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A, N, , R, E, C, S, , T, T, , A, L, E, , A, G, N
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CHILDREN! THESE DIINSTRUCTIONS FOR YOU..., , Graph, , ♦, ♦, ♦, ♦, ♦, ♦, , ♦, ♦, ♦, , T, T, , R, E, C, S, , A, G, N, , A, L, E, ♦, , ♦, ♦, ♦, ♦, , A, N, , For each and every conceptual understanding, a real life context with appropriate illustrations are given in, the textbook. Try to understand the concept through keen reading of context along with observation of, illustration., While understanding the concepts through activities, some doubts may arise. Clarify those doubts by through, discussion with your friends and teachers, understand the mathematical concepts without any doubts., ''Do this/Do these" exercises are given to test yourself, how far the concept has been understood. If you are, facing any difficulty in solving problems in these exercises, you can clarify them by discussing with your, teacher., The problems given in "Try this/try these", can be solved by reasoning, thinking creatively and extensively., When you face difficulty in solving these problems, you can take the help of your friends and teachers., The activities or discussion points given "Think & disicuss" have been given for extensive understanding of, the concept by thinking critically. These activities should be solved by discussions with your fellow students, and teachers., Different typs of problems with different concepts discussed in the chapter are given in an "Exercise" given, at the end of the concept/chapter. Try to solve these problems by yourself at home or leisure time in school., The purpose of "Do this"/do these", and "Try this/try these" exercises is to solve problems in the presence of, teacher only in the class itself., Where ever the "project works" are given in the textbook, you should conduct them in groups. But the, reports of project works should be submitted indivedually., Try to solve the problems given as homework on the day itself. Clarify your doubts and make corrections, also on the day itself by discussions with your teachers., Try to collect more problems or make new problems on the concepts learnt and show them to your teachers, and fellow students., Try to collect more puzzles, games and interesting things related to mathematical concepts and share with, your friends and teachers., Do not confine mathematical conceptual understanding to only classroom. But, try to relate them with your, suroundings outside the classroom., Student must solve problems, give reasons and make proofs, be able to communicate mathematically, connect, concepts to understand more concepts & solve problems and able to represent in mathematics learning., Whenever you face difficulty in achieving above competencies/skills/standards, you may take the help of, your teachers.
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© Government of Telangana, Hyderabad., , All rights reserved., , AN, A, , First Published 2012, New Impressions 2013, 2014, 2015, 2016, 2017, 2018, , LA, N, , G, , No part of this publication may be reproduced,, stored in a retrieval system, or transmitted, in, any form or by any means without the prior, permission in writing of the publisher, nor be, otherwise circulated in any form of binding or, cover other than that in which it is published, and without a similar condition including this, condition being imposed on the subsequent, purchaser., , TE, , The copy right holder of this book is the, Director of School Education, Hyderabad,, Telangana., , ER, , T, , This Book has been printed on 70 G.S.M. SS Maplitho, Title Page 200 G.S.M. White Art Card, , SC, , Free distribution by T.S. Government 2018-19, , Printed in India, at the Telangana Govt. Text Book Press,, Mint Compound, Hyderabad,, Telangana., , –– o ––, , ( ii )
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FOREWORD, , AN, A, , State Curriculum Frame Work (SCF-2011) recommends that childrens’ life at schools must be linked to, their life outside the school. The Right To Education Act (RTE-2009) perceives that every child who enters the, school should acquire the necessary skills prescribed at each level upto the age of 14 years. Academic standards, were developed in each subject area accordingly to maintain the quality in education. The syllabi and text, books developed on the basis of National Curriculum Frame work 2005 and SCF-2011 signify an attempt to, implement this basic idea., , G, , Children after completion of Primary Education enter into the Upper Primary stage. This stage is a crucial, link for the children to continue their secondary education. We recognise that, given space, time and freedom,, children generate new knowledge by exploring the information passed on to them by the adults. Inculcating, creativity and initiating enquiry is possible if we perceive and treat children as participants in learning and not, as passive receivers. The children at this stage possess characteristics like curiosity, interest, questioning,, reasoning, insisting proof, accepting the challenges etc., Therefore the need for conceptualizing mathematics, teaching that allows children to explore concepts as well as develop their own ways of solving problems in a, joyful way., , LA, N, , We have begun the process of developing a programme which helps children understand the abstract, nature of mathematics while developing in them the ability to construct own concepts. The concepts from the, major areas of Mathematics like Number System, Arithmetic, Algebra, Geometry, Mensuration and Statistics are, provided at the upper primary stage. Teaching of the topics related to these areas will develop the skills, prescribed in academic standards such as problem solving, logical thinking, expressing the facts in mathematical, language, representing data in various forms, using mathematics in daily life situations., , T, , TE, , The textbooks attempt to enhance this endeavor by giving higher priority and space to opportunities for, contemplation and wondering, discussion in small groups and activities required for hands on experience in the, form of ‘Do This’ , ‘Try This’ and ‘Projects’. Teachers support is needed in setting of the situations in the, classroom. We also tried to include a variety of examples and opportunities for children to set problems. The, book attempts to engage the mind of a child actively and provides opportunities to use concepts and develop, their own structures rather than struggling with unnecessarily complicated terms and numbers. The chapters are, arranged in such a way that they help the Teachers to evaluate every area of learning to comperehend the, learning progress of children and in accordance with Continuous Comprehensive Evaluation (CCE)., , ER, , The team associated in developing the textbooks consists of many teachers who are experienced and, brought with them view points of the child and the school. We also had people who have done research in, learning mathematics and those who have been writing textbooks for many years. The team tried to make an, effort to remove fear of mathematics from the minds of children through their presentation of topics., , SC, , I wish to thank the national experts, university teachers, research scholars, NGOs, academicians, writers,, graphic designers and printers who are instrumental to bring out this textbook in present form., I hope the teachers will make earnest effort to implement the syllabus in its true spirit and to achieve academic, standards at the stage., The process of developing materials is a continuous one and we hope to make this book better. As an, organization committed to systematic reform and continuous improvement in quality of its products, SCERT,, welcomes comments and suggestions which will enable us to undertake further revision and refinement., , B. Seshu kumari, , Place: Hyderabad, Date: 28 January 2012, , DIRECTOR, , SCERT, Hyderabad, ( vi )
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Mathematics, VII class, Syllabus to be, covered during, , Contents, Integers, , 2., , Fractions, Decimals and, Rational Numbers, , 3., , Simple Equations, , 4., , Lines and Angles, , 5., , Triangle and Its Properties, , 6., , Ratio - Applications, , 7., , Data Handling, , 8., 9., , 1 - 25, , June, July, , 26 - 58, , July, , 59 - 68, , LA, N, , August, , 69 - 87, 88 - 109, , September, , 110 - 141, , September, , 142 - 162, , Congruency of Triangles, , October, , 163 - 181, , Construction of Triangles, , November, , 182 - 191, , Algebraic Expressions, , November, , 192 - 210, , Powers and Exponents, , December, , 211 - 226, , December, , 227 - 244, , ER, , T, , TE, , August, , SC, , 11., , June, , G, , 1., , 10., , Page No, , AN, A, , S. No., , 12., , Quadrilaterals, , 13., , Area and Perimeter, , January, , 245 - 264, , 14., , Understanding 3D and 2D Shapes, , February, , 265 - 276, , 15., , Symmetry, , February, , 277 - 289, , Revision, , March, (v), , Free distribution by Government of T.S. 2018-19
