a + (m – 1) d] = n[a + (n – 1)d](1), ⇒ m[a + (md – d)] = n[a + nd – d], , ⇒ am + m2d – md = an + n2d – nd(1), ⇒, am – an + m2d – n2d – md + nd = 0, ⇒, a(m – n) + d(m2 – n2) – d(m –n) = 0, ⇒ a(m – n) + (m + n) (m – n)d – (m – n)d = 0(1), ⇒, (m – n) [a + (m + n) d – d] = 0, ⇒, a + md + nd – d = 0(1), ⇒, a + (m + n –1)d = 0, Since, m ≠ n, it is clear that (m + n)th term of the AP is zero., (1), 31. 3, 5, 7, 9, ..., (5) 32. –13, –8, –3, (5), 33. 3, (5), 34. Let a1, a2, a3, ... an, ... be the AP with its first term a and, common difference d., It is given that, , 4a4 = 18a18(1), , ⇒, 4(a + 3d) = 18(a + 17d)(1), , ⇒, 4a + 12d = 18a + 306d(1), , ⇒, 14a + 294d = 0 ⇒ 14(a + 21d) = 0, (1), , ⇒, a + 21d = 0 ⇒ a + (22 – 1)d = 0, , ⇒, a22 = 0, Thus, 22nd term is 0., (1), Case Study Based Questions, I. 1. (b) 51, 49, 47…., 2. (c) 11, 3. (b) 30, 4. (a) 2, 5. (a) 6, II.1. (b) 5000 TV sets, 2. (d) 20400, 3. (c) 21600, 4. (b) 12th year, 5. (a) 6600, , 3. Sum of First n Terms of an AP, • If first term of an AP be a and its common difference is d, then the sum Sn of the first n terms of an AP is given by, n, n, Sn = [2a + (n – 1) d] or, Sn = (a + an) where, 2, 2, an = nth term of the AP., • If l is the last term of an AP of n terms, then the sum of all ‘n’ terms can also be given by, n, Sn = (a + l). Sometimes Sn is also denoted by S., 2, • The sum of first n positive integers is given by, n (n + 1), Sn =, ., 2, • If Sn is the sum of the first n terms of an AP, then its nth term is given by an = Sn – Sn–1, i.e., the nth term of an AP is the, difference of the sum to first n terms and the sum to first (n – 1) terms of it., Example 1. Find the sum of the given AP: – 5 + (– 8) + (– 11) + ... + (– 230)., [NCERT][CBSE Standard 2020], Solution. We have, a = – 5 and d = –8 + 5 = – 3, So,, an = a + (n – 1)d, ⇒, – 230 = – 5 + (n – 1) (– 3) ⇒ – 230 = – 5 – 3n + 3, 228, ⇒, – 230 + 2 = – 3n ⇒ – 228 = – 3n ⇒ n =, = 76, 3, 76, n, \, Sn =, (a + an) ⇒ S76 =, [– 5 – 230], 2, 2, = 38 [– 235] = – 8930, , Arithmetic Progressions, , 41

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 1, + 14 + ... + 84 [NCERT], 2, Solution. Let a be the first term, d the common difference and an the last term of given AP., 1, 21, 21 − 14 7, = and an = 84, −7=, We have,, a = 7, d = 10 − 7 =, 2, 2, 2, 2, 7, Now,, an = a + (n – 1)d ⇒ 84 = 7 + (n – 1) ×, 2, 7, ⇒, 77 = (n − 1) ×, ⇒ 11 × 2 = (n – 1) ⇒ 22 = n – 1, 2, Example 2. Find the sum of the AP: 7 + 10, , n = 22 + 1 = 23, n, Now,, Sn = [2a + (n − 1) d ], 2, 23 , 7, 2 × 7 + (23 − 1) × , ⇒, S23 =, , 2 , 2, 23, 2093, 1, 23, 23 , 7, [14 + 77] =, × 91 =, = 1046, 14 + 22 ×  =, ⇒, S23 =, 2, 2, 2, 2, 2 , 2, Example 3. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?  , [NCERT] [Imp.], Solution. Let sum of n terms be 636., Then,, Sn = 636, a = 9, d = 17 – 9 = 8, n, n, [2 × 9 + (n − 1) × 8] = 636, ⇒, [2a + (n − 1)d ] = 636 ⇒, 2, 2, n, × 2[9 + (n − 1) × 4] = 636 ⇒ n(9 + 4n – 4) = 636, ⇒, 2, ⇒, n[5 + 4n] = 636 ⇒ 5n + 4n2 = 636 ⇒ 4n2 + 5n – 636 = 0, \, , n=, , \, , −5 ± (5)2 − 4 × 4 × ( −636) −5 ± 25 + 10176, =, 2×4, 8, , −5 ± 10201 −5 ± 101 96 −106, 53, = 12, −, =, = ,, 8, 8, 8, 8, 4, −53, But, nπ, , So, n = 12, 4, Thus, the sum of 12 terms of the given AP is 636., Example 4. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms., [Delhi 2008] [Imp.], Solution. We have,, S7 = 49, 7, ⇒, 49 = [2a + (7 − 1) × d ] ⇒ 7 × 2 = [2a + 6d], 2, ⇒, 14 = 2a + 6d ⇒ a + 3d = 7, ...(i), and, S17 = 289, 289 × 2, 17, = 34, [2a + (17 − 1) d ] ⇒ 2a + 16d =, ⇒, 289 =, 17, 2, ⇒, a + 8d = 17, ...(ii), Now subtracting equation (i) from (ii), we get, 5d = 10 ⇒ d = 2, Putting the value of d in equation (i), we get, a+3×2=7 ⇒ a=7–6=1, Here, a = 1 and d = 2, n, Now,, Sn = [2a + (n − 1)d ], 2, n, n, n, 2, = [2 × 1 + (n − 1 ) × 2] = [2 + 2n − 2] = × 2n = n, 2, 2, 2, =, , 42, , Mathematics–10

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Example 5. The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term of, this AP., [Foreign 2014], n, Solution. We have,, Sn = [2a + (n − 1)d ], 2, 7, 7, \, S7 = [2a + (7 − 1)d ] ⇒ S7 = [2a + 6d ], 2, 2, ⇒, 63 = 7a + 21d, [S7 = 63 (given)], 63 – 21d, ⇒, a=, ...(i), 7, 14, Also,, S14 =, [2a + 13d], 2, ⇒, S14 = 14a + 91d, But according to question, S1–7 + S8 – 14 = S14, ⇒, , 63 + 161 = 14a + 91d, , ⇒, , 2a + 13d = 32, , ⇒, , ⇒ 224 = 14a + 91d,  63 − 21d , 2,  + 13d = 32, , 7, , 126 – 42d + 91d = 224, 49d = 98 ⇒ d = 2, 63 − 21 × 2 63 − 42 21, =, =, =3, \, a=, 7, 7, 7, Thus,, a28 = a + 27d = 3 + 27 × 2, ⇒, a28 = 3 + 54 = 57, Example 6. Find the sum of the integers between 100 and 200 that are:, (i) divisible by 9, (ii) not divisible by 9, Solution. (i) Numbers divisible by 9 between 100 and 200 are 108, 117, 126, ..., 198., Here,, a = 108, d = 9, an = 198, , \, an = a + (n – 1) d fi 198 = 108 + (n – 1) 9, , fi, 198 = 108 + 9 n – 9 fi 198 = 99 + 9 n, 99, , fi, 198 – 99 = 9 n fi, =n, 9, , fi, n = 11, n, Thus,, Sn = [2a + (n – 1) d], 2, 11, 11, , S11 =, [216 + 90] = 1683, [2 × 108 + 10 × 9] =, 2, 2, , ...(ii), , ⇒, ⇒, , [NCERT Exemplar], , (ii) Numbers between 100 and 200 are 101, 102,..., 199., Here,, a = 101, d = 1, an = 199, Now,, an = a + (n – 1)d, , fi, 199 = 101 + (n – 1) 1 fi 199 = 100 + n fi n = 99, n, So,, Sn = (a + l), where l is the last term, 2, 99, 99, =, (101 + 199) =, × 300 = 14850, 2, 2, Sum of the numbers which are not divisible by 9, = Sum of total numbers – sum of numbers which are divisible by 9, = S99 – S11 = 14850 – 1683 = 13167, Example 7. Find the sum:, Solution. The first term,, , a − b 3a − 2b 5a − 3b, +, +, + ... to 11 terms., a+b, a+b, a+b, a−b, a1 =, a+b, , [NCERT Exemplar], , Arithmetic Progressions, , 43

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\ 23-Nov-2021, , Amit, , Common difference, \, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 3a − 2b (a − b), 2a − b, −, =, a+b, a+b, a+b, 11  2(a − b),  2a − b  , + 10 , S11 =,  a + b  , 2  a + b, d=, , =, , 11, 11, [2a – 2b + 20a – 10 b] =, [11a – 6 b], a+b, 2(a + b), , Example 8. The sum of the first n terms of an AP is 3n2 + 6n. Find the nth term of this AP., [Foreign 2014], Solution. We have,, Sn = 3n2 + 6n, Sn–1 = 3(n – 1)2 + 6(n – 1), = 3(n2 + 1 – 2n) + 6n – 6, = 3n2 + 3 – 6n + 6n – 6 = 3n2– 3, The nth term will be an, Sn = Sn–1 + an, fi, an = Sn – Sn – 1, = 3n2 + 6n – 3n2 + 3 = 6n + 3, Example 9. In an AP, the sum of first ten terms is – 150 and the sum of next ten terms is – 550. Find the AP., [Delhi 2010], Solution. Let a be the first term and d the common difference of the AP., We have,, S10 = – 150, 10, ⇒, [2a + 9d] = – 150 fi 2a + 9d = –30, ...(i), 2, and, S20 – S10 = – 550, ⇒, S20 = – 550 – 150 = – 700, R, V, 20, SSSa S = 20 [2a + 19d]WWW, ⇒, [2a + 19d] = – 700, 20, S, W, 2, 2, T, X, ⇒, 2 a + 19 d = –70, ...(ii), From (i) and (ii),, d = – 4 and a = 3, So, the AP is: 3, –1, –5, ..., Example 10. If an = 3 – 4n, show that a1, a2, a3, ... form an AP. Also find S20., [NCERT Exemplar], Solution. We have,, an = 3 – 4n, \, a1 = – 1, a2 = – 5, a3 = – 9, ..., Since, a2 – a1 = – 4 = a3 – a2, So, –1, –5, –9, ... form an AP., 20, S20 =, [–2 + 19 × (–4)] = 10 [– 2 – 76] = 10 × (–78) = –780, 2, Example 11. If the sum of the first p terms of an AP is ap2 + bp, find its common difference., [CBSE 2010], 2, 2, Solution., ap = Sp – Sp – 1 = (ap + bp) – [a(p – 1) + b(p – 1)], = ap2 + bp – (ap2 + a – 2ap + bp – b), = ap2 + bp – ap2 – a + 2ap – bp + b = 2ap + b – a, \, a1 = 2a + b – a = a + b, a2 = 4a + b – a = 3a + b, ⇒, d = a2 – a1 = (3a + b) – (a + b) = 2a, Example 12. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its, common difference ‘d ’., [Delhi 2014], Solution. We have,, a = 5, Tn = 45, Sn = 400, \, Tn = a + (n – 1)d, ⇒, 45 = 5 + (n – 1)d ⇒ (n – 1)d = 40, ...(i), n, Sn = (a + Tn ), 2, , 44, , Mathematics–10

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⇒, , 400 =, , n, (5 + 45), 2, , ⇒ n = 2 × 8 = 16, , Substituting the value of n in (i), we get, (16 – 1)d = 40 ⇒ 15d = 40, 40 8, =, \, d=, 15 3, Example 13. The sum of the first n terms of an AP is given by Sn = 3n2 – 4n. Determine the AP and the 12th term. , [Delhi 2019] [Imp.], 2, Solution. We have,, Sn = 3n – 4n...(i), Replacing n by n – 1, we get, Sn – 1 = 3 (n – 1)2 – 4 (n – 1), ...(ii), 2, 2, Since,, an = Sn – Sn–1 = {3n – 4n} – {3(n – 1) – 4(n – 1)}, = {3n2 – 4n} – {3n2 + 3 – 6n – 4n + 4}, = 3n2 – 4n – 3n2 – 3 + 6n + 4n – 4 = 6n – 7, So,, nth term, an = 6n – 7, ...(iii), Substituting n = 1, 2, 3, ... respectively in (iii), we get, a1 = 6 × 1 – 7 = –1, a2 = 6 × 2 – 7 = 5, and, a3 = 6 × 3 – 7 = 11, Hence, AP is – 1, 5, 11, ..., 12th term,, a12 = 6 × 12 – 7 = 72 – 7 = 65, [From (iii)], 1, 1, th, th, th, Example 14. If the m term of an AP is n and n term is m , then show that its (mn) term is 1., [Delhi 2017], Solution. Let a and d be the first term and the common difference of the AP respectively., 1, 1, Then,, am = n and an = m, 1, ⇒, a + (m – 1)d = n ...(i), 1, and, a + (n – 1)d = m ...(ii), Subtracting (ii) from (i), we get, 1, 1, a + (m – 1)d – [a + (n – 1)d] = n – m, m–n, 1, fi, (m – n) d = mn, fi d = mn, Putting the value of d in (i), we get, 1, 1, 1, a + (m – 1) mn = n fi a – mn = 0, 1, fi, a = mn, 1, 1, \, amn = a + (mn – 1) d = mn + (mn – 1) mn = 1, Example 15. If Sn denotes the sum of the first n terms of an AP, prove that S30 = 3 (S20 – S10)., n, Solution. We have,, Sn = [2a + (n − 1)d ], 2, \, fi, and, , [Foreign 2014], , 30, [2a + (30 − 1)d ], 2, S30 = 15(2a + 29d) = 30a + 435d, 20, S20 =, [2a + (20 – 1) d], 2, S30 =, , ...(i), , S20 = 10(2a + 19d) = 20a + 190d, 10, S10 =, [2a + (10 – 1)d], 2, , Arithmetic Progressions, , 45

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1, (4a + 22d) fi 2(4a + 6d) = 4a + 22d, 2, 2(20 + 6d) = 20 + 22d, 40 + 12d = 20 + 22d fi 20 = 22d – 12d, 20 = 10d fi d = 2, 4a + 6d =, , fi, fi, fi, fi, , [ a = 5 (given)], , Exercise 2.3, , I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) The sum of first five terms of the AP: 3, 7, 11, 15, ... is:, (a) 44, (b) 55, (c) 22, (d) 11, (2) If the first term of an AP is 1 and the common difference is 2, then the sum of first 26 terms is, (a) 484, (b) 576, (c) 676, (d) 625, (3) If the sum to n terms of an AP is 3n2 + 4n, then the common difference of the AP is, (a) 7, (b) 5, (c) 8, (d) 6, (4) If a, b, c are in AP then ab + bc =, 1, (a) b, (b) b2, (c) 2b2, (d), b, (5) The sum of all natural numbers which are less than 100 and divisible by 6 is, (a) 412, (b) 510, (c) 672, (d) 816, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation ofassertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): Sum of the first 10 terms of the arithmetic progression –0.5, –1.0, –1.5,... is 27.5., n, , Reason (R): Sum of first n terms of an AP is given as Sn = [2a + (n – 1)d] where a = first term, d = common, 2, difference., (2) Assertion (A): The sum of the first n terms of an AP is given by Sn = 3n2 – 4n. Then its nth term, an = 6n – 7., , Reason (R): nth term of an AP, whose sum of n terms is Sn, is given by an = Sn – Sn–1., (3) Assertion (A): Sum of first hundred even natural numbers divisible by 5 is 500., n, , Reason (R): Sum of the first n terms of an AP is given by Sn =, [a + l] where l = last term., 2, 3. Answer the following:, (1), (2), (3), (4), (5), (6), (7), , Find the sum of first 10 terms of the AP: 2, 7, 12, ..., If the sum of first m terms of an AP is 2 m2 + 3 m, then what is its second term?, Find the sum of first 10 multiples of 6., What is the sum of five positive integers divisible by 6?, If the sum of the first q terms of an AP is 2q + 3q2, what is its common difference?, If nth term of an AP is (2n + 1), what is the sum of its first three terms?, Find the sum of first 100 natural numbers., , II. Short Answer Type Questions - I, , [NCERT] [Imp.], [Foreign 2010], [AI 2019], [CBSE Sample Paper 2012], [AI 2010], [CBSE SP 2018-19], [CBSE Standard 2020], [2 Marks], , 4. Find the sum of first 8 multiples of 3., [CBSE 2018], 5. Find the number of terms of the AP: 54, 51, 48, ... so that their sum is 513., [Imp.], 6. In an AP, the first term is –4, the last term is 29 and the sum of all its terms is 150. Find its common difference., [Foreign 2016], 7. Find the sum of all three digit natural numbers, which are multiples of 11., [Delhi 2012], 8. The first and the last terms of an AP are 8 and 65 respectively. If sum of all its terms is 730, find its common difference., , [Delhi 2014], , Arithmetic Progressions, , 47

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 9. The sum of the first n terms of an AP is 4n2 + 2n. Find the nth term of this AP., , [Foreign 2013], , 10. How many terms of the AP: 18, 16, 14, ... be taken so that their sum is zero?, [Delhi 2016], 11. In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms., [AI 2015], 12. The sum of first n terms of an AP is given by Sn = 2n2 + 3n. Find the sixteenth term of the AP., III. Short Answer Type Questions - II, [3 Marks], 13. How many multiples of 4 lie between 10 and 250? Also find their sum., [AI 2011], 14. Find the sum of first n terms of an AP whose nth term is 5n – 1. Hence find the sum of first 20 terms., [AI 2011], 15. The sum of first six terms of an AP is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the, thirteenth terms of the AP., [AI 2009], 16. Find the sum of all multiples of 7 lying between 500 and 900., [AI 2010], 17. If M, N and T are in AP, prove that (M + 2N – T) (2N + T – M) (T + M – N) = 4MNT., 18. In an AP, if the 6th and 13th terms are 35 and 70 respectively, find the sum of its first 20 terms., [Foreign 2011], nd, th, th, th, 19. The sum of the 2 and the 7 terms of an AP is 30. If its 15 term is 1 less than twice its 8 term, find the AP., [AI 2014], th, 20. If the ratio of the sum of first n terms of two AP’s is (7n + 1) : (4n + 27), find the ratio of their m terms., [AI 2016], 21. The digits of a positive number of three digits are in AP and their sum is 15. The number obtained by reversing the digits, is 594 less than the original number. Find the number., [AI 2016], 22. The sums of first n terms of three A.Ps’ are S1, S2 and S3. The first term of each AP is 5 and their common differences, are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2., [Imp], 1 , 2 , 3, , 23. Find the sum of n terms of the series  4 −  +  4 −  +  4 −  + ... , , n , n , n, 24. Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287., IV. Long Answer Type Questions, , [Delhi 2017], , [5 Marks], , 25. The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and the third terms is 5, times the common difference, find the three numbers., [Al 2019], 26. 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 AP is zero., [Al 2019], 27. The first and the last term of an AP are 8 and 350 respectively. If its common difference is 9, how many terms are there, and what is their sum?, [AI 2011], (a + c) (b + c − 2a ), ., 2(b − a ), [NCERT Exemplar][CBSE Standard 2020], 1, 1, 1, 29. If the pth term of an AP is, and qth term is , prove that the sum of the pq terms is (pq + 1)., [CBSE 2012], q, p, 2, 30. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio, of the sum of the first five terms to the sum of the first 21 terms., [NCERT Exemplar], 28. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to, , 31. The sum of the first five terms of an AP is 55 and sum of the first ten terms of this AP is 235, find the sum of its first 20, terms., [Imp.], 32. The sums of n terms of two APs are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 25th terms.  , , [Imp.], , 33. Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 3, when divided by 4., Also, find the sum of all numbers on both sides of the middle terms separately., [Foreign 2015], 34. If the ratio of the sum of the first n terms of two APs is (7n + 1) : (4n + 27), then find the ratio of their 9th terms., 35. If the sum of first 14 terms of an AP is 1050 and its first term is 10, find the 20th term., , [AI 2017], , 36. The first term of an AP is 5, the last term is 45 and sum is 400. Find the number of terms and the common difference., 37. How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?, , 48, , Mathematics–10

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Case Study Based Questions, I. Pollution—A Major Problem: One of the major serious problems that the world is facing today is the environmental, pollution. Common types of pollution include light, noise, water and air pollution., , In a school, students thoughts of planting trees in and around the school to reduce noise pollution and air pollution., , Condition I: It was decided that the number of trees that each section of each class will plant be the same as the class in, which they are studying, e.g. a section of class I will plant 1 tree a section of class II will plant 2 trees and so on a section, of class XII will plant 12 trees., , Condition II: It was decided that the number of trees that each section of each class will plant be the double of the class, in which they are studying, e.g. a section of class I will plant 2 trees, a section of class II will plant 4 trees and so on a, section of class XII will plant 24 trees., Refer to Condition I, 1. The AP formed by sequence i.e. number of plants by students is, (a) 0, 1, 2, 3, ..., 12, (b) 1, 2, 3, 4, ..., 12, (c) 0, 1, 2, 3, ..., 15, (d) 1, 2, 3, 4, ..., 15, 2. If there are two sections of each class, how many trees will be planted by the students?, (a) 126, (b) 152, (c) 156, (d) 184, 3. If there are three sections of each class, how many trees will be planted by the students?, (a) 234, (b) 260, (c) 310, (d) 326, Refer to Condition II, 4. If there are two sections of each class, how many trees will be planted by the students?, (a) 422, (b) 312, (c) 360, (d) 540, 5. If there are three sections of each class, how many trees will be planted by the students?  , (a) 468, (b) 590, (c) 710, (d) 620, II. 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:, , 1. The amount paid by him in 30th installment is, (a) ` 3900, (b) ` 3500, (c) ` 3700, 2. The total amount paid by him upto 30 installments is, (a) ` 37000, (b) ` 73500, (c) ` 75300, 3. What amount does he still have to pay after 30th installment?, (a) ` 45500, (b) ` 49000, (c) ` 44500, 4. If total installments are 40, then amount paid in the last installment is, (a) ` 4900, (b) ` 3900, (c) ` 5900, 5. The ratio of the 1st installment to the last installment is, (a) 1 : 49, (b) 10 : 49, (c) 10 : 39, , (d) ` 3600, (d) ` 75000, (d) ` 54000, (d) ` 9400, (d) 39 : 10, , Arithmetic Progressions, , 49

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Amit, , \ 23-Nov-2021, , Proof-3, , Reader’s Sign _______________________, , Answers and Hints, , 1. (1), (3), (5), 2. (1), , (b) 55, (1) (2) (c) 676, (1), (d) 6, (1) (4) (c) 2b2(1), (d) 816, (1), (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (3) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) 245, (1) (2) 9, (1), (3) First 10 multiples of 6 are 6, 12, 18, ....., 60., This is an AP in which a = 6, n = 10 and d = 6., , ∴ Sum of first 10 multiples of 6 = S10  , (½), n, , ⇒ S10 = 62a + ]n – 1g d @, 2, 10, 62×6 + ]10 – 1g 6@, =, 2, = 5 (12 + 54), = 5 × 66 = 330, (½), (4) 90, (1), (5) Given that,, , Sq = 2q + 3q2, , S1 = 2 + 3 = 5 = T1 = First term [put q = 1], , S2 = 4 + 3(4) = 16, [put q = 2], , S3 = 6 + 3(9) = 33, [put q = 3](½), , \ 2nd term,, , T2 = S2 – S1 = 16 – 5 = 11, , \ 3rd term,, , T3 = S3 – S2 = 33 – 16 = 17, Common difference, = T3 – T2 = 17 – 11 = 6, (½), 3, (6) a1 = 3, a3 = 7, S3 = (3 + 7) = 15, [½+½], 2, (7) Natural numbers are 1, 2, 3, 4, ..., The sum of first 100 natural numbers is given by, n(n + 1), 100 × (100 + 1), , Sn =, =,   , (½), 2, 2, =, , 100 × 101, 2, , = 50 × 101 = 5050, 4. , S8 = 3 + 6 + 9 + 12 +....+ 24, = 3(1 + 2 + 3 + ...+ 8), 8×9, = 3 ×, = 108, 2, 5. 18 or 19, 6. , , 150 =, , , fi, 300 =, , \ Then, l =, , fi, 11d =, , 50, , Date __________, , n, ( −4 + 29), 2, 25n fi n = 12, a12 = 29 = –4 + 11d, 33 fi d = 3, , Mathematics–10, , (½), (1), (1), , (2), n, Q Sn = (a + l ), 2, (1), (1), , 7. 3-digit natural numbers which are multiples of 11 are 110,, 121, 132, ..., 990, , nth term, 990 = 110 + (n – 1)11, , fi, n = 81, (1), , \ Sum of ‘n’ terms,, n, , Sn = [a + l ], 2, 81, [110 + 990] = 44550, 2, , \ Sum of all three-digit natural numbers, which are, multiples of 11 is 44550., (1), n, 8. , Sn = (a + an ), 2, 73n, n, , 730 = (8 + 65) fi, = 730, 2, 2, , fi, n = 20, (1), , \ Given a20 = 65, where an = a + (n – 1)d, , fi, a + 19d = 65 fi 8 + 19d = 65, , fi, 19d = 57, Hence, common differences = d = 3., (1), 2, 9. Given,, Sn = 4n + 2n, , So,, Sn – 1 = 4(n – 1)2 + 2(n – 1), = 4(n2 – 2n + 1) + 2n – 2, = 4n2 – 8n + 4 + 2n – 2, = 4n2 – 6n + 2, (1), th, , an = Sn – Sn – 1 = n term, = (4n2 + 2n) – (4n2 – 6n + 2), = 4n2 + 2n – 4n2 + 6n – 2, = 8n – 2, (1), 10. Let the number of terms taken for sum to be zero be n., Then, sum of n terms, , (Sn) = 0, (Given), n, , fi, Sn = [2a + (n − 1) d ] (1), 2, =, , n, [2 × 18 + (n − 1)( −2)], 2, , fi, n = 19, Hence, sum of 19 terms is 0., (1), 11. , S5 + S7 = 167, 5, 7, (2a + 4d ) + (2a + 6d ) = 167, , fi, 2, 2, n, , , Q Sn = [2a + (n − 1)d ], 2, , , , , fi, 5a + 10d + 7a + 21d = 167, , fi, 12a + 31d = 167, ...(i), Also,, S10 = 235, 10, (2a + 9d ) = 235, , fi, 2, , fi, 2a + 9d = 47, ...(ii)(1), Multiplying eq. (ii) by 6, we get, 6(2a + 9d) = 6 × 47, , fi, , 0=

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fi, , \, , , , , 12a + 54d = 282, Subtracting eq. (i) from (iii), we get, 12a + 54d = 282, 12a + 31d = 167, –, –     –, , ...(iii), , , , , 23d = 115, , \, d=5, Putting ‘d’ in (ii) equation, a = 1, , \ Required AP is 1, 6, 11, ..., (1), 12. , Sn = 2n2 + 3n, , S1 = 5 = a1, , S2 = a1 + a2 = 14 fi a2 = 9, (1), , d = a2 – a1 = 4, , a16 = a1 + 15d = 5 + 15(4) = 65, (1), 13. 60, 7800, (3), 14. Given:, an = 5n – 1, , a1 = 4, , \, a2 = 5(2) – 1 = 9, , d = a2 – a1 = 9 – 4 = 5, (1), Now, sum of first ‘n’ terms,, n, , Sn = [2a + (n − 1)d ], 2, n, = [2 × 4 + 5(n − 1)], 2, n, n(5n + 3), = (8 + 5n − 5) =, (1), 2, 2, Now, sum of first 20 terms,, 20(5 × 20 + 3), , S20 =, 2, = 10 × 103 = 1030, (1), 15. 2 and 26, (3) 16. 39, 900, (3), 18. Given that, a6 = 35 fi a + 5d = 35, ...(i), and also a13 = 70 fi a + 12d = 70, ...(ii)(1), On solving the above equations, we get, , a = 10; d = 5, (1), Now, sum of first 20 terms,, 20, [2 × 10 + 19 × 5], , S20 =, 2, n, Q Sn = [2a + (n − 1)d ], 2, , = 1150, (1), 19. Given,, a2 + a7 = 30, , fi, a + d + a + 6d = 30, , fi, 2a + 7d = 30, ...(i)(1), [ an = a + (n – 1)d], Also, given, a15 = 2a8 – 1, , fi, a + 14d = 2(a + 7d) – 1 fi a = 1 (1), Putting the value of a in (i), we get, , 2 + 7d = 30 fi d = 4, , \, a = 1, d = 4, Hence, AP is 1, 5, 9, 13, 17, ..., (1), 20. Let Sn and Sn′ be the sum of n terms of two APs. Let a,, a¢ and d, d¢ be first terms and common differences of two, APs. Then, , n, [2a + (n − 1)d ], Sn, , = 2, n, Sn′, [2a ′ + (n − 1)d ′ ], 2,  n − 1, a+ , d,  2 , 7n + 1, =, =, 4n + 27,  n − 1, a′ + , d′,  2 , Since, , ...(i)(1), , tm, a + (m − 1)d, =, tm′, a ′ + (m − 1) d ′, , [ Let tm, t¢m be mth terms of two APs], n −1, So, replacing, by m – 1, i.e., n = 2m – 1 in, (i), 2, , , a + (m − 1)d, tm, =, (1), a ′ + (m − 1) d ′, tm′, , =, , 7(2m − 1) + 1 14m − 6, =, 4(2m − 1) + 27 8m + 23, , Thus, the ratio of their mth terms is,    14m – 6 : 8m + 23., (1), 21. Let the required numbers in AP are a – d, a, a + d, respectively., Now, a – d + a + a + d = 15 [ Sum of digits = 15], , fi, 3a = 15 fi a = 5, (1), According to question, number is, 100(a – d) + 10a + a + d, i.e. 111a – 99d, Number on reversing the digits is, 100(a + d) + 10a + a – d, i.e. 111a + 99d, Now, as per given condition in question,, (111a – 99d) – (111a + 99d) = 594, (1), , fi, d = –3, , \ Digits of number are [ 5 – (–3), 5, (5 + (–3))], = 8, 5, 2., , \ Required number is 1 11 × (5) – 99(–3), = 555 + 297 = 852., (1), 1, 2, 3, 23. Sn = 4 − + 4 − + 4 − + .... , upto n terms, n, n, n, 1, = (4 + 4 + ... + 4) − (1 + 2 + 3 + ... + n) (1), n, 1 n(n + 1), = 4n − ×, (1), n, 2, 7n − 1, =, (1), 2, 24. Given equation: 1 + 4 + 7 + 10 + ... + x = 287, Here, a = 1, d = 4 – 1 = 7 – 4 = 3, , Sn = 287, n, Bur,, Sn =, [2a + (n – 1)d], 2, n, , 287 =, [2 × 1 + (n – 1)3], 2, , ⇒, 287 × 2 = n(2 + 3n – 3), , ⇒, 574 = n(3n – 1) = 3n2 – n, , ( )( )( ), , Arithmetic Progressions, , 51

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\ 23-Nov-2021, , Amit, , Proof-3, , , ⇒ 3n2 – n – 574 = 0, We know that,, , Reader’s Sign _______________________, , (1), , −b ± b 2 − 4ac, n =, 2a, , Date __________, , 27. 39; 6981, , (5), , 30. 1 : 3, 5 : 49, , (5), , 31. 970, , (5), , 32. 249 : 447, , (5), , 33. List of 3-digit number leaving remainder 3 when divided, by 4, are 103, 107, 111, ..., 999., , =, , −( −1) ± ( −1) 2 − 4 × 3 × ( −574), 2×3, , Now,, , =, , 1 ± 1 + 6888, 1 ± 6889, 1 ± 83, =, =, (1), 6, 6, 6, , Since, number of terms is odd, so there will be only one, middle term, 225 + 1, = 113 (1), Middle term =, 2, , \, a113 = a + 112d, = 103 + 112 × 4 = 551, (1), th, There are 112 numbers before 113 term., , Either n =, , 1 ± 83, 1 − 83, or n =, 6, 6, , 84, −82, or n =, 6, 6, −41, , ⇒, n = 14 or n =, 3, , \, n = 14, n, 14, Now, Sn = (a + l) ⇒ 287 =, (1 + x), 2, 2, , ⇒ 287 = 7(1 + x) ⇒ 287 = 7 + 7x, 280, , ⇒, 7x = 280 ⇒ x =, = 40, (1), 7, 25. Let the three numbers in AP are a – d, a, a + d, Then a – b + a + a + d = 18, , ⇒, 3a = 18 ⇒ a = 6, (1), Given: (a – d) (a + d) = 5d, , ⇒, a2 – d2 = 5d ⇒ a2 = 5d + d2, , ⇒, 36 = 5d + d2, [Q a = 6](1), 2, , ⇒, d + 5d – 36 = 0, , ⇒, d2 + 9d – 4d – 36 = 0, (1), , ⇒ d(d + 9) – 4(d + 9) = 0, , ⇒, (d – 4) (d + 9) = 0, , ⇒, d – 4 = 0 or d + 9 = 0, , ⇒, d = 4 or d = –9, [Reject], , ⇒, d = 4, (1), Thus, three numbers are a – d, a, a + d, = 6 – 4, 6, 6 + 4, = 2, 6, 10, (1), 26. We know that, an = a + (n – 1)d, From the given conditions,, , m[a + (m – 1) d] = n[a + (n – 1)d], , ⇒, m[a + (md – d)] = n[a + nd – d](1), , ⇒, am + m2d – md = an+ n2d – nd, , ⇒, am – an + m2d – n2d – md + nd = 0, (1), , ⇒, a(m – n) + d(m2 – n2) – d(m – n) = 0, , ⇒ a(m – n) + (m + n) (m – n)d – (m – n)d = 0, (1), , ⇒, (m – n) [a + (m + n)d – d = 0, , ⇒, a + md + nd – d = 0, (1), , ⇒, a + (m + n – 1)d = 0, Since, m ≠ n, it is clear that (m + n)th term of the AP is zero., (1), , ⇒, , 52, , n=, , Mathematics–10, , an = 999, , 103 + (n – 1)4 = 999, , fi a + (n – 1)d = 999, fi n = 225, , , \ Sum of all terms before middle term, 112, [2 × 103 + 111 × 4], , S112 =, 2, = 36400, , (1), , (1), , , \ Sum of all terms = S225 = 123975, , \ Sum of terms after middle term, = S225 – (S112 + 551), = 87024, , (1), , 34. Let the first terms be a and a′ and d and d′ be their respective, common differences., n,  2a + ( n − 1) d , Sn, 2, =, n, Sn',  2a ′ + ( n − 1) d ′ , 2, , , , =, , 7n + 1, (1), 4n + 27, ,  n − 1, a+, d, 7n + 1,  2 , , ⇒, =, (1), 4n + 27,  n − 1, a′ + , d′,  2 , To get ratio of 9th terms, replacing, , ⇒, , n −1, = 8   , 2, , n = 17, , Hence,, , t9, a + 8d, 120, 24, =, =, or, t9′, a ′ + 8d ′ 95, 19, , (1), (1), (1), , 35. Let common difference be d., , ⇒, , 14, [2(10) + (n − 1)d] = 1050, 2, , , ⇒, d = 10, , a20 = a + 19d, = 10 + 19 (10) = 200, , (2), (1), (2)

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a=5, an = 45, Sn = 400, , 36. , , , , n, (5 + 45), 2, , 50n, , n, also, an, , 5 + 15d, , 15d, , ⇒, , =, , = 400, , (2), , =, =, =, =, =, , (1), , 800, 16, 45, 45, 40, , 8, , d = (2), 3, 37. AP is 24, 21, 18, ..., Here, a = 24 and d = 21 – 24 = 18 – 21 = –3, (1), Let the sum of n terms of the AP be 78., n, , Sn =, [2a + (n – 1)d], 2, , fi, , 78 =, , , fi, 78 × 2, , fi, 156, , fi, 156, , fi 3n2 – 49n + 156, We know that, , , n=, , −b ±, , =, =, =, =, , n, [2 × 24 + (n – 1)(–3)], 2, n[48 – 3n + 1], n(49 – 3n), 49n – 3n2, 0, , b 2 − 4ac, 2a, , =, =, , 49 ±, , (49) 2 − 4 × 3 × 156, 2×3, , 49 ±, , 2401 − 1872, 6, , 49 ± 529, 6, , =, , 49 ± 23, (1), 6, , Either n =, , 49 + 23, 49 − 23, or n =, 6, 6, , , , n=, , 72, 26 13, or n =, =, 6, 6, 3, , , , n = 12 or n = 4, , Thus,, , 1, 3, , n = 12., , (2), , Case Study Based Questions, , (1), , I. 1. (b) 1, 2, 3, 4, ..., 12, , 2. (c) 156, 3., , 4. (b) 312, 5., II. 1. (a) ` 3900, 2., , 3. (c) ` 44500, 4., 5. (b) 10 : 49, , (a), (a), (b), (a), , 234, 468, ` 73500, ` 4900, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. Finding nth term of given AP., , 2. Finding nth term of given AP from the end., , 3. Finding n when nth term of an AP is given., , 4. Finding AP or nth term or both when its two terms are given., 5. Finding sum of first n terms of an AP., 6. Finding number of terms when sum of first n terms and AP are given., , IMPORTANT FORMULAE, •• The nth term of an AP, an = a + (n – 1) d, •• The nth term of an AP from end, an = l – (n – 1) d, •• Sum of finite terms of an AP, n, n, , Sn = [2a + (n – 1) d] or Sn = (a + an), 2, 2, •• If there are only n terms in an AP, then, , an = l, the last term, n, , Sn = (a + l), 2, Note:, an = Sn – Sn – 1, where, a = first term, n = number of terms, d = common difference, and an = nth term, l = last term., , Arithmetic Progressions, , 53