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1, , 1.0, , Introduction, , We start learning numbers like 1,2,3,4.... for counting objects, around us. These numbers are called counting numbers or natural numbers. Let us think about these., Which is the smallest natural number?, , (ii), , Write five natural numbers between 100 and 10000., , (iii), , Can you find where the sequence of natural numbers, ends?, , (iv), , What is the difference between any two consecutive, natural numbers?, , LA, N, , G, , (i), , AN, A, , INTEGERS, , TE, , By including ‘0’ to the collection of natural numbers, we get a new collection of numbers called, whole numbers i.e., 0, 1, 2, 3, 4, ......, In class VI we also learnt about negative numbers. If we put whole number and negative numbers, together we get a bigger collection of numbers called integers. In this chapter, we will learn more, about integers, their operations and properties., , ER, , T, , Let us now represent some integers on a number line., -6, , -5, , -4, , -3, , -2, , -1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , Which is the biggest integer represented on the above number line?, , (ii), , Which is the smallest integer?, , (iii), , Is 1 bigger than –3? Why?, , (iv), , Is –6 bigger than –3? Why?, , (v), , Arrange 4, 6, –2, 0 and –5 in ascending order., , (vi), , Compare the difference between (0 and 1) and (0 and –1) on the number line., , SC, , (i), , Free distribution by T.S. Government 2018-19, , INTEGERS, , 1
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Exercise - 1, 1., , Some integers are marked on the number line. Which is the biggest and which is the, smallest?, -6, , -4, , -3, , -2, , -1, , 0, , 1, , 2, , 4, , 5, , 6, , Write the integers between the pairs of integers given below. Also, choose the biggest and, smallest intergers from them., (i) –5, –10, , (ii) 3, –2, , (iii) –8, 5, , Write the following integers in ascending order (smallest to biggest)., (i) –5, 2, 1, –8, , (iii) –10, –15, –7, , Write the following integers in descending order (biggest to smallest)., (i) –2, –3, –5, , LA, N, , 4., , (ii) –4, –3, –5, 2, , G, , 3., , (ii) –8, –2, –1, , (iii) 5, 8, –2, , 5., , Represent 6, –4, 0 and 4 on a number line., , 6., , Fill the missing integers on the number line given below, -5, , 0, , TE, , -9, , 7., , 3, , AN, A, , 2., , -5, , 4, , 8, , The temperatures (in degrees celsius/centigrade) of 5 cities in India on a particular day are, shown on the number line below., , -10, , o, , C, , -5 o C, , 0 oC, , Ooty, 5 oC, , 10 o C, , 15 o C, , Banglore, 20 o C, , ER, , -15 o C, , Nainital, , T, , Kasauli Manali, , Write the temperatures of the cities marked on it?, , (ii), , Which city has the highest temperature?, , (iii), , Which city has the lowest temperature?, , SC, , (i), , 1.1, , (iv), , Which cities have temperature less than 0ºC?, , (v), , Name the cities with temperature more than 0ºC?, , Operations of integers, , We have done addition and subtraction of integers in class VI. First we will review our, understanding of the same and then learn about multiplication and division of integers., 2, , VII CLASS MATHEMATICS
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1.1.1, , Addition of integers, 4+3, , = 7, , 4+2, , = 6, , 4+1, , = 5, , 4+0, , = 4, , AN, A, , Observe the additions given below., , 4 + (–1) = 3, 4 + (–2) = 2, 4 + (–3) = 1, , G, , Do you find a pattern in the answers? You will find that when the number being added is decreased, by one (3, 2, 1, 0, –1, –2, –3) then the value of the sum also decreases by 1., , -6, , -5, , -4, , -3, , -2, , LA, N, , On the number line, when you add 3 to 4 you move right on the number line:, , -1, , 0, , 1, , 2, , 3, , 4, , +3, , 5, , 6, , 7, , TE, , Similarly can you now show the additions of 2 and 1 to 4 on the number line drawn above? You, will find that in each case you have moved right on the number line., Now, let us see what is happening when we add –1 to 4. From the above pattern we find that the, answer is 3. Thus, we have moved one step left on the number line:, , -6, , -5, , -4, , -3, , -2, , T, , -7, , -1, , 0, , -1, 1, , 2, , 3, , 4, , 5, , 6, , 7, , ER, , You can now similarly show addition of –2 and –3 to 4 on the number line drawn above? You will, find that in each case you are moving left on the number line., , SC, , Thus, each time you add a positive integer you move right on the number line. On the, other hand, each time you add a negative number you move left on the number line., Try This, , 1., , 1., , 9+7, , =, , 16, , 9+1, , =, , 9+6, , =, , 15, , 9+0, , =, , 9+5, , =, , 9 + (–1), , =, , 9+4, , =, , 9 + (–2), , =, , 9+3, , =, , 9 + (–3), , =, , 9+2, , =, , Free distribution by T.S. Government 2018-19, , INTEGERS, , 3
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Now represent 9 + 2, 9 + (–1) and 9 + (–3) on the number line., , (ii), , When you added a positive integer, in which direction did you move on the, number line?, , (iii), , When you added a negative integer, in which direction did you move on the, number line?, , 2., , Sangeetha said that each time you add two integers, the value of the sum is, greater than the numbers. Is Sangeetha right? Give reasons for your answer., Exercise - 2, , Represent the following additions on a number line., (i), , (ii), , 5+2, , (iii), , Compute the following., , 1.1.2, , 7+4, , (ii), , 8 + (–3), , (iv) 14 + (–6), , (v), , 9 + (–7), , (vii) 13 + (–15), , (viii) 4 + (–4), , (x), , (xi) 225 + (–145), , 100 + (–80), , Subtraction of integers, , (iv), , 5 + (–7), , (iii) 11 + 3, , LA, N, , (i), , 5 + (–2), , (vi) 14 + (–10), (ix) 10 + (–2), , TE, , 2., , 5+7, , G, , 1., , AN, A, , (i), , Now let us study the subtractions given below., = 3, , 6–2, , = 4, , T, , 6–3, , = 5, , ER, , 6–1, 6–0, , = 6, , 6 – (–1) = 7, 6 – (–2) = 8, , SC, , 6 – (–3) = 9, , 6 – (–4) = 10, , Do you find a pattern in the answers? You will find that when the number being subtracted from 6, is decreased by one (3, 2, 1, 0, –1, –2, –3, –4) the value of the difference increased by 1., On the number line when you subtract 3 from 6, you move left on the number line., 6–3, , -5, , 4, , -4, , -3, , -2, , VII CLASS MATHEMATICS, , -1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8
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You can now, similarly show subtraction of 2, 1 from 6 on the number line. You will find that in each, case you have moved left on the number line., Now, let us see what is happening when we subtract –1 from 6. As seen from the above pattern, we find 6– (–1) = 7., Thus, we have moved one step right on the number line ., , -5, , -4, , -3, , -2, , -1, , 0, , 1, , 2, , 3, , AN, A, , 6–(–1), 4, , 5, , 6, , 7, , 8, , You can now, similarly show subtraction of –2, –3, –4 from 6? You will find that in each case you, are moving right on the number line., , G, , Thus, each time you subtract a positive integer, you move left on the number line., , Try This, Complete the pattern., = 2, , 8–5, , = 3, , 8–4, , =, , 8–3, , =, , 8–2, , =, , 8–1, , =, , 8–0, , =, , TE, , 8–6, , T, , 1., , LA, N, , And each time you subtract a negative integer, you move right on the number line., , ER, , 8 – (–1) =, 8 – (–2) =, 8 – (–3) =, , SC, , 8 – (–4) =, , (i) Now show 8 –6, 8–1, 8–0, 8–(–2), 8–(–4) on the number line., (ii) When you subtract a positive integer in which direction do you move on the, number line?, (iii) When you subtract a negative integer, in which direction do you move on the, number line?, , 2., , Richa felt that each time you subtract an integer from another integer, the value of, the difference is less than the given two numbers. Is Richa right? Give reasons, for your answer., , Free distribution by T.S. Government 2018-19, , INTEGERS, , 5
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Exercise - 3, 1., , Represent the following subtractions on the number line., 7–2, , (ii), , 8 – (– 7), , (iii), , 3–7, , (iv), , 15 – 14, , (v), , 5 – (– 8), , (vi), , (–2) - (–1), , Compute the following., (i), , 17 – (–14), , (ii), , 13 – (– 8), , (iv), , 15 – 28, , (v), , 25 – 33, , (vii) 150 – 75, , AN, A, , 2., , (i), , (viii) 32 – (– 18), , (iii), , 19 – (– 5), , (vi), , 80 – (– 50), , Express ‘–6’ as the sum of a negative integer and a whole number., , 1.1.3, , Multiplication of integers, , LA, N, , G, , 3., , Now, let us multiply integers., , We know that 3 + 3 + 3 + 3 = 4 × 3 (4 times 3), On the number line, this can be seen as., 3, 3, 0, , 1, , 2, , TE, , -1, , 3, , 3, , 4, , 5, , 6, , 7, , 8, , 3, 9, , 10, , 11, , 12, , 13, , Thus, 4 × 3 means 4 jumps each of 3 steps from zero towards right on the number line and, therefore 4 × 3 = 12., , T, , Now let us discuss 4 × (–3) i.e., 4 times (–3), , ER, , 4 × (–3) = (–3) + (–3) + (–3) + (–3) = –12, , SC, , On the number line, this can be seen as., –3, –3, -13, , -12, , -11, , -10, , -9, , -8, , –3, -7, , -6, , -5, , –3, -4, , -3, , -2, , -1, , 0, , 1, , Thus, 4 × (–3) means 4 jumps each of 3 steps from zero towards left on the number line and, therefore 4 × (–3) = -12, Similarly, 5 × (–4) = (–4) + (–4) + (–4) + (–4) + (–4) = –20, On the number line, this can be seen as :, –4, –4, –4, -20, , 6, , -16, , -12, , VII CLASS MATHEMATICS, , –4, -8, , –4, -4, , 0, , 4
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Thus, 5 × –4 means 5 jumps each of 4 steps from zero towards left on the number line, and therefore 5 × –4 = –20, Similarly,, , 2 × –5, , = (–5) + (–5) = –10, , 3 × –6 = (–6) + (–6) + (–6) = –18, , Do This, 1. Compute the following., (i), , 2 × –6, , (ii) 5 × –4, , (iii), , 9 × –4, , Now, let us multiply –4 × 3, Study the following pattern-, , G, , 4 × 3 = 12, 3×3 = 9, , LA, N, , 2×3 = 6, 1×3 = 3, 0×3 = 0, , –3 × 3 = –9, –4 × 3 = –12, , TE, , –1 × 3 = –3, –2 × 3 = –6, , AN, A, , 4 × –8 = (–8) + (–8) + (–8) + (–8) = –32, , T, , You see that as the multiplier decreases by 1, the product decreases by 3., Thus, based on this pattern –4 × 3 = –12., , ER, , We already know that 4 × –3 = –12, Thus, –3 × 4 = 3 × –4 = –12, , Observe the symbol of the product as the negative sign differ in the multiplication., , SC, , Using this pattern we can say that, 4 × –5 = –4 × 5, , = –20, , 2 × –5 = –2 × 5, , = –10, , 3 × –2, , =, , 8 × –4, , =, , 6 × –5, , =, , From the above examples you would have noticed that product of positive integer and a, negative integer is always a negative integer., Free distribution by T.S. Government 2018-19, , INTEGERS, , 7
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1.1.3(a) Multiplication of two negative integers, Now, what if we were to multiply –3 and –4., Study the following pattern., = –12, , –3 × 3, , = –9, , –3 × 2, , = –6, , –3 × 1, , = –3, , –3 × 0, , = 0, , AN, A, , –3 × 4, , G, , –3 × –1 = 3, , LA, N, , –3 × –2 = 6, –3 × –3 = 9, –3 × –4 = 12, , TE, , Do you observe any a pattern? You will see that as we multiply –3 by 4, 3, 2, 1, 0, –1, –2,, –3, –4 the product increases by 3., Now let us multiply –4 and –3. Study the following products and fill the blanks., = –16, , –4×3, , = –12, , = –8, , ER, , –4 × 2, , T, , –4×4, , = –4, , –4 × 0, , = 0, , SC, , –4 × 1, , –4 × –1 = ——, –4 × –2 = ——, –4 × –3 = ——, , You will see that as we multiply –4 by 4, 3, 2, 1, 0, –1, –2, –3, the product increases by 4., According to the two patterns given above, –3 × –4 = –4 × –3 = 12, 8, , VII CLASS MATHEMATICS
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You have also observed that., –3 × –1 = 3, , –4 × –1 = 4, , –3 × –2 = 6, , –4 × –2 = 8, , –3 × –3 = 9, , –4 × –3 = 12, , Activity 1, , AN, A, , Thus, every time if we multiply two negative integers, the product is always a positive integer., , 3, , 2, , 1, , 0, , –1, , –2, , 3, , 9, , 6, , 3, , 0, , –3, , –6, , 2, , 6, , 4, , 2, , 0, , –1, , –3, , –2, , –2, , 0, , –3, , –9, , LA, N, , 1, , –3, , G, , ×, , TE, , Fill the grid by multiplying each number in the first column with each number in the first row., , –1, , 0, , 1, , 2, , 3, , Is the product of two positive integers always a positive integer?, , (ii), , Is the product of two negative integers always a positive integer?, , (iii), , Is the product of a negative and positive integer always a negative integer?, , ER, , T, , (i), , 1.1.3(b) Multiplication of more than two negative integers, , SC, , We observed that the product of two negative integers is a positive integer. What will be the, product of three negative integers? Four negative integers? and so on ....., Let us study the following examples., (i), , (–2) × (–3) = 6, , (ii), , (–2) × (–3) × (–4) = [(–2) × (–3)] × (–4) = 6 × (–4) = – 24, , (iii), , (–2) × (–3) × (–4) × (–5) = [(–2) × (–3) × (–4)] × (–5) = (–24) × (–5) = 120, , (iv), , [(–2) × (–3) × (–4) × (–5) × (–6)] = 120 × (–6) = –720, , Free distribution by T.S. Government 2018-19, , INTEGERS, , 9
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From the above products we can infer that., The product of two negative integers is a positive integer., , (ii), , The product of three negative integers is negative integer., , (iii), , The product of four negative integers is a positive integer., , (iv), , The product of five negative integers is a negative integer., , AN, A, , (i), , Will the product of six negative integers be positive or negative? State reasons., Try This, (–1) × (–1) = ——, , (b), , (–1) × (–1) × (–1) = ——, , (c), , (–1) × (–1) × (–1) × (–1) = ——, , (d), , (–1) × (–1) × (–1) × (–1) × (–1) = ——, , LA, N, , G, , (a), , TE, , We further see that in (a) and (c) above, the number of negative integers that are multiplied are, even (two and four respectively) and the products are positive integers. The number of negative, integers that are multiplied in (b) and (d) are odd and the products are negative integers., , 10, , Exercise - 4, , Fill in the blanks., , (i), , (–100), , (ii), , (–3), , × .......... =, , 3, , (iii), , 100, , × (–6), , =, , ....................., , (iv) (–20) × (–10), , =, , ....................., , (v), , =, , ....................., , SC, , 1., , ER, , T, , Thus, we find that if the number of negative integers being multiplied is even, then the, product is a positive integer. And if the number of negative integers being multiplied is, odd, the product is a negative integer., , 15, , × ( –6) =, , × (–3), , VII CLASS MATHEMATICS, , .....................
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Find each of the following products., (i), , 3 × (–1), , (ii), , (–1) × 225, , (iii) (–21) × (–30), , (iv), , (–316) × (–1), , (v) (–15) × 0 × (–18), , (vi), , (–12) × (–11) × (10), , (vii) 9 × (–3) × (–6), , (viii) (–18) × (–5) × (–4), , (ix) (–1) × (–2) × (–3) × 4, , (x), , AN, A, , 2., , (–3) × (–6) × (–2) × (–1), , A certain freezing process requires that the room temperature be lowered from 40ºC at, the rate of 5ºC every hour. What will be the room temperature 10 hours after the process, begins?, , 4., , In a class test containing 10 questions, ‘3’ marks are awarded for every correct answer, and (–1) mark is for every incorrect answer and ‘0’ for questions not attempted., , LA, N, , G, , 3., , (i) Gopi gets 5 correct and 5 incorrect answers. What is his score?, (ii) Reshma gets 7 correct answers and 3 incorrect answers. What is her score?, , 5., , TE, , (iii) Rashmi gets 3 correct and 4 incorrect answers out of seven questions she attempts., What is her score?, A merchant on selling rice earns a profit of `10 per, bag of basmati rice sold and a loss of `5 per bag of, non-basmati rice., , ER, , T, , (i) He sells 3,000 bags of basmati rice and 5,000, bags of non-basmati rice in a month. What is his, profit or loss in a month?, (ii) What is the number of basmati rice bags he must sell to have neither profit nor loss, if, the number of bags of non-basmati rice sold is 6,400., Replace the blank with an integer to make it a true statement., , SC, , 6., , (i), , (–3) × —————— = 27, , (iii) —————— × (–8) = –56, , (ii), , 5 × —————— = –35, , (iv) —————— × (–12) = 132, , 1.1.4 Division of integers, We know that division is the inverse operation of multiplication. Let us study, some examples for natural numbers., Free distribution by T.S. Government 2018-19, , INTEGERS, , 11