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , QUICK REVISION NOTES, •• A succession of numbers or terms formed and arranged according to some rule is called a sequence /, progression., •• An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed, number to the preceding term except the first term. This fixed number is called the common difference, of the AP, which can be positive, negative or zero., •• The sequence a1, a2, a3, a4,..., an is an AP of n terms with common difference ‘d’ iff, , an – an – 1 = an – 1 – an –2 = ..., = a2 – a1, = d, •• General term (nth term) of an AP (from the beginning) is given by, , an = a + (n – 1)d., •• Three numbers a, b, c are in AP if and only if b – a = c – b or 2b = c + a., c+a, fi, b=, 2, Note that b is known as arithmetic mean of a and c., •• If ‘l’ is the last term of an AP, then nth term from the end of an AP, = l + (n – 1)(–d), = l – (n – 1)d., •• Let a be the 1st term, ‘d’ the common difference and ‘n’ the number of terms of an AP, then Sn, the sum, of ‘n’ terms of an AP is given by, n, , Sn = [2a + (n – 1)d], 2, n, Also, Sn = (a + an), 2, where, an is the last term., , COMMON ERRORS, Errors, (i) Finding incorrectly the common difference (d) when (i), the numbers in AP is in descending order or if the, succeeding term is smaller., (ii) When ‘d’ is negative in questions to find ‘n’, (ii), multiplying (n – 1) incorrectly by positive value of, ‘d’., (iii) Incorrectly differentiating an and Sn., (iii), , Corrections, For finding the common difference, we should subtract the, preceding term from the succeeding term, even if the numbers, in AP is in descending order or the succeeding term is smaller., Put ‘d’ with negative sign in bracket so that multiplication, will be taken up in the next step., , Sn represents the sum of n terms whereas an represents nth, term., th, (iv) Trying incorrectly to find n term when sum to first (iv) The nth term of an AP is the difference of the sum to first n, n terms and the sum to first (n – 1) terms are given., terms and the sum to first (n – 1) terms of it., , i.e.,, an = Sn – Sn – 1, , qqq, , 54, , Mathematics–10

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3, , Circles, , Topic Covered, 1. Tangent to a Circle and Number of Tangents from a Point on a Circle, , 1. Tangent to a Circle and Number of Tangents from a Point on a Circle, Understanding Tangent to a Circle, A tangent to a circle is a line that intersects or touches the circle at only one point. The point of, intersection is called the point of contact or the point of tangency., In the given figure, line AB is a tangent and point C is the point of contact., Theorem 1. The tangent at any point of a circle is perpendicular to the radius through the point of, contact., , A, C, O, , B, , Given: A circle (O, r) and a tangent AB to the circle at the point P., To prove: OP ^ AB., Proof: Take a point ‘Q’ (say) other than P on AB. Join OQ. Suppose OQ intersects the circle at R., OP = OR, ...(i) (Radii of the same circle), OQ = OR + RQ, ⇒, OQ > OR, ⇒, OQ > OP, [Using (i)], ⇒, OP < OQ, Thus, OP is shorter than any other line segments joining O to any point of AB, other than P., Therefore, OP is the shortest distance between the point O and the line AB., We know that the shortest distance between a point and line is the perpendicular distance., \, OP ^ AB, Note: 1. Converse of this theorem is also true, i.e., a line drawn through the end point of a radius and perpendicular to it, is a, tangent to the circle., 2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point., , Number of Tangents from a Point on a Circle, The number of tangents that can be drawn to a circle from a point depends upon the position of the point with respect to, the circle., There are three cases possible., Case 1: There is no tangent to a circle passing through a point lying inside the circle (see Fig. 1)., Case 2: There is one and only one tangent to a circle passing through a point lying on the circle (see Fig. 2)., Case 3: There are exactly two tangents to a circle passing through a point lying outside the circle (see Fig. 3)., T1, , P, , P, , P, , P, , P, , Fig. 2, , T1, , P, , P, , T2, , Fig. 1, , T1, , T2, , P, , T2, , Fig. 3, , 55, , P

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Length of the Tangent, The length of the segment of the tangent from an external point (P in Fig. 3 above) and the point of contact with the circle, (T1 or T2 in Fig. 3 above) is called the length of the tangent from the point P to the circle., Theorem 2. The lengths of tangents drawn from an external point to a circle are equal., [CBSE 2018, AI 2017], Given: A circle with centre O. R is an external point. Two tangents RP and RQ are drawn from point R to the circle., To prove: RQ = RP, Proof: Consider ∆RQO and ∆RPO., Q, Here,, OQ = OP , (Radii of a circle), OR = OR, (Common), O, R, ∠OQR = ∠OPR = 90º, , (Tangents are perpendiculars to the radius of a circle at the point of contact.), \, ∆RQO ≅ ∆RPO, (RHS congruency), P, ⇒, RQ = RP, (CPCT), Note: ∠ORQ = ∠ORP. Therefore, OR bisects ∠QRP. Thus, the centre lies on the bisector of the angle between the two tangents., P, Theorem 3. If two tangents are drawn to a circle from an external point, then, (i) they subtend equal angles at the centre, (ii) they are equally inclined to the line segment joining the centre to that point., A, O, Given: A circle with centre O and a point A outside it., AP and AQ are tangents drawn to the circle from point A., Q, To prove: ∠AOP = ∠AOQ and ∠OAP = ∠OAQ, Proof: In DOAP and DOAQ, we have, OP = OQ , (Radii of the same circle), AP = AQ, (Tangents from an external point are equal), OA = OA, (Common), \, DOAP @ DOAQ, (By SSS congruency), ⇒, ∠AOP = ∠AOQ(CPCT), and, ∠OAP = ∠OAQ(CPCT), D, Angles of Alternate Segment of Circle, B, Look at the figure drawn along side., O, Here, chord AB of any circle, is drawn from point of contact A of the tangent PQ which makes, C, ∠BAP and ∠BAQ with PQ. The chord AB divides the circle into two segments ADB and ACB., The segments ADB and ACB of the circle are alternate segments of ∠BAQ and ∠BAP., A, P, Q, Example 1. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then, find the length of each tangent., [CBSE Standard SP 2020-21] [Imp.], Q, 3 cm, 60°, , O, , P, , 3 cm, , Solution. In DQPO and DRPO,, OQ = OR, OP = OP, QP = RP, \, DQPO ≅ DRPO, , 60°, –QPO = –RPO =, = 30°, 2, –OQP = 90°, OQ, 3, 1, =, \, tan 30° =, fi, QP, QP, 3, Thus, QP = RP = 3 3 cm, , R, , (Radii of same circle), (Common), (Tangents), (By SSS congruency), , fi, In DQPO,, , 56, , Mathematics–10, , (Tangent is perpendicular at the point of contact), fi, , QP = 3 3 cm

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Example 2. In the given figure, PA and PB are tangents to the circle drawn from an external point P., CD is the third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of DPCD., A, C, , Q, , P, , D, B, , Solution. Since PA and PB are tangents from same external point P., \, PA = PB = 15 cm, Now,, Perimeter of DPCD = PC + CD + DP = PC + CQ + QD + DP = PC + CA + DB + DP, = PA + PB = 15 cm + 15 cm = 30 cm, Example 3. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which, touches the smaller circle.[NCERT] [Imp.], Solution. Let O be the common centre of two concentric circles and let AB be a chord of larger circle touching the smaller, circle at P. Join OP., Since OP is the radius of the smaller circle and AB is a tangent to this circle at P., \, OP ^ AB, O, Since perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord., fi, AP = PB, Now, in DAOP, right-angled at P,, A, P, B, OA2 = AP2 + OP2 fi 52 = AP2 + 32 fi 25 – 9 = AP2, fi, AP2 = 16 fi AP = 4, Now, AB = 2 × AP = 2 × 4 = 8, [ AP = PB], Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm., Example 4. Prove that the parallelogram circumscribing a circle is a rhombus., [NCERT] [CBSE 2014] [Imp.], Solution. Given: ABCD is a parallelogram., R, The sides AB, BC, CD and DA touch a given circle at the point P, Q, R and S respectively., Proof: Since lengths of the two tangents from an external point to a circle are equal., S, \, AP = AS ...(i) [Tangents from external point A], Q, BP = BQ ...(ii) [Tangents from external point B], CR = CQ ...(iii) [Tangents from external point C], and, DR = DS ...(iv) [Tangents from external point D], P, Now, AB + CD = AP + BP + CR + DR = AS + BQ + CQ + DS, [Using (i), (ii), (iii) and (iv)], = (AS + DS) + (BQ + CQ) fi AB + CD = AD + BC, But, AB = CD and AD = BC, [Q Opposite sides of a parallelogram are equal], \, AB + AB = AD + AD fi 2AB = 2AD fi AB = AD., \, AB = BC = CD = AD, Since all the four sides of a parallelogram ABCD are equal., \ ABCD is a rhombus., Example 5. In the figure given, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB, with point of contact C intersecting XY at A and X′Y′ at B. Prove that –AOB = 90°., , [NCERT] [AI 2017] [CBSE SP 2018] [Imp.], Solution. Given: A circle C(O, r) in which XY and X′Y′ are parallel tangents. AB is another tangent intersecting XY, and X′Y′ at A and B respectively., X, A Y, P, To prove: ∠AOB = 90°., Prove: Join O and C., 1, In ∆AOP and ∆AOC,, OP = OC, (Radii of a circle), O 2, OA = OA, (Common), 3, C, r 4, ∠OPA = ∠OCA, (Each 90°), \, ∆AOP ≅ ∆AOC, (By RHS Congruency), X', Q, B, Y', ⇒, ∠1 = ∠2, ...(i) (CPCT), Similarly,, ∆BOQ ≅ ∆BOC, (By RHS Congruency), ⇒, ∠3 = ∠4, ...(ii) (CPCT), , Circles, , 57

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Now,, ∠1 + ∠2 + ∠3 + ∠4 = 180°, (Angles in a straight line), ⇒, 2∠2 + 2∠3 = 180°, [From (i) and (ii)], ⇒, 2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ⇒ ∠AOB = 90°, Hence Proved., Example 6. In the given figure, OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle., [CBSE 2009], A, , O, , P, , B, , Solution. Join OP and let it meets the circle at point Q., A, Since, OP = 2r, (Diameter of the circle), ⇒, OQ = QP = r, O, P, Q, Consider ∆AOP in which OA ⊥ AP and OP is the hypotenuse., ∴, OQ = AQ = OA, B, , (Mid-point of the hypotenuse is equidistant from the vertices), ⇒ OAQ is an equilateral triangle., ⇒, ∠AOQ = 60°, (Each angle of an equilateral triangle is 60°), Consider right-angled triangle OAP., ∠AOQ = 60°, (Proved above), ∠OAP = 90° ⇒ ∠APO = 30°, ∠APB = 2∠APO = 2 × 30° = 60°, Also, PA = PB, (Tangents to a circle from an external point are equal.), ⇒, ∠PAB = ∠PBA, (Angles opposite to equal sides in ∆PAB), In ∆ABP,, ∠APB = 60°, 180° − 60°, = 60°, ⇒, ∠PAB = ∠PBA =, 2, ⇒ Each angle of DPAB is 60°, ⇒ PAB is an equilateral triangle. Hence Proved., Example 7. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC into which BC is divided by the point of contact D are of, lengths 8 cm and 6 cm respectively. Find the sides AB and AC., [AI 2014], Solution. Let circle touches AC at point E and AB at point F. Join OA, OB, OC, OE and OF., Since, the tangents to a circle from an exterior point are equal, therefore, BD = BF = 8 cm, CD = CE = 6 cm, and, AE = AF = x cm (let), ⇒, AC = (x + 6) cm, AB = (x + 8) cm, and, BC = (6 + 8) cm = 14 cm, AB + BC + AC, x + 8 + 14 + x + 6, Now,, s=, =, 2, 2, = (x + 14) cm, Also, a = BC = 14 cm,, b = AC = (x + 6) cm, and, c = AB = (x + 8) cm, Using Heron’s formula, we get, Area of ∆ABC = s ( s − a) ( s − b) ( s − c), =, , 58, , Mathematics–10, , ( x + 14) ( x + 14 − 14) ( x + 14 − x − 6) ( x + 14 − x − 8), , 4 cm, , x, , x, , 4 cm

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=, , ( x + 14) · x · 8 · 6 = 48 · x( x + 14) ...(i), , Also,, , Area of ABC = Area of ∆OBC + Area of ∆OCA + Area of ∆OAB, 1, 1, 1, =, BC × OD +, AC × OE +, AB × OF, 2, 2, 2, 1, 1, 1, × 14 × 4 +, × (x + 6) × 4 + (x + 8) × 4, 2, 2, 2, 2, = (28 + 2x + 12 + 2x + 16) cm = (4x + 56) cm2...(ii), From (i) and (ii), we get, =, , 48 x( x + 14) = 4x + 56 = 4(x + 14), 48x(x + 14) = 16(x + 14)2, , ⇒, , (Squaring both sides), , 2, , 3x(x + 14) = (x + 14), , ⇒, , ⇒  3x(x + 14) – (x + 14)2 = 0, (x + 14) (3x – x – 14) = 0, , ⇒, ⇒  , , (x + 14) (2x – 14) = 0, x + 14 = 0, , ⇒, , x = –14, , ⇒, , or, , 2x – 14 = 0, or x = 7., , As the length of any side of triangle can never be negative, x ≠ –14., x=7, , \, , ⇒ AB = (x + 8) = 15 cm, and AC = x + 6 = 13 cm, , Exercise 3.1, I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) In the given figure, if TP and TQ are tangents to a circle with centre O, so that ∠POQ = 110°, then ∠PTQ is, , , (a) 110°, (b) 90°, (c) 80°, (d) 70°, (2) In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre, O. If ∠QPR = 90°, then length of PQ is, [CBSE Standard 2020], , , (a) 3 cm, (b) 4 cm, (c) 2 cm, (d) 2 2 cm, (3) From an external point A, two tangents AB and AC are drawn to the circle with centre O. Then OA is the perpendicular, bisector of, (a) BC, (b) AB, (c) AC, (d) none of these, (4) If the radii of two concentric circles are 6 cm and 10 cm, the length of chord of the larger circle which is tangent to, other is, (a) 14 cm, (b) 16 cm, (c) 18 cm, (d) 12 cm, , Circles, , 59

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , (5) In the given figure, the length PB is equal to, (a) 2 cm, (b) 7 cm, (c) 6 cm, (d) 4 cm, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): When two tangents are drawn to a circle from an external point, they subtend equal angles at the, centre., , Reason (R): A parallelogram circumscribing a circle is a rhombus., (2) Assertion (A): If in a cyclic quadrilateral, one angle is 40°, then the opposite angle is 140°., , Reason (R): Sum of opposite angles in a cyclic quadrilateral is equal to 360°., (3) Assertion (A): PA and PB are two tangents to a circle with centre O such that ∠AOB = 110°, then ∠APB = 90°., , Reason (R): The length of two tangents drawn from an external point are equal., 3. Answer the following:, (1) 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., [AI 2019, Foreign 2015], (2) From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB., , [Delhi 2016], (3) 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 Standard SP 2019-20, AI 2017], (4) 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., [CBSE Standard SP 2019-20], (5) If PQ = 28 cm, then find the perimeter, [CBSE Standard SP 2020-21], of DPLM., , (6) PQ is a tangent to a circle with centre O at point P. If DOPQ is an isosceles triangle, then find ∠OQP., [CBSE Standard SP 2020-21], II. Short Answer Type Questions - I, [2 Marks], 4. Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre., [CBSE 2014], T, 5. In the given figure, from an external point, [CBSE 2016], P, two tangents PT and PS are drawn to a, r, circle with centre O and radius r. If OP =, Q, 2r, show that –OTS = –OST = 30°., O, P, r, , 6. In the given figure, AP and BP are tangents, to a circle with centre O, such that AP =, 5 cm and –APB = 60°. Find the length of, chord AB., , 60°, , A, , B, O, , 60, , Mathematics–10, , [CBSE 2016], , P, , S

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7. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, so that OQ = 12 cm., Find the length of PQ., [NCERT], 8. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the, radius of the circle., [NCERT] [Imp.], 9. Prove that the tangents at the extremities of a chord of a circle make equal angles with the chord., [NCERT Exemplar][AI 2017], 10. In the given figure, common tangents AB, [AI 2015], and CD to two circles with centres O1 and, A, O2 intersect at E. Prove that AB = CD., D, , E, O1, , O2, B, C, , 11. If from an external point P of a circle, with centre O, two tangents PQ and PR, are drawn such that –QPR = 120°, prove, that 2PQ = PO., , [Delhi 2014, Foreign 2016], , Q, , 120°, P, , O, , R, , 12. In the given figure, a circle inscribed in, DABC, touches its sides BC, CA and, AB at the points P, Q and R respectively., If AB = AC, then prove that BP = CP., , 13. In the given figure, DABC is circumscribing, a circle. Find the length of BC., , 14. A circle is inscribed in a ∆ABC having, sides 8 cm, 10 cm and 12 cm as shown in, the following figure. Find AD, BE and CF., , [Foreign 2013], , A, , [CBSE 2009], , [CBSE 2009, 2012, 2013] [Imp.], , 15. PA and PB are tangents from P to the circle with centre O. At point M, a tangent is drawn cutting PA at K and PB at N., Prove that KN = AK + BN., 16. PC is a tangent to the circle at C. AOB is, [CBSE 2013], the diameter which when extended meets, the tangent at P. Find ∠CBA, ∠AOC and, ∠BCO, if ∠PCA = 110°., , 17. In the given figure, O is the centre of the, circle. Determine –AQB and –AMB, if, PA and PB are tangents and –APB = 75°, , [CBSE 2012], , Circles, , 61

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\ 23-Nov-2021, , Amit, , Proof-3, , 18. In the given figure, AB is the diameter of a circle, with centre O and AT is a tangent. If ∠AOQ = 58°,, find ∠ATQ., , Reader’s Sign _______________________, , Date __________, , [Delhi 2016], , 19. 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 the line segment PQ., [Delhi 2016], 20. In the given figure, two tangents RQ and, [AI 2016], RP are drawn from an external point R to, the circle with centre O. If ∠PRQ = 120°,, then prove that OR = PR + RQ., , 21. In the given 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., , [Foreign 2016], , 22. 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 = QD = 3 cm, then, find PC + PD., , [Delhi 2017], , 23. A circle touches all the four sides of a quadrilateral ABCD. Prove that, , AB + CD = BC + DA, [CBSE Standard 2020, AI 2017, CBSE 2016, NCERT], 24. In the given figure, find the perimeter of DABC,, [CBSE Standard 2020], if AP = 12 cm., , 25. In the given figure, a circle touches all the, four sides of a quadrilateral ABCD in which, AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD., , 26. In the given figure, two tangents TP and TQ are, drawn to a circle with centre O from an external, point T. Prove that ∠PTQ = 2 ∠OPQ., , 27. Using the given figure, answer the following questions:, (i) Name the alternate segment of circle of ∠BAQ, (ii) Name the alternate segment of circle of ∠DAP., (iii) If B is joined with C then ∠ACB is equal to which angle?, (iv) ∠ABD and ∠ADB is equal to which angles?, , 62, , Mathematics–10, , [Standard 2020]

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27 cm, , C, , [CBSE Standard SP 2020-21], , R, D, S, , 38, cm, , 28. 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., , 10 cm O, Q, , A, , 29. A quadrilateral ABCD is drawn, to circumscribe a circle. Prove, that AB + CD = AD + BC., , B, , P, x cm, D, , R, , Q, , S, , A, , C, , P, , B, , III. Short Answer Type Questions - II, [3 Marks], 30. The incircle of a ∆ABC touches the sides AB, BC and CA at P, Q, R respectively. Show that, 1, AP + BQ + CR = PB + QC + RA = (Perimeter of ∆ABC), [Imp.], 2, 31. If a chord AB of the larger of the two concentric circles is a tangent to the smaller circle at P, prove that PA = PB., 32. In the given figure, PQ is a chord of, [AI 2019], length 8 cm of a circle of radius 5 cm. The, tangents at P and Q intersect at a point T., Find the length of TP., 33. The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the, smaller circle, touching it at D and intersecting the larger circle at P on producing. Find the length of AP., [CBSE SP 2018-19], 34. A circle is touching the side BC of ∆ABC at X and touching AB and AC produced at P and Q respectively. Prove that, 1, AP = AQ = (Perimeter of ∆ABC). Given AP = 10 cm, find the perimeter of ∆ABC., [CBSE 2001, 2002], 2, 35. In the given figure, two circles with centres, X and Y touch externally at P. If tangents, AT and BT meet the common tangent at, T, then prove that AT = BT., , [CBSE 2013] [Imp.], , 36. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle, subtended by the line segment joining the points of contact at the centre., [NCERT], 37. Prove that the tangents drawn at the ends of a diameter of a circle are parallel., [NCERT, CBSE 2014] [Imp.], 38. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point, of contact., [Foreign 2012], 39. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the, circle., [AI 2019, Delhi 2013], 40. In the given figure, a circle is inscribed in, [AI 2013], a triangle PQR with PQ = 10 cm, QR = 8, cm and PR = 12 cm. Find the lengths of, QM, RN and PL., , 41. Two concentric circles are of radii 7 cm and r cm respectively, where r > 7 cm. A chord of the larger circle, of length 48, cm, touches the smaller circle. Find the value of r., [Delhi 2009], , Circles, , 63

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 42. If d1, d2, (d2 > d1) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to, the other circle, prove that d 22 = c 2 + d12 ., [Foreign 2009] [NCERT Exemplar], 43. In a cyclic quadrilateral ABCD, diagonal, AC bisects the angle C (see the given, figure). Then prove that diagonal BD is, parallel to the tangent PQ of a circle which, passes through the points A, B, C and D., , [Delhi 2020], , D, P, C, O, A, B, Q, , IV. Long Answer Type Questions, [5 Marks], 44. Two tangents PA and PB are drawn to a circle with centre O, such that ∠APB = 120°. Prove that OP = 2AP., [CBSE 2016], 45. Tangents AP and AQ are drawn to circle with centre O from an external point A. Prove that ∠PAQ = 2∠OPQ., [CBSE 2017, 2013, 2012, 2011, 2009] [Imp.], 46. A circle touches the side BC of a DABC, [NCERT Exemplar], at P and AB and AC when produced at Q, [CBSE 2012, 2011], and R respectively as shown in the figure., 1, Show that AQ = (Perimeter of DABC), 2, 1, or show that AQ =, (BC + CA + AB)., 2, C, A, 47. In two concentric circles, prove that all, [CBSE 2015], chords of the outer circle which touch the, inner circle are of equal length (see figure)., N, , M, O, , B, , 48. In the given figure, PQ is a chord of, length 8 cm of a circle of radius 5 cm. The, tangents at P and Q intersect at a point T., Find the lengths of TP and TQ., , D, , [Foreign 2015], , 49. If a, b, c, are the sides of a right-angled triangle, where c is the hypotenuse, then prove that the radius r of the circle which, , a+b–c, ., [NCERT Exemplar; CBSE 2012], 2, 50. Two concentric circles of radii 13 cm and 12 cm, are given. Find the length of chord of the larger circle which touches, smaller circle., [CBSE 2012, 2011], 51. In the given figure, PA and PB are, [CBSE 2011], two tangents drawn from an external, point P to a circle with centre O., Prove that OP is the right bisector, of line segment AB., A, 52. In the given figure, common, [NCERT Exemplar], D, tangents AB and CD of two circles, with centres O and O′ intersect at, O, O′, E, E. Prove that the points O, E and O′, B, are collinear., touches the sides of the triangle is given by r =, , C, , 53. If a number of circles pass through the end points P and Q of a line segment PQ, then show that their centres lie on the, perpendicular bisector of PQ., [NCERT Exemplar], , 64, , Mathematics–10

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54. In the given figure, from an external point P, a tangent PT and, a line segment PAB are drawn to a circle with centre O. ON is, perpendicular on the chord AB. Prove that, (a) PA · PB = PN2 – AN2, (b) PN2 – AN2 = OP2 – OT2, (c) PA · PB = PT2, , [NCERT, Exemplar], , T, , O, , N, , B, , A, , P, , Case Study Based Questions, I. A Ferris Wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with, multiple passenger-carrying 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., , 1. In the given figure, ∠ROQ is, (a) 60°, (b) 100°, (c) 150°, 2. ∠RQP equals, (a) 75°, (b) 60°, (c) 30°, 3. ∠RSQ equals, (a) 60°, (b) 75°, (c) 100°, 4. ∠ORP equals, (a) 90°, (b) 75°, (c) 100°, 5. If ∠P = 60° and OQ = OR = 5 m, then the length of PQ is, (a) 5 m, , (d) 90°, (d) 90°, (d) 30°, (d) 60°, , (b) 12 m, , (c) 5 3 m, (d) None of these, II. Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff . The logo design is, given as in the figure and he is working on the fonts and different colours according to the theme. In the 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, , F, A, , 1. The length of AD is, (a) 7 cm, (b) 8 cm, (c) 5 cm, 2. The length of BE is, (a) 8 cm, (b) 5 cm, (c) 2 cm, 3. The length of CF is, (a) 20 cm, (b) 5 cm, (c) 2 cm, 4. If the radius of the circle is 4 cm, then the area of triangle OAB is, (a) 20 sq cm, (b) 36 sq cm, (c) 24 sq cm, 5. The area of triangle ABC is, (a) 50 sq cm, (b) 60 sq cm, (c) 100 sq cm, , O, D, , E, B, , (d) 9 cm, (d) 9 cm, (d) 3 cm, (d) 48 sq cm, (d) 90 sq cm, , Circles, , 65

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Answers and Hints, 1. (1), (3), (5), 2. (1), , (d) 70°, (1) (2) (b) 4 cm, (1), (a) BC, (1) (4) (b) 16 cm, (1), (d) 4 cm, (1), (b) Both assertion (A) and reason (R) are true but, reason (R) is not the correct explanation of, assertion (A). , (1), (2) (c) Assertion (A) is true but reason (R) is false. (1), (3) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) , AC = BC, , ⇒, AB = 2AC, Now in DOCA,, , AO2 = OC2 + AC2, , ⇒, a2 = b2 + AC2, a 2 − b2, , , ⇒, , AC =, , , \, , AB = 2 a 2 − b 2 = 2AC, , 2, 2, Thus, length of chord = 2 a − b (1), (2) Given,, ∠PAB = 50°, ∠PAB + ∠OAB = 90°, , [Q an angle between radius, OA and tangent PA is 90°], , , , ⇒ 50° + ∠OAB = 90°, , ⇒, ∠OAB = 90° – 50° = 40°, , ⇒, ∠OBA = 40°, [Q OA = OB], Now in DAOB, we have, , ∠AOB + ∠ABO + ∠BAO = 180°, , [Q sum of angles in, a triangle is 180°], , ⇒ ∠AOB + 40° + 40° = 180°, , ⇒, ∠AOB = 180° – 80°, = 100°, (1), (3), , , OB, = sin 30°, OP, r, 1, , ⇒, =, OP, 2, , ∴, OP = 2r(1), (4) Length of chord, In DOBP,, , = 2 × 52 − 42, = 2 × 3 cm = 6 cm, (5) , PQ = PT, , PL + LQ = PM + MT, , PL + LN = PM + MN, Perimeter (DPLM) = PL + LM + PM, , = PL + LN + MN + PM, , = 2(PL + LN), , 66, , Mathematics–10, , (1), , (½), , , , (6) In ΔOPQ, , = 2(PL + LQ), = 2 × 28 = 56 cm, , (½), , ∠P + ∠Q + ∠O = 180°, , 2∠Q + ∠P = 180°, (½), , 2∠Q + 90° = 180°, , 2∠Q = 90°, , ∠Q = 45°, (½), 4. Construction: Join OA and OB., , Proof: As we know, OB is perpendicular to PQ., , [Tangent is perpendicular to radius, at the point of contact.], , , Now, given, PQ || RS, , ⇒ BO (produced to RS) is perpendicular to RS. ...(i), [A line perpendicular to one of the two parallel lines is, perpendicular to other line also], Also, OA is perpendicular to RS., , [Q Tangent perpendicular to radius] ...(ii) (1), From (i) and (ii), OA and OB must coincide as only one, line can be drawn perpendicular to the line from a point, outside the line., , \ AOB is straight line., , \ A, O, B are collinear., , ⇒ AB passes through O, the centre of the circle. (1), 6. 5 cm, (2), 7. Since, , PQ =, , OQ 2 − OP 2 = 122 − 52 (1), , = 144 − 25 = 119 cm(1), 8. \ Using Pythagoras theorem, we get, , OQ2 = QT2 + OT2, , ⇒, OT2 = OQ2 – QT2, = 252 – 242, = (25 – 24) (25 + 24), = 1 × 49 = 49, (1)

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 20. OR bisects ∠PRQ, , \, ∠PRO = ∠QRO = 60°, , [Q ∠PRQ = ∠ORP + ∠ORQ = 120°], , , In right DOPR,, , OP ⊥ PR, , [ radius is perpendicular to the, tangent at point of contact], , PR, = cos 60°, OR, , ⇒, OR = 2PR, Similarly, in right DOQR,, , \ cos ∠ORP =, , ...(i) (1), , QR, = cos 60° ⇒ OR = 2QR, ...(ii), OR, Adding (i) and (ii), we get, , 2OR = 2PR + 2QR, , ⇒, OR = PR + RQ, (1), 21. We know that tangent is perpendicular to radius. Hence,, , ∠OTP = ∠OQP = 90°, , ,     , (1), In quadrilateral PQOT,, , ∠QOT + ∠OTP + ∠TPQ + ∠OQP = 360°, , ⇒ ∠TOQ + ∠TPQ = 1 8 0 ° ⇒ ∠ T O Q = 1 1 0 °, Also, ∠TOQ = 2∠TRQ, , ⇒, 110º = 2∠TRQ, , ⇒, ∠TRQ = 55°, (1), 22. , PA = PC + CA = PC + CQ, , fi, 12 = PC + 3, , fi, PC = 9 cm, (1), Similarly,, PD = 9 cm, , \, PC + PD = 18 cm, (1), 23. Here, AP = AS, , BP = BQ, , CR = CQ, , DR = DS, (1), , Date __________, , 24. Since AP and AQ are two tangents drawn from common, external point A., , , , \, AP = AQ = 12 cm, (Given), Similarly, BP = BD, ...(i), and, CQ = CD, ...(ii) (1), Now perimeter of, , DABC = AB + BC + AC, = AB + BD + CD + AC, = AB + BP + CQ + AC, , [From (i) and (ii)], = AP + AQ = 12 + 12, = 24 cm, (1), 25. 3 cm, (2), 27. (i) ∠ADB, (ii) ∠ABD, (iii) ∠BAP, (iv) ∠DAP and ∠BAQ(2), 28. , ∠A = ∠OPA = ∠OSA = 90°, (½), Hence,, ∠SOP = 90°, Also, , AP = AS, Hence, OSAP is a square, , AP = AS = 10 cm, (½), , CR = CQ = 27 cm, , BQ = BC – CQ, , = 38 – 27 = 11 cm, (½), , BP = BQ=11 cm, , x = AB = AP + BP, , = 10 + 11 = 21 cm, (½), 29. To prove: AB + CD = AD + BC, , Proof: AS = AP (Length of tangents from an, , external point to a circle are equal), , BQ = BP, , CQ = CR, , DS = DR, (1),       , ,     , , AS + BQ + CQ + DS = AP + BP + CR + DR,    (AS + DS) + (BQ + CQ), , , , = (AP + BP) + (CR + DR), AD + BC = AB + CD, , 32. Given:, , PQ = 8 cm and PO = 5 cm, , Now,, , , PR = RQ, , (1), , (Perpendicular from the, centre bisects the chord), , 8, = 4 cm (1), 2, In DOPR, OR = OP 2 –PR 2 = 5 2 – 4 2, =, , , Adding (AP + PB) + (CR + RD), = (AS + SD) + (BQ + QC), , ⇒ AB + CD = AD + BC, , 68, , Mathematics–10, , =, Let ∠POR be q., (1), , In DPOR, tan q =, , 25 – 16 = 9 = 3 cm, PR 4, = (1), RO 3

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We know, OP ⊥ TP as point of contact of a tangent is, perpendicular to the line from the centre., OP, 4, 5, 15, In DOTP, tan q =, , & =, & TP =, TP, 3 TP, 4, , ⇒, TP = 3.75 cm, (1), 33. , –APB = 90º, (Angle in semi-circle) (½), , –ODB = 90º, (Radius is perpendicular, , to tangent), , and, ∠AOP, , ⇒, ∠APB, and, ∠AOB, But, ∠AOP, , ⇒, 2∠AOP, , ⇒, ∠AOB, , ⇒ ∠AOB + ∠APB, , =, =, =, =, =, =, =, , ∠BOP, 2 ∠OPA, 2 ∠AOP(1), 90° – ∠OPA, 180° – 2∠OPA, 180° – ∠APB, 180°., (1), , 37. In the figure, we have:, PQ is diameter of the given circle and O is its centre., Let tangents AB and CD be drawn at the end points of the, diameter PQ., ,     , , (1), , , , DABP ~ DOBD, AB, AP, 26, AP, , ⇒, =, ⇒, =, OB, OD, 13, 8, , ⇒, AP = 16 cm, (1½), 34. In the given figure,, , AP = AQ, BP = BX and CX = CQ, , ⇒ AB + BX = AC + CX, ...(i) (1), , ∴ Perimeter of, , ∆ABC = AB + BC + CA, = AB + (BX + XC) + CA, , Q, , , = (AB + BX) + (XC + CA), = 2(AB + BX), using (i).(1), 1, Hence, AP =, × (perimeter of DABC) = AQ., 2, If AP = 10 cm, then perimeter of ∆ABC = 20 cm., (1), 36. In right D OAP and right D OBP, we have, , PA = PB, [Tangents to circle from, , an external point P], , OA = OB, [Radii of the same circle], , OP = OP, [Common], , \ By SSS congruency,, , DOAP @ DOBP(1), , , Since the tangent at a point to a circle is perpendicular to, the radius through the point., (1), , \, PQ ⊥ AB ⇒ ∠APQ = 90°, and, PQ ⊥ CD ⇒ ∠PQD = 90°, (1), , ⇒, ∠APQ = ∠PQD, But they form a pair of alternate angles., , \, AB || CD., (1), 38. Given: Let O be the centre of two concentric circles C1, and C2., Let AB be the chord of larger circle C2 which is a tangent, to the smaller circle C1 at D., , , , To prove: Now we have to prove that the chord AB is, bisected at D that is AD = BD., (1), , Construction: Join OD., , Proof: Now since OD is the radius of the circle C1 and, AB is the tangent to the circle C1 at D., So,, , OD ⊥ AB, [Radius of the circle is perpendicular, to tangent at any point of contact] (1), , Since AB is the chord of the circle C2 and OD ⊥ AB., , \, , , \ Their corresponding parts are equal., , \, ∠OPA = ∠OPB, , AD = DB, [Perpendicular drawn from the centre to, the chord always bisects the chord] (1), , Circles, , 69

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 39. Given: ABCD is a quadrilateral circumscribing the circle, with centre O touching it at P, Q, R, S., , , , To prove:, ∠AOB + ∠DOC = 180°, ∠AOD + ∠BOC = 180°, , Construction: Join AO, PO, BO, QO, CO, RO, DO, SO., (1), , Proof: In DAOS and DAOP,, , AO = AO, (Common), , AS = AP, , (Tangents from external point), , OS = OP (Radii of same circle), By SSS congruence rule,, , DAOS ≅ DAOP, , ∠1 = ∠2, ...(i) (by CPCT) (1), Similarly,, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8 ...(ii), Now,, , ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°, , [Q ASP of quadrilateral], , ⇒ ∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360°, , [By (i) and (ii)], , ⇒2[∠2 + ∠3 + ∠6 + ∠7] = 360°, , ∠AOB + ∠COD = 180°, Similarly,, ∠AOD + ∠BOC = 180°, (1), 40. Let, QM = x = QL, , MR = y = RN, and, PL = z = PN, Now, PQ = 10 cm,, , QR = 8 cm,, , PR = 12 cm, , ⇒, x + y = 8, y + z = 12, z + x = 10, (1), , ⇒ 2x + 2y + 2z = 8 + 12 + 10 = 30, , ⇒, x + y + z = 15 ⇒ 8 + z = 15, , ⇒, z = 7, (1), , ⇒, x + 12 = 15 ⇒ x = 3, , ⇒, y + 10 = 15 ⇒ y = 5, Hence, QM = 3 cm, RN = 5 cm and PL = 7 cm. (1), 41. Given:, OP = 7 cm, OA = r cm, , AB = 48 cm, Now, OP ⊥ AB, (as radius makes an angle of 90° with, the tangent at point of contact), Also,, AP = PB, (Perpendicular drawn from centre, to the chord bisects the chord), So,, AP = 24 cm, (1), , 70, , Mathematics–10, , , In DOPA, ∠P = 90°, By Pythagoras theorem in DOPA,, , OA2 = AP2 + OP2, , r2 = 242 + 72, = 576 + 49 = 625, , ⇒, r = 25 cm, 42. Q Diameter of bigger circle = d2, , (1), , (1), , , 1, d 2 = OB, 2, Diameter of smaller circle = d1(1), 1, So, radius of smaller circle = d1 = OA, 2, c, , AB =, 2, In right DOAB,, , ∠A = 90° [Q Radius is perpendicular to the, tangent at point of contact] (1), By Pythagoras theorem, , OB2 = AB2 + OA2, So, radius of bigger circle =, , 2, , 2, 1 , 1 , 1 , , ⇒  d 2  =  c +  d1 , 2 , 2 , 2 , 1 2, 1, 1, d 2 = c 2 + d12, , ⇒, 4, 4, 4, , , ⇒, , 2, , d 22 = c 2 + d12 (1), , 46. We have BQ = BP, CP = CR and AQ = AR  , , , Now, 2AQ =, =, =, =, =, , (1), , AQ + AR, (AB + BQ) + (AC + CR), (1), AB + BP + AC + CP, (BP + CP) + AC + AB, (1), BC + CA + AB, (1), 1, i.e., AQ = (BC + CA + AB) (1), 2, 48. Join OT intersecting PQ at R., OT bisects ∠PTQ, , \, ∠PTO = ∠QTO, , \, ∠PTR = ∠QTR...(i) (1)

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In DPTR and DQTR,, , PT = QT, , , , \, , ⇒, , RT, ∠PTR, DPTR, PR, , =, =, ≅, =, , [Length of tangents drawn from, common external point are equal], RT, [common] (1), ∠QTR, [Q from (i)], DQTR, [By SAS], RQ, [Q By CPCT], ,     , , ⇒ R is mid-point of PQ, , OR ⊥ PQ, In right triangle ORP,, , , ⇒, , , (1), , OP2 = PR2 + OR2, [Q Given OP = 5 cm,, PQ = 8 cm, PR = QR = 4 cm], 25 = 16 + OR2, OR = 3 cm, , (1), , In DORQ and DOQT, , , ∠ORQ = ∠OQT, , (Each 90°), , , , ∠ROQ = ∠ROQ, , (Common), , , , DORQ ~ DOQT, , (By AA criterion), , , ⇒, , OR, RQ, =, OQ, QT, 4, 3, =, QT, 5, , (By C.P. of similar triangles), , 20, cm, 3, 20, Also, PT = QT ⇒ PT =, cm, (1), 3, 49. Let the circle touches the sides BC, CA, AB of the right, triangle ABC at D, E and F respectively, where BC = a,, CA = b and AB = c (see figure). Then AE = AF and BD, = BF. Also CE = CD = r.(2), , ⇒, , ⇒ QT =, , and, , , , ⇒, , AP = BP, [Tangents from P], DPAO ≅ DPBO, (SSS congruence rule) (2), ∠1 = ∠2[CPCT], , , In DAPC and DBPC,, , ∠1 = ∠2[Proved], , AP = BP and PC = PC, (1), , \, DAPC ≅ DBPC, (SSS congruence rule), , ⇒, AC = BC, [CPCT], , and, ∠ACP = ∠BCP, [CPCT] (1), Also, ∠ACP + ∠BCP = 180°, , ⇒, ∠ACP = ∠BCP = 90°, , \ OP is the right bisector of AB., (1), 52. In DAEO and DCEO,, , OE = OE, [Common], , OA = OC, [Radii of same circle], , EA = EC [Tangents from an external point, to a circle are equal in length], \, DOEA  DOEC, , [By SSS criterion of congruence] (1), ⇒, OEA = OEC, [CPCT], \, 1 = 2, [CPCT] (1), Similarly, 5 = 6, and, 3 = 4, [Vertically opposite angles] (1), , Since sum of angles at a point = 360°, \, 1 + 2 + 3 + 4 + 5 + 6 = 360°, ⇒, 2(1 + 3 + 5) = 360°, A, D, 3, 1, 2, , O, , E, 4, , 5, 6, , O, B, , C, , ,      , (1), i.e., b – r = AF, a – r = BF, or, AB = c = AF + BF = b – r + a – r(1), a+b−c, This gives r =, .(1), 2, 50. 10 cm, (5), 51. Join OA and OB., In DPAO and DPBO,, , OA = OB, [Radii], , OP = OP, [Common], , (1), , ⇒, 1 + 3 + 5 = 180°, , ⇒, OEO = 180°, \ OEO is a straight line., Hence, O, E and O are collinear., (1), 53. Centre of any circle passing through the end points P and, Q of a line segment are equidistant from P and Q., , \, , A1P = A1Q, , , , A2P = A2Q, , , , A3P = A3Q(2), , Circles, , 71

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, ,      , (1), As we know that any point on perpendicular bisector of a, segment is equidistant from the end points of the segment., Hence, A1, A2, A3 points are the centres of circles passing, through the end points P and Q of a segment PQ or the, centres of circles lie on the perpendicular bisector of, PQ.(2), , Date __________, , 54. (a) PA · PB = (PN – AN) (PN + BN), = (PN – AN)(PN + AN) (As AN = BN), , ⇒ PA.PB = PN2 – AN2...(i) (1), (b) PN2 – AN2 = (OP2 – ON2) – AN2 (As ON ⊥ PN), = OP2 – (ON2 + AN2), (1), 2, 2, = OP – OA , (As ON ⊥ AN), , ⇒ PN2 – AN2 = OP2 – OT2(As OA = OT) ...(ii) (1), (c) From (i) and (ii), PA . PB = OP2 – OT2 = PT2(As ∠OTP = 90°) (1), Case Study Based Questions, I. 1. (c) 150°, 2. (a) 75°, , 3. (b) 75°, 4. (a) 90°, , 5. (c) 5 3 m, II. 1. (a) 7 cm, 2. (b) 5 cm, , 3. (d) 3 cm, 4. (c) 24 sq cm, , 5. (b) 60 sq cm, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. To prove and use the theorem "The tangent at any point of a circle is perpendicular to the radius through the point of, contact.", 2. To prove and use the theorem "The lengths of tangents drawn from an external point to a circle are equal.", , QUICK REVISION NOTES, •• A tangent to a circle is a line that intersects the circle at only one point., •• There is only one tangent at a point of the circle., •• The tangent to a circle is a special case of the secant., •• There is no tangent to a circle passing through a point lying inside the circle., •• There is one and only one tangent to a circle passing through a point lying on the circle., •• There are exactly two tangents to a circle through a point lying outside the circle., •• The tangent at any point of a circle is perpendicular to the radius through the point of contact., •• The lengths of tangents drawn from an external point to a circle are equal., , COMMON ERRORS, Errors, , Corrections, , (i) Interpreting incorrectly that all secants are tangents., , (i) Remember that the tangent to a circle is a special case of secant,, when the two end points of its corresponding chord coincide., , (ii) Interpreting incorrectly that many tangents can be, drawn to a circle passing through a point lying on, the circle., , (ii) There is one and only one tangent to a circle passing through, a point lying on the circle., , (iii) Distinguishing incorrectly the terms ‘inscribed’ and (iii) Understand the difference between terms ‘inscribed’ and, ‘circumscribed’., ‘circumscribed’ clearly., , , , Inscribed, , Circumscribed, , (iv) Not recognising corresponding parts of congruent (iv) Write corresponding parts of congruent triangles correctly, triangles to find the values of angles., which are equal., (v) Do not using angle sum property of triangle and, quadrilateral to find the unknown angles., , (v) Must use angle sum property of triangle and quadrilateral to, find the unknown angles when sum are given., , qqq, , 72, , Mathematics–10

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\ 23-Nov-2021, , 4, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Constructions, , Topics Covered, 1. Division of a Line Segment, 2. Construction of Tangents to a Circle, , 1. Division of a Line Segment, Division of a Line Segment in a Given Ratio, Construction 1: To divide a given line segment in a given ratio, Let us divide a line segment, say AB = 6 cm in the ratio 2 : 3., We have to locate a point P on AB such that AP : PB = 2 : 3., Steps:, (i) Draw a line segment AB = 6 cm., (ii) Draw ray AX making a suitable angle with AB., (iii) Since, AP : PB = 2 : 3 and 2 + 3 = 5; cut 5 equal segments AA1, A1A2,, A2A3, A3A4 and A4A5 on AX., (iv) Join A5 with B., (v) Through A2, draw A2P parallel to A5B by making corresponding angles, AA2P and AA5B equal., (vi) The line through A2 and parallel to A5B will meet the given line segment, AB at point P., Then P is the required point which divides AB in the ratio 2 : 3,, i.e.,, AP : PB = 2 : 3., Justification: Since, in a triangle the line drawn parallel to one side of the triangle divides other two sides proportionally, (Basic Proportionality Theorem), therefore, in DAA5B; A2P || A5B, AP, AA 2, ⇒, =, A 2 A5, PB, AP, 2, ⇒, =, PB, 3, Alternative Method:, Steps:, (i) Draw a line segment AB = 6 cm., (ii) Draw ray AX making any suitable angle with AB., (iii) Draw ray BY parallel to ray AX by making alternate angles BAX, and ABY equal., (iv) From AX, cut five equal segments, AA1, A1A2, A2A3, A3A4 and, A4A5., (v) From BY, cut five equal segments [of same size as taken in step (iv)], BB1, B1B2, B2B3, B3B4 and B4B5., (vi) Join A2 and B3., Line A2B3 cuts given line segment AB at point P., i.e.,, AP : PB = 2 : 3., Then P is the required point such that, AP : PB = 2 : 3, Justification: DAA2P and BB3P are similar by AAA, AA 2 2, AP, =, \, =, BB3, 3, BP, i.e.,, AP : BP = 2 : 3, , 73

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 1. Draw a line segment of length 7.8 cm and divide it in the ratio 5 : 8. Measure the two parts., Solution. Steps:, (i) Draw a line segment AB = 7.8 cm., A13, A, A11 12, (ii) Draw ray AX making any suitable angle with AB., A10, A9, (iii) Draw ray BY parallel to ray AX by making alternate angles BAX, A8, A7, A6, and ABY equal., A5, A, (iv) Since 5 + 8 = 13, from AX cut 13 equal segments i.e., AA1, A1A2,, A3 4, A2, A1, A2A3, ..., A12A13., (v) Also, from BY cut 13 equal segments [of the same size as taken A, P, in step (iv)] i.e., BB1, B1B2, B2B3, ..., B12B13., B, B3 2, B4, (vi) Join A5 with B8., B, B6 5, Line joining A5 and B8 cuts line segment AB at point P., B7, B8, B, \ P is the required point such that AP : PB = 5 : 8., B10 9, B11, B12, On measuring we find: AP = 3 cm and BP = 4.8 cm., B, Y, , X, , B, B1, , 13, , Exercise 4.1, I. Very Short Answer Type Questions, , [1 Mark], , 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn such that ∠BAX is an acute angle and then at, equal distances points are marked on the ray such that the minimum number of these points is, (a) 8, (b) 10, (c) 11, (d) 12, (2) To divide line segment AB in the ratio m : n, draw a ray AX so that ∠BAX is an acute angle and then mark points, on ray AX at equal distance such that the minimum number of these points is, (a) m + n, (b) m – n, (c) m × n, (d) m ÷ n, (3) To divide a line segment AB in the ratio 5 : 6, we draw a ray such that ∠BAX is an acute angle, locate points A1,, A2, A3, ... at equal distances on the ray AX and join point B to, (a) A12, (b) A11, (c) A10, (d) A9, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): By geometrical construction, it is possible to divide a line segment in the ratio 3 : 1 ., 3, , Reason (R): To divide a line segment in the ratio p : q, p and q must be positive integers., (2) Assertion (A): When a line segment is divided in the ratio 2 : 3, the number of parts in which the line is divided is 5., , Reason (R): To divide a line segment in the ratio p : q, we divide it into (p + q) parts., 3. Answer the following:, , (1) Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4., , [Delhi 2011], AP 3, = ., (2) Draw a line segment AB of length 6.5 cm. Find a point P on it such that, [Foreign 2011], AB 5, (3) Geometrically divide a line segment of length 8.4 cm in the ratio 5 : 2., [CBSE 2015], (4) Draw a line segment of length 7.6 cm and divide it in the ratio 3 : 2., , [Foreign 2011], , (5) In the figure given on the next page, if B1, B2, B3,…... and A1, A2, A3,….. have been marked at equal distances. In, what ratio C divides AB?, [CBSE Standard SP 2020-21], , 74, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , (6) To divide a line segment BC internally in the ratio 3 : 5, we draw a ray BX such that ∠CBX is an acute angle. What, will be the minimum number of points to be located at equal distances, on ray BX?, II. Short Answer Type Questions - I, [2 Marks], 4. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP : PB = 3 : 5., , [NCERT Exemplar] [CBSE 2011], 5. Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5., [Delhi 2017], 6. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.  , [NCERT], III. Short Answer Type Questions - II, 7. Draw a line segment of length 8 cm and divide it in the ratio 2 : 3., 8. Draw a line segment of length 6 cm and divide it in the ratio 3 : 2., 9. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3 : 5., , [3 Marks], , Answers and Hints, 1. (1) (d) 12, (1) (2) (a) m + n, (1), (3) A11(1), 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), 3. (1) Steps of construction:, (i) Draw a line segment AB = 6 cm., (ii) Draw any ray AX making an acute angle XAB, with AB., , , (iii) Along AX mark 7 (3 + 4) points A1, A2, A3, A4,, ......, A7 at equal distances such that AA1 = A1A2, = ..... = A6A7., (iv) Join A7B., (v) From A3, draw A3P parallel to A7B (by making, an angle equal to ∠AA7B at A3), to meet AB at, point P., Then AP : PB = 3 : 4., (1), (5) 8 : 5, (1), (6) 8, (1), 4. Steps of construction:, (i) Draw a line segment AB = 7 cm., (ii) Draw AX || BY such that ∠A and B are acute angles., , (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 intersects AB at P and divides AP :, PB = 3 : 5., , , Hence, P is the required point on AB which divides it in, 3 : 5., (2), 6. Steps of construction:, (i) Draw a line segment AB = 7.6 cm., (ii) Draw a ray AX making an acute angle with AB., (iii) Mark 13 (8 + 5) equal points on AX, and mark them, as X1, X2, X3, ........, X13., (iv) Join ‘point X13’ and B., , (v) From ‘point X5’, draw X5C || X13B, which meets AB, at C., Thus, C divides AB in the ratio 5 : 8, On measuring the two parts, we get:, AC = 2.9 cm and BC = 4.7 cm., (2), , Constructions, , 75

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 2. Construction of Tangents to a Circle, Construction 1: To draw a tangent to a given circle through a given point on its circumference, (using centre of the given circle), Let ‘O’ be the centre of given circle of radius 4 cm and P be a point on its circumference., Steps:, O, A, (i) Draw the given circle with centre ‘O’ and radius 4 cm., (ii) Mark a point ‘P’ anywhere on the circumference of the circle drawn., (iii) Join P with centre O., P, (iv) Through point P, draw a perpendicular to radius. Let the perpendicular drawn be AP., Extend AP to B., B, , \ AB is the required tangent., Construction 2: To draw a tangent to a circle at a point on it (without using the centre of the circle), Let P be a point on the circumference of the given circle of radius 3 cm., Steps:, (i) Draw a circle of radius 3 cm and mark a point ‘P’ anywhere on its circumference., (ii) Draw chord PQ of any suitable length., (iii) Mark a point R on the circumference of the given circle, then join RP and RQ., (iv) Construct ∠QPB = ∠R, , \ PB i.e., AB is the required tangent., Construction 3: To draw a tangent to a circle from a point outside the circle (using its centre), Let the radius of the given circle (with centre O) be 2.5 cm and the point P is outside the circle at a distance of 5 cm, from the centre O., T, Steps:, (i) Draw a circle with centre O and radius 2.5 cm., (ii) Mark a point P such that OP = 5 cm., P, A, O, (iii) Draw the perpendicular bisector of the line segment PO which meets PO at point A., (iv) With A as centre and AP (AO) as radius, draw a semicircle which meets the given circle, at point T., (v) Join P and T., T, , \ PT is the required tangent., We know that two tangents can be drawn to a circle through a point outside it. And, both, of these tangents are always equal in length., P, O, A, In order to draw both the tangents from the exterior point P, complete the circle with centre, A and radius PA. This circle will meet the given circle at two points say, T and Q. Then, PT, Q, and PQ are the required two tangents. Also PT = PQ., Construction 4: To draw a tangent to a circle from a point outside the circle (without using its centre), Let the radius of the given circle be 3.6 cm and P be a point outside the circle at a distance of 8 cm from its centre., Steps:, D, (i) Draw a circle with centre O and radius 3.6 cm., E, (ii) Mark a point P outside the circle such that OP = 8 cm., (iii) Through point P, draw a secant PAB which meets the given circle at points A and B., O, (iv) By drawing perpendicular bisector of PB, find the mid-point of PB. Let C be the midpoint of PB., P, A, C, B, (v) With C as centre and PC as radius, draw a semicircle., (vi) Through point A, draw a line perpendicular to PB which meets the semicircle [drawn, in step (v)] at point D., (vii) With P as centre and PD as radius, draw an arc which meets the given circle at point E., (viii) Join PE., , \ PE is the required tangent., , 76, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the, circle and measure its length. , [NCERT] [Imp.], Solution. Steps:, (i) With point O as centre, draw a circle of radius = 6 cm., (ii) Mark a point A such that OA = 10 cm., (iii) Join O and A. Then, draw the perpendicular bisector of OA which cuts OA at point B., (iv) With B as centre and AB = BO as radius, draw a circle which cuts the circle drawn in step (i) at points P and Q., (v) Join PA and QA., PA and QA are the required tangents., On measuring, we find PA = QA = 8 cm., Example 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure, its length. Also, verify the measurement by actual calculation.  , [NCERT] [Delhi 2013], Solution. Steps:, (i) Mark a point as O., (ii) With O as centre, draw two circles with radii 4 cm and 6 cm respectively., (iii) On the bigger circle, mark a point A., O, (iv) Join O and A., P, (v) Draw perpendicular bisector of OA to get its mid-point P., (vi) With P as centre and PA = PO as radius, draw a semicircle which meets smaller circle A, B, at point B., C, (vii) Join A and B, and produce it to meet outer circle at point C., ABC is the required tangent., The length of AB = 4.5 cm (approx.), Verification: Join OB., Since, angle between radius and tangent is 90°., ⇒, ∠ABO = 90°., 2, In right DABO,, AB + OB2 = OA2, ⇒, AB2 = 62 – 42 = 36 – 16 = 20, and, AB = 20 cm = 4.5 cm (approx.), Example 3. ABC be a right triangle in which AB = 6 cm, BC = 4 cm and, ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is, drawn. Construct the tangents from A to this circle., Solution. Steps:, (i) Draw perpendicular bisectors of BC and CD which meet at point O., O is the centre of the circle through B, C and D., (ii) Join A and O., (iii) Draw a circle with AO as diameter which meets the given circle at points P, and B., (iv) Join AP and AB., AP and AB are the required tangents., , Exercise 4.2, , I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) To draw a pair of tangents to a circle which are inclined to each other at an angle of 30°, it is required to draw tangents, at end points of those two radii of the circle, the angle between them should be, (a) 150°, (b) 90°, (c) 60°, (d) 120°, (2) If you draw a pair of tangents to a circle C(O, r) from point P such that OP = 2r, then the angle between the two, tangents is, (a) 90°, (b) 30°, (c) 60°, (d) 45°, , Constructions, , 77

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 2. Assertion-Reason Type Question, , In the following question, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): If two tangents AB and AC are drawn from a common point A to a circle of radius 4 cm, then, AB = AC., , Reason (R): Tangents drawn from an external point to a circle are equal., 3. Answer the following:, (1) Is it possible to construct a pair of tangents from point P to circle of radius 5 cm situated at a distance of 4.9 cm from, the centre?, (2) Is it possible to construct a pair of tangents from point P lying on circle of radius 4 cm and centre O?, II. Short Answer Type Questions - I, , [2 Marks], , 4. Draw a circle of radius 3 cm. Take a point P at a distance of 4.5 cm from the centre of the circle and from the point P,, draw two tangents to the circle., 5. Draw two tangents to a circle of radius 4 cm from a point P at a distance of 6 cm from its centre., 6. 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., [CBSE Standard SP 2020-21], III. Short Answer Type Questions - II, , [3 Marks], , 7. Draw a circle of radius 3 cm. From a point 6 cm away from its centre, construct the pair of tangents to the circle and, measure their lengths., [CBSE 2014], 8. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. [Foreign 2014], 9. Draw a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°, , [CBSE 2016, AI 2013], 10. Draw a circle of radius 4 cm. From the point 7 cm away from its centre, construct the pair of tangents to the circle., 11. 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., [CBSE Standard 2020], IV. Long Answer Type Questions, , [5 Marks], , 12. Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and, draw a circle passing through the points B, C and D. Construct tangents from A to this circle., [Delhi 2014], 13. 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 of the other circle., [AI 2014], 14. DABC is a right-angled triangle, in which AB = 3 cm, BC = 4 cm and ∠B = 90°. BD is perpendicular from B on AC., The circle through B, C and D is drawn. Construct the tangents from A to the given circle., [CBSE 2015], 15. Draw two concentric circles C1 and C2 of radii 3 cm and 5 cm. Taking a point on outer circle C2, construct the pair of, tangents to the other. Measure the length of a tangent and verify it by actual calculation., [NCERT Exemplar], 16. Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle., Also measure its length., [Delhi 2016], 17. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance, of 7 cm from its centre. Draw tangents to the circle from these two points P and Q., [NCERT]], 18. Construct a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of 60°., [CBSE Standard SP 2019-20], 19. 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 circle., [CBSE Standard 2020] [Imp.], , 78, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Case Study Based Questions, , 6m, , I. The management of a school decided to arouse interest of their students in Mathematics. So they want to construct some, geometrical shapes in one corner of the school premises. They showed a rough sketch of a right triangular structure on a, plane sheet of paper with sides AB = 6 m, BC = 8 m and ∠B = 90°. The diagram shows a perpendicular from the vertex, B to the front side AC. They want to build a circular wall through B, C and D but they had certain problems in doing so., So they called on some students of class X to solve this problem. They made some suggestions., 1. To find centre of the circle, the students made some suggestions which are as follows:, (a) Draw perpendicular BD on AC, A, (b) Draw perpendicular bisectors of BC and CD., (c) The intersecting point of perpendiculars of BC and CD are the centre of the circle., D, E, (d) All of the above, 2. Referring to the above, what will be the length of AD?, (a) 3.6 m, (b) 3.8 m, (c) 4.8 m, (d) 5.6 m, 3. Referring to the above, what is the length of perpendicular drawn on side AC from vertex, B, C, O, B?, 8m, (a) 2.6 m, (b) 3.0 m, (c) 4.8 m, (d) 4.0 m, 4. Referring to the above, the length of tangent AE is, (a) 10 m, (b) 8 m, (c) 12 m, (d) 6 m, 5. Referring to the above, sum of angles ∠BAE and ∠BOE is, (a) 120°, (b) 180°, (c) 90°, (d) 60°, II. The construction of a road is in progress. A road already exists through a forest that goes over a circular lake., , The engineer wants to build another road through the forest that connects this road, but does not go through, the lake., , T, , Lake, , , , B, , Road to be, built through, the forest, , dge, Bri Road, g, stin, Exi, , P, , A, , , As it turns out, the road the engineer will be building and the road it will connect to both represent characteristics, of a circle that have their own name. The road/bridge that already exists is called a secant of the circular lake,, and the road the engineer is going to build is called the tangent of the circular lake., 1. Refer to the above, if PT = 12 km and PA = 9 km, then the length of existing bridge is, (a) 7 km, (b) 9 km, (c) 12 km, (d) 16 km, 2. Refer to the above if the length of existing bridge is 5 km and the length of the existing road outside the lake is 4 km,, then the length of the road under construction is, (a) 4 km, (b) 6 km, (c) 10 km, (d) 14 km, 3. Refer to the question (2) if the road under construction, PT is 6 km and it is inclined at an angle of 30° to the line joining, the centre, the radius of the lake is, (a) 3 3 km, (b) 4 3 km, (c) 2 3 km, (d) 5 3 km, 4. Refer to the question (3) above, the circumference of the lake is, (a) 2 3 π km, (b) 3 3 π km, (c) 4 3 π km, 5. Refer to the question (3) above, the area of the lake is, (a) 12π km2, (b) 16π km2, (c) 18π km2, , (d) 5 3 π km, (d) 9π km2, , Constructions, , 79

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Answers and Hints, 1. (1) (a) 150°, (1) (2) (c) 60°, (1), 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), 3. (1) No, (1) (2) No, (1), 7. 5.20 m (approx), (3), 8. Steps of construction:, (i) A circle, with centre O and radius 5 cm is drawn., (ii) As tangents are inclined at 60°., , \ ∠TOS = 120°., T, , 90°, O, , 120°, , P, , 60°, , (ii) With centre A, a circle of radius 4 cm is drawn and, with centre B, a circle of radius 3 cm is drawn., (iii) With AB as diameter, a circle is drawn meeting circle, with centre A at S and T respectively and circle with, centre B at P and Q respectively., (iv) Then AP and AQ are tangents from A to circle with, centre B and BS and BT are tangents from B to circle, with centre A., (5), 15. 4 cm, (5), 16. Steps of construction:, (i) Take point O. Draw two concentric circles of radii 3, cm and 5 cm respectively., (ii) Locate point P on the circumference of larger circle., , 90°, , , , S, , (iii) Two radius OT and OS, inclined at an angle of 120°, are drawn., (iv) Tangents are drawn to the circle at T and S meeting, at P., Then PT and PS are the required tangents., (3), 9. Proceed same as Q8., , (3), , 12. Steps of construction:, (i) Draw a right angle triangle ABC, right angled at B,, AB = 6 cm and BC = 8 cm., , , (iii) Join OP and bisect it. Let M be mid-point of OP., (iv) Taking M as centre and MP as radius, draw an arc, intersecting smaller circle at A and B., (v) Join PA and PB. Thus, PA and PB are required, , (ii) Draw BD ⊥ AC., , tangents.(5), , (iii) Draw a circumcircle of DBDC., (iv) From point A draw a pair of tangents AB and AP., , 17. Steps of construction:, (i) Join P and O., (ii) Bisect PO such that M be its mid-point., (iii) Taking M as centre and MO as radius, draw a circle., Let it intersects the given circle at A and B., (iv) Join PA and PB., Thus, PA and PB are the two required tangents from P., (v) Now, join O and Q., (vi) Bisect OQ such that N is its mid-point., (vii) Taking N as centre and NO as radius, draw a circle., , Then AB and AP are the required tangents., 13. Steps of construction:, , (5), , Let it intersects the given circle at C and D., (viii) Join QC and QD., A, , C, , S, P, , P, A, , , , 4 cm, , 3 cm, , Q, T, , (i) AB = 8 cm is taken., , 80, , B, , Mathematics–10, , O, N, , M, , B, , Q, , D, , Thus, QC and QD are the required tangents to the given, circle.(5)

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , II. 1. (a) 7 km, , Case Study Based Questions, , 2. (b) 6 km, , , 3. (c) 2 3 km, , I. 1. (d) All of the above, , 2. (a) 3.6 m, , 3. (c) 4.8 m, , , 4. (d) 6 m, , 5. (b) 180°, , Date __________, , 4. (c) 4 3 π km, , 2, , , 5. (a) 12π km, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. To divide a line segment internally in the given ratio., 2. To construct a pair of tangents to a circle from a point outside it., , QUICK REVISION NOTES, •• Dividing a line segment in the given ratio means to determine a point on the given line segment which divides it in the, given ratio., •• A tangent to a circle is a straight line which touches the circle at a point. This point is called the point of contact and, the radius through the point of contact is perpendicular to the tangent., •• Tangents drawn from an external point to a circle are equal., , COMMON ERROR, Error, , Correction, , (i) Constructing incorrect tangent(s) when centre of, the circle is not given., , (i) If centre of the circle is not given, first locate its, centre and then construct the tangent(s) correctly., , qqq, , Constructions, , 81

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5, , Some Applications of Trigonometry, , Topic Covered, 1. Heights and Distances, , 1. Heights and Distances, So far, we have studied the trigonometric ratios. Knowledge of these ratios can be used to calculate the height or length, of an object or the distance between two distant objects. But, let’s first define a few important terms., Line of Sight: The line from our eyes to the object we are viewing is known as the line of sight or line of vision., When an object is being viewed, it may be, (i) at the eye level, , (ii) above the eye level, , (iii) below the eye level, , Horizontal level, E (Eye), O, (Object) Line of sight, , Angle of Elevation: If a person is looking up at an object, the acute angle so formed from, the horizontal level to a line of sight, is called the angle of elevation., Let EA be the horizontal level at the eye level. If a person looks at an object P situated, above the eye level, then –PEA is the angle of elevation of P as seen from E., Angle of Depression: If a person is looking down at an object,, the acute angle measured from horizontal level to a line of sight is, called the angle of depression., Let EA be the horizontal level at the eye level. If a person looks, at an object P situated below the eye level, the –AEP is the angle, of depression of P as seen from E., Note: Angle of depression of P as seen from O = Angle of elevation of O as seen from P i.e., –AOP = –OPH, Example 1. Find the values of x and y in the following figure:, Solution. DABC is a right-angled triangle, in which ∠B = 90° and ∠C = 45°, BC, 6, 6, 1, Now,, cos 45° =, = y fi, = y fi y= 6 2m, AC, 2, x, x, AB, Again, tan 45° =, =, fi 1=, fi x = 6m, 6, 6, BC, Thus, x = 6 m and y = 6 2 m, Example 2. The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres, towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.  [NCERT Exemplar], Solution. Let AB be the tower and angle of elevation from point C = 30°, Then, angle of elevation from point D = 30° + 15° = 45°, , 82

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In DADB,, ⇒, In DACB,, ⇒, ⇒, ⇒, ⇒, \, , A, AB, AB, tan 45° =, ⇒ 1=, DB, DB, DB = AB, ...(i) [ tan 45° = 1], AB, tan 30° =, 45°, 30°, B, BC, C, 20 m, D, AB, 1, 1, =, ( tan 30º =, ), CD + DB, 3, 3, AB, 1, =, [ DB = AB from (i) and CD = 20 m (given)], 20, + AB, 3, , 20 + AB =, , 3 AB ⇒ 20 = ( 3 – 1) AB, 10 ( 3 + 1) = AB, Height of the tower = 10 ( 3 + 1) metres., , ⇒, , 20 ( 3 + 1), = AB, 3 −1, , Example 3. At a point on level ground, the angle of the elevation of the top of a vertical tower is found to be such that, 5, its tangent is, . On walking 192 metres towards the tower, the tangent of the new angle of elevation is found to be., 12, Find the height of the tower., , P, , Solution. Let PQ be the tower of height h metres, point A be the first point of, observation and B be the second point of observation towards the towar PQ such, that AB = 192 m., Also, let, ∠PAQ = x and ∠PBQ = y, 5, 5, PQ, Given, tan x =, ⇒, =, 12, 12, AQ, i.e., , 5, h, =, 12, AQ, , Also,, , tan y =, , 3, 4, , ⇒ AQ =, ⇒, , Since,, , h, A, , y, , x, , Q, , B, , 192 m, , 12h, ...(i), 5, , 3, PQ, =, 4, BQ, , ⇒, , h, 3, =, BQ, 4, , AB = 192 m ⇒ AQ – BQ = 192 m, 36 h – 20 h, 12h 4h, From (i) and (ii),, = 192 ⇒, = 192 ⇒, −, 15, 5, 3, \ The height of the tower = 180 m, , ⇒ BQ =, 16 h, = 192, 15, , 4h, ...(ii), 3, ⇒ h = 180 m, , Example 4. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5 m. From a point on, the ground, the angles of elevation of the top and bottom of the flagstaff are 60° and 30° respectively. Find the height of, the tower and the distance of the point from base of the tower., [Take 3 = 1.732] [CBSE 2016] [Imp.], Solution. Let height of tower (TR) be x m, distance (RP) of a point (P) from the base of tower be y m, height of the flagstaff, (QT) be 5 m., Q, x, x, 1, Then in the DTRP, tan 30° = y fi, = y fi y = 3 x, ...(i), 3, Now, in DQRP,, , QR = (5 + x) m, , QR, 5+ x, and, tan 60° =, fi, fi 5+x= 3y, 3 = y, PR, fi, 5 + x = 3x , [using (i)], 5, = 2.5 m, fi, x=, 2, and from (i),, y = 3 x = 2.5 × 1.732 = 4.33 m, Thus, the height of the tower = 2.5 m and the distance of the observation point from the base, of the tower = 4.33 m., , 5m, , T, 60°, , x, , 30°, y, , P, , Some Applications of Trigonometry, , 83, , R

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Example 5. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between, the two towers and also the height of the other tower., [NCERT Exemplar], Solution. Let AB be the tower of height 30 m and CD be another tower of height h m., Then,, –ADB = 60° and –CBD = 30°, A, 30, AB, In DABD,, tan 60° =, ⇒, 3 = BD, BD, ⇒, , BD =, , 30 × 3, 3× 3, , =, , 30 3, = 10 3 m ...(i), 3, , 30 m, , CD, 1, h, C, In DCBD,, tan 30° =, ⇒, =, , [From (i)], BD, 3 10 3, h, 10 3, ⇒, = h ⇒ h = 10 m, 30°, 60°, 3, D, B, \, CD = 10 m, Thus, height of the other tower = 10 m and the distance between two towers = BD = 10 3 m., Example 6. A man standing on the deck of a ship, which is 10 m above the water level, observes the angle of elevation, of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from, the ship and height of the hill., [CBSE 2006], A, Solution. Let AB be the hill (A its top and B its base), P be the position of the man on the deck, C be a point on the water level and PD be the eye level of the man., Clearly,, PC = 10 m = DB, ∠APD = 60° and ∠BPD = 30° = ∠PBC, We have to find AB and BC or PD., DB, 10, 1, In DPDB,, tan 30° =, ⇒, =, PD, PD, 3, PD = 10 3 m...(i), 60°, P, D, AD, AD, 30°, In DAPD,, tan 60° =, ⇒, =, , [From (i)], 3, 10 m, PD, 10 3, 30°, ⇒, AD = 10 3 × 3 = 30 m, C Water Level B, \ Distance of the hill from the ship = BC = PD = 10 3 m, And, the height of the hill = AB = AD + DB = 30 m + 10 m = 40 m, Example 7. A tree breaks down due to storm and the broken part bends, so that the top of the tree touches the ground, making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8, metres. Find the height of the tree before it was broken., [NCERT] [Imp.], Solution. Let AB be the height of the tree that was broken, AC be the height of broken part of the tree and P be the point, at which the top of the broken part of the tree touches., Clearly,, ∠CPB = 30° and PB = 8 m, A, Let CB be x m and AC = PC be y m., Therefore, the height of the tree before it was broken = (x + y) m, 8 3, x, BC, 1, x, y, =, In DCPB,, tan 30° =, =, ⇒, ⇒ x=, m ...(i), PB, 8, 3, 3 8, ⇒, , Again,, ⇒, , cos 30° =, y=, , PB, 8, =, PC, y, 16, 3, , ⇒, , 3 8, =, 2, y, , C, ym, , m=, , 16 3, m, 3, , ...(ii), , \ The height of the tree before it was broken, = x + y =, , 84, , Mathematics–10, , 24 3, 8 3 16 3, = 8 3 = 13.86 m, =, +, 3, 3, 3, , P, , 30°, 8m, , x, , B, , [From (i) and (ii)]

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Exercise 5.1, I. Very Short Answer Type Questions, 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, , [1 Mark], , (1) A pole of 6 m high casts a shadow 2 3 m long, then sun’s elevation is, (a) 60°, (b) 45°, (c) 30°, (d) 90°, (2) At some time of the day, the length of the shadow of a tower is equal to height. Then the sun’s altitude at that time, is, (a) 30°, (b) 60°, (c) 90°, (d) 45°, (3) The angle of depression of a car standing on the ground, from the top of a 75 m high tower is 30°. The distance of, the car from the base of the tower (in m) is:, (a) 25 3, , (b) 50 3, , (c) 75 3, , (d) 150, , (4) In the given figure, the angles of depressions from the observing, positions O1 and O2 respectively of the object A respectively are, (a) 30°, 45°, (b) 45°, 60°, (c) 60°, 75°, (d) 60°, 30°, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): If the angle of elevation of Sun, above a perpendicular line (tower) decreases, then the shadow of, tower increases., , Reason (R): It is due to decrease in slope of the line of sight., (2) Assertion (A): When we move towards the object, angle of elevation decreases., , Reason (R): As we move towards the object, it subtends large angle at our eye than before., 3. Answer the following:, (1) A kite is attached to a string. Assuming that there is no slack in the string, find the height of the kite above the level, of the ground, if the length of the string is 54 m and it makes an angle of 30° with the ground., [Imp.], (2) In the given figure, 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., [CBSE Standard 2020], (3) The ratio of the length of a vertical rod and the length of its shadow is 1 : 3 . Find, the angle of elevation of the sun at that moment?, [CBSE Standard 2020], (4) In the given figure, the angle of elevation of the top of a tower AC from a point, B on the ground is 60°. If the height of the tower is 20 m, find the distance of the, point from the foot of the tower., , (5) A tower stands vertically on the ground. From a point on the ground, which is 15m away from the foot of the tower,, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower., , Some Applications of Trigonometry, , 85

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II. Short Answer Type Questions - I, , [2 Marks], , 4. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the, ground. Find the height of the pole, if the angle made by the rope with the ground level is, 30° (see figure given alongside)., [NCERT], 5. From the top of a 60 m high building, the angles of depression of the top and the bottom of, a tower are 45° and 60° respectively. Find the height of the tower., [Take, , 3 = 1.73] [AI 2015], , 6. The shadow of a tower standing on a level ground is found to be 30 m longer when the sun’s altitude is 30° than when it, is 60°. Find the height of the tower., 7. An electrician has to repair an electric fault on a pole of height 4 m. He needs to reach a point, , A, , 1.3 m below the top of the pole to undertake the repair work (see figure given along side)., What should be the length of the ladder that he should use which, when inclined at an angle, of 60° to the horizontal would enable him to reach the required position?, [Imp.], , B, , 60°, , D, , C, , 8. The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is, supporting the disc as shown in the figure. If AC = 1.5 m long and CD = 3 m, find, (a) tan q, (b) sec q + cosec q, [CBSE Standard 2020], , III. Short Answer Type Questions - II, , [3 Marks], , 9. The angle of depression of the top and bottom of a tower as seen from the top of a 60 3m high cliff are 45° and 60°, respectively. Find the height of the tower., [CBSE 2013], 10. 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 PQ is 45°. Find the height of the tower PQ and the distance, PX.[Use 3 = 1.73], 11. From the top of a tower 50 m high, the angle of depression of the top of a pole is 45° and from the foot of the pole,, the angle of elevation of the top of the tower is 60°. Find the height of the pole if the pole and tower stand on the same, plane., [Foreign 2013], 12. The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 m from, the point A towards the foot of the tower to the point B, the angle of elevation of the top of the tower from the point B is, 60°. Find the height of the tower and its distance from the point A., [Foreign 2013], 13. A flag-staff stands on the top of a 12 m high tower. From a point on the ground, the angles of elevation of the top and, bottom of the flag-staff are observed to be 45° and 30° respectively. Find the height of the flag-staff., [Imp.], 14. The given figure shows a statue, 1.6 m tall, standing on the top of pedestal., , From a point on the ground, the angle of elevation of the top of the statue, is 60° and from the same point, the angle of elevation of the top of the, pedestal is 45°. Find the height of the pedestal., [Imp.], , Statue = 1.6 m, , Pedestal, 45°, , 60°, , 15. As observed from the top of a 60 m high lighthouse from the sea-level, the angles of depression of the two ships are 30°, and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two, ships.[Use, , 86, , Mathematics–10, , 3 = 1.732]