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We know that 3 × 5 = 15, Therefore, 15 ÷ 5 = 3 or 15 ÷ 3 = 5, Similarly, 4 × 3 = 12, Therefore, 12 ÷ 4 = 3 or 12 ÷ 3 = 4, , AN, A, , Thus, we can say that for each multiplication statement of natural numbers there are two, corresponding division statements., Can we write a multiplication statement and its corresponding division statements for integers?, Study the following and complete the table., Multiplication statement, , Division statements, (–12) ÷ (–6) = 2, , ,, , (–12) ÷ 2 = (–6), , (–4) × 5 = (–20), , (–20) ÷ (5) = (–4), , ,, , (–20) ÷ (–4) = 5, , (–8) × (–9) = 72, , 72 ÷ (–8) = (–9), , ,, , 72 ÷ (–9) = (–8), , (–3) × (–7) = _______, , ______ ÷ (–3) = _____, , ,, , __________________, , (–8) × 4 = __________, , __________________, , ,, , __________________, , 5 × (–9) = __________, , __________________, , ,, , __________________, , (–10) × (–5) =, , __________________, , ,, , __________________, , TE, , LA, N, , G, , 2 × (–6) = (–12), , We can infer from the above that when we divide a negative integer by a positive integer, or a positive integer by a negative integer, we divide them as whole numbers and then, negative (–) sign for the quotient. We thus, get a negative integer as the quotient., , T, , Do This, , ER, , 1. Compute the following., (i) (–100) ÷ 5, , (ii), , (–81) ÷ 9, , (v) 125 ÷ (–25), , (vi) 80 ÷ (–5), , (iii), , (–75) ÷ 5, , (iv) (–32) ÷ 2, , (vii) 64 ÷ (–16), , SC, , Try This, , Can we say that (–48) ÷ 8 = 48 ÷ (–8)?, Check whether(i) 90 ÷ (–45) and (–90) ÷ 45, , (ii), , (–136) ÷ 4 and 136 ÷ (–4), , We also observe that, (–12) ÷ (–6) = 2; (–20) ÷ (–4) = 5; (–32) ÷ (–8) = 4; (–45) ÷ (–9) = 5, So, we can say that when we divide a negative integer by a negative integer, we get a, positive number as the quotient., 12, , VII CLASS MATHEMATICS
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Do This, 1., , Compute the following., , (i), , –36 ÷ (–4), , (–201) ÷ (–3), , (iii) (–325) ÷ (–13), , Properties of integers, , AN, A, , 1.2, , (ii), , 1.2.1, , Properties of integers under addition, , (i), , Closure property, , Study the following., Statement, , Conclusion, The sum is a whole number, , LA, N, , 5 + 8 = 13, , G, , In class VI we have learnt the properties of whole numbers. Here we will learn the properties of, integers., , 6+3=, 13 + 5 =, 10 + 2 =, 2+6=8, , The sum is a whole number, , TE, , Is the sum of two whole numbers always a whole number? You will find this to be true. Thus, we, say that whole numbers follow the closure property of addition., , Statement, , Conclusion, , The sum is an integer, , ER, , 6+3=9, , T, , Do integers satisfy closure property of addition ? Study the following additions and complete the, blanks., , –10 + 2 =, –3 + 0 =, , –5 + 6 = 1, , SC, , (–2) + (–3) = –5, 7 + (–6) =, , The sum is an integer, , Is the sum of two integers always an integer?, Can you give an example of a pair of integers whose sum is not an integer? You will not be able to, find such a pair. Therefore, integers are also closed under addition., In general, for any two integers a and b, a + b is also an integer., , Free distribution by T.S. Government 2018-19, , INTEGERS, , 13
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(ii), , Commutative property, , Study the following and fill in the blanks., Statement 2, , Conclusion, , 4+3=7, , 3+4=7, , 4+3=3+4=7, , 3+5=, , 5+3=, , 3+1=, , 1+3=, , AN, A, , Statement 1, , Similarly, add as many pairs of whole numbers, as you wish. Did you find any pair of whole, numbers for which the sum is different, when the order is changed. You will not find such a pair., Thus, we say that the addition of whole numbers is commutative., , Statement 2, , 5 + (–6) = –1, , (–6) + 5 = –1, , –9 + 2 =, , 2 + (–9) =, , –4 + (–5) =, , (–5) + (–4) =, , Conclusion, , LA, N, , Statement 1, , G, , Is addition of integers commutative? Study the following and fill in the blanks., , 5 + (–6) = (–6) + 5 = –1, , TE, , Did you find any pair of integers for which the sum is different when the order is changed? You, would have not. Therefore, addition is commutative for integers., In general, for any two integers a and b, a + b = b + a, Associative property, , T, , (iii), , ER, , Let us study the following examples(2 + 3) + 4, =5+4, =9, (ii) (–2 + 3) + 5, =1+5, =6, (iii) (–2 + 3) + (–5), = 1 + (–5), , SC, , (i), , = –4, , 2 + (3 + 4), = 2+7, = 9, –2 + (3 + 5), = –2 + 8, = 6, (- 2) + [3 + (–5)], = (- 2) + (-2), = –4, , (iv) [(–2) + (–3)] + (-5) –2 + [(–3) + (–5)], , 14, , = –5 + (–5), , = –2 + (–8), , = –10, , = –10, , VII CLASS MATHEMATICS
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Is the sum in each case equal? You will find this to be true., Therefore, integers follow the associative property under addition., , 1., , (i), , (2 + 5) + 4, , =, , 2 + (5 + 4), , (ii) (2 + 0) + 4, , =, , 2 + (0 + 4), , AN, A, , Try This, , Does the associative property hold for whole numbers? Take two more examples, and write your answer., , Additive identity, , LA, N, , (iv), , G, , In general, for any three integers a, b and c, (a + b) + c = a + (b + c), , Carefully study the following., –2 + 0 = –2, 5 + 0 = 5, , TE, , 8 + 0 =, –10 + 0 =, , On adding zero to integers, do you get the same integer? Yes., , T, , Therefore, ‘0’ is the additive identity for integers., , ER, , In general, for any integer a, a+0 = 0 + a= a, , Try This, , Add the following, (i), , 2+0=, , (ii), , 0+3=, , (iii), , 5+0=, , SC, , 1., , 2., , Similarly, add zero to as many whole numbers as possible., Is zero the additive identity for whole numbers?, , Free distribution by T.S. Government 2018-19, , INTEGERS, , 15
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(v), , Additive Inverse, , What should be added to 3 to get its additive identity ‘0’?, , 3 + (–3), , =0, , 7 + (–7), , =0, , (–10) + 10 = 0, Check whether we get similar pairs for other integers., , AN, A, , Study the following-, , In each pair given above, one integer is called the additive inverse of the other integer., , G, , In general, for any integer ‘a’ there exists an integer (–a) such that a + (–a) = 0., Both the integers are called additive inverse of each other., , (i), , LA, N, , 1.2.2 Properties of integers under multiplication, Closure property, , Study the following and complete the table, , 9 × 8 = 72, 10 × 0 =, , The product is an integer, , T, , –15 × 2 =, , Conclusion, , TE, , Statement, , ER, , –15 × 3 = –45, –11 × –8 =, 10 × 10 =, , SC, , 5 × –3 =, , Is it possible to find a pair of integers whose product is not an integer? You will not find this to be, possible., Note: Do you remember fractions and decimals are not Integers., Therefore, integers follow the closure property of multiplication., In general, if a and b are two integers, a × b is also an integer., 16, , VII CLASS MATHEMATICS
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Try This, , 2., , Multiply the following, (i), , 2 × 3 = _______, , (ii), , 5 × 4 = _______, , (iii), , 3 × 6 = _______, , Similarly, multiply any two whole numbers of your choice., , AN, A, , 1., , Is the product of two whole numbers always a whole number?, (ii), , Commutative property, , Statement 1, , Statement 2, (–2) × 5, , = –10, , (–3) × 6 =, , 6 × (–3), , =, , –20 × 10 =, , 10 × (–20) =, , Conclusion, , 5 × (–2) = (–2) × 5 = –10, , LA, N, , 5 × (–2) = –10;, , G, , We know that multiplication is commutative for whole numbers. Is it also commutative, for integers?, , TE, , Therefore, multiplication of integers follows the commutative property., In general, for any two integers a and b, a × b = b × a, (iii), , Associative property, , Consider the multiplication of 2, –3, –4 grouped as follows., , ER, , We see that-, , and, , 2 × [(–3) × (–4)], , T, , [2 × (–3)] × (–4), [2 × (–3)] × (–4), , and, , 2 × [(–3) × (–4)], = 2 × 12, , = 24, , = 24, , SC, , = (–6) × (–4), , In first case 2, –3 are grouped together and in the second –3, –4 are grouped together. In both, cases the product is the same., Thus, [2 × (–3)] × [(–4)] = 2 × [(–3) × (–4)], Does the grouping of integers affect the product of integers? No, it does not., The product of three integers does not depend upon the grouping of integers. Therefore, the, multiplication of integers is associative., In general, for any integers, a, b and c, (a × b) × c = a × (b × c), Free distribution by T.S. Government 2018-19, , INTEGERS, , 17