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16. The angles of elevation of the top of a tower from two points at a distance of 6 m and 13.5 m from the base of the tower, and in the same straight line with it are complementary. Find the height of the tower., [Foreign 2013], 17. Two ships are there in the sea on either side of a lighthouse in such a way that the ships and the lighthouse are in the, same straight line. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45°. If, the height of the lighthouse is 200 m, find the distance between the two ships., [Use 3 = 1.73] [Delhi 2014], 18. The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation, becomes 30°. If the aeroplane is flying at a constant height of 3000 3 m, find the speed of the aeroplane. [AI 2014], 19. From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flagstaff is fixed at the, top of the building and the angle of elevation of the top of the flagstaff from the point P is 45°. Find the length of the, flagstaff and the distance of building from the point P., [Delhi 2013], 20. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression, of its foot is 30°. Determine the height of the tower., [Foreign 2013], 21. A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards him., If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this, will the car reach the, observation tower? , [CBSE SP 2018-19] [CBSE 2006 (C)], 22. Two men on either side of a 75 m high building and in line with base of building observe the angles of elevation of the, top of the building as 30° and 60°. Find the distance between the two men., (Use 3 = 1.73). [Foreign 2015], 23. As observed from the top of a lighthouse, 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 observation., [Use 3 = 1.73] [AI 2016], 24. 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., , IV. Long Answer Type Questions, [5 Marks], 25. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and, 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships., [Use 3 = 1.732] [CBSE 2018], 26. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively., Find the height of the tower and the horizontal distance between the tower and the building., [Use 3 = 1.73] [Delhi 2015], 27. A 7 m long flagstaff is fixed on the top of a tower standing on the horizontal plane. From a 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 to, one place of decimal., [Use 3 = 1.73] [Foreign 2016], 28. An aeroplane, when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant, when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the, vertical distance between the aeroplanes at that instant., [Take 3 = 1.73] [Foreign 2016], 29. The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of, cloud in the lake is 60°. Find the height of the cloud.  , [CBSE SP 2018-19, 2008, 2010], 30. A round balloon of radius r subtends an angle a at the eye of the observer while the angle of elevation of its centre is b., α, Prove that the height of the centre of the balloon is r sin b cosec . , [CBSE 2010], 2, 31. A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top, of the lighthouse from 60° to 30°. Find the speed of the boat in metres per minute., [Use 3 = 1.732] [Delhi 2019], , Some Applications of Trigonometry, , 87

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32. Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a, point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the, height of the poles and the distances of the point from the poles., [Delhi 2019], 33. 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] [Delhi 2019], 34. The angles of elevation and depression of the top and bottom of a lighthouse from the top of a 60 m high building are, 30° and 60° respectively. Find, (i) the difference between the heights of the lighthouse and the building., (ii) the distance between the lighthouse and the building., [AI 2014], 35. From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically, above the first point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and, its horizontal distance from the point of observation.  , [AI 2016], 36. The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 10 seconds, the angle of, elevation changes to 30°. If the jet is flying at a speed of 900 km / hour, find the constant height at which the jet is flying., [Use, , 3 = 1.732] [CBSE 2008], , 37. The angle of elevation of the top of a tower as observed from a point on the ground is a and on moving a metres, a tan α tan β, towards the tower, the angle of elevation is b. Prove that the height of the tower is:, ., tan β − tan α, [NCERT] [CBSE 2006], 38. From a window, x metres high above the ground in a street, the angles of elevation and depression of the top and foot of, the other house on the opposite side of the street are a and b respectively. Show that the height of the opposite house is, x(1 + tan a tan b) metres., [CBSE 2006], 39. The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of, elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff., [Use, , 3 = 1.73] [AI 2017], , 40. At the foot of a mountain, the angle of elevation of its summit is 45°. After ascending 1 km towards the mountain up an, incline of 30°, the elevation changes to 60°. Find the height of the mountain., , [Use, , 3 = 1.732] [CBSE 2010], , 41. The angles of depression of the top and bottom of a building 50 metres 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. , [CBSE Standard SP 2020-21, SP 2018], 42. As observed from the top of a 75 m highlight house above the sea level, the angles of depression of two ships are 30°, and 45° respectively. If one ship is exactly behind the other on the same side of the lighthouse and in the same straight, line, find the distance between the two ships., [Use 3 = 1.732], 43. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the, statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal., [Use 3 = 1.73] [CBSE Standard 2020], 44. A verticle tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 6 m. At a point on the, plane, the angle of elevation of the bottom and top of the flag-staff are 30° and 45° respectively. Find the height of the, 45., 46., 47., 48., , 88, , tower.[Take 3 = 1.73] [CBSE Standard 2020], From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20 m high, building are 45° and 60° respectively. Find the height of the tower., [CBSE Standard 2020], From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and, 45°, respectively. If the bridge is at a height of 10 m from the banks, then find the width of the river. [Use 3 = 1.73], The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the, tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building., 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., [CBSE Standard SP 2020-21], , Mathematics–10

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49. The angles of depression of the top and bottom of a 8 m tall building from the top of a multi-storied building are 30° and, 45°, respectively. Find the height of the multi-storied building and the distance between the two buildings., 50. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height 88.2 m from the ground. The angle, of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces, 30°. Find the distance travelled by the balloon during the interval., , , , Case Study Based Questions, I. Application of Trigonometry—Height of Tree/Tower: Mr. Suresh is an electrician. He receives a call regarding a fault, on a pole from three different colonies A, B and C. He reaches one-by-one to each colony to repair that fault., He needs to reach a point 1.3 m below the top of each pole to undertake the repair work. Observe the following diagrams., , A, , B, , C, , Refer to Diagram A, 1. What should be the length of ladder DQ that enable him to reach the required position if the height of the pole is 4 m?, 7 2, 4 3, 5 3, 9 3, m, m, m, m, (b), (c), (d), 5, 5, 7, 5, 2. What is the distance of the point where the ladder is placed on the ground if the height of pole is 4 m?, (a) 2.5 m, (b) 3.8 m, (c) 1.56 m, (d) 5.3 m, Refer to Diagram B, 3. Given that the length of ladder is 4 2 m . What is height of pole?, 1, (a) 4 m, (b) 4 5 m, (c) 5 5 m, (d) 5.3 m, 2, 4. The distance of the point where the ladder lies on the ground is, (a), , (a) 3 5 m, , (b) 4 2 m, , (c) 4 m, , (d) 4 7 m, , Refer to Diagram C, 5. The angle of elevation of reaching point of ladder at pole, i.e., H, if the height of the pole is 8.3 m and the distance GF, is 7 3 m, is, (a) 30°, , (b) 60°, , (c) 45°, , (d) None of these., , Some Applications of Trigonometry, , 89

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II. A group of students of class X visited India Gate on an educational 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, Kingsway), is about 138 feet (42 metres) in height., , 1. 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°, 2. They want to see the tower at an angle of 60°. The distance where they should stand will be, (a) 25.24 m, (b) 20.12 m, (c) 42 m, (d) 24.25 m, 3. 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, 15, 20, (a) 20 3 m, (b), (c), (d) 15 3 m, m, m, 3, 3, 4. 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°, 5. 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, III. A satellite flying at a height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, they are, 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 mid-point of the distance between the two mountains., 1. The distance of the satellite from the top of Nanda Devi is, (a) 1118.29 km , (b) 577.52 km, (c) 1937 km , (d) 1025.36 km, 2. 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, 3. The distance of the satellite from the ground is, (a) 1139.4 km , (c) 1937 km , , (b) 566.96 km, (d) 1025.36 km, , 4. What is the angle of elevation if a man is standing at a distance of 7816 m away, from Nanda Devi?, (a) 30°, (b) 45°, (c) 60°, (d) 0°, 5. If a mile stone very far away from, makes 45° to the top of Mullayangiri mountain. So, find the distance of this mile, stone from the mountain., (a) 1118.327 km, (b) 566.976 km, (c) 1937 km, (d) 1025.36 km, , 90, , Mathematics–10

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Answers and Hints, 1. (1) (a) 60°, , (1) (2) (d) 45°, , (1), , (3) (c) 75 3, (1) (4) (a) 30°, 45°, (1), 2. (1) (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (d) Assertion (A) is false but reason (R) is true. (1), 3. (1) 27 m, (1)  (2) 10 3 m (1), 20, m (1), (3) 30°, (1)  (4), 3, (5) In right DABC, tan 60° =, A, , h, 15, , h, B, , , ⇒, , 15 m, , 60°, , C, , 3 = h (½), 15, , , , ⇒, h = 20[3 − 3 ] = 20[3 – 1.73], = 20 × 1.27 = 25.4 m, , \ Height of the tower is 25.4 m., 6. Let AB be the tower of height h., Let BC = x, , (1), , \ , h = 15 3 m (½), 4. In the given figure, let AC be the rope and AB be the pole., , , , In right D ABC, we have:, AB, , = sin 30°, AC, AB, 1, , fi, = [Q AC = 20 m] (1), 20, 2, 1, , fi, AB = 20 × = 10 m, 2, Thus, the required height of the pole is 10 m.  , (1), 5. Let AB be 60 m high building and CD be the tower of, height h., , \, ∠ACE = 45° and ∠ADB = 60°, , (using alternate angles), Let, BD = CE = x, , BE = CD = h fi AE = 60 – h, , In right-angled triangle ABD,, x, 1, BD, , = cot 60° fi, =, 60, 3, AB, 60 60, 3, , x=, ...(i), =, ×, = 20 3, 3, 3, 3, (1), In right-angled triangle AEC,, AE, , = tan 45°, CE, 60 − h, , ⇒, = 1 ⇒ 60 – h = x, x, , ⇒, h = 60 − 20 3, [using (i)], , In DABC,, , h, =, x, , , fi, In DADB,, , fi, , fi, , fi, , AB, = tan 60°, BC, 3, , fi, , x=, , h, 3, , ...(i) (½), , AB, = tan 30°, BD, , 1, h, =, fi h 3 = x + 30, x + 30, 3, h, + 30, [From (i)] (½), h 3 =, 3, h + 30 3, h 3 =, fi 3h = h + 30 3, 3, , , fi, , 2h = 30 3 fi h = 15 3, Hence, the height of the tower is 15 3 m.(1), , 5.4, , m (2), 3, 8. DACD is a right angled triangle., 7., , , , Some Applications of Trigonometry, , 91

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So,, , ⇒, , ⇒, , ⇒, , CD2 = AC2 + AD2, (3)2 = (1.5)2 + AD2, AD2 = 9 – 2.25 = 6.75, AD =, , 6.75 × 4, =, 4, , 27, 3 3, =, 4, 2, , AC, 1.5, 1.5 × 2, 1, =, =, =,  (1), AD, 3 3, 3 3, 3, 2, 2, CD, 3, (b) sec q =, =, =, AD, 3 3, 3, 2, CD, 3, 30, cosec q =, =, =, =2, AC 1.5, 15, (a) tan q =, , , \ sec q + cosec q =, , 9. In DABC,, , =, , 2, 3, , 2+2 3, , AB, = tan 60° fi, BC, , BC = 60 m, Now, in DAED,, AE, , = tan 45°, ED, , , +2, , 3, , (1), , 60 3, =, BC, , 3, ...(i) (1), , 60 3 − h, = 1, BC, , [ AE = AB – BE and ED = BC], , , fi 60 3 − h = 60, , [From (i)] (1), , , fi, , , In DQRY,, , h − 40, QR, = tan 45° fi, =1, z, YR, , fi, h – 40 = z, ...(ii) (1), From (i) and (ii), we get, h, , = h – 40 fi h = h 3 − 40 3, 3, 40 3, h 3 − h = 40 3 fi h =, , fi, (1), 3 −1, , , 40 3 ( 3 + 1), 2, = 20(3 + 3 ) = 20(3 + 1.73), , fi, h = 20 × 4.73 = 94.6 m, ...(iii), So, height of the tower PQ = 94.6 m and the distance PX, = 94.6 – 40 = 54.6 m., [From (ii) and (iii)] (1), 11. Let the height of the pole is h., In right DEDC,, 50 − h, ED, , tan 45° =, fi 1=, DC, DC, , DC = 50 – h = AB, ...(i) (1), , fi, , h=, , , , h = 60 3 – 60 fi h = 60( 3 − 1), , h = 60(1.73 – 1), = 60 × 0.73 = 43.8 m, PQ, h, 10. In DQPX,, = tan 60° fi, = 3, PX, z, h, , fi, =z, 3, , (1), , ...(i), , , In right DEAB,, , , fi, , fi, , fi, , fi, , 92, , Mathematics–10, , EA, fi, AB, 50, 3 =, , 50 − h, , tan 60° =, , 50, AB, [From (i)] (1), , 3, ×, 3, 3, 50 × 1.73, (50 – h) =, 3, 3(50 – h) = 86.50 fi 150 – 3h = 86.50, 50 – h =, , 50, , 3 =

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fi 150 – 86.50 = 3h fi 63.50 = 3h, 63.50, , fi, h=, fi h = 21.16 m  , 3, 12. In right DDCB,, h, DC, , tan 60° =, fi, 3 =, BC, BC, h, , fi, BC =, 3, , (1), , 16. In DABC, tan q =, , h, 6, , ...(i) (1), , ...(i) (1), , In DABD,, , , In right DDCA,, , , fi, , fi, , fi, , DC, tan 30° =, AC, , h, =, BC+AB, 3, , 1, , fi, , BC + 20 = 3h, h, + 20 = 3h, 3,  3 − 1, h, = 20 fi,  3 , , [From (i)] (1), h = 10 3 = 17.30 m, , On putting h = 10 3 in equation (i), we get, , , 10 3, , BC =, , 3, , tan (90° – q) =, , h, h, fi cot q =, 13.5, 13.5, , , ⇒, , 13.5, h, , ...(ii) (1), , , ∴, , 13.5, h, =, h, 6, , [From (i) and (ii)], , , , h2 = 13.5 × 6 = 81.0, , tan q =, , , fi, , h = 9 m, , (1), , 17. In DABD,, , , AB, = tan 45° fi, BD, , , fi, , BD = 200 m, , 200, =1, BD, ...(i) (1), , = 10 m, , So, the height of the tower is 17.30 m and its distance from, point A = 20 + 10 = 30 m., (1), 13. 12( 3 – 1) m, , (3), , 14. (0.8) ( 3 + 1) m, 15. In DABP,, , (3), , , , tan 45° =, , , fi, , AB, BP, , fi, , 1=, , (1), , A, , = BC + BD =, , 45°, , 60 m, 45°, , , In DABQ,, , , ⇒, , B, , tan 30° =, , AB, BQ, , 30°, Q, , P, , ⇒, , AB, = tan 60° fi, BC, 200, , fi, BC =, m, 3, , \ Distance between the ships, , , 60, BP, , BP = 60 m, 30°, , , Now, in DABC, , 1, 3, , =, , AB, BQ, , BQ = AB 3 = 60( 3 ), , = 103.92 m, Distance between two ships, = BQ – BP = 103.92 – 60, , (1), , = 43.92 m., , (1), , 200, =, BC, , 3, ...(ii) (1), , 200, 3, , + 200, , [From (i) and (ii)], 200 3, =, + 200, 3, 200 × 1.73, 346, =, + 200 =, + 200, 3, 3, = 115.33 + 200 = 315.33 m, (1), 18. In right-angled triangle OLA,, OL, , = cot 60°, AL, 1, , fi, OL = 3000 3 ×, 3, = 3000 m, ...(i) (1), , Some Applications of Trigonometry, , 93

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fi, , AD =, , CD, 3, , In DADE,, , , , , fi, , In right-angled triangle OMB,, OM, = cot 30°, BM, OM = 3000 3 × 3 = 9000 m, , , , fi, , \, , , fi, ...(ii), , AB = LM = OM – OL, , = (9000 – 3000), = 6000 m  , , (From (i) and (ii)) (1), , Now, distance covered in 30 s = 6000 m, , \ Distance covered in 1 hour (3600 s), 6000 3600, km = 720 km, ×, 30, 1000, , \ Speed of the aeroplane is 720 km/h., =, , (1), , tan 30° =, , DE, AD, , 1, , DE, AD, , 3, , =, , ...(i) (1), , AD = DE ( 3 ) ...(ii) (1), , From (i) and (ii), we get, CD, , = DE( 3 ), 3, , fi, , CD, , = 7( 3 ), [ DE = AB = 7 m], 3, , fi, CD = 21 m, Total height of the cable tower, = CD + DE = 21 + 7 = 28 m, (1), 21. 16.39 mins., (3), 22. Let C and D be the positions of two men., Let CB = y and BD = x, , 19. Let height of flagstaff be h and the distance of the building, from the point P be x., A, flagstaff h, B, , , In DABC,, , 10 m, 30°, 45°, C, , , In DBCP, BC, CP, , fi, x, AC, In DACP,, PC, , fi, h + 10, , , , fi, , fi, , = tan 30° =, , P, , x, , 1, 3, , fi, , 10, 1, =, x, 3, , = 10 3 m, = tan 45° fi, , ...(i) (1), h + 10, =1, x, , =x, , h + 10 = 10 3, , [From (i)] (1), , h = 10( 3 − 1) = 10(1.73 – 1), , = 10 × (0.73) = 7.3 m, , \ Height of the flagstaff is 7.3 m., The distance of the building from the point P, = 10 3 = 10 × 1.73 = 17.3 m  , 20. In DADC,, CD, , tan 60° =, AD, , fi, , 94, , 3 =, , CD, AD, , Mathematics–10, , AB, 75, = tan 60° fi, = 3 (1), BC, y, 75 75 3, =, = 25 3 m, , fi, y=, (1), 3, 3, Now, in DABD,, 75, 1, 75, ⇒, =, , tan 30° =, x, x, 3, , , , fi, x = 75 3, Hence, distance between two men is x + y, = 25 3 + 75 3 = 100 3, = 100 × 1.732 = 173.20 m, AB, 23. In DABC,, = tan 60°, BC, , fi, , (1), , , In DABD,, , 100, =, y, , 3 ⇒ y=, , 100, 3, , (1), , ...(i) (1)

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AB, = tan 30° fi, BD, , 100, 1, =, y+x, 3, , fi x = 100 3 − y, 100 300 − 100 200, =, =, , fi, x = 100 3 −, 3, 3, 3, , [From (i)] (1), 200 3 200 × 1.73, =, , fi, x=, = 115.33 m, 3, 3, The distance travelled by the ship during the period of, observation is 115.33 m., (1), , fi, , x + y = 100 3, , , , 24., , Let AB = candle, C and D are two coins, , AB h, =, BC b, , , , tan 60° =, , , , 3 =, , , , h= b 3, , 1, AB, =, (1), AB + CD, 3, , , fi, , fi, , AB + CD =, , 3 AB, , fi, CD = AB ^ 3 − 1 h, = 100 × (1.732 – 1) = 73.2 m  , 26. In DAED,, , y, = tan 45° fi, x, , (1), (1), , y, =1, x, , (½), , (½), , , , fi, In DABC,, , , h, b, , , fi, ...(i) (½), , AB h, =, , tan 30° =, BD a, 1, h, , =, a, 3, a, , h=, ...(ii) (½), 3, Multiplying (i) and (ii), we get, a, , h2 = b 3 ×, (½), 3, , h2 = ba, , h = ab m , (½), 25. Let AB be the tower and ships are at points C and D., , , fi, , fi, , , y=x, , ...(i) (1), , AB, = tan 60°, BC, AE + EB, = 3 fi, BC, x + 50 = 3x, 3x − x = 50, x=, , 50, , y + 50, = 3, x, [ y = x, using (i)] (1), ( 3 − 1) x = 50, , fi, =, , 50( 3 + 1), ( 3 − 1)( 3 + 1), , 3 −1, 50(1.732 + 1), =, (1), 2, , x = 68.49 m, (1), Height of the tower = 50 + y = 50 + 68.49, ( x = y), = 118.49 m, Horizontal distance between the tower and the building, = x = 68.49 m., (1), AB, 27. In DABC,, = tan 45°, BC, h, , =1 fi h=x, ...(i) (1), x, , , , , tan 45° =, , AB, (1), BC, , AB, = 1 fi AB = BC, BC, 1, AB, =, Also tan 30° =, 3 BC + CD, , fi, , (1), , , Now, in DDBC,, DB, , = tan 60° fi, BC, , h+7, = 3 (1), x, , Some Applications of Trigonometry, , 95

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fi, , fi, , h+7=, ( 3 − 1)h = 7, , =, , 3h, fi, , h=, , 7( 3 + 1), , [ h = x, using (i)] (1), 7, 3 −1, , (1), ( 3 − 1)( 3 + 1), 7(1.732 + 1), 7 × 2.732, =, =, 2, 2, = 9.5 m, So, height of the tower is 9.5 m., (1), AB, 4000 − y, 28. In DABC,, = tan 45° fi, =1, BC, x, , fi, x = (4000 – y) m, ...(i) (1), , , fi, , AE =, , h + 60, , ...(ii) (1), 3, From (i) and (ii), we get, h + 60, (h − 60) 3 =, (1), 3, , fi, 3h – 180 = h + 60, (1), , fi, 2h = 240 fi h = 120 m, , \ Height of the cloud above the lake is 120 m., (1), 30. According to the given statement, the diagram will be as, shown:, A, , r, O, r, B, , α/2, α, , β, , P, , , Now, in DDBC, DB, 4000, , = tan 60° fi, = 3   , (1), BC, x, 4000, , x=, ...(ii) (1), 3, From (i) and (ii),, 4000, 4000, , 4000 – y =, ⇒ y = 4000 −, (1), 3, 3, , ⇒, y = 4000 – 2312.14 = 1687.86 m, So, distance between the aeroplanes is 1687.86 m. (1), 29. In right DAEC,, AE, , = cot 30°, EC, , fi, AE = (h − 60) 3, ...(i) (1), , C,    Eye, (1), In the diagram, O is the centre of the balloon, P is the, eye of the observer. If PA and PB are tangents, angle, α, APB = a and ∠APO = ∠BPO = .(1), 2, Since the angle of elevation of the centre is b, , ⇒ ∠OPC = b, , α, (1), 2, OP, α, α, In DOAP, cosec, =, ⇒ OP = r cosec, ., OA, 2, 2, , ( OA = r) (1), OC, In DOPC, sin b =, ⇒ OC = OP sin b, OP, α, α, , i.e., OC = r cosec, sin b = r sin b cosec   , (1), 2, 2, 31. Let BD = x m and CD = y m, Required to prove: OC = r sin b cosec, , , , AB = 100 m, In triangle ABD,, 100, , tan 60° =, x, 100, 100, , ⇒, ⇒ x=, (1), 3 =, x, 3, In triangle ABC,, 1, AB, 100, , tan 30° =, ⇒, =, x+ y, x+ y, 3, , In right DAED,, AE, , = cot 60°, ED, , 96, , Mathematics–10, , , ⇒, , ⇒, , ⇒, , x + y = 100 3 (1), 100, y = 100 3 – x = 100 3 –, 3, 200, y=, (1), 3

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200, Distance, 3, =, Now, speed =, Time, 2, 200 100 100 3, =, =, =, (1), 3, 2 3, 3, , ⇒ speed of boat, 100×1.732 173.2, =, =, 3, 3, = 57.7 m/min, (1), 32. Let AB and CD be the two poles of equal height standing, on the opposite sides of the road BD., , , AB = CD, From figure in right ∆ABE, AB, , tan 60° =, & AB = BE 3 ...(i) (1), BE, , , , tan 60° =, , , ⇒, , AB, BC, , ⇒, , 3=, , h, x, ...(i) (1), , h= x 3 , , In right ∆ABD,, , , tan 30° =, , 1, h, AB, =, ...(ii) (1), &, BD, x + 40, 3, , Using (i) in (ii), we get, , , 1, x 3, =, & ^ x 3 h 3 = x + 40, x + 40, 3, , , ⇒, , 3x – x = 40, , , ⇒, , 2x = 40, , ⇒ x = 20, , (1), , So,, , h = 20 3 , , [From (i)], , , ⇒, , h = 20 × 1.732 ⇒ h = 34.64, , Therefore, the height of the tower is 34.64 m.  , , (1), , 34. In right DABD,,    , (1), Also, in right ∆CDE,, CD, AB, 1, =, , tan 30° =, &, ED, 80 – BE, 3, 80 – BE, ⇒, AB =, 3, , BE = 80 – 3 AB ...(ii) (1), Using (ii) in (i),, , AB = ^80 – 3 AB h 3, , , , BD, = cot 60° fi, AB, , , fi, , BD =, , , \, , AE = 20 3 m, , 60, 3, , =, , 60, 3, , ×, , 1, BD, =, (1), 3, 60, 3, 3, , = 20 3 m, ( BD = AE) (1), , = 80 3 – 3 AB , , ⇒, , 4 AB = 80 3, , ⇒ AB = 20 3 m, , , ∴ Height of poles = 20 3 m, Now,, , BE = 80 –, , = 80 –, , (1), , 3 AB, 3 × 20 3, ,   , , = 80 – 60 = 20 m, , , ED = 80 – BE = 80 – 20 = 60 m, , ∴ Point E is 20 m and 60 m away from both the poles. (1), 33. Let AB be a tower of height h m, and BC its shadow when, sun’s altitude is 60° and BD also its shadow when sun’s, altitude is 30°., , , , (1), , Let, , BC = x m, , Then,, , BD = (x + 40) m, , In right ∆ABC,, , [ CD = 40 m, given], , (1), , Now, in right DCEA,, CE, , tan 30° =, AE, , fi, , 1, , =, , CE, , fi CE = 20 m, (1), 20 3, 3, (i) Difference between the heights of the lighthouse and, the building = CE = 20 m, (ii) The distance between the lighthouse and the building, = BD = 20 3 m., (1), 35. Let h be the height of the tower and x be the horizontal, distance from the point of observation., , \, CB = ED = x and CE = BD = 40 m, In right DABC,, 1, AB, AB, , tan 30° =, fi, =, x, 3, BC, , Some Applications of Trigonometry, , 97

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fi, , x = AB 3 ...(i) (1), , , fi, , h = 120[ 3 − 1] (1), , , fi, , h = 120[1.73 – 1] m, , , fi, , h = 120 × 0.73 = 87.6 m, , , \ Height of the flagstaff is 87.6 m., , (1), , 40. 1.366 km, , (5), , 41. In right DBTP,, , fi,      , , (1), , Now, in right DAED, , , tan 60º =, , , fi, , 3 =, , h, x, , fi, , h, , x=, , 3, , h, , AB 3 =, , 3, , =, , TP, BP, , BP = TP 3, , ...(i) (1), ,      , , 3, , ⇒, , fi, , 2h = 120, , 3, AB = h – 40], , 60, , ⇒ h = 60 m, , From (ii),, , x=, , , ⇒, , x = 34.641 m, , 3, , (1), , h, , 3 (h − 40) =, , 3(h – 40) = h, , , ⇒, , 1, , fi, , ...(ii) (1), , [ AB + 40 = h, , , , , TP, BP, , AD, DE, , From equations (i) and (ii), we get, , , tan 30° =, , (1), , In right DGRT,, , , tan 60° =, , , fi, , = 20 3, (1), , 36. 2165 m, , (5), , 39. Let BC = x be the tower and BD be the flagstaff of height h., AC = 120 m, ∠BAC = 45° and ∠DAC = 60°, , GR =, , Now,, , fi, , TP 3 =, , TR, GR, TR, 3, , 3 =, , fi, , TR, GR, , ...(ii) (1), , TR, , (∵ BP = GR), 3, 3TP = TP + PR fi 2TP = BG, 50, m = 25 m, 2, , , fi, , TP =, , (1), , Now,, , TR = TP + PR = (25 + 50) m, , Height of tower = TR = 75 m, Distance between building and tower, = GR =, , fi,      , , (1), , GR =, , 75, , 42., , 3, , TR, 3, , ., , m = 25 3 m .(1), , In right-angled triangle ACB,, , , fi, , AC, = cot 45° fi, BC, , 120, =1, x, , x = 120, , ...(i) (1), , In right-angled triangle ACD,, , , fi, , CD, = tan 60° fi, AC, , h+ x, = 3, 120, , h + x = 120 3 (1), , , fi, , h = 120 3 − 120 , , 98, , Mathematics–10, , [using (i) x = 120], ,   , In right DADC,, , , ⇒, , 75, CD, CD = 75, , tan 45° =, , (1), ⇒ 1=, , 75, CD, , (1½)

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In right DADB,, 1, , 75, tan 30° =, BD, , 75, , ⇒, =, BD, 3, , ⇒, BD = 75 3 (1½), , ⇒ Distance between two ships, = BC = 75( 3 − 1) m = 54.9 m, (1), 43. Let AD be 1.6 m tall statue, BD the pedestal and C the, point of observation such that ∠ACB = 60° and ∠DCB =, 45°, In right DABC,, AB, , tan 60° =, BC, AD + BD 1.6 + BD, , ⇒, 3 =, =, BC, BC, 1.6 + BD, , ⇒, BC =, ...(i) (1), 3, , , ⇒, BC = 6 + h, In right DBCD,, CD, , tan 30° =, ⇒, BC, , ⇒, BC = h 3, From equations (i) and (ii), , 6+h= h 3 ⇒, , ⇒, , ⇒, , h( 3 − 1) = 6, , BD, BD, ⇒ 1=, BC, BC, , ⇒, BC = BD, From (i) and (ii),, 1.6 + BD, , = BD  , 3, , ⇒  1.6 + BD = 3 BD, , , , ⇒, , tan 45° =, , ...(ii) (1), , (1), , 3 BD – BD = 1.6 ⇒ BD( 3 − 1) = 1.6, , 3 +1, ×, 3 −1, 3 +1, 1.6(1.732 + 1), =, = 0.8 × 2.732, 2, = 2.1856 ≈ 2.19 m, (1), 44. Let AD be a flagstaff of height 6 m, CD = h be the tower, and B be the point of observation., , ⇒, , BD =, , 1.6, ,    , In right DABC,, , , tan 45° =, , 3, , =, , h, BC, ...(ii) (1), , h 3−h =6, 6, 3 −1, , 6( 3 + 1), = 3(1.732 + 1), 3 −1, h = 3 × 2.732 ⇒ h = 8.196, h = 8.2 m, , (1), , , ⇒, , ⇒, (1), 45. 14.64 m, (5), 46. Let BC be the river and AD = 10 m be the height of bridge,, , (1), , AD, 1, 10, ⇒, =, BD, BD, 3, , ⇒, BD = 10 3 m ...(i) (1), In right DADC,, AD, 10, , tan 45° =, ⇒ 1=, CD, CD, , ⇒, CD = 10 m, ...(ii) (1), From (i) and (ii),, , BC = BD + CD = 10 3 + 10 (1), = 10 × 1.732 + 10, = 17.32 + 10 = 27.32 m, So, the width of the river is 27.32 m., (1), 47. Let AB be the building and CD be the tower of height 60 m., In right DBCD,, CD, 60, , tan 60° =, ⇒, 3 =, BC, BC, 60, 3, ×, ,    BC =, = 20 3 m ...(i) (1), 3, 3, , , (1), , 1, , h=, ,     , In right DABD,, ,      , In right DDCB,, , ⇒ h=, , ...(i) (1), , tan 30° =, , (1), AC, BC, , ⇒ 1=, , AD + CD, BC, ,     , In right DABC,, , (1), , Some Applications of Trigonometry, , 99

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tan 30° =, , , ⇒, , AB =, , AB, BC, BC, , ⇒, =, , 1, , 20 3, , =, , 3, , AB, (1), BC, , , ⇒, , tan 45° =, , [From (i)] (1), , , ⇒, , 1=, , , , 3, 3, , ⇒, AB = 20 m, So, the height of building = 20 m, 48., , (1), , h, , , , , ⇒, , C, , A, , B, , 30°, O, , D, 80 m, , (1), , Let BD = river, (½), , , AB = CD = palm trees = h, , BO = x(½), , OD = 80 – x, h, In DABO, tan 60° = (½), x, h, , 3 =, x, , h= 3x, ...(i) (½), h, In DCDO, tan 30° =, (80 − x ), 1, h, , =, ...(ii) (½), 3, (80 − x ), Solving (i) and (ii), we get, , x = 20, , h = 3 x = 34.6, (½), The height of the trees = h = 34.6 m, , BO = x = 20 m, (½), , DO = 80 – x, , = 80 – 20 = 60 m, (½), 49. Let AB be the multi-storied building of height h m and CD, the building at a distance x m., F, , h, x, x=h, , Perpendicular , , Q tan θ =, , Base, , ...(i) (1), , BE, ED, h−8, 1, =, x, 3, , tan 30° =, , , ⇒, , h, , 60°, , , ⇒, In right DBDE,, , AB, , AC, , , ⇒, x = 3 (h − 8), From (i) and (ii), we get,, , , h=, , , ⇒, , 3h − h = 8 3, , , ⇒, , h ( 3 − 1) = 8 3, , ...(ii) (1), , 3h − 8 3, , , \, , h=, , 8 3, , , , h=, , 8 3, , , , h = 4 3 ( 3 + 1), , 3 −1, 3 −1, , ×, , 3 +1, 3 +1, , (1), , , h = 12 + 4 3 m, Distance between the two building, ,     x = (12 + 4 3 ) m [From (i)](½), 50., , , , 30°, , E, , C, , 45°, (h – 8) m, , E, , hm, , 30°, , D, , A, , 8m, 45°, xm, , (1), , AE = CD = 8 m , [Given], , BE = AB – AE = (h – 8) m, and, AC = DE = x m, [Given], Also,, ∠FBD = ∠BDE = 30°, (Alternate angles), , ∠FBC = ∠BCA = 45° (Alternate angles) (½), Now,, In right DACB,, , 100, , Mathematics–10, , D, , B, , From the figure, the angle of elevation for the first position, of the balloon ∠EAD = 60° and for second position ∠BAC, = 30°. The vertical distance, , ED = CB = 88.2 – 1.2 = 87 m., (1), Let, AD = x m and AB = y m, DE, Then in right DADE, tan 60° =, AD, 87, 3, , ⇒, =, x

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\, , x=, , 87, , ... (i) (1), , 3, , BC, In right DABC, tan 30° =, AB, 1, 87, , ⇒, =, y, 3, , \, y = 87 3, , y – x = 87 3 −, , , ⇒, , y–x=, , 87, 3, , ...(ii) (1), , , , , (1), , 87 3 × 3 − 87, , 3, 87 × 2 × 3, =, 3, y – x = 58 3 m, , , , = BD = y – x, , , , = 58 3 m (1), , Case Study Based Questions, , Subtracting (i) from (ii), we get, , , Hence, the distance travelled by the balloon, , =, , 87 × 2, , 9 3, m, 5, , 3. (d) 5.3 m, I. 1. (b), , 2. (c) 1.56 m, 4. (c) 4 m, , 5. (a) 30°, II. 1. (b) 45°, , 2. (d) 24.25 m, , , 3. (a) 20 3 m, , 3, , 4. (b) 45°, , 5. (c) angle of depression, III. 1. (a) 1118.29 km, , 2. (c) 1937 km, , , 3. (b) 566.96 km, , 4. (b) 45°, , , 5. (c) 1937 km, , IMPORTANT FORMULAE, •• The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric, ratios., A, •• In a right triangle ABC, right-angled at B,, BC, AC, , (i) sin C =, , AB, AC, , (ii) cos C =, , (iii) tan C =, , AB, BC, , (iv) cosec C =, , AC, 1, =, cos C BC, , (vi) cot C =, , (v) sec C =, , AC, 1, =, sin C AB, , BC, 1, =, tan C AB, , B, , C, , QUICK REVISION NOTES, •• One of the important application of trigonometry is to find the height or length of an object or the, distance between two distant objects., •• Line of sight: The line of sight is the line drawn from the eye of an observer to the point of the object, viewed by the observer., , •• Angle of elevation: The angle of elevation of an object viewed, is the angle formed by the line of sight, with the horizontal when it is above the horizontal level., , Some Applications of Trigonometry, , 101

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•• Angle of depression: The angle of depression of an object viewed, is the angle formed by the line of, sight with the horizontal level when it is below the horizontal level., , ••, •• Values of trigonometric ratios of standard angles:, –q, , 0°, , 30°, , 45°, , 60°, , 90°, , sin q, , 0, , 1, 2, , 1, 2, , 3, 2, , 1, , cos q, , 1, , 3, 2, , 1, 2, , 1, 2, , 0, , tan q, , 0, , 1, 3, , 1, , 3, , Not defined, , cosec q, , Not defined, , 2, , 2, , 2, 3, , 1, , sec q, , 1, , 2, 3, , 2, , 2, , Not defined, , cot q, , Not defined, , 3, , 1, , 1, 3, , 0, , COMMON ERRORS, Errors, , Corrections, , (i) Drawing incorrect figures for word problems and, so getting wrong solution or no solution., , (i) Read 2-3 times the statement given in the word, problems and then draw correct figure for the, given situation., , (ii) Taking incorrect value of trigonometric ratios., 1, For example, tan 60° =, and tan 30º = 3 ., 3, , (ii) Be careful when taking value of trigonometric, ratios. In this case, the correct value of, 1, tan 60° = 3 and tan 30° =, ., 3, , (iii) Do not using given values of, word problems., , 2 , 3 , etc. in (iii) Values of 2 , 3 , etc. if given in the questions,, have to be used as such., , (iv) Drawing untidy construction., , (iv) Draw construction with proper labelling and, neatly., , (v) Doing incorrect calculation and do not writing, the units in the answer., , (v) Be careful during calculation and must write units, in the answer., , (vi) Interpreting incorrectly about the length of (vi) If the angle of elevation of sun, above a, shadow of a perpendicular line when the angle of, perpendicular line (tower) decreases, then the, elevation of sun above it decreases or increases., length of shadow of a tower increases and viceversa., , qqq, , 102, , Mathematics–10

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\ 23-Nov-2021, , 6, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Surface Areas and Volumes, , Topics Covered, 1. Surface Areas and Volumes of Some Important Solids, 2. Surface Areas and Volumes of Combinations of Solids, , 1. Surface Areas and Volumes of Some Important Solids, Surface Area: Surface area of a solid body is the area of all of its surfaces taking together. It is measured in square unit., Volume: Space occupied by a solid body is called the volume of that particular solid body. It is always measured in cube, unit., Let as know important formulae related to surface areas and volumes of solid figures., 1. Cuboid: Let l, b and h denote respectively the length,, breadth and height of a cuboid. Then, , , , , (i) Total surface area = 2(lb + bh + hl) (unit)2, (ii) Lateral surface area = 2h (l + b) (unit)2, (iii) Diagonal of a cuboid, , 4. Right circular hollow cylinder: Let R and r be the, external and internal radii respectively of a hollow, cylinder of height h. Then, , = l 2 + b 2 + h 2 units, (iv) Volume of a cuboid = lbh (unit)3, 2. Cube: Let the length of each edge of a cube is a unit., Then, , , , , (i) Total surface area = 6a2 (unit)2, (ii) Lateral surface area = 4a2 (unit)2, (iii) Diagonal of a cube = 3 a units, (iv) Volume of a cube = a3 (unit)3, 3. Right circular cylinder: Let base radius and height, of circular cylinder be r and h respectively. Then, (i) Area of each end or base = pr2 (unit)2, (ii) Curved surface area = 2prh (unit)2, (iii) Total surface area = 2pr (h + r) (unit)2, (iv) Volume = pr2h (unit)3, , (i) Area of each end or base, = p (R2 – r2) (unit)2, (ii) Curved surface area, = External surface area + internal surface area, = 2pRh + 2prh, = 2ph (R + r) (unit)2, (iii) Total surface area, = 2pRh + 2prh + 2(pR2 – pr2), = 2ph (R + r) + 2p (R + r) (R – r), = 2p (R + r) (R + h – r) (unit)2, (iv) Volume of material, = External volume – Internal volume, = pR2h – pr2h = ph (R2 – r2) (unit)3, , 103

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , 5. Right circular cone: Let h, l and r denote the height,, slant height and base radius respectively of a cone., Then, θ, l, , h, , r, , , , Date __________, , (iv), , (v), , , Curved surface area of hemisphere, = 2pr2 sq. units, Total surface area of hemisphere, = 2pr2 + pr2 = 3pr2 sq. units, 2, (vi) Volume of hemisphere = pr3 cu. units, 3, 7. Spherical shell: Let R and r be the outer and inner, radii respectively of a spherical shell. Then, , r, Cone, , (i) Slant height, l = r 2 + h 2 units, (ii) Curved surface area = prl sq. units, (iii) Total surface area, = Curved surface area + Area of the base, = prl + pr2 = pr (l + r) sq. units, 1, (iv) Volume =, (Area of the base) × Height, 3, 1, , = pr2h cube units, 3, 6. Sphere and hemisphere: Let r be the radius of the, sphere and hemisphere both. Then, , , (i) Outer surface area = 4pR2 sq. units, (ii) Inner surface area = 4pr2 sq. units, (iii) Volume of material, 4, = p (R3 – r3) cu. units, 3, 8. Hemispherical shell: Let R and r be the outer and, inner radii respectively of hemispherical shell. Then, O, , r, , O, r, , r, , r, , r, , r, , R, , r, , Hemispherical shell, , (i) External curved surface area, = 2pR2 sq. units, Sphere, Hemisphere, , (ii) Internal curved surface area, (i) Curved surface area of sphere, = 2pr2 sq. units, = 4pr2 sq. units, (iii) Total surface area, (ii) Total surface area of sphere, = 2pR2 + 2pr2 + p(R2 – r2), = 4pr2 sq. units, = p(3R2 – r2) sq. units, 4 3, (iv) Volume of material, (iii) Volume of sphere = pr cu. units, 3, 2, = p(R3 – r3) cu. units, 3, Example 1. If the number of square centimetres on surface of a sphere is equal to the number of cubic centimetres in its, volume, what is the diameter of the sphere?, 4 3, Solution. Given:, 4pr2 = πr, 3, 2, ⇒, 3r = r3 ⇒ 3 = r, \ Diameter of the sphere = 2r = 6 cm., Example 2. Two right circular cones X and Y are made. X having three times the radius of Y and Y having half the volume, of the X. Calculate the ratio of the heights of X and Y., Solution. Let radius (r1) of cone Y be r, then the radius (r2) of cone X = 3r, Let h1 and h2 be the heights of X and Y respectively., 1, Since, Volume of cone Y =, × volume of cone X, 2, 1 2, 1 1, 1 2, 1 JK 1 2 NO, πr h2 = × π(3r ) 2 h1, ⇒, πr1 h2 = KK πr 2 h1OO ⇒, 3, 3, 2L3, 2 3, P, , 104, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , h1, 9, 2, h1 ⇒, = = 2:9, h2, 9, 2, Example 3. The internal and the external diameters of a hollow hemispherical vessel are 42 cm and, 45.5 cm respectively. Find its capacity and outer curved surface area., O, Solution. Let D = 45.5 cm be the external and d = 42 cm be the internal diameters of a hollow, hemisphere., 42, = 21 cm, Then, internal radius,, r=, 2, 45.5, = 22.75 cm, and external radius,, R=, 2, \ Capacity of hollow hemispherical vessel = Internal volume of hollow hemispherical vessel, 2 3, 2 22, × (21)3 cm3 = 19404 cm3, = πr = ×, 3 7, 3, 22, × (22.75) 2 cm 2 = 3253.25 cm2, Outer curved surface area = 2pR2 = 2 ×, 7, Example 4. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top which is open, is 5, cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the, vessel, one-fourth of water flows out. Find the number of lead shots dropped into the vessel. [NCERT] [CBSE 2015], , ⇒, , h2 =, , 15 cm, , 8 cm, , Solution. We have, height of the conical vessel, h = 8 cm, O 5 cm, and radius of the conical vessel, r = 5 cm, 200, 1 2, 1, 2, π cm3, \ Volume of water filled in the vessel cone = π r h = π × 5 × 8 =, 3, 3, 3, Also, we have radius of a spherical lead shot = 0.5 cm, 4 3 4, 3, 3,  Volume of each lead shot = π r = π (0.5) cm, 3, 3, \ Volume of lead shots dropped = Volume of water that overflows, 1 200, 50, π cm3 =, π cm3, = ×, 3, 4, 3, 50, π, 3, = 100, \ Number of lead shots dropped =, 4, π × 0.5 × 0.5 × 0.5, 3, Hence required number of lead shots is 100., Example 5. The sides of a right angled triangle are 15 cm and 8 cm. If the triangle is revolved around the hypotenuse,, find the volume and curved surface area of the double cone generated., Solution. Let ABC be the given triangle in which ∠ABC = 90°, AB = 15 cm, BC = 8 cm, \, AC2 = 152 + 82 = 289 ⇒ AC = 17 cm, 1, 1, A, \, Area of DABC = × AC × OB = × 17 × OB, 2, 2, 1, 1, 2, Also,, area of DABC = × BC × AB = × 8 × 15 cm, 2, 2, 120, 1, 1, cm, × 8 × 15 ⇒ OB =, × 17 × OB =, So,, 17, 2, 2, Volume of the double cone generated, O, 1, 1, 2, 2, = × π × (OB) × OA + × π × (OB) × OC, 3, 3, B, C, 8 cm, 1, 2, = × π × (OB) × (OA + OC), 3, 2, , =, , 1 22  120 , ×, ×,  × 17, 3 7  17 , , [Q OA + OC = AC = 17], , Surface Areas and Volumes, , 105

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Amit, , Proof-3, , Reader’s Sign _______________________, , A, , 15, cm, , 105600, cm3, =, 119, Curved surface area of the double cone generated, = p × OB × AB + p × OB × BC, 22 120, = pOB (AB + BC) =, ×, × 23, 7, 17, 60720, cm 2, =, B, 119, Example 6. A copper wire of diameter 8 mm is evenly wrapped on a cylinder of length, 24 cm and diameter 49 cm to cover the whole surface. Find:, (i) the length of the wire, (ii) the volume of the wire., Solution. Given: The thickness of wire = its diameter = 8 mm = 0.8 cm., And, the length of the cylinder = 24 cm, , \ Number of turns of the wire required to cover the whole surface of the cylinder, 24, Length of the cylinder, = 30, =, =, 0.8, Diameter of the wire, , Date __________, , 15, cm, , \ 23-Nov-2021, , O, 8c, , B′, , m, , 8c, , m, , C, , Since, diameter of the cylinder = 49 cm, 49, , \ Radius of the cylinder, r =, cm, 2, 22 49, (i) Length of wire wrapped in 1 round = Circumference of the cylinder = 2πr = 2 ×, = 154 cm, ×, 7, 2, Length of wire wrapped in 30 rounds = 30 × 154 cm = 4620 cm, 0.8, = 0.4 cm, (ii) Since radius (r) of wire =, 2, and its length or height (h) = 4620 cm, 22, × (0.4) 2 × 4620 = 2323.2 cm3, , \, Volume of the wire = pr2h =, 7, , Conversion of a Solid from one Shape to Another, Sometimes, we need to convert a solid into another solid of different shape or size or both. In such situations,, (i) Total volume of the solids to be converted, = Total volume of the solids into which the given solid(s) is (are) to be converted, (ii) Number of solids of a given shape in which a given solid is to be converted, Total volume of the solid to be converted, =, Volume of one converted solid, Example 7. How many coins of 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid of 11 cm × 10 cm, × 7 cm?, [Imp.], Solution. Let n coins to be melted., 1.75, Since, for each coin, r=, cm = 0.875 cm, 2, and, height (h) = 2 mm = 0.2 cm, \, Volume of each coin = pr2h = p(0.875)2 × 0.2 cm3, 22, × (0.875) 2 × 0.2 × n cm3, ⇒, Volume of n coins =, 7, As per the statement:, Volume of n coins to be melted = Volume of the cuboid to be formed, 22, × 0.875 × 0.875 × 0.2 × n = 11 × 10 × 7, ⇒, 7, 11 × 10 × 7 × 7, ⇒, n=, = 1600, 22 × 0.875 × 0.875 × 0.2, Thus, 1600 coins are required for the given purpose., , 106, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 8. A well is to be dug with 4 m inside diameter and 10 m in depth. Find the quantity of earth to be excavated., The earth taken out is spread all round to a width of 4 m to form an embankment. Find the height of the embankment., 4, Solution. We have, inside radius (r) of the well = m = 2 m, [Imp.], 2, \ The quantity of earth excavated = Volume of the well, 22, × (2) 2 × 10 m3 = 880 m3, = pr2h =, 7, 7, Let height of the embankment = h m, h, R, Its internal radius (r) = 2 m, fi, External radius (R) = 2 + 4 = 6 m, Now, the embankment is of the form of a hollow cylinder., \ Volume of the embankment = Volume of the well, 880, 880, 22 2, fi, p (R2 – r2) h =, fi, (6 – 22) h =, 7, 7, 7, 880, 22, fi, × 32 × h =, 7, 7, 880 × 7, fi, h=, = 1.25 m, 7 × 22 × 32, , 4 cm, , Example 9. Two cubes each of volume of 64 cm3 are joined end to end. Find the surface area of the resulting cuboid., Solution. Since, the volume of cube = (edge)3 fi 64 = (Edge)3, [Imp.], 3, 3, fi, (Edge) = 4, \, Edge = 4 cm, As two cubes are joined end to end, so for the resulting cuboid, length = 2 × 4 = 8 cm, breadth = 4 cm, and, height = 4 cm, 4, So,, required surface area = 2 [l × b + b × h + h × l], cm, 2, = 2(32 + 16 + 32) = 160 cm, 8 cm, , Exercise 6.1, I. Very Short Answer Type Questions, [1 mark], 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) The radius of a sphere (in cm) whose volume is 12p cm3, is, [CBSE Standard 2020], 2/3, 1/3, (a) 3, (b) 3 3, (c) 3, (d) 3, (2) If the surface area of a sphere is 144p, then its radius is, (a) 6 cm, (b) 8 cm, (c) 12 cm, (d) 10 cm, (3) The edge of a cube whose volume is equal to that of a cuboid of dimensions 8 cm × 4 cm × 2 cm is, (a) 6 cm, (b) 4 cm, (c) 2 cm, (d) 8 cm, (4) If the radii of two spheres are in the ratio 2 : 3, then the ratio of their respective volumes is, (a) 8 : 27, (b) 3 : 5, (c) 7 : 24, (d) 5 : 14, (5) The edge of a cube whose volume is 8x3 is, (a) x, (b) 2x, (c) 4x, (d) 8x, (6) If the volume of a 7 cm high right circular cylinder is 448p cm3, then its radius is equal to, (a) 2 cm, (b) 4 cm, (c) 6 cm, (d) 8 cm, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., , Surface Areas and Volumes, , 107

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , (1) Assertion (A): Total surface area of the cylinder having radius of the base 14 cm and height 30 cm is 3872 cm2., , Reason (R): If r be the radius and h be the height of the cylinder, then total surface area = (2prh + 2pr2)., (2) Assertion (A): If the height of a cone is 24 cm and diameter of the base is 14 cm, then the slant height of the cone, is 15 cm., , Reason (R): If r is the radius and h the height of the cone, then slant height = h 2 + r 2 ., 3. Answer the following., (1) Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?, , [Delhi 2017], (2) Total surface area of a cube is 216 cm2. Find its volume., [CBSE SP 2012], (3) If a solid right-circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, find, the radius of the sphere., [CBSE SP 2012], (4) Find the curved surface area of a right-circular cone of height 15 cm and base diameter 16 cm. , [CBSE 2011], (5) Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?, , [CBSE Standard 2020], (6) Find the radius of the sphere whose surface area is 36 p cm2., (7) 12 solid spheres of the same radii are made by melting a solid metallic cylinder of base diameter 2 cm and height 16, cm. Find the diameter of the each sphere., [CBSE Standard SP 2020-21], II. Short Answer Type Questions - I, [2 Marks], 4. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the, cylinder.[NCERT], 5. Three solid metal cubes of edges 6 cm, 8 cm and 10 cm are melted and recasted into a single solid cube. Find the length, of the edge of the cube so obtained., [Imp.], 6. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness, of the wire., [NCERT] [Imp.], 7. Two spheres of same metal weigh 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., [CBSE 2015], 8. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by, 14 m. Find the height of the platform., [NCERT], 9. 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., Aisha, a juice seller, was serving juice to her customers in two types of glasses. Both the glasses had inner radius 3 cm., The height of both the glasses was 10 cm., , First type: A glass with hemispherical raised, Second type: A glass with conical raised, bottom., 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?, , [CBSE Standard SP 2019-20], 10. A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their, volumes., [CBSE Standard 2020], 1, 11. The volume of a right circular cylinder with its height equal to the radius is 25 cm3 . Find the height of the cylinder., 7, 22, [Use p =, ] [CBSE Standard 2020], 7, III. Short Answer Type Questions - II, [3 Marks], 12. A solid sphere is melted and recasted into a hollow cylinder of uniform thickness. If the external radius of the base of the, cylinder is 4 cm, its height 24 cm and thickness 2 cm; find the radius of the sphere., [Imp.], , 108, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 13. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much, canvas cloth is required to just cover the heap?, [CBSE 2018], 14. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere., Find the radius of the sphere, [NCERT] [Imp.], 15. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5, cm × 10 cm × 3.5 cm?, [CBSE 2011], , 3, th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a, 4, cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel., [Delhi 2017], 17. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30, minutes, if 8 cm standing water is needed?, [NCERT] [Delhi 2019, SP 2018], 18. In a hospital, used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is, used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will, be the height of standing water used for irrigating the park?, [Delhi 2017], 19. In a rain-water harvesting system, the rain-water from a roof of 22 m × 20 m drains into a cylindrical tank having diameter, of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm. , 20. Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted to form a solid sphere. Find the radius of the, resulting sphere., IV. Long Answer Type Questions, [5 Marks], 21. A well, whose diameter is 3 m, has been dug 21 m deep and the earth dug out is used to form an embankment 4 m wide, around it. Find the height of the embankment., [Imp.], 22. A farmer connects a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in, diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?, [NCERT], 23. Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from an underground tank, which is in the shape of a cuboid. The underground tank has dimensions 1.57 m × 1.44 m × 95 cm. The overhead tank, has its radius of 60 cm and its height is 95 cm. Find the height of the water left in the underground tank after the overhead, tank has been completely filled with water from the underground tank which had been full. Compare the capacity of the, overhead tank with that of the underground tank., [Use p = 3.14] [NCERT], 24. In a cylindrical vessel of radius 10 cm, containing some water, 9000 small spherical balls are dropped which are completely, immersed in water which raises the water level. If each spherical ball is of radius 0.5 cm, then find the rise in the level, of water in the vessel., [CBSE Standard 2020], 25. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the, rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour?, [CBSE Standard SP 2020-21], 16. The, , Case Study Based Questions, I. Adventure camps are the perfect place for the children to practise, decision making for themselves without parents and teachers guiding, them 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 while 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., , d = 7 cm, h = 10.5 cm, , d = 7 cm, , Surface Areas and Volumes, , 109

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , Date __________, , 2, , Area = 551 m, , r=7m, , 1. The volume of cylindrical cup is, (a) 295.75 cm3, (b) 7415.5 cm3, (c) 384.88 cm3, (d) 404.25 cm3, 2. The volume of hemispherical cup is, (a) 179.67 cm3, (b) 89.83 cm3, (c) 172.25 cm3, (d) 210.60 cm3, 3. Which container had more juice and by how much?, (a) Hemispherical cup, 195 cm3, (b) Cylindrical glass, 207 cm3, 3, (c) Hemispherical cup, 280.85 cm, (d) Cylindrical glass, 314.42 cm3, 4. The height of the conical tent prepared to accommodate four students is, (a) 18 m, (b) 10 m, (c) 24 m, (d) 14 m, 5. How much space on the ground is occupied by each student in the conical tent, (a) 54 m2, (b) 38.5 m2, (c) 86 m2, (d) 24 m2, , Answers and Hints, 1. (1), (3), (5), 2. (1), , 2, 33, , (c), (1) (2) (a) 6 cm, (1), (b) 4 cm (1) (4) (a) 8 : 27, (1), (b) 2x, (1) (6) (d) 8 cm, (1), (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A)., Total surface area = 2prh + 2pr2, , = 2pr(h + r), 22, , = 2×, × 14(30 + 14), 7, , = 88(44) = 3872 cm2, (2) (d) Assertion (A) is false but reason (R) is true., 2, , Slant height =, , , =, , , , =, ,  14 , 2,   + (24), 2, , 9, 2 3, pr = 3pr2 fi r = units, 2, 3, , \, d = 9 units, (1), (2) , 6l2 = 216 ⇒ l2 = 36, , ⇒, l=6, , \ Volume of cube = l 3 = (6) 3 = 216 cm 3 (1), (3) Volume of cone = volume of sphere, 1, 4, , \ p62 × 24 = pr3, 3, 3, , ⇒, 864 = 4r3, , ⇒, r3 = 216, , ⇒, r = 6 cm, So, radius of sphere = 6 cm, (1), , 110, , 8 + 15, (Q Diameter = 16 cm), , Mathematics–10, , r1, 3, =, 1, r2, , Now,, 1 2, πr1 h1, V1 (Volume of first cone), , = 3, 1 2, V2 (Volume of second cone), πr h, 3 2 2, , ⇒, , 3. (1) , , (4) Slant height of cone, l =, , , Then,, , 2, , 49 + 576, 625 = 25, , 2, , , ⇒, l = 17 cm, , \ CSA of cone = prl = p × 8 × 17, , = 136 p cm2(1), (5) Let h1 and h2 be the heights of two cones., h1, 1, Then,, =, 3, h2, Also, let r1 and r2 be the radii of two cones., , 2, , r , h , V1, r2 h, = 12 × 1 =  1  ×  1 , V2, h, r,  2,  h2 , r2, 2, 2, , 9 1,  3,  1, =   ×   = ×,  1,  3, 1 3, 3, =3:1, 1, (6) Let r be the radius of sphere., =, , Then,, , ⇒, , ⇒, (7) , , 4pr2 = 36p, r2 =, , 36π, =9, 4π, , r = 3 cm, pR2H = 12 ×, , 4 3, πr, 3

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 4, 1 × 1 × 16 = × r 3 × 12 (½), , r3 = 13, , r=1, , d = 2 cm, (½), 4. Radius of the sphere (r1) = 4.2 cm, , \ Volume of the sphere, 4 22 42 42 42,  4 3, =  πr1  = × × × ×, cm3, , 3, 3 7 10 10 10, Radius of the cylinder (r2) = 6 cm, Let ‘h’ be the height of the cylinder, , \ Volume of the cylinder = pr22h, 22, =, × 6 × 6 × h cm3(1), 7, Since, Volume of the metallic sphere, = Volume of the cylinder., 4 22 42 42 42, × × × ×, 3 7 10 10 10, 22, =, ×6×6×h, 7, 4 22 42 42 42 7 1 1, , ⇒ h= × × × ×, × × × cm, 3 7 10 10 10 22 6 6, 2744, 4 × 7 × 7 × 14, =, =, 1000, 10 × 10 × 10, , ⇒, , = 2.744 cm, 5. Volume of single cube so obtained, , (1), , 6. Volume of rod = volume of wire, 2,  1, , \ p ×   × 8 = p × r2 × 1800,  2, ⇒ r2 =, , 1, 900, , 1, cm(1), 30, So, thickness of wire, = diameter of cross-section of wire, 1, 1, =, ×2=, cm = 0.67 mm (approx), 30, 15, (1), , ⇒, , r=, , 7. 12 cm, , (2), , 22 7 7, × × × 20, 7 2 2, = 22 × 7 × 5 m3, , ⇒ Volume of the earth taken out = 22 × 7 × 5 m3(1), Now this earth is spread out to form a cuboial platform, having length = 22 m, breadth = 14 m, Let ‘h’ be the height of the platform., , \ Volume of the platform = 22 × 14 × h m3, , \ 22 × 14 × h = 22 × 7 × 5, 22 × 7 × 5 5, = = 2.5 m, , ⇒, h=, 22 × 14, 2, 8. Volume of well = pr2h =, , Thus, the required height of the platform is 2.5 m., (1), 9. Capacity of first glass, 2, = πr 2 H − πr 3, 3, = p × 9(10 − 2) = 72p cm3(1), Capacity of second glass, 1, = πr 2 H − πr 2 h = p × 3 × 3(10 − 0.5), 3, = 85.5p cm3, , ∴ Suresh got more quantity of juice., (1), 10. Let the radii of a cone and a cylider be r, And, let the height of the cylinder be h, Then, the height of the cone = 3h, 1, Now, volume of cone (V1) = πr 2 (3h) = pr2h, 3, and volume of cylinder (V2) = pr2h(1), πr 2 h, V1, =, πr 2 h, V2, , ⇒, V1 : V2 = 1 : 1, (1), 11. Let the height and radius of cylinder be r, , [Q Height of cylinder = radius of cylinder], Then, Volume of cylinder = pr2h, 176, 1, , ⇒, pr2 × r = 25, ⇒ pr3 =, (1), 7, 7, 22, 176, 176, , ⇒, × r3 =, ⇒ r3 =, 7, 7, 22, , ⇒, r = 2 cm, (1), 12. Volume of sphere = volume of hollow cylinder, 4, , ⇒, p × r3 = p × 24 × (42 – 22)(1), 3, 4πr 3, 288π × 3, , ⇒, = 288 p ⇒ r3 =, , (1), 3, 4π, , ⇒, r3 = 216 ⇒ r = 6 cm, (1), , \, , = (63 + 83 + 103) cm3, , ⇒ l3 = 216 + 512 + 1000 = 1728 ⇒ l = 12 cm(2), , , ⇒ 2p = pr2 × 1800, , Date __________, , 13. Radius of conical heap = 12 m, 1, 22, Volume of rice =, ×, × 12 × 12 × 3.5 m2, 7, 3, = 528 m3(1), , Area of canvas cloth required = prl, , , l=, , 122 + (3.5) 2 = 12.5 m, , , \ Area of canvas required =, , (1), , 22, × 12 × 12.5, 7, , = 471.4 m2(1), 14. Volume of sphere = volume of cone, (1), 4 3 1, , ⇒, pr = × p × 6 × 6 × 24, (1), 3, 3, , ⇒, r3 = 6 × 6 × 6 ⇒ r = 6, , \ Radius of sphere = 6 cm, , Surface Areas and Volumes, , (1), , 111

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 2, , 15. Volume of coin =, , , 22  175 , 2, 3, ×,  × cm (1), 7  200  10, |Q A coin is like a cylinder, , Let the number of coins need to melt be ‘n’, 55 35   22 175 175 2 , , ÷, ×, ×, ×, ×, , \ n = 10 ×, (1), 10 10   7 200 200 10 , , 55 35 7 200 200 10, ×, ×, × = 400, = 10 × × ×, 10 10 22 175 175 2, Thus, the required number of coins = 400., 3, 16., × Volume of conical vessel, 4, , (1), , = Volume of cylindrical vessel, , (1), , Let the height of cylindrical vessel be h, 6, 3 1, (1), × × π × 5 × 5 × 24 = p × 10 × 10 × h  , 4 3, 3, , fi, h = cm or 1.5 cm (1), 2, 17. Let the area that can be irrigated in 30 minute be A m2, Water flowing in canal in 30 minutes, 1, , = 10000 ×  m = 5000 m, , 2, Volume of water flowing out in 30 minutes, = (5000 × 6 × 1.5) m3, = 45000 m3...(i) (1), Volume of water required to irrigate the field, 8, = A ×, m3, ...(ii) (1), 100, Equating (i) and (ii), we get, 8, , A×, = 45000, 100, , ⇒, A = 562500 m2.(1), 18. Volume of water in cylindrical tank, = Volume of water in park (1), 22, fi, × 1 × 1 × 5 = 25 × 20 × h(1), 7, where h is the height of standing water, 11, 22, m or, cm (1), fi, h=, 350, 7, 19. Volume of rain water on the roof, = Volume of cylindrical tank, (1), 22, , i.e. 22 × 20 × h =, × 1 × 1 × 3.5, (1), 7, , , fi, , , ⇒, 20. r1 = 6 cm, , r2 = 8 cm, , r3 = 10 cm, , h=, , 1, m = 2.5 cm, 40, , (1), , , ⇒, , r=, , , ⇒, , 4 3 4 3 4 3, πr1 + πr2 + πr3 (1), 3, 3, 3, 4, 3, 3, 3, π (r1 + r2 + r3 ), 3, 63 + 83 + 103, 1728, 3, , 1728, , ⇒, r = 12 cm, Therefore, the radius of the resulting sphere is, 12 cm., (1), 21. 1.69 m, (5), 22. Diameter of the pipe = 20 cm, 20, , ⇒ Radius of the pipe (r) =, = 10 cm, (1), 2, Since, the water flows through the pipe at 3 km/hr., , \ Length of water column per hour (h), = 3 km = 3 × 1000 m = 3000 × 100 cm, (1), = 300000 cm., , \ Volume of water = pr2h, , = p × 102 × 300000 cm3, , = p × 30000000 cm3(1), Now, for the cylindrical tank,, , Diameter = 10 m, 10, = 5 × 100 cm = 500 cm, 2, Height (H) = 2 m = 2 × 100 cm = 200 cm, , \ Volume of the cylindrical tank, = pR2H = p × (500)2 × 200 cm3(1), Now, time required to fill the tank, [ Volume of the tank ], =, [ Volume of water flown in 1 hour ], , ⇒ Radius (R) =, , =, , π × 500 × 500 × 200, hrs, π × 30000000, , 5×5×2, 5, 5, hrs = hrs = × 60 minutes, 30, 3, 3, = 100 minutes., =, , (1), , 23. The volume of water in the overhead tank equals the, volume of the water removed from the sump., Now, the volume of water in the overhead tank (cylinder), = pr2h = 3.14 × 0.6 × 0.6 × 0.95 m3(1), The volume of water in the sump when full, = l × b × h = 1.57 × 1.44 × 0.95 m3(1), The volume of water left in sump after filing tank, = [(1.57 × 1.44 × 0.95), , 4, Volume of sphere = πr 3 (1), 3, Volume of the resulting sphere, = Sum of the volumes of the smaller spheres, , 112, , , ⇒, , ⇒, , 4 3, πr =, 3, 4 3, πr =, 3, r3 =, r3 =, , , , Date __________, , Mathematics–10, , , , – (3.14 × 0.6 × 0.6 × 0.95)] m3, , = (1.57 × 0.6 × 0.6 × 0.95 × 2) m2(1), So, the height of water left in the sump

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\ 23-Nov-2021, , Amit, , Proof-3, , Volume of water left in the sump, l×b, 1.57 × 0.6 × 0.6 × 0.95 × 2, =, = 0.475 m, 1.57 × 1.44, = 47.5 cm, , Reader’s Sign _______________________, , =, , Capacity of tank, Capacity of sump, 1, 3.14 × 0.6 × 0.6 × 0.95, =, =, 2, 1.57 × 1.44 × 0.95, Therefore, the capacity of the tank is half the capacity of, the sump., (1), 24. Let radius of spherical ball, r1 = 0.5 cm, Then volume of one spherical ball,, 4, , V1 = × πr13, 3, 4 22, , ⇒, V1 = ×, × 0.5 × 0.5 × 0.5 (1), 3 7, 4 22 5, 5, 5, , ⇒, V1 = ×, ×, ×, ×, 3 7 10 10 10, , ⇒, , 11, V1 =, cm3, 21, , So, volume of 9000 spherical balls, 99000, 11, = 9000 ×, =, cm3 (1), 21, 21, Radius of cylindrical vessel = 10 cm, Let height of water raised = h, , Q Volume of 9000 balls, = Volume of water raised, , , ⇒, , 99000, = pr2h(1), 21, , , ⇒, , 99000, 22, × 10 × 10 × h (1), =, 21, 7, , (1), , Also,, , Date __________, , , ⇒, , h=, , 99000 × 7, 21 × 22 × 10 × 10, , , ⇒, h = 15 cm, 25. For pipe, r = 1 cm, , (1), (½), , Length of water flowing in 1 sec,,  , ,    h, , = 0.7 m = 70 cm, , (½), , Cylindrical Tank, R = 40 cm, rise in water level = H (½), Volume of water flowing in 1 sec,    = pr2h = p × 1 × 1 × 70 = 70p,                                                                  , , (½), , Volume of water flowing in 60 sec,     = 70p × 60, , (1), , Volume of water flowing in 30 minutes,     = 70p × 60 × 30, , (½), , Volume of water in Tank,     = pr2H = p × 40 × 40 × H, , (½), , Volume of water in Tank, , , = Volume of water flowing in 30 minutes, ,   p × 40 × 40 × H = 70p × 60 × 30, , (½), ,        H = 78.75 cm, , (½), , Case Study Based Questions, I. 1. (d) 404.25 cm3, 2. (b) 89.83 cm3, , 3. (d) Cylindrical glass, 314.42 cm3, , 4. (c) 24 m, 5. (b) 38.5 m2, , 2. Surface Areas and Volumes of Combinations of Solids, , Sometimes, we come across some complex solid figures which are combination of two or more similar or different solid, figures., For example:, • A circus tent which is a combination of a cylindrical portion surmounted by a conical or hemispherical roof., • Many toys which are in the form of a cone mounted on a hemisphere., In the same way, a solid can be composed of a cylinder with one hemispherical end or with both hemispherical ends., Examples of combination of solids, (i) This is a cube surmounted by a cone., (a) Total surface area = Total surface area of cube – Base area of the cone, , + Curved surface area of the cone, = 6a2 – pr2 + prl (unit)2, (b) Total volume = Volume of cube + Volume of cone, (i), 1 2, l, 3, = (a) + πr h, 3, r, (ii) This is a cylinder surmounted by a cone., (a) Total surface area = Curved surface area of cone, (ii), + Curved surface area of cylinder, h, h, + Area of circular base, = prl + 2prh + pr2 sq. units, r, (b) Total volume = Volume of cylinder + volume of cone, , Surface Areas and Volumes, , 113, , (ii)

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 1, = pr2h + πr 2 h, 3, 4 2, = πr h, 3, (iii) This is a cylinder surmounted by a hemisphere., (a) Total surface area = Curved surface area of hemisphere, + Curved surface area of cylinder + Area of circular base, = 2pr2 + 2prh + pr2 = 3pr2 + 2prh, = pr(3r + 2h) sq. units, (b) Total volume = Volume of hemisphere + Volume of cylinder, 2, = πr3 + pr2h, 3, JK 2, NO, = pr2 KK r + hOO, l, L3, P, (iv) This is a hemisphere surmounted by a cone as in some toys., r, (a) Total surface area = Curved surface area of cone, r, + Curved surface area of hemisphere, = prl + 2pr2 sq. units, (iv), (b) Total volume = Volume of cone + Volume of hemisphere, 1, 2, = πr 2 h + πr3, 3, 3, 1, = πr 2 (h + 2r), 3, (v) A cylindrical pipe, tunnel, factory shed are few examples of figures with uniform cross-section., , r, , h, , (iii), , Cross-section, Cross-section, , Length, , Length, In case of a figure with uniform cross-section, its, (a) Surface area (excluding cross-section) = Perimeter of cross-section × length, (b) Volume (capacity) = Area of a cross-section × length, Example 1. A solid is in the form of a cone mounted on a hemisphere with both their radii being equal to 1 cm and the, height of the cone is equal to its radius. Find the volume of the solid in terms of p.[NCERT], Solution. Let the radius of the hemisphere, r = 1 cm, Then, the height of the cone, h = 1 cm, r, The volume of the solid, = Volume of the cone + Volume of the hemisphere, r, 1, 2, = π r 2 h + π r 3, 3, 3, 1, 2, 1, 2, = π × (1)2 × (1) + π (1)3 = π + π = p cm3, 3, 3, 3, 3, Example 2. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of, the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid., [NCERT], Solution. Given: The diameter of the hemispherical depression = l, l, \, r=, 2, The surface area of the remaining solid, = Total S.A. of the cube – Area of the circle, , + Curved S.A. of the hemisphere, = 6 × (side)2 – pr2 + 2pr2 = 6 × (side)2 + pr2, 2, l, = 6 × l 2 + π  ,  2, , 114, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 1, π l 2 24l 2 + π l 2, =, = l 2 (24 + π) (unit)2, 4, 4, 4, Example 3. A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20, m. The height of the cylindrical and conical portions are 4.2 m and 2.1 m respectively., Find the volume inside the tent., Solution. Let h1 = 4.2 m be the height of cylindrical portion, h2 = 2.1 m be the height of conical portion and r = 20 m be, the radius of tent., 2.1 m, Then, the volume inside the tent, = Volume of cylindrical part + Volume of conical part, 1, = pr2h1 + pr2h2, 4.2 m, 3, 1 22, 22, × (20) 2 × 2.1 m3, =, × (20)2 × 4.2 m3 + ×, 20 m, 3 7, 7, 3, 3, 3, = 5280 m + 880 m = 6160 m ., Example 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter hemisphere can, have? Find the surface area of the solid., [Imp.], Solution. The figure shows the solid as per the question:, 7 cm, Now the greatest diameter the hemisphere can have, = size of the cube = 7 cm, 7 cm, Surface area of the solid, 7 cm, = Total S.A. of the cube – Area of the circle common to the hemisphere and the cube + Curved, S.A. of the hemisphere, = 6 × (side)2 – pr2 + 2pr2 = 6 × (side)2 + pr2, = 6 l 2 +, , 2, , 22  7 , 7, ×   sq. cm, ( Diameter = 7 cm ⇒ radius =, cm), 7  2, 2, = (294 + 38.5) sq. cm = 332.5 sq. cm, Example 5. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at, its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a, height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions, of the model to be nearly the same.), [NCERT] [Imp.], Solution. The figure shows the model as per the given question:, Now, for the cylinder, 3, 2 cm, radius,, r=, cm, 2, and height,, h1 = (12 – 2 × 2) cm, = 8 cm, 8 cm, 3 cm, Also, for each conical portion, 3, radius (r) =, cm, 2, 2 cm, and, h2 = 2 cm, Volume of the air contained in this model, = Volume of the cylindrical part + 2 × Volume of conical part, 1 2, = pr2h1 + 2 × π r h2, 3, = 6 × 7 2 +, , 2, , 2, , = 22 ×  3  × 8 cm3 + 2 × 22 ×  3  × 2 cm3, 7  2, 3 7  2, 2, , , 3, 8 + 3 × 2 = 66 cm, Example 6. A playing top (lattu) is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and, the diameter of top is 3.5 cm. Find its whole surface area., [NCERT] [Imp.], =, , 22 9, ×, 7 4, , Surface Areas and Volumes, , 115