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1., , Is [(–5) × 2)] × 3 = (–5) × [(2 × 3)]?, , 2., , Is [(–2) × 6] × 4 = (–2) × [(6 × 4)]?, , AN, A, , Do This, , Try This, (5 × 2) × 3= 5 × (2 × 3), , Distributive property, , LA, N, , (iv), , G, , Is the associative property true for whole numbers? Take many more examples and, verify., , We know that, 9 × (10 + 2) = (9 × 10) + (9 × 2), , Thus, multiplication distributes over addition is true for whole numbers., , (i), , –2 × (1 + 3), , = [(–2) x 1] + [(–2) × 3], , –2 × 4, , = –2 + (–6), , = –8, , T, , –8, , –1 × [3 + (–5)] = [(–1) × 3] + [(–1) × (–5)], , ER, , (ii), , TE, , Let us see, is this true for integers-, , –1 × (–2), 2, , = –3 + (+5), =2, , SC, , Verify –3 × (–4+2) = [(–3) × (–4)] + [–3 × (2)], , You will find that in each case, the left hand side is equal to the right hand side., Thus, multiplication distributes over addition of integers too., In general, for any integers a, b and c, a × (b + c) = a × b + a × c, , 18, , VII CLASS MATHEMATICS
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Multiplicative identity, 2×1, , = 2, , –5 × 1, , = –5, , –3 × 1, , = _____, , –8 × 1, , = _____, , 1 × –5, , = _____, , 1 is the multiplicative identity of, integers, , AN, A, , (v), , You will find that multiplying an integer by 1 does not change the integer. Thus, 1 is called the, multiplicative identity for integers., , (vi), , G, , In general, for any integer ‘a’, a × 1 = 1 × a = a, Multiplication by zero, , =, , 0, , 0 × (–8), , =, , ——, , 9×0, , =, , ——, , TE, , (–3) × 0, , LA, N, , We know that any whole number when multiplied by zero gives zero. What happens in case of, integers? Study the following-, , This shows that the product of an integer and zero is zero., , 3., , Exercise 5, , Verify the following., (i), , 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)], , (ii), , (–21) × [(–4) + (–6)] = [(–21) × (–4)] + [(–21) × (–6)], , (i), , For any integer a, what is (–1) × a equal to?, , (ii), , Determine the integer whose product with (–1) is 5, , SC, , 2., , a×0=0×a=0, , ER, , 1., , T, , In general for any integer a,, , Find the product, using suitable properties., (i), , 26 × (–48) + (–48) × (–36), , (ii), , 8 × 53 × (–125), , (iii) 15 × (–25) × (–4) × (–10), , (iv) (–41) × 102, , (v), , (vi) 7 × (50 – 2), , 625 × (–35) + (–625) × 65, , (vii) (–17) × (–29), Free distribution by T.S. Government 2018-19, , (viii) (–57) × (–19) + 57, INTEGERS, , 19
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1.2.3, , Properties of integers under subtraction, , (i), , Closure under subtraction, , Do we always get an integer, when subtracting an integer from an integer?, 9–7, , =, , _____, , 7 – 10, , =, , _____, , 2–3, , =, , _____, , –2 – 3, , =, , _____, , –2 – (–5) =, , _____, , 0–4, , _____, , =, , AN, A, , Do the following., , G, , What did you find? Can we say that integers follow the closure property for subtraction?, , LA, N, , Therefore, for any integers a and b, a – b is also an integer., (ii), , Commutativity under subtraction, , Let us take an example. Consider the integers 6 and –4, 6 – (–4) = 6 + 4 = 10 and, , TE, , – 4 – (6) = – 4 – 6 = –10, 6 – (– 4) ≠ – 4 – (6), , Therefore,, , Thus, subtraction is not commutative for integers., , T, , Try This, , ER, , Take atleast 5 different pairs of integers and see if subtraction is commutative., , Properties of integers under division, , (i), , Closure Property, , SC, , 1.2.4, , Study the following table and complete it., Statement, , Inference, , Statement, , (–8) ÷ (–4) = 2, , Result is an integer, , (–8) ÷ 4 =, , −8, = –2, 4, , Result is not an integer 4 ÷ (–8) =, , 4 −1, =, −8 2, , (–4) ÷ (–8) =, , 20, , −4 1, =, −8 2, , VII CLASS MATHEMATICS, , Inference
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What can you infer? You will infer that integers are not closed under division., Try This, Take atleast five pairs of integers and check whether they are closed under division., , Commutative Property, , AN, A, , (ii), , We know that division is not commutative for whole numbers. Let us check it for integers also., You can see from the table given above that (–8) ÷ (–4) ≠ (–4) ÷ (–8)., Is (–9) ÷ 3 equal to 3 ÷ (–9)?, Is (–30) ÷ (6) equal to (–6) ÷ (–30)?, , G, , Thus, we can say that division of integers is not commutative., , LA, N, , Try This, , Take atleast 5 pairs of integers and observe whether the division of integers is, commutative or not?, (iii), , Division by Zero, , TE, , Like whole numbers, any integer divided by zero is meaningless and zero divided by a nonzero integer is equal to zero., For any integer a, a ÷ 0 is not defined but 0 ÷ a = 0 for a ≠ 0., (iv), , Division by 1, , ER, , T, , When we divide a whole number by 1 it gives the same whole number. Let us check whether, this is true for negative integers also., Observe the following(–8) ÷ 1 = (–8), , (11) ÷ 1 = +11, , (–13) ÷ 1 = _____, , (–25) ÷ 1 = _______, , SC, , Thus, a negative integer or a positive integer divided by 1 gives the same integer as quotient., , In general, for any integer a, a ÷ 1 = a., , What happens when we divide any integer by (–1)? Complete the following table(–8) ÷ (–1) = 8, , 11 ÷ (–1) = –11, , 13 ÷ (–1) = _____, , (–25) ÷ (–1) = _____, , We can say that if any integer is divided by (–1) it does not give the same integer, but gives its, additive identity., Free distribution by T.S. Government 2018-19, , INTEGERS, , 21
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Try This, 1., , For any integer a, is, (i), , a ÷ 1 = 1?, , (ii) a ÷ (–1) = –a?, Take different values of ‘a’ and check., Associative property, , AN, A, , (iii), , Is [(–16) ÷ 4] ÷ (–2) = (–16) ÷ [4 ÷ (–2)]?, [(–16) ÷ 4] ÷ (–2) = (–4) ÷ (–2) = 2, , Therefore,[(–16) ÷ 4] ÷ (–2) ≠ (–16) ÷ [4 ÷ (–2)], , Try This, , LA, N, , Thus, division of integers is not associative., , G, , (–16) ÷ [4 ÷ (–2)] = (–16) ÷ (–2) = 8, , Take atleast five more examples and check whether division is associative for integers., Exercise - 6, , Fill the following blanks., (i), , –25 ÷ ........ = 25, , (ii), , ........ ÷ 1, , (iii) 50 ÷ 0, , = –49, , =, , ....................., , = ....................., , T, , (iv) 0 ÷ 1, , Some practical problems using negative numbers, , ER, , 1.3, , TE, , 1., , In a test (+5) marks are given for every correct answer and (–2) marks are given, for every incorrect answer. (i) Radhika answered all the questions and scored 30, marks through 10 correct answers. (ii) Jaya also answered all the questions and, scored (–12) marks through 4 correct answers. How many incorrect answers had, both Radhika and Jaya attempted?, , SC, , Example 1 :, , Solution :, , (i) Marks given for one correct answer, , =5, , So marks given for 10 correct answers = 5 × 10 = 50, Radhika’s score, , = 30, , Marks obtained for incorrect answers, , = 30 – 50 = –20, , Marks given for one incorrect answer = (–2), Therefore, number of incorrect answers = (–20) ÷ (–2) = 10, 22, , VII CLASS MATHEMATICS
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(ii) Marks given for 4 correct answers, , = 5 × 4 = 20, , Jaya’s score, , = –12, , Marks obtained for incorrect answers, , = –12 – 20 = –32, , Marks given for one incorrect answer, , = (–2), , Example 2 :, , AN, A, , Therefore number of incorrect answers = (–32) ÷ (–2) = 16, A shopkeeper earns a profit of ` 1 by selling one pen and incurs a loss of 40 paise, per pencil while selling pencils of her old stock., (i) In a particular month she incurs a loss of ` 5. In this period, she sold 45 pens., How many pencils did she sell in this period?, , Solution :, , LA, N, , G, , (ii) In the next month she earns neither profit nor, loss. If she sold 70 pens, how many pencils did, she sell?, (i) Profit earned by selling one pen ` 1, , Profit earned by selling 45 pens = ` 45, which we denote by 45, Total loss given, , = ` 5, which we denote by –5., , TE, , Profit earned + Loss incurred = Total loss, Therefore, Loss incurred, , = Total loss-Profit earned, , = –5 – (45) = (–50) = – ` 50 = – 5000 paise, , Loss incurred by selling one pencil = 40 paise which we write as –40 paise, , T, , So, number of pencils sold, , = (–5000) ÷ (–40) = 125 pencils., , ER, , (ii) In the next month there is neither profit nor loss., So, profit earned + loss incurred, , =0, , i.e., Profit earned, , = – Loss incurred., , SC, , Now, profit earned by selling 70 pens = ` 70, Hence, loss incurred by selling pencils = – ` 70 or –7000 paise., Total number of pencils sold, , 1., , = (–7000) ÷ (–40) = 175 pencils., , Exercise - 7, , In a class test containing 15 questions, 4 marks are given for every correct answer and, (–2) marks are given for every incorrect answer. (i) Bharathi attempts all questions but, only 9 answers are correct. What is her total score? (ii) One of her friends Hema attempts, only 5 questions and all are correct. What will be her total score?, , Free distribution by T.S. Government 2018-19, , INTEGERS, , 23
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2., , A cement company earns a profit of ` 9 per bag of white cement sold and a loss of, ` 5 per bag of grey cement sold., (i), The company sells 7000 bags of white cement and 6000 bags of grey cement in a, month. What is its profit or loss?, , AN, A, , (ii), What is the number of white cement bags it must sell to have neither profit nor, loss, if the number of grey bags sold is 5400., The temperature at 12 noon was 10o C above zero. If it decreases at the rate of 2oC per, hour until midnight, at what time would the temperature be 8oC below zero? What would, be the temperature at midnight?, , 4., , In a class test (+3) marks are given for every correct answer and (–2) marks are given for, every incorrect answer and no marks for not attempting any question. (i) Radhika scored, 20 marks. If she has got 12 correct answers, how many questions has she attempted, incorrectly? (ii) Mohini scores (–5) marks in this test, though she has got 7 correct answers., How many questions has she attempted incorrectly?, , 5., , An elevator descends into a mine shaft at the rate of 6 meters per minute. If the descent, starts from 10 m above the ground level, how long will it take to reach – 350 m., Looking Back, , LA, N, , G, , 3., , TE, , 1. N (natural numbers) = 1, 2, 3, 4, 5 . . ., , W (whole numbers) = 0, 1, 2, 3, 4, 5 . . ., Z (Integers) = ..., –4, –3, –2, –1, 0, 1, 2, 3, 4 . . ., , T, , also 0, ±1, ±2, ±3 (set of integers also represented as I.), , ER, , 2. (i) Each time you add a positive integer, you move right on the number line., (ii) Each time you add a negative integer, you move left on the number line., , 3. (ii) Each time you subtract a positivie integer, you move left on the number line., , SC, , (iii) Each time you subtract a negative integer, you move right on the number line., , 4. (i) Each time you multiply a negative integer by a positive integer or a positive, integer by a negative integer, the product is a negative integer., (ii) Each time you multiply two negative integers, the product is a positive integer., (iii) Product of even number of negative integers is positive (+ve), product of odd, number of negative integers is negative (–ve)., , 24, , VII CLASS MATHEMATICS
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5., , (i) Each time you divide a negative integer by a positive integer or a positive integer by, a negative integer the quotient is negative integer., (ii) Each time you divide negative integer by a negative integer the quotient is positive, integer., , 6., , AN, A, , (iii) When you multiply or divide two integers of same sign the result is always positive;, if they are of opposite signs the result is negative., The following are the propeties satisfied by addition and subtraction of integers-, , (i) Integers are closed for addition and subtraction both. i.e., a + b and a–b are integers,, where a and b are any integers., (ii) Addition is commutative for integers, i.e., a + b = b + a, for all integers a and b., , LA, N, , G, , (iii) Addition is associative for integers, i.e., (a+b) + c = a + (b + c), for all integers a, b,, and c., (iv) Integer 0 is the identity under addition, i.e., a + 0 = 0 + a = a, for every integer a., 7., , Integers show some properties under multiplication., , (i) Integers are closed under multiplication. i.e., a × b is an integer for any two integers, a and b., , TE, , (ii) Multiplication is commutative for integers. i.e., a × b = b × a for any integers a and b., (iii) The integer 1 is the identity under multiplication, i.e., 1 × a = a × 1 = a, for any integer a., , T, , (iv) Multiplication is associative for integers, i.e., (a×b) × c = a × (b×c) for any three, integers a, b, and c., In integers multiplication distributes over addition. i.e., a × (b+c) = a ×b + a × c for any, three integers a, b and c. This is called distributive property., , 9., , The properties of commutativity and associativity under addition and multiplication and, the distributive property help us make our calculations easier., , 10., , For any integer a, we have, , SC, , ER, , 8., , (i) a ÷ 0 is not defined or meaningless, (ii) 0 ÷ a = 0 (for a ≠ 0), (iii) a ÷ 1 = a, , Free distribution by T.S. Government 2018-19, , INTEGERS, , 25
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FRACTIONS, DECIMALS, AND RATIONAL NUMBERS, Introduction, , AN, A, , 2.0, , 2, , G, , We come across many examples in our day-to-day life where we use fractions. Just try to recall, them. We have learnt how to represent proper and improper fractions and their addition and, subtraction in the previous class. Let us review what we have already learnt and then go further to, multiplication and division of fractional numbers as well as of decimal fractions. We will conclude, by an introduction to a bigger set of numbers called rational numbers., , Figure 1, , TE, , LA, N, , The shaded portion of the figures given below have been represented using fractions. Are the, representations correct?, , Figure 2, , Figure 3, , 1, 2, , 1, 3, , Y/N, , Y/N, , Y/N, , Reason ......................, , Reason ......................, , Reason ......................, , ER, , T, , 1, 2, , While considering the above you must have checked if parts of each figure are equal or not?., Make 5 more such examples and give them to your friends to verify., , SC, , Here is Neha’s representation of, , 1, in different figures., 2, , 1, Do you think that the shaded portions correctly represent ? Then what fractions are represented, 2, by unshaded portions?, 26, VII CLASS MATHEMATICS
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Try This, 3 1, , in different ways using different figures. Justify your representation., 4 4, , Represent, , Share, and check it with your friends., , AN, A, , Proper and Improper fractions, You have learnt about proper and improper fractions. A proper fraction is a fraction that represents, a part of a whole. Give five examples of proper fractions., Is, , 3, a proper fraction? How do you check if it is a proper fraction or not?, 2, , LA, N, , G, , What are the properties of improper fractions? One of them is that in improper fractions the, numerator is more than or equal to the denominator. What else do we know about these fractions., We can see that all improper fractions can be written as mixed fractions. The improper fraction, , 3, 2, , 1, can be written as 1 . This is a mixed fraction. This contains an integral part and a fractional part., 2, , Write five examples, each of proper, improper and mixed fractions?, Try This, , T, , 1., , TE, , The fractional part should be a proper fraction., Do This, , 1, pictorially. How many units are needed for this., 4, , ER, , Represent 2, , SC, , Comparison of fractions, , Do you remember how to compare like fractions, for e.g., , 1, 3 3, 1, and ? is bigger than . Why?, 5, 5 5, 5, , Can you recall how to compare two unlike fractions, for e.g., , 5, 3, and ?, 7, 4, , We convert these into like fractions and then compare them., 5, 4, 20, ×, =, 7, 4, 28, , and, , 3 7, 21, × =, 4 7 28, , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 27
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Since, , 5 20, 3 21, =, and =, 7 28, 4 28, , Thus,, , 5 3, <, 7 4, , 1. Write five equivalent fractions for (i), 2. Which is bigger, , 3, 5, , (ii), , 5, 3, or ?, 8, 5, , 4, ., 7, , AN, A, , Do These, , 3, 375, and, 8, 1000, , (ii), , (iii), , 6, 600, and, 10, 1000, , (iv), , 18, 23, and, 54, 69, , LA, N, , (i), , G, , 3. Determine if the following pairs are equal by writing each in their simplest form., , 17, 25, and, 27, 45, , Example 1 :, , TE, , You have already learnt about addition and subtraction of fractions in class VI. Let us, solve some problems now., Razia completes, , 3, 4, part of her homework while Rekha completed of it. Who, 7, 9, , has completed the least part?, , 3, 4, and ., 7, 9, 3 27 4 28, , =, Converting them to like fractions we have =, 7 63 9 63, , T, , To find this we have to compare, , ER, , Solution :, , 27 28, <, 63 63, , SC, , Thus,, , 27 28, 3 4, 〈, and so <, 63 63, 7 9, , Razia has completed a least part of her homework than Rekha., , Example 2 :, , Shankar’s family consumed 3, next 15 days they consumed 3, the whole month?, , 28, , VII CLASS MATHEMATICS, , 1, kg sugar in the first 15 days of a month. For the, 2, , 3, kg sugar. How much sugar did they consume for, 4