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Amit, , Proof-3, , Reader’s Sign _______________________, , 5 cm, , \ 23-Nov-2021, , Date __________, , 3.5 cm, , Solution. For the hemispherical part,, 3.5, 1.75 cm, radius (r) =, cm = 1.75 cm, r, 2, For the conical part, r = 1.75 cm, and, height (h) = 5 cm – 1.75 cm = 3.25 cm., h, 3.25 cm, If slant height of conical part be l cm, l2 = h2 + r2, ⇒, l2 = (3.25)2 + (1.75)2 = 13.625 ⇒ l = 13.625 cm = 3.7 cm (approx.), \ Whole surface area, = Curved surface area of the conical part + C.S.A. of hemisphere, = prl + 2pr2, 22, ,  22, × (1.75) 2  cm2 = 39.6 cm2., =  × 1.75 × 3.7 + 2 ×, 7, 7, , Example 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical, part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making, the tent. Also, find the cost of the canvas of the tent at the rate of ` 500 per m2.[NCERT], 4, Solution. Radius (r) of the cylindrical part = Radius of the conical part =, m=2m, 2, Area of the canvas used, 2.8 m, = Curved S.A. of the cylindrical part + Curved S.A. of the conical, part, = 2prh + prl, 2.1 m, 22, 22, × 2 × 2.8 m 2, × 2 × 2.1 m 2 +, = 2 ×, 72, 7 2 + 17.6 m2 = 44 m, = 26.4 m, 4m, Cost of the canvas = Rate per m2 × area of canvas, = ` 500 × 44 = ` 22,000, Example 8. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each, 5 mm, of its ends (see along side). The length of the entire capsule is 14 mm and the diameter of, the capsule is 5 mm. Find its:, 14 mm, (i) surface area, , (ii) volume, [NCERT], 5, Solution. Since, diameter of the capsule = 5 mm, its radius =, mm = 2.5 mm, 2, r, r, 5, 5 mm, Length (height) of the cylindrical portion = 14 mm – 2 × mm = 9 mm, 2, 9 mm, (i) Surface area of the capsule, = Curved S.A. of its cylindrical part + 2 (curved S.A. of a hemispherical end), 22, 22, × 2.5 × 9 + 4 ×, × (2.5) 2 mm 2 = 220 mm2, = 2prh + 2(2pr2) = 2 ×, 7, 7, (ii) Volume of the capsule, = Volume of cylindrical part + 2 (volume of a hemispherical end), 22, 4 22, 2, × (2.5) 2 × 9 + ×, × (2.5)3 mm3 = 242.26 mm3, = π r 2 h + 2 × πr 3 =, 7, 3 7, 3, Example 9. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part, is 8.5 cm. By measuring the amount of water it holds, Rama finds its volume to be 345 cm3. Is she correct by taking the, above as the inside measurements?, [Take p = 3.14] [NCERT], 2, Solution. Radius (r) of the cylindrical neck = cm = 1 cm, its height = 8 cm and radius (R) of the spherical part, 2, , 116, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 2 cm, , 8.5, cm = 4.25 cm., 2, Volume of water in the vessel, = Volume of the whole vessel, = Volume of cylindrical neck + Volume of spherical part, 4, = π r 2 h + πR 3, 3, 4, = 3.14 × (1)2 × 8 cm3 +, × 3.14 × (4.25)3 cm3 = 346.51 cm3., 3, Since, Rama finds the volume 345 cm3, therefore Rama is not correct., Example 10. A pen stand (as shown in the figure) made of wood is in the shape of a, 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. Find the volume of wood in the entire stand., [NCERT] [Imp.], Solution. Volume of the wood in the entire stand, =, , r 8 cm, , R, 25, , cm, , 4., , = Volume of cuboid – 4 × Volume of one conical depression, = 15 × 10 × 3.5 cm3 – 4 ×, , 1, × p × (0.5)2 × 1.4 cm3 = 523.53 cm3, 3, , r = 8 cm, , Example 11. 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 g mass. [Use p = 3.14] [NCERT [Imp.], Solution. Volume of the pole, = Volume of the bigger cylinder + Volume of the smaller cylinder, = pR2H + pr2h, = 3.14 × (12)2 × 220 cm3 + 3.14 × (8)2 × 60 cm3, = 111532.8 cm3, Since, 1 cm3 of iron has approximately 8 g mass, , 60 cm = h, , 24 cm, 220 cm = H, , R = 12 cm, , \ The mass of the pole, = 8 × 111532.8 g, = 892262.4 g, = 892.262 kg, Example 12. A storage building is in the form as shown in the given figure. The vertical cross-section parallel to the width, side of the building is a rectangle of size 7 m × 3 m mounted by a semicircle of radius 3.5 m. The inner measurements of, the cuboid are 10 m × 7 m × 3 m. Find the volume of the building and the total internal surface area excluding the floor., Solution. Since the storage building is of uniform cross-section, So,, its Volume = Its area of cross-section × its length., 2, , 1 22  7  , = 7 × 3 + ×, ×    × 10 m3, 2 7  2  , , m, cm, 3, 10, = (21 + 19.25) × 10 m3 = 402.5 m3., 7 cm, For total internal surface area excluding the floor;, Perimeter of cross-section excluding floor, 2π r, 22 7, +3m = 6+, × m = 17 m, = 3 +, 2, 7 2, \ Required internal surface area, = Perimeter of cross-section × length + 2 × Area of cross-section, = 17 m × 10 m + 2 × (21 + 19.25) m2 = 250.5 m2, , Surface Areas and Volumes, , 117

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 13. A rectangular metal block has length 15 cm, breadth 10 cm and height 5 cm. From this block, a circular hole, of diameter 7 cm is drilled out. Find:, (i) the volume of the remaining solid, , (ii) the surface area of the remaining solid., , Solution. (i) The volume of the remaining solid, , [Delhi 2015], , 7 cm, , = Volume of rectangular block – Volume of the circular hole, 2, , 22  7 , ×   × 5 cm3, 7  2, 3, = 750 cm – 192.5 cm3 = 557.5 cm3, (ii) The surface area of the remaining solid, = Total surface area of the block – 2 (area of circle of the hole), + curved surface of circular hole (cylinder), = 2(l × b + b × h + h × l) – 2(pr2) + 2prh, 2, 22  7 , 22 7, ×  +2×, × ×5, = 2(15 × 10 + 10 × 5 + 5 × 15) – 2 ×, , , 7, 2, 7 2, = 550 cm2 – 77 cm2 + 110 cm2 = 583 cm2., = 15 × 10 × 5 cm3 –, , 5 cm, , 10 cm, , 15 cm, , Example 14. A metallic cylinder has radius 3 cm and height 5 cm. It is made of metal X. To reduce its weight, a conical, hole is drilled in the cylinder as shown and it is completely filled with a lighter metal Y. The conical hole has a radius, 3, 8, of, cm and its depth is cm. Calculate the ratio of the volume of the metal X to the volume of metal Y in the solid., 2, 9, Solution. Volume of metal X in the solid, = Volume of the whole cylinder – Volume of conical hole, = p × (3)2 × 5 cm3 –, , 2, , 1, 8,  3, × π ×   × cm3,  2, 3, 9, , 3 cm, , 2, , 133, = π  45 −  cm3 =, p cm3., 3, 3, Volume of metal Y = Volume of conical hole, =, , 8 cm, 9, 5 cm, , 2, 2π, 1  3, 8, cm3, π   × cm3 =, , , 3, 3, 2, 9, , \ Ratio of the volume of the metal X to the volume of metal Y, , 3 cm, , 133π, Volume of metal X, 3 = 133, =, =, = 133 : 2, 2π, 2, Volume of metal Y, 3, Example 15. 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:, 0.7 cm, , (i) the curved surface area of the cavity formed., , (i) If l is the slant height of the conical cavity hollowed out, , , ⇒, , l2 = h2 + r2 = (2.4)2 + (0.7)2 = 5.76 + 0.49 = 6.25, l=, , 6.25 cm = 2.5 cm, , , \ The curved surface area of the conical cavity, , 118, , Mathematics–10, , cm, , 2.4 cm, , 1.4, cm = 0.7 cm, 2, , 2.5, , Solution. Since, diameter = 1.4 cm, radius =, , 2.4 cm, , (ii) the total surface area of the remaining solid to the nearest cm2.

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 22, × 0.7 × 2.5 cm2 = 5.5 cm2, 7, (ii) Total surface area of the remaining solid, = Curved S.A. of the conical cavity + external curved S.A. of the cylinder, 22, 22, × 0.7 × 2.4 cm2 +, × (0.7)2, = 5.5 cm2 + 2prh + pr2 = 5.5 cm2 + 2 ×, 7, 7, = 17.6 cm2 = 18 cm2, = πrl =, , Exercise 6.2, , I. Very Short Answer Type Questions, [1 Mark], 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) A cylindrical pencil sharpened at one edge is the combination of, (a) a cone and a cylinder, (b) frustum of a cone and a cylinder, (c) a hemisphere and a cylinder, (d) two cylinders, (2) A surahi is the combination of, (a) a sphere and a cylinder, (b) a hemisphere and a cylinder, (c) two hemispheres, (d) a cylinder and a cone, (3) In a right circular cone, the cross-section made by a plane parallel to the base is a, (a) Triangle, (b) Circle, (c) Square, (d) None of these, (4) If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this, new solid is, (a) 4pr2, (b) 3pr2, (c) 2pr2, (d) pr2, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): The number of coins 1.75 cm in diameter and 2 mm thick is formed from a melted cuboid 10 cm ×, 5.5 cm × 3.5 cm is 400., , Reason (R): Volume of a cylinder = pr2h cubic units, , and volume of cuboid = (l × b × h) cubic units., (2) Assertion (A): Number of spherical balls that can be made out of a solid cube of 1ead whose edge is 44 cm, each, ball being 4 cm in diameter is 2541., , Reason (R): Number of balls =, , Volume of one ball, ., Volume of lead, , 3. Answer the following., (1) What is the capacity of cylindrical vessel with the hemispherical bottom portion, raised upwards?, (2) A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas, of two parts are equal, then what is the ratio of its radius and the slant height of the conical part?, II. Short Answer Type Questions - I, [2 Marks], 4. A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the, total surface area of the toy., [NCERT][CBSE 2012], 5. A solid is in the form of a right circular cylinder with hemispherical ends. The total height of the solid is 58 cm and the, 22, diameter of the cylinder is 28 cm. Find the total surface area of the solid., [Use p =, ] [CBSE 2006], 7, 6. Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2 : 1. Find, the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.[NCERT Exemplar], , Surface Areas and Volumes, , 119

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , , III. Short Answer Type Questions - II, , [3 Marks], , 7. A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical, portion is 6 cm and its height is 4 cm. Determine the surface area of the toy., [Take p = 3.14], 8. From a solid cylinder of height 12 cm and diameter of the base 10 cm a conical cavity of the same height and same, diameter is hollowed out. Find the surface area of the remaining solid., [Imp.], 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown:, , 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.[NCERT], 10. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The, height and the radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical, and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if height of the conical, part is 12 cm., [Imp.], 11. A gulab jamun when completely ready for eating contains sugar syrup upto about 30% of its volume. Find approximately, how much syrup would be found in 45 gulab jamuns shaped like a cylinder with two hemispherical ends, if the complete, length of each of the gulab jamuns is 5 cm and its diameter is 2.8 cm., [NCERT][CBSE 2008], 12. A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the, bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass, was 10 cm, find its actual capacity and its apparent capacity.(Use p = 3.14) [NCERT][CBSE 2009][Imp.], IV. Long Answer Type Questions, , [5 Marks], , 13. The following figure shows a model which is the combination of a cone and, cylinder., , 12 cm, 6 cm, , The total length of the model is 52 cm, the length of conical portion is 12 cm,, the base of the conical portion has diameter of 10 cm and the base of the, 52 cm, cylindrical portion is of diameter 6 cm. If the conical portion is to be painted, orange and the cylindrical portion yellow, find the area of the model painted with each of these colours., , 10 cm, , 14. In the given 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, (a) the total surface area of the block., 22, ] [CBSE 2019 (AI)], 7, 15. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius, as that of cylinder. The diameter and height of cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical, portion is 5 cm, then find the total surface area and volume of rocket., (Use p = 3.14) [NCERT Exemplar], (b) the volume of the block formed, , [Take p =, , 16. A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height, 180 cm such that it touches the bottom. Find the volume of water left in cylinder, if the radius of the cylinder is equal to, the radius to the cone., [NCERT Exemplar], 17. A solid is in the shape of a cone surmounted on a hemisphere. The radius of each of them being 3.5 cm and the total, height of the solid is 9.5 cm. Find the volume of the solid., , 120, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Case Study Based Questions, I. 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 is 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, 22, dome with a cuboidal structure mounted on it. (Take =, ), 7, Dome, Torana, , Charan, Harmika, Balutade, , Stairs, , 1. The volume of the hemispherical dome if the height of the dome is 21 m, is, (a) 19404 cu. m, (b) 2000 cu. m, (c) 15000 cu. m, (d) 19000 cu. m, 2. The formula to find the volume of sphere is, 2 3, 4, (a) πr, (b) πr 3, (c) 4 πr2, (d) 2 πr2, 3, 3, 3. The cloth required to cover the hemispherical dome if the radius of its base is 14 m is, (a) 1222 sq. m, (b) 1232 sq. m, (c) 1200 sq. m, (d) 1400 sq. m, 4. 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, 5. The volume of the cuboidal shaped top with dimensions mentioned in question 4, is, (a) 182.45 m3, (b) 282.45 m3, (c) 292 m3, (d) 192 m3, , Answers and Hints, 1. (1), (2), (3), 2. (1), , (a) a cone and a cylinder, (1), (a) a sphere and a cylinder, (1), 2, (b) Circle, (1) (4) (a) 4pr (1), (a) Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of, assertion (A)., Volume of cuboid, Number of coins =, Volume of one coin, 10×5.5×3.5, 1.75 1.75, π×, ×, × 0.2, 2, 2, 10×5.5×3.5, =, 22 1.75 1.75, ×, ×, × 0.2, 7, 2, 2, = 400, =, , (2) (c) Assertion (A) is true but reason (R) is false., We have,, Volume of lead, Number of balls =, Volume of one ball, , 44×44×44, 4 22, ×, ×2×2×2, 3 7, 44×44×44×3×7, =, 4 × 22 × 2 × 2 × 2, =, , = 2541, , (1), , πr 2, [3h − 2r ] (1) (2) 1 : 2, 3, 4. Here,, r = 3.5 cm, , \, h = (15.5 – 3.5) cm, = 12.0 cm, Surface area of the conical part = prl, 3. (1), , (1), , Surface area of the hemispherical part = 2pr2 , (1), , , \ Total surface area of the toy, , , Q, , 12.0, , =prl + 2pr2 = pr(l + 2r) cm2 cm, 2, , 2, , l = (12) + (3.5) = 156.25, , , ⇒, , 15.5, cm, , 2, , l = 12.5 cm, , cm2, (1), , Surface Areas and Volumes, , 3.5, cm, , 121

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , , \ TSA of the toy, , Date __________, , 22, 7, = 103.62 cm2(1½), = 18p + 15p = 33p = 33 ×, , 22 35, =, (12.5 + 2 × 3.5) cm2, ×, 7 10, , 8. Surface area of remaining solid, , = 11 × (12.5 + 7) cm2, , = 2p × 5 × 12 + p × 5 × 5 + p × 5 ×, , = 11 × 19.5 cm2, = 214.5 cm2.(1), 5. Total surface area of solid = Curved surface area, of cylindrical portion + 2(Curved surface area of, hemispherical portion), 22, 22, = 2 × × 14 × 30 + 2 (2 ×, × 14 × 14), (1), 7, 7, = 2640 + 2464 = 5104 cm2(1), 6. As the ratio of volumes of cones A and B is 2 : 1, their, 6, radii are same equal to r = = 3 cm ., 2, , 52 + 122, , (1), , = 120p + 25p + 65p(1), = 210p = 210 ×, , 22, = 660 cm2(1), 7, , 9. Radius of the cylinder (r) = 3.5 cm, Height of the cylinder (h) = 10 cm, , \ Total surface area, = 2prh + 2pr2 + 2pr2(1), = 2pr (h + r + r) = 2pr (h + 2r), = 2 ×, , 22, × 3.5 (10 + 2 × 3.5), 7, , (1), , 22, × 3.5 × 17 = 374 cm2(1), 7, 10. 770 cm2(3), , = 2 ×, 1 2, πr1 h1, V1, = 3, V2, 1 2, πr h, 3 2 2, , \, , 2, (3) h1, =, 1, (3)2 h2, 2, , ⇒, , 11. 337.88 cm3(3), , h1 = 2h2, , ⇒, , ...(i), , Also, h1 + h2 = 21 cm, , ⇒, , 3h2 = 21, , [Using (i)], , ⇒, , h2 = 7 cm, , Now,, , h1 = 21 cm – 7 cm = 14 cm, , ...(ii), , Hence, height of cone A = 14 cm and height of cone B =, 7 cm., (1), Volume of cone A =, , 1, 22, ×, × 9 × 14 = 132 cm3, 3, 7, , 1, 22, ×, × 9 × 7 = 66 cm3, 3, 7, Volume of remaining portion of tube, Volume of cone B =, , = Vol. of cylinder – Vol. of cone A – Vol. of cone B, = pr2h – 132 cm3 – 66 cm3, =, , = 22 × 27 – 198 = 594 – 198 = 396 cm3, 3, , Hence, the required volume is 396 cm .(1), 7. Surface area of toy, , 122, , , ⇒ length (h) of cylindrical part, Volume of each gulab jamun, 2, p × (1.4)3 cm3, 3, Volume of syrup found in 45 gulab jamuns, = p(1.4)2 × 2.2 + 2 ×, , = 45 × 30% of volume of each gulab jamun., 12. Since the inner diameter of the glass = 5 cm and height, = 10 cm, the apparent capacity of the glass = pr 2h, = 3.14 × 2.5 × 2.5 × 10 cm3 = 196.25 cm3 , (1), , But the actual capacity of the glass is less by the volume, of the hemisphere at the base of the glass., i.e., it is less by, =, , 22, × 3 × 3 × 21 – 198, 7, , = 2p × 3 × 3 + p × 3 ×, , 2.8, cm = 1.4 cm, 2, , = 5 cm – 2 × 1.5 cm = 2.2 cm, , 2h2 + h2 = 21, , ⇒, , Hint: Radius (r) =, , 32 + 42 (1½), , Mathematics–10, , 2 3, pr, 3, , 2, × 3.14 × 2.5 × 2.5 × 2.5 cm3 = 32.71 cm3 (1), 3, , , So, the actual capacity of the glass, = apparent capacity of glass, , – volume of the hemisphere, = (196.25 – 32.71) = 163.54 cm3(1)

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 13. orange: 254.34 cm2, yellow: 781.86 cm2(5), , Hint: For conical portion, h = 12 cm,, 10, r =, cm = 5 cm, 2, and slant height (l) = h 2 + r 2 = 13 cm;, whereas for cylindrical portion,, 6, h = (52 – 12) cm = 40 cm and r = cm = 3 cm., 2, , \ Area to be painted orange, = C.S.A. of the cone + Area of its base, = p × 5 × 13 + p (52 – 32) sq. cm, Area to be painted yellow, = C.S.A. of the cylinder + Area of its base, = 2p × 3 × 40 + p32 sq. cm, 14. Diameter of the hemisphere = 4.2 cm, , ∴ Radius of the hemisphere = 2.1 cm, Edge of the cube (l) = 6 cm, (1), (a) Total surface area of the block, = Total surface area of the cube, , – Base area of hemisphere, , + Curved surface area of hemisphere, = 6 × (edge)2 – pr2 + 2pr2, 22 21 21, = 6 l2 + pr2 = 6 × 36 +, × ×, 7 10 10, = 216 + 13.83 = 229.86 cm2(2), , , , Date __________, , [Q h =, , 52 − 32 = 4 cm], ,  40 , = 3.14 × 9   = 3.14 × 3 × 40 = 376.8 cm 3, 3, , \ Volume of Rocket = 376.8 cm3(1), Total surface area of rocket, = Curved surface area of cylinder, , + Curved surface area of cone, , + Area of base of cylinder, , [As it is closed (Given)], = 2prH + prl + pr2 = pr [2H + l + r](1), = 3.14 × 3[2 × 12 + 5 + 3] = 3.14 × 3 × 32, = 301.44 cm2(1), Hence, the surface area of rocket is 301.44 cm2., 16. The water left in cylinder, = Volume of cylinder – Volume of cone, Volume of water left after immersing the cone into cylinder, full of water, = Volume of cylinder – Volume of cone (1), 1 2, 2, = πR H − πr h, 3, \ Required volume of water in cylinder, , 15. Volume of rocket, , 1 2, 2, = πr H − πr h [Q R = r] (1), 3, 1 , 22, 120 , , 2, × 60 × 60 180 −, = πr  H − h  =, 3 , , 7, 3 , , 22, =, × 60 × 60 × 140 cm3 (1), 7, 22 × 60 × 60 × 140 22 × 72 1584, =, =, =, 7 × 100 × 100 × 100, 1000, 1000, \ Vol. of water in cylinder = 1.584 m3, , = Volume of cylinder + Volume of cone, , Hence, required volume of water left = 1.584 m3.(1), , (b) Volume of the block, = Volume of cube + Volume of hemisphere, 2 22 21 3, 2, = (edge)3 + πr 3 = (6)3 + × × c m, 10, 2 322 21 21 321 7, = 216 + × × × ×, 3 7 10 10 10, = 216 + 19.40 = 235.40 cm3(2), , 17. Let the radius of hemisphere or a cone, r = 3.5 cm, Also, the height of the cone,, , l, , h, , h = 9.5 – 3.5 = 6 cm, Volume of solid, , r, , = Volume of hemisphere + volume of cone, 2, 1, = πr 3 + πr 2 h (1), 3, 3, 2 22, 1 22, × (3.5)3 + × × (3.5) 2 × 6, = ×, 3 7, 3 7, = 89.83 + 77 = 166.83 cm3(2), , H, , r, ,       , , (2), , (1), , 1 , 1, 2, = πr 2 H + πr 2 h = πr  H + h , 3 , , 3, 1 , , = 3.14 × 3 × 3 12+ × 4 (1), 3 , , , Case Study Based Questions, I. 1. (a) 19404 cu. m, , 3. (b) 1232 sq. m, , 2. (b), , 4 3, πr, 3, , 4. (c) 1392 sq. m, , , 5. (d) 192 m3, , Surface Areas and Volumes, , 123

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. To find surface area, volume, height, radius, etc. when a solid (or solids) of one shape is converted into solid (or solids), of another shape., 2. To find surface areas and volumes of a combination of solids., , IMPORTANT FORMULAE, •• If l, b and h denote respectively the length, breadth and height of a cuboid, then, (i) Total surface area of the cuboid = 2 (lb + bh + lh) square units., (ii) Volume of the cuboid = Area of the base × height = lbh cubic units., (iii) Diagonal of the cuboid = l 2 + b 2 + h 2 units., (iv) Area of four walls of a room = 2 (l + b) h sq. units., •• If the length of each edge of a cube is ‘a’ units, then, (i) Total surface area of the cube = 6 a2 sq. units. (ii) Volume of the cube = a3 cubic units, (iii) Diagonal of the cube =, , 3 a units., , •• If r and h denote respectively the radius of the base and height of a right circular cylinder, then, (i) Area of each end = pr2, (ii) Curved surface area = 2prh, (iii) Total surface area = 2pr (h + r) sq. units., (iv) Volume = pr2h = Area of the base × height, •• If R and r denote respectively the external and internal radii of a hollow right circular cylinder, then, (i) Area of each end = p (R2 – r2), (ii) Curved surface area of hollow cylinder = 2p (R + r) h, (iii) Total surface area = 2p (R + r) (R + h – r), (iv) Volume of material= ph (R2 – r2), •• If r, h and l denote respectively the radius of base, height and slant height of a right circular cone, then, (i) l2 = r2 + h2, (ii) Curved surface area = prl, 1, (iii) Total surface area = pr2 + prl, (iv) Volume = πr 2 h, 3, •• For a sphere of radius r, we have, 4, (i) Surface area = 4pr2, (ii) Volume = πr 3, 3, •• For a hemisphere of radius r, we have, (i) Lateral/curved surface area = 2pr2, (ii) Total surface area = 3pr2, 2, (iii) Volume = πr 3, 3, •• For a hemispherical shell of outer radius R and inner radius r, we have, (i) Lateral / curved surface area = 2p (R2 + r2), (ii) Total surface area = 3pR2 + pr2, 2, (iii) Volume = π(R 3 − r 3 ), 3, , COMMON ERRORS, Errors, (i) In calculating the surface area of combination of solids, some, students add the surface areas of the two individual solids., , Corrections, , (i) It is not correct. In combination of solids, some parts, of surface area disappeared. So, find the surface area, of visible portion of individual solids only and add, them., (ii) Taking incorrectly diameter in place of radius and vice versa. (ii) Be careful while substituting radius in the formula., (iii) Some students do not identify whether to find the volume or (iii) Read the problem 2-3 times and identify what is to be, surface area for the given information., find correctly., , 124, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , QUICK REVISION NOTES, •• Surface areas and volumes of various solids are obtained by using these formulae., S.No., , (i), , Name of the Solid, , Figure, , Cube, , a, , a, , Lateral/Curved, Surface Area, , Total Surface, Area, , Volume, , 4a2, , 6a2, , a3, , 2 (l + b) h, , 2 (lb + bh + hl), , lbh, , 2prh, , 2pr (r + h), , pr2h, , prl, , pr (l + r), , 1 2, pr h, 3, , 4pr2, , 4pr2, , 4 3, pr, 3, , 2pr2, , 3pr2, , 2 3, pr, 3, , a, , (ii), , Cuboid, , h, b, , (iii), , Right circular cylinder, , l, , h, , r, , (iv), , Right circular cone, , l, , h, r, , r, , (v), , Sphere, , r, , (vi), , Hemisphere, , r, , •• When a solid or solids of one shape is (are) converted into solid or solids of another shape, then total volume of the solid(s), to be converted is (are) same to the total volume of the solid(s) into which the given solid(s) is (are) converted., •• To find the number of solids of a given shape in which a given solid is to be converted, we divide the total volume of the, solid(s) to be converted by volume of one converted solid., •• To find the surface area of a complex solid, i.e., a solid formed by combination of other solids, we add the curved surface, area of only visible portion of individual solid., •• To find the volume of a complex solid, we add the volume of individual solid., , qqq, , Surface Areas and Volumes, , 125

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\ 23-Nov-2021, , 7, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Statistics, , Topics Covered, 1. Mean of Grouped Data, , 2. Mode and Median of Grouped Data, , 1. Mean of Grouped Data, We know that mean (or average) of a number of observations is the sum of the values of all the observations divided by, the total number of observations., The mean of the values x1, x2, x3, ..., xn of the variable x is denoted by x and is given by, n, , ∑ xi, x1 + x2 + ... + xn i = 1, x=, =, n, n, The mean of grouped data can be found by the following methods:, (i) Direct method: The mid-points x1, x2, x3, ..., xn of the class intervals are taken as values of the variate whose, respective frequencies are f1, f2, f3, ..., fn. Then mean of the data is given by, n, , , , Mean x =, , f1 x1 + f 2 x2 + ... + f n xn, f1 + f 2 + ... + f n, , fi, , x=, , ∑ fi xi, i =1, , Σf i, , (ii) Assumed mean method: In this method, first of all obtain the frequency distribution and prepare the frequency, table in such a way that its first column consists of class intervals of the variable and the second column consists, of the corresponding frequencies., 1, For each class, find the mid-point i.e., class-mark xi =, (lower limit + upper limit). Now choose an arbitrary, 2, number preferably among the middle values of xi, in the third column. We call it assumed mean and denote it by, A. Find the deviations di = xi – A., Σf d, Thus,, Mean x = A + i i where, A is the assumed mean and di = xi – A, Σf i, Example 1. Find the arithmetic mean of 1, 2, 3, ..., n, Solution. We know that,, , x=, , x1 + x2 + ... + xn Σxi, 1 n, 1 + 2 + 3 + ... + n, , =, =, =  (2 + n − 1) , n, n, n, 2, n, , , , [Q 1 + 2 + 3 + ... + n is an A.P. and sum of this A.P. =, 1  n(n + 1) , n +1, =, , , n 2 , 2, Example 2. If the mean of the following distribution is 6.4, then find the value of ‘p’., , n, (2 + n – 1)], 2, , =, , 2, 3, , x, f, , Solution., , x=, , fi, , 64 =, , fi, , 4, p, , Σfi xi, Σf, , fi, , 6.4 =, , (92 + 4 p ) × 10, 14 + p, , 24p = 24, , fi, , p=, , 6, 5, , 8, 3, , 10, 2, , [Imp.], 12, 1, , 6 + 4 p + 30 + 24 + 20 + 12, 92 + 4 p, =, 3+ p + 5 + 3+ 2 +1, 14 + p, , fi, , 920 + 40 p = 896 + 64 p, , 24, =1, 24, , 126

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Example 3. The following data gives the information on the observed life-times (in hours) of 225 electrical components., Life-time (in hours), Frequency, , 0–20, , 20–40, , 40–60, , 60–80, , 10, , 35, , 52, , 61, , 80–100 100–120, 38, , 29, , Determine the mean of the above data., Solution., , [NCERT], , Class interval, , fi, , xi, , di = xi – A, , fidi, , 0–20, , 10, , 10, , –60, , –600, , 20–40, , 35, , 30, , –40, , –1400, , 40–60, , 52, , 50, , –20, , –1040, , 60–80, , 61, , 70 = A, , 0, , 0, , 80–100, , 38, , 90, , 20, , 760, , 100–120, , 29, , 110, , 40, , 1160, , Sfi = 225, , Sfidi = –1120, , Σf i d i, 1120, = 70 −, = 70 – 4.98 = 65.02, Σf i, 225, x = 65.02 approximately., x=A+, , \, fi, , Exercise 7.1, I. Very Short Answer Type Questions, , [1 Mark], , 1. Multiple Choice Questions (MCQs), , Choose the correct answer from the given options:, (1) While computing mean of grouped data, we assume that the frequencies are, (a) evenly distributed over all the classes, (b) centered at the classmarks of the classes, (c) centered at the upper limits of the classes, (d) centered at the lower limits of the classes, (2) If xi are the mid-points of the class intervals of grouped data, fis are the corresponding frequencies and x is the mean,, then S( fixi – x) is equal to, (a) 0, (b) –1, (c) 1, (d) 2,  Σf i d i , (3) In the following x = A + , , for finding the mean of grouped frequency distribution, di =,  Σfi , A − xi, (a) xi + A, (b) A – xi, (c) xi – A, (d), fi, (4) If the arithmetic mean of n numbers of a series is x and the sum of first (n – 1) numbers is k, the value of the last, number is, x+k, (a) nx – k, (b) nx + k, (c), (d) n( x + k), n, (5) Arithmetic mean of all factors of 20 is, (a) 5, (b) 6, (c) 7, (d) 8, (6) The mean of 5 numbers is 27. If one number is excluded their mean is 25. The excluded number is, (a) 30, (b) 35, (c) 32, (d) 36, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): The arithmetic mean of the following given frequency distribution table is 13.81., x, f, , 4, 7, , 7, 10, , 10, 15, , 13, 20, , 16, 25, , 19, 30, , Statistics, , 127

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\ 23-Nov-2021, , Amit, , , Reason (R): x =, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Σf i xi, Σf i, , Σf i d i, where a is the assumed mean and di the deviation., Σf i, , Reason (R): To find deviation, we use di = a – xi where a is the assumed mean and xi is the class mark., 3. Answers the following:, (1) Find the class-mark of class 25–35., (2) Find the mean of first ten odd natural numbers., (3) If the mean of the first n natural number is 15, then find n., [CBSE Standard 2020], (4) Find the class-marks of the classes 10-25 and 35-55., [CBSE Standard 2020], (2) Assertion (A): To find mean of a grouped data, we use x = a +, , II. Short Answer Type Questions - I, 4. If the mean of the following data is 20.6, find the value of p., x, , 10, , 15, , p, , 25, , 35, , f, , 3, , 10, , 25, , 7, , 5, , [2 Marks], , 5. Find the mean of the following distribution:, Class, , [CBSE Standard 2020], , 3–5, , 5–7, , 7–9, , 9–11, , 11–13, , 5, , 10, , 10, , 7, , 8, , Frequency, , 6. Find the mean of the following distribution:, Class, , 5–15, , 15–25, , 25–35, , 35–45, , 2, , 4, , 3, , 1, , Frequency, , III. Short Answer Type Questions - II, [3 Marks], 7. The mean of the following frequency distribution is 62.8 and sum of all frequencies is 50. Find the missing frequencies, f1 and f2., [Imp.], Class, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , 100–120, , 5, , f1, , 10, , f2, , 7, , 8, , Frequency, , 8. The arithmetic mean of the following frequency distribution is 53. Find the value of k., Class, Frequency, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , 12, , 15, , 32, , k, , 13, , [Delhi 2019], , 9. The table below shows the daily expenditure on grocery of 25 households in a locality., Daily expenditure (in `), , 100–150 150–200 200–250 250–300 300–350, 4, , No. of households, , 5, , 12, , 2, , 2, , Find the mean daily expenditure on food by a suitable method., [NCERT], 10. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is, ` 18. Find the missing frequency f., Daily pocket allowance (in `), , 11–13, , 13–15, , 15–17, , 17–19, , 19–21, , 21–23, , 23–25, , 7, , 6, , 9, , 13, , f, , 5, , 4, , Number of children, , 11. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of, days a student was absent., [AI 2019], Number of days:, , 0–6, , 6–12, , 12–18, , 18–24, , 24–30, , 30–36, , 36–42, , Number of students:, , 10, , 11, , 7, , 4, , 4, , 3, , 1, , IV. Long Answer Type Questions, [5 Marks], 12. The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are as follows:, Mileage (km / l), , 10–12, , 12–14, , 14–16, , 16–18, , 7, , 12, , 18, , 13, , No. of Cars, , Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km / l. Do you agree with this, claim?, [NCERT Exemplar] [Imp], , 128, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 13. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given as follows:, 100–104 104–108 108–112 112–116 116–120, , No. of seats, , 15, , Frequency, , 20, , 32, , 18, , 15, , Determine the mean number of seats occupied over the flights., [NCERT Exemplar], 14. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ` 18., Find the missing frequency k., [CBSE Standard 2020, 2018], Daily pocket allowance (in `), , 11–13, , 13–15, , 15–17, , 17–19, , 19–21, , 21–23, , 23–25, , 3, , 6, , 9, , 13, , k, , 5, , 4, , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , 100–120, , 20, , 35, , 52, , 44, , 38, , 31, , Number of children, , 15. Find the mean of the following data:, Classes, Frequency, , Case Study Based Questions, I. Student-Teacher Ratio: Student-teacher ratio expresses the relationship between the number of students enrolled in, a school and the number of teachers in that school. It is important for a number of reasons. For example, it can be an, indicator of the amount of individual attention any child is likely to receive, keeping in mind that not all class size are, going to be the same., The following distribution gives the state-wise student-teacher ratio in higher secondary schools of India (28 states and, 7 UTs only)., , Number of students, per teacher, , Number of States/, UTs, , Number of students, per teacher, , Number of States/, UTs, , 15-20, , 3, , 35-40, , 3, , 20-25, , 8, , 40-45, , 0, , 25-30, , 9, , 45-50, , 0, , 30-35, , 10, , 50-55, , 2, , 1. In order to find the mean by direct method, we use the formula, n, , (a), 2., 3., , 4., 5., , Σ f i xi, , i =1, , n, , (b), , n, n, , n, , (c) n × Σ f i xi, i =1, , Σ f i xi, The mean of the above data is, (a) 29.2, (b) 30.5, (c) 38.3, The formula for assumed mean method to find the mean is, Σf d, Σf i, Σf d, (a) A − i i, (b) A +, (c) A × i i, Σf i, Σf i d i, Σf i, The sum of class marks of 25-30 and 45-50 is, (a) 62, (b) 70, (c) 75, The sum of the upper and lower limits of modal class is, (a) 55, (b) 65, (c) 85, , n, , (d) n + Σ f i xi, i =1, , i =1, , (d) 40.1, (d) A +, , Σf i d i, Σf i, , (d) 85, (d) 75, , Answers and Hints, 1. (1), (2), (4), (6), 2. (1), , (b) centered at the class-marks of the classes, (1), (a) 0, (1) (3) (c) xi – A, (1), (a) nx – k, (1) (5) (c) 7, (1), (b) 35, (1), (a) Both assertion (A) and reason (R) are true and, reason (R) is the correct explanation of assertion, (A).(1), (2) (c) Assertion (A) is true but reason (R) is false. (1), 25 + 35 60, =, 3. (1) Class-mark =, = 30(1), 2, 2, , (2) 10(1), 1 + 2 + 3 + ... + n, (3) , 15 =, n, , ⇒, , , ⇒, , 1 n, , (2 + n − 1) , n  2, , [Q 1 + 2 + 3 + ... + n is an AP, n, and sum of this AP =, (2 + n – 1)], 2, 1  n(n + 1) , n +1, 15 = , =, , n 2 , 2, 15 =, , Statistics, , 129

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , , ⇒ n + 1 = 30 ⇒ n = 29, (4) Class-mark of class 10 – 25, =, , (1), , 7., , Class, interval, , xi, , fi, , fi xi, , 0-20, , 10, , 5, , 50, , 20-40, , 30, , f1, , 30 f1, , 40-60, , 50, , 10, , 500, , 60-80, , 70, , f2, , 70 f2, , 10 + 25, 35, =, = 17.5, 2, 2, , Class-mark of class 35–55, =, 4., , fi, , fi xi, , 80-100, , 90, , 7, , 630, , 10, , 3, , 30, , 100-120, , 110, , 8, , 880, , 15, , 10, , 150, , p, , 25, , 25 p, , Sfi = 30, + f1 + f2, , Sfixi = 2060, + 30f1 + 70f2, , 25, , 7, , 175, , 35, , 5, , 175, , Total, , Sfi = 50, , Sfixi = 530 + 25p, , Mean x =, , , ⇒, 5., , (1), , xi, , , , , 35 + 55, 90, =, = 45, 2, 2, , 20.6 =, , 6., , 130, , , , So,, , , Sfi = 30 + f1 + f2 but Sfi = 50, 50 = 30 + f1 + f2 ⇒ f1 + f2 = 20, Sfixi = 2060 + 30f1 + 70 f2, Σf x, Now, mean = i i, Σf i, 2060 + 30 f1 + 70 f 2, 50, , ⇒ 3f1 + 7 f2 = 108, Solving (i) and (ii), we get f1 = 8 and f2 = 12, 8. Calculation of mean., , ⇒, , 530 + 25 p, 50, , ⇒ p = 20, , (1), , xi, , fi, , fi xi, , 3-5, , 4, , 5, , 20, , 5-7, , 6, , 10, , 60, , 7-9, , 8, , 10, , 80, , 0-20, , 12, , 10, , 120, , 20-40, , 15, , 30, , 450, , 9-11, , 10, , 7, , 70, , 40-60, , 32, , 50, , 1600, , 11-13, , 12, , 8, , 96, , 60-80, , k, , 70, , 70k, , Sfi = 40, , Sfixi = 326, , 80-100, , 13, , 90, , 1170, , Mean x =, , 326, Sfi xi, =, = 8.15, 40, Sf i, , (1), , Class, , xi, , fi, , fi xi, , 5-15, , 10, , 2, , 20, , 15-25, , 20, , 4, , 80, , 25-35, , 30, , 3, , 90, , 35-45, , 40, , 1, , 40, , Sfi = 10, , Sfixi = 230, , Mean x =, , Sfi xi, 230, =, = 23, Sf i, 10, , Mathematics–10, , Class Frequency Class, Interval, (fi), Mark (xi), , (1), , Sfi = 72 + k, , , , ∴, , Mean x =, , , , ⇒ 3340 + 70k =, , ⇒ 3340 + 70k =, , ⇒ 70k – 53k =, , ⇒, 17k =, 9., , (1), (1), , (1), (given), ...(i), , 62.8 =, , Class, , , , , (1), , Σfi xi, Σf i, , , , , Date __________, , Daily, expenditure xi, (in `), , Σfi xi, Σf i, , ...(ii) (1½), (½), , fi xi, , Sfixi = 3340 + 70k, , (1), , 3340 + 70k, 72 + k, , (1), , ⇒ 53 =, , 53 (72 + k), 3816 + 53k, 3816 – 3340, 476 ⇒ k = 28, , (1), , fi, , di= (xi – a), , fi di, , 100-150, , 125, , 4, , –100, , 4 × (–100) = –400, , 150-200, , 175, , 5, , –50, , 5 × (–50) = –250, , 200-250, , 225, , 12, , 0, , 12 × 0 = 0

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , 250-300, , 275, , 2, , 50, , 50 × 2 = 100, , 300-350, , 325, , 2, , 100, , 100 × 2 = 200, , Sf i = 25, , , , \, , Sfi di = –350, ,  Σf i d i ,  −350 , x= a+,  = 225 + ,  25 ,  Σf i , , = 225 –14 = 211, Thus, the mean daily expenditure of food is ` 211., 10., , Class, x, interval i, , fi di = (xi – a), , 12 7, , –6, , 7 × (–6) = –42, , 13-15, , 14 6, , –4, , 6 × (–4) = –24, , 15-17, , 16 9, , –2, , 9 × (–2) = –18, , 17-19, , 18 13, , 0, , 13 × 0 = 0, , 19-21, , 20, , f, , 2, , f × 2 = 2f, , 21-23, , 22 5, , 4, , 5 × 4 = 20, , 23-25, , 24 4, , 6, , 4 × 6 = 24, , Sfi di, = 2f – 40, , Sfi, = f + 44, , , Since, , \, ⇒, , 12. xi = class-mark and a = assumed mean., , (1), , 0-6, , 10, , (1), fidi, , 3, , –18, , –180, , 6-12, , 11, , 9, , –12, , –132, , 12-18, , 7, , 15, , –6, , –42, , 18-24, , 4, , 21, , 0, , 0, , 24-30, , 4, , 27, , 6, , 24, , 30-36, , 3, , 33, , 12, , 36, , 36-42, , 1, , 39, , 18, , 18, , Sfi = 40, , Sfidi = –276, , fi, 7, , fidi, –14, , 0, , 12, , 0, , 2, 4, ,  , , A = 13, , , , x=A+, , (1½), , ⇒, , No. of Class, days students Mark di = (xi – a), (CI), (fi), (xi), , di = (xi – a), –2, , 12-14 13 = a, 14-16, 15, 16-18, 17, , 13., ,  2 f − 40 , 18 = 18 +  f + 44  (½), , , , 11. No. of, , xi, 11, , 18, 36, 13, 52, Sfi = 50 Sfidi = 74, , (2½), , Σf i d i, 74, = 13 +, Σf i, 50, = 13 + 1.48 = 14.48 km L–1, Hence, mean mileage of car is 14.48 km/litre., So, the manufacturer’s statement is wrong that mileage is, 16 km L–1.(2½), , x = 18, a = 18,  Σf d , x= a+ i i,  Σf i , ,  2 f − 40 , 0= , ,  f + 44 , ⇒, 0 = 2f – 40, ⇒, 2f = 40, , ⇒, f = 20, Thus, the missing frequency is 20., , C.I., 10-12, , (2), , fi di, , 11-13, , Date __________, , , (2), Let assumed mean a = 21 and class size h = 6, Σf i d i, −276, Mean,, x= a+, = 21 +, Σf i, 40, = 21 – 6.9 = 14.1, Hence, mean number of days a student was absent is 14.1., (1), , C.I., , xi, , di = (xi – a), , fi, , fidi, , 100-104, 104-108, 108-112, 112-116, 116-120, , 102, 106, 110 = a, 114, 118, , –8, –4, 0, 4, 8, , 15, 20, 32, 18, 15, Sfi = 100, , – 120, – 80, 0, 72, 120, Sfidi = –8, , (2½), Here,, a = 110, Σf d, , x=a+ i i, Σf i, −8, = 110 +, 100, = 110 – 0.08, = 109.92, but, seat cannot be in decimal., , ⇒, x = 110., Hence, the mean number of seats occupied over the flights, is 110., (2½), 14. k = 8, (5), 15., , , , , Class, 0-20, 20-40, 40-60, 60-80, 80-100, 100-120, Total, , xi, 10, 30, 50, 70, 90, 110, , fi, fi xi, 20, 200, 35, 1050, 52, 2600, 44, 3080, 38, 3420, 31, 3410, Sfi = 220 Sfixi = 13760, , Σfi xi, 13760, Mean x = Σf =, i, 220, = 62.55 (approx), , (2½), , (2½), , Case Study Based Questions, n, , I. 1. (a), , Σ f i xi, , i =1, , n, , 3. (d) A +, , Σf i d i, Σf i, , 2. (a) 29.2, , 4. (c) 75, , 5. (b) 65, , Statistics, , 131

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , Date __________, , 2. Mode and Median of Grouped Data, Mode, The mode or modal value is that value of the variate which occurs most frequently., To find the mode of a grouped data, we proceed as follows:, (i) Obtain the grouped data., (ii) Locate the class having maximum frequency. This class is called modal class., (iii) Mode of a grouped data is given by the formula, , f1 − f 0 , ×h, Mode = l + ,  2 f1 − f 0 − f 2 , , , , where, l = lower limit of the modal class, , f1 = frequency of the modal class, f0 = frequency of the class preceding the modal class, , f2 = frequency of the class succeeding the modal class, , h = size of the modal class, Example 1. The marks distribution of 30 students in a science examination are as follows. Find the mode of this data., , [Imp.], 10, 1, , Marks obtained (xi), Number of students (fi), , 20, 1, , 36, 3, , 40, 4, , 50, 3, , 56, 2, , 60, 4, , 70, 4, , 72, 1, , 80, 1, , 88, 2, , 92, 3, , 95, 1, , Solution. First, we will make the class interval with class size of 15., Class interval, 10–25, 25–40, 40–55, 55–70, 70–85, 85–100, Total, , Number of students, 2, 3, 7, 6, 6, 6, Sfi = 30, , f0, Model class, f2, , f1 = 7, , Since the maximum number of students (7) have got marks in the interval 40–55, the model class is 40–55., So, , lower limit of the model class, l = 40, class size, h = 15, frequency, f1 of the model class = 7, , frequency, f0 of the class preceding the model class = 3, frequency, f2 of the class succeeding the model class = 6, Now, using the formula, , f1 − f 0 , ×h, Mode = l + ,  2 f1 − f 0 − f 2 , ,  7−3 , × 15 = 52, = 40 + ,  14 − 3 − 6 , \, Mode = 52, , Median, The median is the middle value of a distribution i.e., median of a distribution is the value of the observation which divides, it into two equal parts., • Median of ungrouped data:, (i) Arrange the data in ascending order., , 132, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , n + 1, (ii) If n (number of observations) is odd, then median = ,  2 , , th, , Date __________, , observation., , th,  n  th, , n ,   observation +  + 1 observation , 2 ,  2 , , • Median of grouped data: Median of a grouped data or continuous frequency distribution can be found by using the, formula:, n, , − cf, 2, , , Median = l + ,  ×h,  f , where, l = lower limit of the median class, , n = number of observations, , f = frequency of the median class, , h = size of the median class (assuming all class sizes to be equal), , cf = cumulative frequency of the class preceding the median class, , (iii) If n (number of observations) is even, then median =, , 1, 2, , Example 2. Find the median of the following data:, , Solution., , xi, , 22, , 24.50, , 28, , 31.50, , 34, , 36.50, , fi, , 10, , 23, , 32, , 28, , 12, , 05, , xi, 22, 24.50, 28, 31.50, 34, 36.50, Total, , fi, 10, 23, 32, 28, 12, 05, 110, , cf, 10, 33, 65, 93, 105, 110, , Since Sfi = n = 110, which is even., R, VW, Jn, Nth, 1 SJ n Nth, \, Median = SSSKKK OOO observation + KKK + 1OOO observationWWW, W, 2 SL 2 P, L2, P, T, X, 1, 1, = ^55 th observation + 56 th observation h = (28 + 28) = 28, 2, 2, (Q Both 55 and 56 lies in cf 65, therefore 55th observation = 56th observation = 28), Example 3. The distribution below gives the marks of 30 students of a class in mathematics. Find the median marks of, the students., [Imp.], Marks, Number of students, , 40–45, 2, , 45–50, 3, , 50–55, 8, , 55–60, 6, , 60–65, 6, , 65–70, 3, , 70–75, 2, , Solution., Marks, , Number of students (fi), , Cumulative frequency (cf), , 40–45, , 2, , 2, , 45–50, , 3, , 5, , 50–55, , 8, , 13, , 55–60, , 6, , 19 → median class, , 60–65, , 6, , 25, , 65–70, , 3, , 28, , 70–75, , 2, , 30, , Total, , Sfi = 30, , Statistics, , 133

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\ 23-Nov-2021, , Proof-3, , Amit, , Reader’s Sign _______________________, , Sfi = n = 30,, , Date __________, , n, = 15, 2, , n, = 15 is 19., 2, \ The corresponding class is 55–60 which is the median class., n, Now, we have, = 15, l = 55, c f = 13, f = 6, h = 5, 2, n, , − (cf ), 2, ,  15 − 13 , \, Median = l + , × 5 = 55 + 1.67 = 56.67,  × h = 55 + ,  6 , , , f, Since cf just greater than, , Example 4. If the median of the distribution given below is 28.5, find the values of x and y., Class interval, Frequency, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , Total, , 5, , x, , 20, , 15, , y, , 5, , 60, , Solution. Here, median = 28.5, n = 60, Class interval, , Frequency (fi), , Cumulative frequency (cf), , 0–10, , 5, , 5, , 10–20, , x, , 5+x, , 20–30, , 20, , 25 + x, , 30–40, , 15, , 40 + x, , 40–50, , y, , 40 + x + y, , 50–60, , 5, , 45 + x + y, , Total, , Sfi = 60, , Since the median = 28.5, therefore, median class is 20–30, n, \, = 30, l = 20, h = 10, cf = 5 + x, f = 20, 2, , n, − cf, 2, ,  30 − (5 + x) , \, Median = l + ,  × 10,  × h ⇒ 28.5 = 20 + , 20,  f , ⇒, ⇒, Also,, , 25 − x, 25 − x, × 10 ⇒ 28.5 = 20 +, 2, 20, 57 = 65 – x ⇒ x = 65 – 57 = 8, 45 + x + y = 60 ⇒ y = 7, 28.5 = 20 +, , ⇒ 57 = 40 + 25 – x, , Exercise 7.2, I. Very Short Answer Type Questions, , [1 Mark], , 1. Multiple Choice Questions (MCQs), Choose the correct answer from the given options:, (1) Consider the following frequency distribution of the heights of 60 students of a class, Height (in cm), No. of students, , 150–155 155–160 160–165 165–170 170–175 175–180, 15, , 13, , 10, , The upper limit of the median class in the given data is, (a) 165, (b) 155, (c) 160, , 134, , Mathematics–10, , 8, , 9, , 5, , (d) 170

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , (2) For the following distribution:, Marks, , Number of students, , Marks, , Number of students, , Below 10, , 3, , Below 40, , 57, , Below 20, , 12, , Below 50, , 75, , Below 30, , 28, , Below 60, , 80, , The modal class is, (a) 0-20, (b) 20-30, (c) 30-40, (d) 50-60, (3) The cumulative frequency of a given class is obtained by adding the frequencies of all the classes, (a) preceding it, (b) succeeding it, (c) Both (a) and (b), (d) None of these, 2. Assertion-Reason Type Questions, , In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct, choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., (1) Assertion (A): If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0,, 52, 36, 27, then median is 30.,  n + 1, , Reason (R): Median = , value, if n is odd.,  2 , (2) Assertion (A): If the value of mode and mean is 60 and 66 respectively, then the value of median is 64., , Reason (R): Median = (Mode + 2 Mean), 3. Answer the following., (1) Write the modal class of the following frequency distribution:, th, , Class interval, Frequency, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , 33, , 38, , 65, , 52, , 19, , 48, , (2) Write the median of the following data: 3, 5, 2, 9, 7, 11, II. Short Answer Type Questions - I, , [2 Marks], , 4. Find the median class of following data:, Class interval, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 8, , 10, , 12, , 22, , 30, , 18, , Frequency, , 5. Find the mode of the following data:, Class, , [CBSE Standard 2020], , 0–20, , 20–40, , 40–60, , 60–80, , 80–100, , 6, , 8, , 10, , 12, , 6, , Frequency, , 100–120 120–140, 5, , 3, , 6. The following distribution shows the transport expenditure of 100 employees:, Expenditure (in `), , 200–400, , 400–600, , 600–800, , 800–1000, , 1000–1200, , No. of employees, , 21, , 25, , 19, , 23, , 12, , Find the mode of the distribution., III. Short Answer Type Questions - II, 7. The weight of tea in 70 packets are as follows:, Weight (in kg), No. of packets, , [3 Marks], , 200–201 201–202 202–203 203–204 204–205 205–206, 12, , 26, , 20, , 4, , 2, , 1, , Determine the modal weight., , [Imp.], , Statistics, , 135

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , 8. The annual rainfall record of a city of 100 days is given in the following table:, Rainfall (in cm), , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , 8, , 8, , 14, , 22, , 30, , 8, , 10, , No. of days, , Calculate the median rainfall., , [Imp.], , 9. The following table shows the ages of the patients admitted in a hospital during a year:, Age (in year), , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , No. of patients, , 22, , 10, , 8, , 15, , 5, , 6, , Find the median of the data given above. , , [Imp.], , 10. The table below shows the salaries of 280 persons:, Salary (In thousand `), , No. of Persons, , 5 – 10, , 49, , 10 – 15, , 133, , 15 – 20, , 63, , 20 – 25, , 15, , 25 – 30, , 6, , 30 – 35, , 7, , 35 – 40, , 4, , 40 – 45, , 2, , 45 – 50, , 1, , Calculate the median salary of the data., , [CBSE 2018], , 11. The following data gives the information on the observed life-times (in hours) of 25 electrical components. Determine, the model life-time of the components., [Imp.], Life-time (in hrs), , 0–50, , 50–100, , 2, , 3, , No. of components, , 100–150 150–200 200–250 250–300 300–350, 5, , 6, , 5, , 3, , 1, , 12. The table shows the daily expenditure on grocery of 25 households in a locality. Find the modal daily expenditure on, grocery by a suitable method. , [CBSE SP 2018-19], Daily Expenditure (in `), , 100-150 150-200 200-250 250-300 300-350, 4, , No. of households, , 5, , 12, , 2, , 2, , 13. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70., , [CBSE Standard SP 2020-21], Class, Frequency, , 0-5, 12, , 5-10, a, , 10-15, 12, , 15-20, 15, , 20-25, b, , 25-30, 6, , 30-35, 6, , 35-40, 4, , IV. Long Answer Type Questions, , [5 Marks], , 14. If the median of the following frequency distribution is 32.5. Find the values of f1 and f2., Class, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , 50–60, , 60–70, , Total, , f1, , 5, , 9, , 12, , f2, , 3, , 2, , 40, , Frequency, , 15. Compare the modal age of two groups of students A and B appearing for an entrance test., Class interval, , 136, , Frequency, Group A, , Group B, , 16–18, , 50, , 54, , 18–20, , 78, , 89, , Mathematics–10, , [HOTS]

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 20–22, , 46, , 40, , 22–24, , 28, , 25, , 24–26, , 23, , 17, , Date __________, , 16. The median of the following data is 525. Find the values of x and y if the total frequency is 100., Class Interval, , 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, , 17. Daily wages of 110 workers, obtained in a survey, are tabulated below:, Daily Wages (in `), Number of Workers, , [CBSE Standard SP 2019-20], , 100–120 120–140 140–160 160–180 180–200 200–220 220–240, 10, , 15, , 20, , 22, , 18, , 12, , 13, , Compute the mean daily wages and modal daily wages of these workers., 18. 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., [CBSE Standard 2020], Number of wickets, No. of bowlers, , 20–60, , 60–100, , 100–140, , 140–180, , 180–220, , 220–260, , 7, , 5, , 16, , 12, , 2, , 3, , 19. The mode of the following data is 67. Find the missing frequency x., Class, Frequency, , [CBSE Standard SP 2020-21], , 40-50, , 50-60, , 60-70, , 70-80, , 80-90, , 5, , x, , 15, , 12, , 7, , 20. A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was, obtained. Find the median height and the mean using the formulae., Height (in cm), , Number of Girls, , Less than 140, , 4, , Less than 145, , 11, , Less than 150, , 29, , Less than 155, , 40, , Less than 160, , 46, , Less than 165, , 51, , Statistics, , 137

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Case Study Based Questions, I. Females’ Literacy: The education of women helps to remove the social stigma that surrounds it. It is the key to eliminating, social evils such as female infanticide, dowry, child marriage, harassment, etc. This will not just help the women of today, but of the future generations who can live in a world where gender equality exists which ultimately raises the literacy, rate., The following distribution shows the number of literate females in the age group 0 to 60 years of a particular area., Age (in years), , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , No. of literate females, , 350, , 1100, , 900, , 750, , 600, , 500, , 1. The class marks of class 40-50 is, (a) 70, (b) 90, (c) 10, (d) 45, 2. The number of literate females whose ages are between 20 years and 50 years is, (a) 1350, (b) 1650, (c) 2000, (d) 2250, 3. The modal class of the above distribution is, (a) 0-10, (b) 10-20, (c) 20-30, (d) 30-40, 4. The number of literate females whose ages are less than 40 years is, (a) 1450, (b) 2350, (c) 3100, (d) 3700, 5. The upper limit of modal class is, (a) 10, (b) 20, (c) 30, (d) 60, II. 100 m Race, , A stopwatch was used to find the time that it took a group of students to run 100 m., 0-20, 20-40, 40-60, 60-80, Time (in sec.), 8, 10, 13, 6, No. of students, , 80-100, 3, [CBSE Standard SP 2020-21], , 1. The estimated mean time taken by a student to finish the race is, (a) 54, (b) 63, (c) 43, (d) 50, 2. What will be the upper limit of the modal class?, (a) 20, (b) 40, (c) 60, (d) 80, 3. The construction of cumulative frequency table is useful in determining the, (a) mean, (b) median, (c) mode, (d) All of the above, 4. The sum of lower limits of median class and modal class is, (a) 60, (b) 100, (c) 80, (d) 140, 5. How many students finished the race within 1 minute?, (a) 18, (b) 37, (c) 31, (d) 8, III. 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 among humans., , The following tables shows the age distribution of case admitted during a day in two different hospitals., Table 1, Age (in years), No. of cases, , 5-15, , 15-25, , 25-35, , 35-45, , 45-55, , 55-65, , 6, , 11, , 21, , 23, , 14, , 5, , Table 2, Age (in years), No. of cases, , 138, , 5-15, , 15-25, , 25-35, , 35-45, , 45-55, , 55-65, , 8, , 16, , 10, , 42, , 24, , 12, , Mathematics–10

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Refer to Table 1, 1. The average age for which maximum cases occurred is, (a) 32.24 years, (b) 34.36 years, (c) 35.91 years, 2. The upper limit of modal class is, (a) 15, (b) 25, (c) 35, 3. The mean of the given data is, (a) 26.2, (b) 32.4, (c) 33.5, Refer to Table 2, 4. The mode of the given data is, (a) 41.4, (b) 48.2, (c) 55.3, 5. The median of the given data is, (a) 32.7, (b) 40.2, (c) 42.3, , Date __________, , (d) 42.24 years, (d) 45, (d) 35.4, , (d) 64.6, (d) 48.6, , Answers and Hints, 1. (1) (a) 165, (1) (2) (c) 30-40 , (1), (3) (a) preceding it, (1), 2. (1) (d) Assertion (A) is false but reason (R) is true., Arrange the terms in ascending order,, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52, th, 11 + 1, , Median value = ,  2 , = 6th value = 27, , , , (1), , (2) (c) Assertion (A) is true but reason (R) is false., 1, , Median =, (Mode + 2 Mean), 3, 1, , =, (60 + 2 × 66) = 64, (1), 3, 3. (1) Frequency of the class interval 30-40 is maximum,, i.e., 65. So, the modal class is 30-40., (1), (2) Median, th, 1  6 , =   observation, 2  2 , th, , 6 , +  + 1 observation , , 2 , , [Q n = 6 (even)], 1, = [3rd observation + 4th observation], 2, 1, = [5 + 7] = 6, (1), 2, 4., , , , , Class interval, , f, , cf, , 0–10, , 8, , 8, , 10–20, , 10, , 18, , 20–30, , 12, , 30, , 30–40, , 22, , 52, , 40–50, , 30, , 82, , 50–60, , 18, , 100, , n = 100, n, , ⇒, = 50, 2, , ⇒ Median class is 30-40., , 5. Maximum frequency = 12, , \ Modal class = 60–80,  f1 − f 0 , Now, Mode = l + h , ,  2 f1 − f 0 − f 2 , , , ,  12 − 10, = 60 + 20 , (1),  2 × 12 − 10 − 6 , , , ,  2 , = 60 + 20 ,  24 − 16 , , , , = 60 +, , 20 × 2, 8, , , = 60 + 5 = 65, (1), 6. Maximum frequency = 25, So, Modal class = 400–600, , , f1 − f 0, , \, Mode = l + h ,  (1),  2 f1 − f 0 − f 2 ,  25 − 21 , = 400 + 200 , ,  50 − 21 − 19 , = 400 + 200 ×, , 4, 10, , = 400 + 80, = 480, (1), 7. Modal class = 201–202 as its frequency is maximum., , \ Modal weight, , , f1 − f 0, = l + h × ,  (1),  2 f1 − f 0 − f 2 , 26 − 12, , , = 201 + 1 × , (1),  2 × 26 − 12 − 20 , = 201 +, , (1), , 14, 52 − 32, , 14, 20, = 201.7 kg., 8. 39.09 cm, 9. 21.25 years, = 201 +, , (1), , (1), (3), (3), , Statistics, , 139

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\ 23-Nov-2021, , 10., , Proof-3, , Amit, , Reader’s Sign _______________________, , Salary, No. of Persons, (In thousand `), (f), 5–10, 49, 10–15, 133, 15–20, 63, 20–25, 15, 25–30, 6, 30–35, 7, 35–40, 4, 40–45, 2, 45–50, 1, , cf, , , , N, 280, =, 2, 2, = 140, , , Median = l +, , 1=, , , , , , , a=, 55 + a + b =, 55 + 8 + b =, b=, , (½), , (½), , Frequency (fi), , cf, , 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, , 60–70, , 2, , f1 + f2 + 31, , Total, , Sfi = 40, , 5, (140 − 49) (1), 133, , 11. Modal life-time = 175 hrs., , (3), , 12. 200 – 250 is the modal class., f1 − f 0, × h (1), 2 f1 − f 0 − f 2, , = 200 +, , 12 − 5, × 50 (1), 24 − 5 − 2, , = 200 + 20.59, = `220.59 (1), 13., , 8, 70, 70, 7, , Class Interval, , , Median salary is `13.42 thousand or `13420 (approx.)(1), , Mode = l +, , 3, , Now, we have, , 5 × 91, = 10 +, 133, = 13.42, , , , (11 − a ), , 14. Here, median = 32.5 and n = 40, , h N, ,  − C (1), f 2, , = 10 +, , 35 − 24 − a, ×5, 15, , 16 = 15 +, , , , Median class is 10–15, , , N, − cf, Median = l + 2, × h (½), f, , , , 49, 182, 245, 260, 266, 273, 277, 279, 280, , Class, , Frequency, , Cumulative, Frequency, , 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, , N = 70, , , , 55 + a + b = 70, , , , a + b = 15, , (1), (½), , , , Mathematics–10, , (1), , Since the median is given to be 32.5, thus the median class, is 30-40., n, , − cf, 2, , Median = l + , × h (1),  f , , ⇒, , 32.5 = 30 +, , 20 − f1 − 14, × 10, 12, , , ⇒, , 32.5 = 30 +, , 6 − f1, × 10 (1), 12, , , ⇒, , 2.5 =, , 6 − f1, × 10, 12, , , ⇒, , 2.5 =, , 6 − f1, ×5, 6, , , ⇒, , 15, = 6 – f1, 5, , , ⇒, , 3 = 6 – f1, , ⇒ f1 = 3, , (1), , Now, f1 + f2 + 31 = 40, , ⇒, , f2 = 9 – 3, , , ⇒, , f2 = 6, , Thus, f1 = 3 and f2 = 6, , 140, , Date __________, , (1)

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , 15. For group A, modal class is 18-20, , , ⇒ (14 – x) × 5 = 25, , ⇒, x = 9, from (i), y = 15, , , \ Mode of group A, , f1 − f 0 , = l + , × h (1),  2 f1 − f 0 − f 2 , , 17. Daily Wages, , , , 78 − 50, = 18 + ,  ×2, 2, ×, 78, −, 50, −, 46, , , 28, = 18 +, ×2, 156 − 96, 56, = 18 +, 60, = 18.93 years, , (1), , For group B, modal class is 18-20, , \ Mode of group B, , f1 − f 0 , = l + , × h (1),  2 f1 − f 0 − f 2 , , , 89 − 54, = 18 + , × 2, ,  2 × 89 − 54 − 40 , = 18 +, , (1), , 35, × 2 = 18.83 years, 84, , Since 18.93 > 18.83, , (1), , (in `), , Number of, Workers (fi), , 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, , 0, , 0, , 180–200, , 18, , 190, , 1, , 18, , 200–220, , 12, , 210, , 2, , 24, , 220–240, , 13, , 230, , 3, , 39, , Total, , 110, , 1, , , , Mean daily wages, 1, × 20, 110, = `170.19 (approx.), , (1½), , 22 − 20, × 20, 44 − 20 − 18, = `166.67 (approx.), 18. Mean:, , , Mode = 160 +, , (1½), , CI, , xi, , fi, , fixi, , 20-60, , 40, , 7, , 280, , Class Interval, , Frequency, , cf, , 60–100, , 80, , 5, , 400, , 0–100, , 2, , 2, , 100–140, , 120, , 16, , 1920, , 100–200, , 5, , 7, , 140–180, , 160, , 12, , 1920, , 200–300, , x, , 7+x, , 180–220, , 200, , 2, , 400, , 300–400, , 12, , 19 + x, , 220–260, , 240, , 3, , 720, , 400–500, , 17, , 36 + x, , Sfi = 45, , Sfixi = 5640, , 500–600, , 20, , 56 + x, , 600–700, , y, , 56 + x + y, , , \ Mean x =, , 700–800, , 9, , 65 + x + y, , , Median:, , 800–900, , 7, , 72 + x + y, , 900–1000, , 4, , 76 + x + y, , (2), , , N = 100 ⇒ 76 + x + y = 100, (½), , ⇒, x + y = 24, ...(i) (½), Median = 525, , ⇒ 500 – 600 is median class., n, − cf, Median = l + 2, × h (1), f,  50 − 36 − x , , ⇒ 500 + ,  × 100 = 525, , 20, , (2), , = 170 +, , , \ The modal age of students of group A > modal age of, students of groups B., (1), 16., , Date __________, , Total, , , , Σfi xi, 5640, =, = 125.33, Σf i, 45, , Number, of wickets, (CI), , Frequency, (fi), , Cumulative, frequency, (cf), , 20–60, , 7, , 7, , , Q, , 60–100, , 5, , 12, , 100–140, , 16, , 28, , 140–180, , 12, , 40, , 180–220, , 2, , 42, , 220–260, , 3, , 45, , Total, , Sf = 45, , x = 45, , (1), (1), , ← Median class, , (1), , n = 45, , Statistics, , 141

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , n, 45, =, 2, 2, , Date __________, , = 22.5, , N, ,  − C, 2, Median = l +, ×h, f, , n, Since, the cumulative frequency just greater than, i.e.,, 2, 22.5 is 28., (1), , , , C = C.F. of the class preceding the median class, = 11, (½), , , , h = higher limit – lower limit, , , \, , So, the median class is 100-140., Now, Median, n,  2 − cf, = l + h ,  f, , , , , , , Here, l = lower limit of median class = 145, , , , ,  22.5 − 12 , = 100 + 40 , ,  16, , = 150 – 145 = 5, f = frequency of median class = 18, , , \ Median = 145 +, , , 10.5, = 100 + 40 ×, = 126.25, 16, , (1), , (25.5 − 11), ×5, 18, , = 149.03, , (1), , Calculating mean, Height, (in cm), , f, , xi, , fxi, , below 140, , 4, , 137.5, , 550, , 140 – 145, , 7, , 142.5, , 997.5, , 145 – 150, , 18, , 147.5, , 2655, , 150 – 155, , 11, , 152.5, , 1677.5, , 7 × (18 – x) = 10(15 – x)(½), , 155 – 160, , 6, , 157.5, , 945, , , , 160 – 165, , 5, , 162.5, , 812.5, , Mode = l +, , 19. , , f1 − f 0, × h (½), 2 f1 − f 2 − f 0, , , , 15 − x, 67 = 60 +, × 10 (½), 30 − 12 − x, , , , 15 − x, 7=, × 10 (½), 18 − x, , 126 – 7x = 150 – 10x, , , , 3x = 150 – 126, , , , 3x = 24, , , , x = 8, , (1½), (1½), , Mean =, , , , =, , , , = 149.75, , f, , cf, , Below 140, , 4, , 4, , 140 – 145, , 7, , 11, , I. 1. (d) 45, , 2. (d) 2250, , 145 – 150, , 18, , 29, , 150 – 155, , 11, , 40, , , 3. (b) 10-20, , 4. (c) 3100, , 155 – 160, , 6, , 46, , 160 – 165, , 5, , 51, , Case Study Based Questions, , , 5. (b) 20, , N 51, =, = 25.5, 2, 2, As 29 is just greater than 25.5, therefore median class, is 145 – 150., ⇒, , II. 1. (c) 43, , 3. (b) median, , Mathematics–10, , 2. (c) 60, 4. (c) 80, , , 5. (c) 30, III. 1. (c) 35.91 years, , 2. (d) 45, , , 3. (d) 35.4, , 4. (a) 41.4, , , 5. (b) 40.2, , 142, , 7637.5, 51, , Height (in cm), , N = 51, , (1), , Σfx, N, , , , 20. Calculating Median, , (1), , (½), , (1)

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , Experts’ Opinion, Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly., 1. To find the mean of grouped data by direct method., 2. To find the mode of grouped data., 3. To find the median of grouped data., , IMPORTANT FORMULAE, n, , x1 + x2 + ... + xn, =, n, Σf x, (i) Using direct method: x = i i, Σf i, (ii) Using assumed mean method:, Σf d, , x = A + i i , di = xi – A, Σf i, ••, , Mean x =, , , f1 − f 0 , •• Mode = l + , ×h,  2 f1 − f 0 − f 2 , , ∑ xi, , i =1, , n, , n, − cf, 2, •• Median = l + ,  f, , , ,  ×h, , , •• 3 median = mode + 2 mean, •• Note: Symbols used above have their usual meanings., , COMMON ERRORS, Errors, (i) Choosing not a suitable method to obtain the mean of, grouped data., , Corrections, (i) The choice of method to be used depends on the numerical, values of xi and fi. If xi and fi are sufficiently small, the, direct method is an appropriate choice. If xi and fi are, numerically large numbers, choose the assumed mean, method., (ii) Interpreting incorrectly that mode of a given data is (ii) For a given data, mode may be equal, more than or less, always less than the mean of that data., than the mean of that data., (iii) Interpreting incorrectly that mode can be calculated for (iii) The mode can also be calculated for grouped data with, grouped data of equal size only., unequal class sizes., , QUICK REVISION NOTES, •• There are three measures of central tendency: Mean, Mode and Median., •• Mean: The mean x of an ungrouped data is obtained by dividing the sum of numbers by the number of data. The mean, of ‘n’ numbers of data x1, x2, x3, ..., xn denoted by x is defined as, n, , , , x=, , x1 + x2 + ... + xn, =, n, , ∑ xi, , i =1, , n, , •• The mean of grouped data can be found by following methods:, (i) Direct Method: If the variates observations x1, x2, x3, ..., xn have frequencies f1, f2, ..., fn respectively, then the mean, is given by, f x + f 2 x2 + ... + f n xn, , Mean x = 1 1, ,, f1 + f 2 + ... + f n, n, , , , x=, , ∑ fi xi, , i =1, , Σf i, , (ii) Assumed Mean Method: Mean x = A +, , Σf i d i, , where A is the assumed mean and di = xi – A, Σf i, , Statistics, , 143

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\ 23-Nov-2021, , Amit, , Proof-3, , Reader’s Sign _______________________, , Date __________, , In this method, first find the class mark or mid-value of each class, as:,  lower limit + upper limit , xi = class mark = , , , 2, Further, take an approximate mean, called assumed mean. This assumed mean is taken preferably near the middle,, say A and the deviation di = xi – A for each variate xi., •• Mode: The mode of a distribution is the value of the observation for which the frequency is maximum. The mode for, grouped data can be found by using the formula:, , , , f1 − f 0 , Mode = l + , ×h, 2, f,  1 − f 0 − f 2 , , ••, , •• where, l = lower limit of the model class, ••, f1 = frequency of the model class, ••, f0 = frequency of the class preceding the model class, ••, f2 = frequency of the class succeeding the model class, ••, h = size of the model class, •• Median: The median is the middle value of a distribution i.e., the value of the observation which divides it into two equal, parts., •• In the case of ungrouped data, first the data is arranged in ascending order. Then if n i.e., the number of observations,  n + 1, is odd, then median is the value of ,  2 , observation is the median., , th, ,  n, observation, and if n is even, then the average of  ,  2, , th, , n , and  + 1, 2 , , th, , •• Median of a grouped or continuous frequency distribution can be found by using the formula:, , Median, where,, , , n, , − cf, 2, , =l + ,  ×h,  f , l = lower limit of the median class, n = number of observations, , , , f = frequency of the median class, , , , h = size of the median class (assuming class size to be equal), , , , cf = cumulative frequency of the class preceding the median class, , qqq, , 144, , Mathematics–10

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Sample Paper - 1 (Solved), Time Allowed: 2 Hours, , Maximum Marks: 40, , General Instructions:, 1. This question paper contains two parts, A and B., 2. Both Part A and Part B have internal choices., Part-A:, 1. It consists of two sections, I and II., 2. Section I has 7 questions of 1 mark each. Internal choice is provided in 2 questions., 3. Section II has 2 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to, attempt any 4 out of 5 sub-parts., Part-B:, 1. Question No. 10 to 12 in section III are Short Answer Type-I questions of 2 marks each., 2. Question No. 13 to 15 in section IV are Short Answer Type-II questions of 3 marks each., 3. Question No. 16 and 17 in section V are Long Answer Type questions of 5 marks each., 4. Internal choice is provided in 1 question of 2 marks, 1 question of 3 marks and 1 question of 5 marks., , PART-A, SECTION-I, 9x2, , 1. For what values of k, the equation, + 6kx + 4 = 0 has equal roots?, 2, Sol. , 9x + 6kx + 4 = 0, , , (6k)2 – 4 × 9 × 4 = 0, , , 36k2 = 144   ⇒ k2 = 4, , , k = ±2, 2. For what value(s) of ‘a’ quadratic equation 3ax2 – 6x + 1 = 0 has no real roots?, , 1, , 1, , OR, x2, , , Find the roots of the equation + 7x + 10 = 0, Sol. , 3ax2 – 6x + 1 = 0, , (–6)2 – 4(3a)(1) < 0, , 12a > 36   ⇒ a > 3, OR, 2, , x + 7x + 10 = 0, , x2 + 5x + 2x + 10 = 0, , (x + 5)(x + 2) = 0, , x = –5, x – 2, 3. In the figure, if B1, B2, B3,…... and A1, A2, A3,….. have been marked at equal distances, in what ratio C divides AB?1, , Sol. 8 : 5, , 145

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , Date __________, , 4. 12 solid spheres of the same radii are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm., Find the diameter of the each sphere., 1, Sol. , , pR2H = 12 ×, , 4 3, πr, 3, 4, × r 3 × 12 ⇒ r3 = 1 ⇒ r = 1, 3, d = 2 cm, , , 1 × 1 × 16 =, , , , 5. Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?, , 1, , OR, , Find the radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm., Sol. Let h1 and h2 be the heights of two cones and r1 and r2 be their radii., , Then,, , , , r1, h1, 1, 3, =, and, = [Given], r2, h2, 3, 1, 1 2, 2, 2, πr1 h1,  3,  1,  r1 , 1 3, h1, 3, Ratio of their volumes =, =   ×, =   ×   = 9 × = = 3 : 1., 1, 3, 1 2, 3 1, r, h,  2, 2, πr2 h2, 3, OR, , Q, , Edge of the cube = 4.2 cm, , 4.2, 1, (Edge of the square) =, = 2.1 cm, 2, 2, , Directions (Q6 and Q7): In the following questions, a statement of assertion (A) is followed by a statement of reason, (R). Mark the correct choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., 6. Assertion (A) : 4x2 – 12x + 9 = 0 has repeated roots., Reason (R) : The quadratic equation ax2 + bx + c = 0 have repeated roots if discriminant D > 0., 1, Sol. (c) Assertion (A) is true but reason (R) is false., Assertion, 4x2 – 12x + 9 = 0, , D = b2 – 4ac = (–12)2 – 4(4)(9), , = 144 – 144 = 0, Roots are repeated., 7. Assertion (A) : If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52,, 36, 27, then median is 30., , \ Radius of the largest right circular cone =, , n + 1, Reason (R) : Median = ,  2 , , th, , value, if n is odd., , Sol. (d) Assertion (A) is false but reason (R) is true., Arranging the terms in ascending order,, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52, , , 146, ,  11 + 1, Median value = ,  2 , , Mathematics–10, , th, , = 6th value = 27, , 1