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Solution :, , The total weight of the sugar for the whole month, 3⎞, ⎛ 1, = ⎜ 3 + 3 ⎟ kg, 4⎠, ⎝ 2, , =, , Example 3 :, , ⎛ 14 15 ⎞, ⎜ + ⎟, ⎝ 4 4⎠, , AN, A, , ⎛ 7 15 ⎞, = ⎜ + ⎟ kg =, ⎝2 4 ⎠, , 29, 1, kg = 7 kg., 4, 4, , At Ahmed’s birthday party,, , 5, part of the total cake was distributed. Find how, 7, , Total cake is one = 1 or, Cake distributed =, , 5, 7, , TE, , The cake left = 1 −, , 5, 7, , 1, 1, , LA, N, , Solution :, , G, , much cake is left?, , 7 5 2, = − =, 7 7 7, , 2, part of the total cake is left now., 7, , 2., , Exercise - 1, , Compute and express the result as a mixed fraction?, 3, 4, , (i), , 2+, , (iv), , 2 1, 2 +, 3 2, , SC, , 1., , ER, , T, , Thus,, , (ii), , 7 1, +, 9 3, , (iii), , 1−, , 4, 7, , (v), , 5 1, −, 8 6, , (vi), , 2, 1, 2 +3, 3, 2, , Arrange the following in ascending order., (i), , 5 5 1, , ,, 8 6 2, , (ii), , 2 1 3, , ,, 5 3 10, , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 29
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Check in the following square, whether in this square the sum of the numbers in each row, and in each column and along the diagonals is the same., 6, 13, , 13, 13, , 2, 13, , 3, 13, , 7, 13, , 11, 13, , 12, 13, , 1, 13, , 8, 13, , AN, A, , 3., , 1, 2, A rectangular sheet of paper is 5 cm long and 3 cm wide. Find its perimeter., 3, 5, , 5., , The recipe requires 3 cups of flour. Radha has 1 cups of flour. How many more cups, , G, , 4., , 1, 4, , 3, 8, , 6., , LA, N, , of flour does she need?, , Abdul is preparing for his final exam. He has completed, , 5, part of his course content., 12, , Find out how much course content is left?, , B, , C, , TE, , Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in, this figure. Which figure has greater perimetre and by how, much?, , 7., , Multiplication of fractions, , 2.1.1, , Multiplication of a fraction by a whole number, , T, , 2.1, , 1, 3 cm, 3, 1, 2 cm, 5, D, , 2, 1 cm, 3, , E, , A, , 2, 4 cm, 3, , ER, , When we multiply whole numbers we know we are repeatedly adding, a number. For example 5 x 4 means adding 5 groups of 4 each or 5 times 4., 1, 1, 1, it means adding twice or 2 times ., 4, 4, 4, , SC, , Thus, when we say 2 ×, , Let us represent this pictorially. Look at Figure 1. Each shaded part is, shaded parts together will represent 2 ×, , +, Figure 1, , 30, , VII CLASS MATHEMATICS, , 1 1 1 2, = + = ., 4 4 4 4, , =, Figure 2, , 1, part of a square. The two, 4
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1, 1, Let us now find 3 × . This means three times or three halves., 2, 2, 3×, , We have, , 1 1 1 1 3, = + + =, 2 2 2 2 2, , 1. Find (i) 4 ×, , 2, 7, , (ii) 4 ×, , (iii) 7 ×, , 1, 3, , 1 2 2, 3, , , and are proper fractions., 2 3 7, 5, , 5, 5, .and how to multiply 2 ×, 3, 3, , LA, N, , Let us consider some improper fractions like, , 3, 5, , G, , The fractions that we have considered till now, i.e.,, , AN, A, , Do This, , 5 5 5, 10, 1, 2× = + =, =3, 3 3 3, 3, 3, , Pictorially, , 5, 3, , 5, 3, , =, , 10, 3, , T, , +, , =, , TE, , +, , 1. Find, , (i) 5 ×, , 3, 2, , (ii) 4 ×, , 7, 5, , (iii) 7 ×, , 8, 3, , SC, , We know the area of a rectangle is equal to length × breadth. If the length, and breadth of a rectangle are 6 cm and 5 cm respectively, then what will be, its area? Obviously the area would be 6 × 5 = 30 cm2., If the length and breadth of a rectangle are 6 cm, 2, , 6 cm, , 5 cm, , ER, , Do This, , 1, cm respectively, what, 3, , would be the area of that rectangle., Area of a rectangle is the product of its length and breadth. To multiply a mixed fraction with a, whole number, first convert the mixed fractions into an improper fraction and then multiply., , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 31
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6 cm, , 1, Therefore, area of a rectangle = 6 × 2, 3, , = 6×, , 1, 2 cm, 3, , 7 42, =, cm2 = 14cm2, 3, 3, , 1., , Find the following., (i) 3 × 2, , 2, 7, , (ii) 5 × 2, , 1, 3, , (iii) 8 × 4, , 2, 9, , (v) 5 × 1, , 1, 3, , 1, × 5 . How do you understand it?, 2, , TE, , Consider, , (iv) 4 × 1, , LA, N, , 1 2, Represent pictorially 2 × =, 5 5, , 2., , 1, 7, , G, , Do These, , AN, A, , You might have realised by now that to multiply a whole number with a proper or an, improper fraction, we multiply the whole number with the numerator of the fraction, keeping, the denominator the same., , 1, 5, 1, × 5 means half of 5, which is, or 2, 2, 2, 2, , T, , 1, 1, 5, of 5 = × 5 =, 2, 2, 2, , ER, , Thus,, , Similarly =, , 1, 1, 3, 1, of 3 = × 3 = or 1, 2, 2, 2, 2, , SC, , Here onwards ‘of’ represents multiplication., , So what would, , 1, of 16 mean? It tells us that the whole (16) is to be divided into 4 equal parts and, 4, , one part out of that has to be taken. When we make 4 equal parts of 16, each part will be 4. So, 1, of 16 is 4., 4, , 32, , VII CLASS MATHEMATICS
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This can be illustrated with marbles as :, , 1, 1, 16, of 16 = ×16 = = 8 ., 2, 2, 2, , Example 4 :, , Nazia has 20 marbles. Reshma has, , 1, of the number of marbles that Nazia has., 5, , How many marbles does Reshma have?, , LA, N, , 1, × 20 = 4 marbles., 5, , G, , Similarly,, , AN, A, , 1, 1, 16, = 4, of 16 or, × 16 =, 4, 4, 4, , Solution :, , Reshma has, , Example 5 :, , In a family of four persons 15 chapaties were consumed in a day., , 3, were consumed by the children and the, 5, , TE, , were consumed by the mother and, , 1, of the chapaties, 5, , remaining were eaten by the father., , (i) How many chapaties were eaten by the mother?, , T, , (ii) How many chapaties were eaten by the children?, , ER, , (iii) What fraction of the total chapaties has been eaten by the father?, Solution :, , Total number of chapaties = 15, 1, × 15 = 3 chapaties, 5, , SC, , (i) Number of chapaties eaten by mother, (ii), , 3, 3, of the total number is eaten by children, × 15 = 9 chapaties, 5, 5, , (iii) The chapaties left for father = 15 – 3 – 9 = 3 chapaties, Fraction of chapathies eaten by father =, , Free distribution by T.S. Government 2018-19, , 3 1, =, 15 5, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 33
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Exercise - 2, Multiply the following. Write the answer as a mixed fraction., , Shade :, , (i), , 1, ×4, 3, , (iii), , 6, ×2, 7, , 1, of the circles in box (a), 2, , (iii), , (ii), , 3, of the rectangles in box (c), 5, , (a), , (b), 1, of 12, 3, , (ii), , (v) 15 ×, , 2, 5, , 2, of the triangles in box (b), 3, , (iv), , 3, of the circles in box (d), 4, , (d), , 2, of 15, 5, , Find (i), , 2.1.2, , Multiplication of a fraction with a fraction, , TE, , 1 1, 1, 1, × mean? From the above we can understand that it means of ., 2 4, 2, 4, , 1, 4, , T, , Consider, , 2, ×5, 9, , (c), , 3., , What does, , (iv), , G, , 2., , (ii), , AN, A, , 3, ×10, 6, , (i), , LA, N, , 1., , ⎛1 ⎞, 1, of this shaded part? We can divide this one-fourth ⎜ ⎟ shaded part into, 2, ⎝4 ⎠, , ER, , How will we find, , SC, , two equal parts (Figure 1). Each of these two parts represents, , 1, 1, of ., 2, 4, , Let us call one of these parts as part ‘A’. What fraction is ‘A’ of the whole circle?, we divide the remaining parts of the circle into two equal parts each, we get a, total of eight equal parts. ‘A’ is one of these parts., So, ‘A’ is, , 34, , 1, 1, 1 1 1 1, of the whole. Thus, of, = × =, 8, 2, 4 2 4 8, , VII CLASS MATHEMATICS, , Figure 1, , A, , Figure 2, , I f
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Find, , 1 1, 1 1, × and × ., 3 2, 2 3, , =, , 1, 1, 1 1 1 1, thus of = × =, 6, 3, 2 3 2 6, , 1, 1, of is, 2, 3, , =, , 1, 1, 1 1 1 1, of = × =, thus, 6, 2, 3 2 3 6, , Do These, , TE, , Fill in these boxes:, 1 1 1× 1, × =, =, 5 7 5× 7, , (i), , Find, , 1 1, × =, 2 6, , =, , 1 1, 1 1, 1 1 1 1, × and × using diagram, check whether × = ×, 2 5, 5 2, 2 5 5 2, , 2, 2, 2, 2 2, of . We have shown in Figure 1 and × in Figure 2., 3, 5, 5, 3 5, , ER, , 2., , (ii), , T, , 1., , G, , 1 1 1 1, × = ×, 3 2 2 3, , LA, N, , We can see that, , AN, A, , 1, 1, of is, 3, 2, , SC, , Consider one more example, , 2, 3, 2, 5, , 2, 5, , Figure 1, , Figure 2, , The cross hatched portion in figure (2) represents, Free distribution by T.S. Government 2018-19, , 2, 2, 2 2 4, of, or × =, 3, 5, 3 5 15, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 35