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SECTION-II, , Case Study based questions Q8 and Q9 are compulsory. Attempt any four sub-parts of each question., 8. Case Study-1, , Adventure Camp, , Adventure camps are the perfect place for children to practice, decision making for themselves without parents and teachers, guiding their every move., , Some students of a school reached for adventure at Mukteshwar., 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 above., , After that they went for the 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., , Refer to Glass and Cup, 1, , (i) The volume of cylindrical glass is, (a) 295.75 cm3, , (b) 7415.5 cm3, , (c) 384.88 cm3, , (d) 404.25 cm3, , cm3, , cm3, , 1, , (ii) The volume of hemispherical cup is, (a) 179.67, , cm3, , (b) 89.83, , cm3, , (c) 172.25, , (d) 210.60, , 1, , (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, , Refer to Tent, 1, , (iv) The height of the conical tent prepared to accommodate 4 students is, (a) 18 m, , (b) 10 m, , (c) 24 m, , (d) 14 m, , (v) How much space on the ground is occupied by each student in the conical tent?, (a) 54 m2, Sol. (i), , d = 7 cm, r=, , , ∴ , , , (b) 38.5 m2, , (c) 86 m2, , 1, , (d) 24 m2, , and h = 10.5 cm, , 7, cm, 2, , Volume of cylindrical glass = pr2h, , , , ∴ Correct, , =, option is (d), , , (ii), , 22 7 7, × × × 10.5 = 404.25 cm3, 7 2 2, , d = 7 cm ∴ R =, , 7, cm, 2, , Volume of hemispherical cup =, , 2 3, πR, 3, , , , 2 22 7 7 7, ×, × × × = 89.83 cm3, 3 7 2 2 2, , , ∴ Correct, , =, option is (b), , Sample Paper - 1 (Solved), , 147

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , Date __________, , , (iii) Vol. of cylindrical glass (= 404.25 cm3) > Vol. of hemispherical cup (= 89.83 cm3), , , ∴ Vol. of cylindrical glass is more than the vol. of hemispherical cup., , , Difference in volume = 404.25 cm3 – 89.83 cm3 = 314.42 cm3, ∴ Correct option is (d), , (iv) , Net area of canvas = 551 m2 – 1 m2 = 550 m2; radius (r) = 7 m, , , Net area of canvas = curved surface area of cone, , , 550 = prl, 22, × 7 × h2 + r 2, 7, , , , , 550 =, , , , ⇒, , 550, =, 22, , h2 + 72, , , , ⇒, , 25 m =, , h 2 + 49, , ⇒ 625 – 49 = h2, , , , ⇒, 576 = h   ∴ h = 24 m, ∴ Correct option is (c), , (v) Area of space on the ground of conical tent = pr2, 22, , , =, × 7 × 7 = 154 m2., 7, , There are 4 students in the tent., , So, Area of space for 1 student =, ∴ Correct option is (b), , 154 2, m = 38.5 m2., 4, , 9. Case Study-2, , 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, 8, , 20-40, 10, , 40-60, 13, , 60-80, 6, , 80-100, 3, 1, , (i) The estimated mean time taken by a student to finish the race is, (a) 54, , (b) 63, , (c) 43, , (d) 50, 1, , (ii) What will be the upper limit of the modal class?, (a) 20, , (b) 40, , (c) 60, , (d) 80, , (iii) The construction of cumulative frequency table is useful in determining the, (a) mean, , (b) median, , (c) mode, , (d) All of these, 1, , (iv) The sum of lower limits of median class and modal class is, (a) 60, , 148, , (b) 100, , Mathematics–10, , (c) 80, , 1, , (d) 140

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1, , (v) How many students finished the race within 1 minute?, (a) 18, Sol., , (i), , (b) 37, , Class Interval, (CI), 0-20, 20-40, 40-60, 60-80, 80-100, , (c) 31, , Frequency (fi), , Mid-value (xi), , 8, 10, 13, 6, 3, Sfi = 40, , 10, 30, 50 = A, 70, 90, , Let the assumed mean be, , (d) 8, di = xi – A, (A = 50), –40, –20, 0, 20, 40, , fidi, –320, –200, 0, 120, 120, Sfidi = –280, , A = 50, , Sfi = 40 and Sfidi = –280, , , Hence,, , Σf i d i, −280, = 50 +, = 50 – 7 = 43, Σf i, 40, So, option (c) is the correct answer., \, , Mean x = A +, , (ii) Since frequency of 40-60 is maximum. So, the modal class is 40-60 and its upper limit is 60., So, option (c) is the correct answer., (iii) (b) median, (iv), , Class Interval, (CI), 0-20, 20-40, 40-60, 60-80, 80-100, Total, , Sfi = n = 40,, , Frequency (fi), 8, 10, 13, 6, 3, Sfi = 40, , Cumulative Frequency, (cf), 8, 18, 31, 37, 40, , n, = 20, 2, , Since cf just greater than 20 is 31., \ The corresponding class is 40 = 60 which is the median class., Form part (ii), modal class is also 40 = 60., Now, sum of lower limits of median class and modal class = 40 + 40 = 80, So, option (c) is the correct answer., (v) The number of students who finished the race within 1 minute = 8 + 10 + 13 = 31., So, option (c) is the correct answer., , PART-B, SECTION-III, 10. How many two-digit numbers are divisible by 3?, 2, Sol. Numbers divisible by 3 are 3, 6, 9, 12, 15, ..., 96 and 99. Lowest two-digit number divisible by 3 is 12 and highest twodigit number divisible by 3 is 99., , Hence, the sequence starts with 12, ends with 99 and common difference is 3., , So, the AP is 12, 15, 18, ..., 96, 99., , Hence,, , a = 12, d = 3 and an = 99, , Sample Paper - 1 (Solved), , 149

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , , , an = a + (n – 1)d, , , , 99 = 12 + (n – 1)3, , Date __________, , 87, = 29 ⇒ n = 30, 3, , Therefore, there are 30 two-digit numbers divisible by 3., n–1=, , , , 11. 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., 2, , OR, , In the given figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T., , , , Prove that: ∠PTQ = 2∠OPQ., Sol., , ∠A = ∠OPA = ∠OSA = 90°, , , Hence,, , ∠SOP = 90°, , , Also,, , AP = AS, , , Hence, OSAP is a square, , , AP = AS = 10 cm, , , , CR = CQ = 27 cm, , , , BQ = BC – CQ = 38 – 27 = 11 cm, , , , BP = BQ = 11 cm, , , , x = AB = AP + BP = 10 + 11 = 21 cm, OR, , , We know that, tangents drawn from same external point are equal., , So,, , TP = TQ, , , ⇒, , ∠TQP = ∠TPQ...(i), , , Now,, , OP ^ TP, , , ⇒, , ∠OPT = 90°, , , ⇒, , ∠TPQ = 90° – ∠OPQ...(ii), , ⇒ ∠OPQ + ∠TPQ = 90°, , , In DPTQ,, , , ∠TPQ + ∠TQP + ∠PTQ = 180°, , , ⇒, , ∠TPQ + ∠TPQ + ∠PTQ = 180°, , 150, , Mathematics–10, , [From (i)]

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⇒, , 2∠TPQ + ∠PTQ = 180°, , , ⇒, , 2(90° – ∠OPQ) + ∠PTQ = 180°, , , ⇒, , 180° – 2∠OPQ + ∠PTQ = 180°, , , ⇒, , [From (ii)], , ∠PTQ = 2∠OPQ, , , Hence proved., 12. A kite is flying at a height of 90 m above the ground. The string attached to the kite is temporarily tied to a point on the, ground. The inclination of the string with the ground is 60°. Find the length of the string assuming that there is no slack in, the string., 2, Sol. As per given in question we have drawn figure below., , A, , , In right DABC, we have, , , , , sin 60° =, , AB, AC, , 3, 90, =, 2, x, x=, , 60°, B, , 90 × 2, 3, , , , x, , 90 m, , , , =, , 180, 3, , =, , C, , 3 × 60, 3, , = 60 3 = 60 × 1.732, , , Hence, length of string is 103.92 m., , SECTION-IV, 13. In the figure, ABCD is a square of side 14 cm. Semicircles are drawn with each side of square as diameter. Find the area, of the shaded region., 3, , Sol. , , Area of 1 segment = area of sector – area of triangle, , , , 90°  2 1, = , πr − × 7 × 7,  360° , 2, , , , =, , , , = 14 cm2, , , , 1 22, 1, ×, × 72 − × 7 × 7, 4 7, 2, , Area of 8 segments = 8 × 14 = 112 cm2, , Area of the shaded region = 14 × 14 – 112, , , = 196 – 112 = 84 cm2, , , (each petal is divided into 2 segments), , Sample Paper - 1 (Solved), , 151

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , Date __________, , 14. 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., 3, Sol. Steps of construction:, , 1. Draw a line segment AB of length 7 cm., , 2. With A as centre, draw a circle of radius 3 cm., , 3. With B as centre, draw a circle of radius 2 cm., , 4. Draw the perpendicular bisector of AB. Let P be the mid-point of segment, AB., , 5. With P as centre and radius PA draw a circle which intersects the circle, with centre A at M and the circle with centre B at R and S., , 6. Join BM and BN. Also join AR and AS. Then, BM, BN, AR and AS are required tangents., 15. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70., Class, Frequency, , 0-5, 12, , 5-10, a, , 10-15, 12, , 15-20, 15, , 20-25, b, , 25-30, 6, , 30-35, 6, , 3, , 35-40, 4, , OR, , The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is, ` 18. Find the missing frequency k., Daily pocket, allowance (in `), Number of children, Sol., , , , , , , 11–13, , 13–15, , 15–17, , 17–19, , 19–21, , 21–23, , 23–25, , 3, , 6, , 9, , 13, , k, , 5, , 4, , Class, 0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, Total, , Frequency, 12, a, 12, 15, b, 6, 6, 4, N = 70, 55 + a + b = 70, a + b = 15, , , , , Median =, , , , , 16 =, , , , , 1=, , , , , a=, , N, − cf, l+ 2, ×h, f, 35 − 24 − a, 15 +, ×5, 15, (11 − a ), 3, 8, , , , , 55 + a + b = 70, , , , , 55 + 8 + b = 70, , , , , 152, , b=7, , Mathematics–10, , OR, , Cumulative Frequency, 12, 12 + a, 24 + a, 39 + a, 39 + a + b, 45 + a + b, 51 + a + b, 55 + a + b

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Daily pocket, allowance (in `), , Number of, children (fi), , Mid-point (xi), , di = xi – 18, , fidi, , 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25, , 3, 6, 9, 13, k, 5, 4, , 12, 14, 16, 18 = a, 20, 22, 24, , –6, –4, –2, 0, 2, 4, 6, , – 18, – 24, – 18, 0, 2k, 20, 24, , Σfi = 40 + k, , Σfi di = 2k – 16, , , By assumed mean method,, , , Mean = x = a +, , , , 18 = 18 +, , , ⇒, , 2k = 16, , Sfi di, Sfi, , 2k –16, 40 + k, , ⇒, ⇒, , 2k –16, =0, 40 + k, k=8, , SECTION-V, 16. Solve for x :, Sol. We have, , , , 1, 2, 4, +, =, x ≠ − 1, − 2, − 4 , x +1 x + 2 x + 4, 1, 2, 4, +, =, x +1 x + 2, x+4, x + 2 + 2( x + 1), 4, =, ( x + 1)( x + 2), x+4, 3x + 4, x 2 + 3x + 2, , =, , 4, x+4, , , , (3x + 4)(x + 4) = 4(x2 + 3x + 2), , , , 3x2 + 16x + 16 = 4x2 + 12x + 8, x2 – 4x – 8 = 0, , , , Now, , , , Hence,, , 5, , −( −4) ± ( −4) 2 − 4(1)( −8), −b ± b 2 − 4ac, x=, =, 2 ×1, 2a, =, , 4 ± 16 + 32, 4 ± 48 4 ± 4 3, =, =, =2± 2 3, 2, 2, 2, , x = 2 + 2 3 and 2 – 2 3, , 17. 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., 5, , OR, , The angles of depression of the top and bottom of a building 50 metres 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., Sol. Let, BD = river, , AB = CD, , = palm trees = h, , BO = x, , Sample Paper - 1 (Solved), , 153

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , Date __________, , , , OD = 80 – x, h, , In DABO,, tan 60° = , x, h, 3 =, , x, , h = 3 x...(i), h, , In DCDO,, tan 30° =, (80 − x ), 1, h, , =, ...(ii), 3, (80 − x ), , Solving (i) and (ii), we get, , x = 20, , h = 3 x = 34.6, , The height of the trees = h = 34.6 m, , BO = x = 20 m, , DO = 80 – x = 80 – 20 = 60 m, OR, , Let AB = Building of height 50 m, , , , , , RT = height of tower = 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), (, , =, ...(i), 3, x, RT, , In DRBT,, tan 60° =, BT, h, , 3 = ...(ii), x, , Solving (i) and (ii), we get, , h = 75 m, h, 75, , From (ii), x=, =, = 25 3 m, 3, 3, , Hence, height of the tower = h = 75 m, , Distance between the building and the tower = 25 3 = 43.25 m, , qqq, , 154, , Mathematics–10

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Sample Paper - 2 (Solved), Time Allowed: 2 Hours, , Maximum Marks: 40, , General Instructions: Same as Sample Paper-1, , PART-A, SECTION-I, 1. If PQ = 28 cm, then find the perimeter of DPLM., , 1, , Sol., PQ = PT, , PL + LQ = PM + MT, , PL + LN = PM + MN, , Perimeter (DPLM) = PL + LM + PM, , = PL + LN + MN + PM, , = 2(PL + LN) = 2(PL + LQ), , = 2 × 28 = 56 cm, 2. If two tangents inclined at 60° are drawn to a circle of radius 3 cm, then find length of each tangent., , OR, PQ is a tangent to a circle with centre O at point P. If DOPQ is an isosceles triangle, then find ∠OQP., , tan 30˚ =, , AO, PA, , , , 1, , 3, PA, , , , PA = 3 3 cm, , Sol. In DPAO,, , 3, , =, , 155, , 1

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\ 23-Nov-2021, , Amit, , Proof-2, , OR, In ΔOPQ,, , ∠P + ∠Q + ∠O = 180°, , 2∠Q + ∠P = 180°, , 2∠Q + 90° = 180°, , 2∠Q = 90°, , ∠Q = 45°, 3. In the given figure, in what ratio does P divides AB internally?, , Reader’s Sign _______________________, , Date __________, , 1, , Sol. Point P divides AB internally in the ratio 4 : 4, i.e., 1 : 1, 4. Find the number of spherical lead shots each 4.2 cm in diameter that can be obtained from a rectangular solid lead with, dimensions 66 cm, 42 cm and 21 cm., 1, OR, , A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively is melted and recast into the, form of a cone of base diameter 8 cm. Find the height of the cone., Sol. Given: Diameter,, D = 4.2 cm, 4.2, , \ Radius,, r=, cm = 2.1 cm, h = 21 cm, 2, , Volume of rectangular solid = n × volume of one spherical shot, b = 42 cm, l = 66 cm, 4 3, , l × b × h = n × pr, 3, 4 22, #, , ⇒, 66 × 42 × 21 = n ×, × (2.1)3, 3, 7, 66 # 42 # 21 # 3 # 7, 4.2 cm, , ⇒, =n, 4 # 22 # 2.1 # 2.1 # 2.1, , \, n = 1500, , OR, , , , , ⇒, , ⇒  , , ⇒  , , \, , 156, , Volume of spherical shell = Volume of cone, 4, 1, p(r23 – r13) = pR2h, 3, 3, 4[(4)3 – (2)3] = (4)2 × h, 4[64 – 8] = 16h, 4 # 56, =h, 16, h = 14 cm, , Mathematics–10

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5. Find the median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3., Sol. Arranging data in ascending order: 3, 4, 4, 5, 6, 7, 7, 7, 12., No. of observations (n) = 9, , 1, , n +1, th term, 2, 9 +1, , =, = 5th term, 2, , = 6., , Directions (Q6 and Q7): In the following questions, a statement of assertion (A) is followed by a statement of reason, (R). Mark the correct choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., 6. Assertion (A): The equation x2 + 3x + 1 = (x – 2)2 is a quadratic equation., Reason (R): Any equation of the form ax2 + bx + c = 0 where a π 0, is called a quadratic equation., 1, Sol. (d) Assertion (A) is false but reason (R) is true., We have,, x2 + 3x + 1 = (x – 2)2 = x2 – 4x + 4, ⇒, x2 + 3x + 1 = x2 – 4x + 4, ⇒, 7x – 3 = 0, it is not of the form ax2 + 6x + c = 0, So, A is incorrect but R is correct., 7. Assertion (A): Two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre., Reason (R): A parallelogram circumscribing a circle is a rhombus., 1, Sol. (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., , \, , Median =, , SECTION-II, , Case Study based questions Q8 and Q9 are compulsory. Attempt any four sub-parts of each question., 8. Case Study-1, , Clinometer, , A clinometer is a tool that is used to measure the angle of elevation. We can use a clinometer to measure the height of, tall things, i.e., flag poles, towers, buildings, tree, etc., , Study some results after using clinometer., , R1: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the, tower is 30°., , R2: From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of, a 20 m high building are 45° and 60° respectively., , R3: The elevation of the Sun is 30°., , Refer Result R1, (i) The height of the tower is, 1, (a) 8 2 m, (b) 10 2 m, (c) 8 3 m, (d) 10 3 m, (ii) In case, the angle of elevation is 60°, then the height of tower is, 1, (a) 16 2 m, (b) 30 2 m, (c) 30 3 m, (d) 16 3 m, , Refer Result R2, (iii) The height of the tower is, 1, (a) 12 3 m, , (b) 20, , (, , ), , 3 −1 m, , (c) 18, , (, , ), , 3 −1 m, , (d) None of these., , Sample Paper - 2 (Solved), , 157

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , (iv) The distance between the observation point and the bottom of building is, (a) 8 m, (b) 12 m, (c) 20 m, (d) 15 m, , Refer Result R3, (v) The length of the shadow cast by a tower of 150 m height is, (a) 150 3 m, (b) 125 2 m, (c) 125 3 m, (d) 120 3 m, BC, Sol. (i) In rt DABC, tan 30° =, AB, h, 1, 30°, , \, =, A, 30, 3, 30, 3, #, , ⇒, h=, = 10 3 m, 3, 3, , \ Correct option is (d), EF, (ii) In rt DDEF, tan 60° =, DE, h, , ⇒, 3 =, 30, , ⇒, h = 30 3 m, , \ Correct option is (c), (iii) In DABC,, BC, , tan 45° =, AB, 20, , ⇒, 1=, ⇒ x = 20 m, x, CD + BC, In DABD,, tan 60° =, AB, h + 20, , 3 =, x, , ⇒, [From (i)], 3 x = h + 20, , ⇒, 20 3 = h + 20 ⇒ h = 20 3 – 20, , ⇒, h = 20( 3 – 1), , \ Correct option is (b), , Date __________, , 1, 1, , C, , h, 30 m, , B, , F, , h, , D, , 60°, 30 m, , E, , (i), , (iv) Distance between observation point and bottom of building = x = 20 m, \ Correct option is (c), (v) In rt DABC,, , tan 30° =, , BC, AB, , Sun, C, , 150 m, , 150, 1, =, y, 3, , , ⇒, , , \, y = 150 3 m, , \ Correct option is (a), , 30°, A, , B, , y, 9. Case Study-2, , 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 among humans., , The following tables shows the age distribution of case admitted during a day in two different hospitals., Table 1, , Age (in years), No. of cases, , 5-15, , 15-25, , 25-35, , 35-45, , 45-55, , 55-65, , 6, , 11, , 21, , 23, , 14, , 5, , Table 2, Age (in years), No. of cases, , 158, , 5-15, , 15-25, , 25-35, , 35-45, , 45-55, , 55-65, , 8, , 16, , 10, , 42, , 24, , 12, , Mathematics–10

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Refer to Table 1, 1, , (i) The average age for which maximum cases occurred is, (a) 32.24, , (b) 34.36, , (c) 40, , (d) 42.24, 1, , (ii) The upper limit of modal class is, (a) 15, , (b) 25, , (c) 35, , (d) 45, 1, , (iii) The mean of the given data is, (a) 26.2, , (b) 32.4, , (c) 33.5, , (d) 35.4, , Refer to Table 2, 1, , (iv) The mode of the given data is, (a) 41.4, , (b) 48.2, , (c) 55.3, , (d) 64.6, 1, , (v) The median of the given data is, (a) 32.7, , (b) 40.2, , (c) 42.3, , (d) 48.6, , Sol., , (i) In Table 1, maximum cases are 23, for which age (in years) lies in 35-45., 35 + 45 80, =, = 40, , \, Average age =, 2, 2, , \ Correct option is (c), (ii) In Table 1, highest frequency is 23, so modal class will be 35-45., , \ Upper limit of modal class is 45., , \ Correct option is (d), (iii), , C.I., , Frequency ( fi ), , Class mark (xi), , fi xi, , 5-15, , 6, , 10, , 60, , 15-25, , 11, , 20, , 220, , 25-35, , 21, , 30, , 630, , 35-45, , 23, , 40, , 920, , 45-55, , 14, , 50, , 700, , 55-65, , 5, , 60, , 300, , Sfi = 80, , , Mean =, , , \ Correct option is (d), , Sfi xi, Sfi, , =, , Sfixi = 2830, 2830, 80, , = 35.4 (approx), , (iv) In table-2, highest frequency is 42, so modal class will be 35-45., Here,, l = 35, h = 10, f0 = 10, f1 = 42, f2 = 24, , , Mode = l + f, , f1 − f0, 2f1 − f0 − f2, , p× h, , 42 − 10, , , , = 35 + f, , , , = 35 +, , , , \ Correct option is (a), , = 35 + 6.4 = 41.4, , 2 (42) − 10 − 24, , 32 × 10, 50, , = 35 +, , p× 10, , 32, 5, , Sample Paper - 2 (Solved), , 159

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\ 23-Nov-2021, , (v), , Amit, , Proof-2, , Reader’s Sign _______________________, , C.I., , fi, , c.f, , 5-15, , 8, , 8, , 15-25, , 16, , 24, , 25-35, , 10, , 34 = c, , (35-45), , 42 = f, , (76), , 45-55, , 24, , 100, , 55-65, , 12, , 112, , Date __________, , n = 112, , \, , n = 112 fi, , n, 2, , = 56, , 56 lies in c.f column at 76., So, median class is 35-45. Here l = 35, f = 42, c = 34, h = 10, JK n, N, KK − c OOO, , Median = l + KK 2, O ×h, K f OO, , L, , , , = 35 + d, , P, , 56 − 34, 42, , n × 10, , 220, , = 35 +, = 35 + 5.2 = 40.2, 42, \ Correct option is (b), , PART-B, SECTION-III, x2, , 10. Find the value of k such that the polynomial – (k + 6) x + 2(2k + 1) has sum of its zeroes equal to half of their product., , 2, Sol. Let a and b be the roots of given quadratic equation, , x2 – (k + 6)x + 2(2k + 1) = 0, − ( k + 6), , Now sum of toots,, a+b= –, =k+6, 1, 2(2k + 1), = 2(2k + 1), , Product of roots,, ab =, 1, , According to given condition,, 1, , a + b = αβ, 2, 1, , k + 6 = [2(2k + 1)], 2, , k + 6 = 2k + 1 ⇒ k = 5, 11. Which term of the AP 3, 15, 27, 39, ... will be 120 more than its 21st term?, 2, OR, 1, 1, , Find the 21st term of the AP −4 , − 3, − 1 , ..., 2, 2, Sol. Given AP is 3, 15, 27, 39 ..., , Here, first term, a = 3 and common difference, d = 12, , Now, 21st term of AP is, , an = a + (n – 1)d, , a21 = 3 + (21 – 1) × 12 = 3 + 20 × 12 = 243, , Therefore, 21st term is 243., , Now we need to calculate term which is 120 more than 21st term i.e., it should be 243 + 120 = 363, , Therefore,, an = a + (n – 1)d, , 160, , Mathematics–10

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363 = 3 + (n – 1)12, 360 = 12(n – 1), n – 1 = 30 ⇒ n = 31, OR, 1, 1, 9, 3, , Given AP is −4 , − 3, − 1 , ... or − , − 3, − , ..., 2, 2, 2, 2, −9, , First term,, a=, 2, 9,  9, , Common difference,, d = – 3 – −  = − 3 +,  2, 2, −6 + 9 3, , =, =, 2, 2, , Now,, an = a + (n – 1)d,  9,  3, , a21 =  −  + (21 − 1)  ,  2,  2, , , , 9, 9, 3, + 20 × = − + 30, 2, 2, 2, −9 + 60 51, 1, =, = 25, =, 2, 2, 2, = −, , 1, 2, 12. In the given figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point, of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°., 2, , , Hence, 21st term of given AP is 25, , Sol. Join OC, , In DOPA and DOCA, OP = OC, (Radii of same circle), , PA = CA, (Length of two tangents), , AO = AO, (Common), , \, DOPA @ DOCA, (by SSS congruency criterion), , Hence,, ∠1 = ∠2(CPCT), , Similarly,, ∠3 = ∠4, , Now,, ∠PAB + ∠QBA = 180°, (Co-interior angles), , ⇒, 2∠2 + 2∠4 = 180° ⇒ ∠2 + ∠4 = 90°, , Now in DAOB, ∠AOB + ∠2 + ∠4 = 180°, , ⇒, ∠AOB + 90° = 180°, , ⇒, ∠AOB = 90° Hence proved, , (Angle sum property), , SECTION-IV, 13. Draw a line segment of length 6 cm. Using compass and ruler, find a point P on it which divides it in the ratio 3 : 4. 3, Sol. Steps of construction:, , 1. Draw a line segment AB = 6 cm., , 2. Draw any acute angle ∠BAX., , 3. Along AX take 7 points, such that, ,    AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7., , 4. Join BA7, 5. Through A3 draw A3P || A7B which meets AB at P., , , Sample Paper - 2 (Solved), , 161

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\ 23-Nov-2021, , Proof-2, , Amit, , Reader’s Sign _______________________, , Date __________, , , 6. AP : PB = 3 : 4 and P is the required point., 14. 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., 3, , Sol. Let, AB = candle., , C and D are two coins, then in rt. ∆ABC,, AB h, , tan 60° =, =, BC b, h, 3 =, , b, , h = b 3 ...(i), , Again, in rt DABD,, AB h, =, BD a, 1, h, , =, 3, a, a, , h=, ...(ii), 3, , Multiplying (i) and (ii), we get, a, , h2 = b 3 ×, 3, , h2 = ba, , h = ab m, , , tan 30° =, , 15. The mode of the following data is 67. Find the missing frequency x., Class, Frequency, , 40-50, 5, , 50-60, x, , 3, , 60-70, 15, , 70-80, 12, , 80-90, 7, , Sol. , , , 200–300, , 300–400, , 400–500, , 500–600, , 600–700, , 700–800, , 800–900, , 900–1000, , Frequency, , 100–200, , Class, , 0–100, , OR, , The median of the following data is 525. Find the values of x and y, if total frequency is 100., , 2, , 5, , x, , 12, , 17, , 20, , y, , 9, , 7, , 4, , Mode = l +, , f1 − f 0, ×h, 2 f1 − f 2 − f 0, , 67 = 60 +, , 15 − x, × 10, 30 − 12 − x, , 15 − x, × 10, 18 − x, , , , 7=, , , , 7 × (18 – x) = 10(15 – x), , 162, , Mathematics–10

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126 – 7x =, 3x =, 3x =, x=, , 150 – 10x, 150 – 126, 24, 8, , OR, , We prepare the cumulative frequency table, as given below:, Class, , Frequency, , cf, , 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, , n = Sfi = 76 + x + y = 100, , ⇒, x + y = 24, , Median is 525, so it lies in the class 500-600., , Here, l = 500, cf = 36 + x, f = 20 and h = 100., , , , , ⇒, , ⇒, , \, , ⇒, , Hence,, , n, − cf, , Median = l +  2,  f, , , ,  × h, , 25 = (50 – 36 – x) 5, x = 50 – 41 = 9, 9 + y = 24, y = 15, x = 9, y = 15, , ...(i), ,  100, , − 36 − x, , , 2, × 100, ⇒ 525 = 500 +, 20, , ⇒, , 5 = 50 – 36 – x, [using (i)], , SECTION-V, 16. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows her mother’s, present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both, Asha and Nisha., 5, , OR, 3, 4, 29, 1, +, =, ; x ≠ − 1, 1,, , Solve for x :, x + 1 x − 1 4x − 1, 4, Sol. Let the present age of Nisha be x years., , Then, according to the question, we have, , Asha’s present age = x2 + 2, , Again, when Nisha grows her mother’s present age, Asha’s age would be one year less than 10 times the present age, of Nisha., , So, when Nisha’s age = x2 + 2,, , Asha’s age = 10x – 1, , Now, we have, (x2 + 2) + (x2 + 2 – x) = 10x – 1, ⇒, 2x2 – x + 4 = 10x – 1, ⇒, 2x2 – 11x + 5 = 0, 2, ⇒, 2x – 10x – x + 5 = 0, ⇒, 2x(x – 5) –1(x – 5) = 0, , Sample Paper - 2 (Solved), , 163

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\ 23-Nov-2021, , Amit, , Proof-2, , Reader’s Sign _______________________, , (x – 5)(2x – 1) = 0, , ⇒, , ⇒, , x–5=0, , or, , Date __________, , 2x – 1 = 0, , 1, 1, ; Rejecting x = , we get x = 5., 2, 2, , Hence, Nisha’s present age is 5 years and her mother’s present age = (5)2 + 2 = 25 + 2 = 27 years., , OR, 29, 3, 4, , We have, =, +, 4x − 1, x +1 x −1, , ⇒, , , , , x=5, , or x =, , 3x − 3 + 4 x + 4, x −1, 2, , 7x + 1, x −1, 2, , =, , 29, 4x − 1, , =, , 29, 4x − 1, , , (7x + 1)(4x – 1) = 29x2 – 29, , 28x2 – 7x + 4x – 1 = 29x2 – 29, , –3x = x2 – 28, , x2 + 3x – 28 = 0, 2, , x + 7x – 4x – 28 = 0, , x(x + 7) – 4(x + 7) = 0, , (x + 7)(x – 4) = 0, , Hence, x = 4, –7, 17. Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the, rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour?, 5, Sol. For pipe, r = 1cm, , Length of water flowing in 1 sec, h = 0.7 m = 70 cm, , Cylindrical Tank, R = 40 cm, rise in water level = H, , Volume of water flowing in 1 sec = pr2h = p × 1 × 1 × 70 = 70p, , Volume of water flowing in 60 sec = 70p × 60, , Volume of water flowing in 30 minutes = 70p × 60 × 30, , Volume of water in Tank = pr2H = p × 40 × 40 × H, , Volume of water in Tank = Volume of water flowing in 30 minutes, , p × 40 × 40 × H =70p × 60 × 30, , H = 78.75 cm, , qqq, , 164, , Mathematics–10

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Sample Paper - 3 (Unsolved), Time Allowed: 2 Hours, , Maximum Marks: 40, , General Instructions: Same as Sample Paper-1, , PART-A, SECTION-I, 1. Find the value(s) of k, if the quadratic equation 3 x 2 − k 3 x + 4 = 0 has equal roots., , 1, , OR, , If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots,, then find the value of k., 2. Find the whole number roots of the equation 2x2 – 7x + 6 = 0., 1, 3. To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn such that ∠BAX is an acute angle and then at equal, distances points are marked on the ray. Find the minimum number of these points., , , 1, 4. A metallic sphere of radius 10.5 cm is melted and then recast into small cones each of radius 3.5 cm and height 3 cm,, then find the number of such cones., 1, OR, , , The volume of a hemisphere is 2425, , 1, cm3 . Find its curved surface area., 2, , 5. If the mean of numbers 27 + x, 31 + x, 89 + x, 107 + x, 156 + x is 82, then find the mean of 130 + x, 126 + x, 68 + x,, 50 + x and 1 + x, 1, , Directions (Q6 and Q7): In the following questions, a statement of assertion (A) is followed by a statement of reason, (R). Mark the correct choice as:, (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)., (c) Assertion (A) is true but reason (R) is false., (d) Assertion (A) is false but reason (R) is true., 8, 6. Assertion (A): The equation 8x2 + 3kx + 2 = 0 has equal roots, then the value of k is ± ., 3, , Reason (R): The equation ax2 + bx + c = 0 has equal roots if D = b2 – 4ac = 0., 1, 7. Assertion (A): If the height of a cone is 24 cm and diameter of the base is 14 cm, then the slant height of the cone is 15 cm., , Reason (R): If r be the radius and h be the height of the cone, then slant height =, , h 2 + r 2 ., , 1, , SECTION-II, , Case Study based questions Q8 and Q9 are compulsory. Attempt any four sub-parts of each question., 8. Case Study-1, , Bird-Bath, , A bird-bath is a small ‘container’ which is placed in a garden and filled with water for birds to bath in., , Mayank and Swati both made two bird-baths as their school projects. The shape of Mayank’s bird-bath’s is a cylinder, with a hemispherical dipression at one end as shown in figure (A), whereas the shape of Swati’s bird-bath’s is a cylinder, with a conical dipression at one end as shown in figure (B)., , 165

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Ved_Goswami, , \ 23-Nov-2021, , Proof-1, , Reader’s Sign _______________________, , Bird-Bath (i), , Date __________, , Bird-Bath (ii), , Figure (A), Figure (B), Bird-Bath (i): The height of hollow cylinder used in making this bird-bath is 1.45 m and radius is 30 cm., Bird-Bath (ii): The height of the entire bird-bath is 2.6 m, while the height of the conical dipression is 36 cm., The radius of the bird-bath is 28 cm., Refer to Bird-Bath (i), 1, , (i) The curved surface area of the cylinder is, (a) 1.4, , m2, , (b) 2.7, , m2, , (c) 3.9, , m2, , (c) 1.5, , m2, , (d) 4.3, , m2, 1, , (ii) The curved surface area of the hemisphere is, (a) 0.6, , m2, , (b) 3.5, , m2, , (d) None of these., 1, , (iii) The volume of hemisphere is, (a) 4.3, , m3, , (b) 2.8, , m3, , (c) 0.057, , m3, , (d) 6.9, , m3, , Refer to Bird-Bath (ii), 1, , (iv) The total surface area of bird-bath is approximately, (a) 5, , m2, , (b) 0.6, , m2, , (c) 0.7, , m2, , (d) None of these., 1, , (v) The volume of conical dipression is approximately, m3, , m3, , m3, , m3, , (a) 2, (b) 0.03, (c) 4, (d) 6, 9. Case Study-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)., , 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, 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, (in units), No. of families, 16, 12, 18, 6, 4, 0, , The similar survey is conducted for 80 families of Colony B and the data is recorded as below., Weekly consumption, (in units), No. of families, , 166, , 0-10, , 10-20, , 20-30, , 30-40, , 40-50, , 50-60, , 0, , 5, , 10, , 20, , 40, , 5, , Mathematics–10

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Refer to data received from Colony A, 1, , (i) The median weekly consumption is, (a) 12 units, , (b) 16 units, , (c) 20 units, , (d) None of these., 1, , (ii) The mean weekly consumption is, (a) 19.64 units, , (b) 22.5 units, , (c) 26 units, , (d) None of these., 1, , (iii) The modal class of the above data is, (a) 0-10, , (b) 10-20, , (c) 20-30, , (d) 30-40, , Refer to data received from Colony B, 1, , (iv) The modal weekly consumption is, (a) 38.2 units, , (b) 43.6 units, , (c) 26 units, , (d) 32 units, 1, , (v) The mean weekly consumption is, (a) 15.65 units, , (b) 32.8 units, , (c) 38.75 units, , (d) 48 units, , PART-B, SECTION-III, 10. For what values of k, the roots of the equation x2 + 4x + k = 0 are real?, OR, , 2, , , Find the roots of the quadratic equation 15x2 – 10 6x + 10 = 0., 11. If 3k – 2, 4k – 6 and k + 2 are three consecutive terms of AP, then find the value of k., , 2, , 12. The shadow of a tower is 30 m long when the Sun’s elevation is 30°. What is the length of the shadow, when Sun’s elevation, is 60°?, 2, , SECTION-IV, 2, , then find the value of p and the other root of the equation.3, 3, OR, , 13. If one root of the quadratic equation 3x2 + px + 4 = 0 is, , , The roots a and b of the quadratic equation x2 – 5x + 3(k – 1) = 0 are such that a – b = 1. Find the value k., 14. Draw a line segment PQ = 8.4 cm by using ruler and compass only. Find a point R on PQ such that PR =, , 3, RQ. , 4, , 15. Find the mean and median for the following data:, Class, Frequency, , 3, 3, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , Total, , 8, , 16, , 36, , 34, , 6, , 100, , SECTION-V, 16. Two hoardings on cleanliness are kept on two poles of equal heights standing opposite to each other on either side of, the road, which is 80 m wide. From a point between them on the road, the angle of elevation of the top of the poles are, 60° and 30° respectively. Find the height of the pole and the distance of the point from the poles., 5, OR, , A tree breaks down due to storm and the broken part bends, so that the top of the tree touches the ground making an, angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 metres., Find the height of the tree before it was broken., , Sample Paper - 3 (Unsolved), , 167

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Ved_Goswami, , \ 23-Nov-2021, , Proof-1, , Reader’s Sign _______________________, , Date __________, , 17. In the figure, PQ and RS are two parallel tangents to a 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°., 5, , ANSWERS, 1. k = ± 4 OR, , k=, , 7, 4, , 2. 2, , 3. 12, , 4. 126 OR 693 cm2, 5. 75, 6. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)., 7. (d) Assertion (A) is false but reason (R) is true., 8. (i) (b) 2.7 cm2 (ii) (a) 0.6 m2, (iii) (c) 0.057 m3, 9. (i) (c) 20 units (ii) (a) 19.64 units (iii) (c) 20-30, 10. k ≤ 4 OR x =, 13. p = –8, x = 2 OR, , 2, 3, , ,, , 2, , 3, k=3, , 15. Mean = 26.4, meadian = 27.2, , 11. k = 3, , (iv) (a) 5 m2, (v) (b) 0.03 m3, (iv) (b) 43.6 units (v) (c) 38.75 units, 12. 10 m, , 14. Point R is 3.6 cm away from P., 16. 20 3 ; 20 m and 60 m OR, , 13.86 m, , Visit https://telegram.me/booksforcbse for more books., , 168, , Mathematics–10