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To find the, , 2, 2, 2, of , we have made three equal parts of and then selected 2 out of the 3 parts., 3, 5, 5, , This represent 4 parts out of a total 15 parts so, , 2, 2 2 2 4, of, = × =, ., 3, 5 3 5 15, , AN, A, , Product of Numerators, Here, we can observe that Product of two fractions = Product of Denominators ., , 1, 1, cm and 3 cm respectively., 2, 2, , 1, cm, 2, , G, , are 6, , 6, , 1, 3 cm, 2, , Now, what will be the area of the rectangle be if its length and breadth, , Example 6 :, , Narendra reads, , Solution :, , 1, of a short novel in 1 hour. What part of the book will he have, 4, , 1, hours?, 2, , TE, , read in 2, , LA, N, , 1, 1 13 7, 91, 3, = 22 cm 2 ., The area = 6 × 3 = × cm2. =, 2, 2 2 2, 4, 4, , The part of the novel read by Narendra in 1 hour =, , T, , So the part of the novel read by him in 2, , 1, 1 1, hours = 2 ×, 2, 2 4, , =, , 5 1 5, × =, 2 4 8, , So Narendra would read, , 5, 1, part of the novel in 2 hours., 8, 2, , A swimming pool is filled, , 3, 1, part in half an hour. How much will it be filled in 1, 10, 2, , ER, , Example 7 :, , 1, 4, , SC, , hour?, , Solution :, , The part of the pool filled in half an hour =, , So, the part of pool which is filled in 1, , 3, ., 10, , 1, hour is 3 times the pool filled in half an, 2, , hour., = 3×, , 36, , 3, 9, =, 10 10, , VII CLASS MATHEMATICS, , Thus,, , 9, 1, part of the pool will be filled in 1 hours., 10, 2
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Try This, You have seen that the product of two natural numbers is one or more than one or, bigger than each of the two natural numbers. For example, 3 × 4 = 12, 12 > 4 and 12, > 3. What happens to the value of the product when we multiply two proper fractions?, Fill the following table and conclude your observations., 2 4 8, × =, 3 5 15, , 8 2 8 4, < ,, <, 15 3 15 5, , AN, A, , Eg:, , Product is less than each, of the fractions, , G, , 1 2, × =−−−−−, 5 7, , 5, , LA, N, , 3, 21, ×, =, 5 2 10, 4 20, × =, 3 6, , 1., , Find each of the following products., 6×, , 1, 5, , (iii), , 1 1, 2 ×3, 3 5, , ER, , Multiply and reduce to lowest form., (i), , 2, 1, ×5, 3, 5, , (ii), , 2 1, ×, 7 3, , (iii), , 9 5, ×, 3 5, , Which one is greater?, , SC, , 3., , 5 7, ×, (ii), 6 11, , T, , (i), 2., , TE, , Exercise - 3, , 2 4, 3, 1, 1, 4, 2, 3, of, or, of, of or, of, (ii), 5 7, 4, 2, 2, 7, 3, 7, 1, Rehana works 2 hours each day on her embroidery. She completes the work in 7 days., 2, How many hours did she take to complete her work?, , (i), , 4., , 5., , A truck runs 8 km using 1 litre of petrol. How much distance will it cover using 10, , 2, litres, 3, , of petrol?, Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 37
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1, meters in 1 second. How much distance will he walk in 15 minutes?, 2, , 6., , Raja walks 1, , 7., , Provide the number in the box, , 2.2, , 2, ×, 3, , =, , 20, ., 21, , (ii), , 5, 3, ×, =, 7 5, , AN, A, , (i), , to make the statement true., , Division of fractions, , Imagine you have 15 meters length of cloth and you want to make pieces of 1, , 1, metres length, 2, , G, , 1, 1, each from it. How many 1 meter pieces will you get? Here we will successively subtract 1, 2, 2, , LA, N, , meters from 15 meters and see how many times we can do this, till we have no cloth left., Look at one more example. A paper strip of length, , 21, cm has to be cut into smaller strips of length, 2, , by, , 3, 21 3, ÷ ., i.e.,, 2, 2 2, , TE, , 3, 3, 21, cm each. How many pieces would we get? Clearly, we can cut cm each time or divide, 2, 2, 2, , Let us recall division with whole numbers. In15 ÷ 3 , we find out how many 3’s are there in 15. The, , ER, , T, , answer to this is 5. Similarly, to find the number of 2’s in 18, we divide 18 by 2 or 18 ÷ 2 . The, answer to this is 9. Now correlate the same process in dividing whole numbers by fractions and, fractions by fractions., 2.2.1, , Division of whole number by a fraction, , SC, , 1, Let us find 3 ÷ ., 2, , 1, ⎛1⎞, Kiran says we have to find how many halves ⎜ ⎟ are there in 3. To find the number of in 3, we, 2, ⎝2⎠, , draw the following., 1, 2, , 1, 2, , 1, 2, , 1, 2, , The figure above suggests that there are 6 halves in 3., 38, , VII CLASS MATHEMATICS, , 1, 2, , 1, 2
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We can therefore say 3 ÷, Think about 2 ÷, , 1, =6, 2, , 1, 3, , AN, A, , ⎛1⎞, This means finding how many one-thirds ⎜ ⎟ are there two wholes. How would you find these?, ⎝3⎠, 1, We can see that there are 6 one-thirds in two wholes or 2 ÷ = 6, 3, , 2.2.1a), , 1, 4, , (ii) 7 ÷, , 1, 2, , Reciprocal of a fraction, , (iii) 3 ÷, , 1, 5, , LA, N, , Find (i) 2 ÷, , G, , Do This, , are divided into, , TE, , 1, 1, Now consider 3 ÷ . This means the number of parts obtained, when each of the three wholes,, 4, 4, 1, equal parts, 4, , 1, 4, 1, 4, , 1, 4, 1, 4, , ER, , T, , 1, 4, 1, 4, , SC, , The number of one-fourths is 12, or 3 ÷, We also see that, 3 ÷, , 1, 4, = 3 × = 12 ., 4, 1, , This suggests that 3 ÷, , 1, 4, = 3×, 4, 1, , Free distribution by T.S. Government 2018-19, , 1, 4, 1, 4, , 1, 4, 1, 4, , 1, 4, 1, 4, , 1, = 12, 4, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 39
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1, Also examine 2 ÷ ., 3, , We already found that 2 ÷, , 1, =6, 3, , Similarly, 4 ÷, , 3, 1, can be obtained by interchanging the numerator and denominator of or by, 1, 3, , 1, 4, 1, . Similarly, is obtained by inverting ., 3, 1, 4, , LA, N, , inverting, , 1, 4, = 16 and 4 × = 16 ., 4, 1, , G, , The number, , AN, A, , 1, 3, As inthe above example 2 ÷ = 2 × = 6, 3, 1, , Observe these products and fill in the blanks:, , 2 3 2×3 6, × =, = =1, 3 2 3× 2 6, , 1, =1, 7, , 1, × 9 = ............., 9, , 2, × ............. = 1, 7, , TE, , 7×, , 5 4, × = ............., 4 5, , 5, =1, 9, , T, , ......... ×, , Multiply five more such pairs., , ER, , Any two non-zero numbers whose product is 1, are called reciprocals of one another. So the, 4 7, 7 4, is and the reciprocal of is ., 7 4, 4 7, , SC, , reciprocal of, , What is the reciprocal of, , 5, 2, and ?, 9, 5, , Try This, , 40, , 1., , Will the reciprocal of a proper fraction be a proper fraction?, , 2., , Will the reciprocal of an improper fraction be an improper fraction?, , VII CLASS MATHEMATICS
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Therefore, we can say that,, , 1, 4, 1, = 3 × reciprocal of = 3 × ., 4, 1, 4, 1, = ........ = ......................, 2, , So, 2 ÷, , 5÷, , of 1, , 3, 3, 4, = 2 × reciprocal of = 2 × ., 4, 4, 3, , 1, 2, is 1 . Is he right? Check., 2, 1, , AN, A, , 3÷, , Raju applied this inverting procedure to, mixed fractions and said that the reciprocal, , G, , 3÷, , 1, 2, 1, = 1 × reciprocal of = 1 × ., 2, 1, 2, , 2, = 5 × ........... = 5 × ......................, 4, , LA, N, , 1÷, , Thus dividing by a fraction is equivalent to multiplying the number by the reciprocal of, that fraction., , Find (i) 9 ÷, , 2, 5, , TE, , Do This, , (ii) 3 ÷, , 4, 7, , (iii) 2 ÷, , 8, 9, , ER, , T, , For dividing a whole number by a mixed fraction, we first convert the mixed fraction into, an improper fraction and then solve it., , SC, , 2, 17, 5 20, Example 4 ÷ 3 = 4 ÷ = 4 × =, 5, 5, 17 17, , 1, 10, Find , 11 ÷ 3 = 11÷ = ?, 3, 3, , Do This, , Find, , (i) 7 ÷ 5, , 1, 3, , Free distribution by T.S. Government 2018-19, , (ii) 5 ÷ 2, , 4, 7, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 41
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Division of a fraction by a whole number, 3, ÷ 3 be equal to?, 4, , What will, , Based on our earlier observations we have :, , So,, , 2, 2 1, ÷5 = × = ?, 3, 3 5, , What is, , 3, 3 3 3 1 3 1, ÷3 = ÷ = × =, =, 4, 4 1 4 3 12 4, , 5, 2, ÷ 6 and ÷ 8?, 7, 7, , AN, A, , 2.2.2, , For dividing mixed fractions by whole numbers, we convert the mixed fractions into improper, fractions. For example, , Division of a fraction by another fraction, , We can now find, , LA, N, , 2.2.3, , 3, 2 ÷ 2 = .......... = ........, 5, , G, , 1, 7, 7 1 7, 2, 2 ÷ 5 = ÷ 5 = × = . Find 4 ÷ 3 = .......... = ........;, 3, 3, 3 5 5, 5, , 1 5, ÷ ., 4 6, , 8 2 8, 2, ÷ = × reciprocal of = ......... = ........ and 1 ÷ 3 = ............ = ........., 5 3 5, 3, 2 4, , 3 1, ÷, 5 2, , ER, , Do This, , T, , Similarly,, , TE, , 1 5 1, 5 1 6 6, 3, ÷ = × reciprocal of = × =, = ., 4 6 4, 6 4 5 20 10, , Find (i), , (ii), , 1 3, 1 3, ÷, (iii) 2 ÷, 2 5, 2 5, , SC, , Example 8 : An empty swimming pool is to be filled up to, hour to fill, , 1 9, (iv) 5 ÷, 6 2, 9, of its capacity. A pump takes half an, 10, , 3, 9, of the pool, how long will it take to fill, of the pool?, 10, 10, , Solution : We need to find how many, , 3, 9, 9 3, ÷ ., ’s are there in , solve the division problem, 10, 10, 10 10, , 1, 9, 9 10, × = 3 Thus, it would take 3 half an hours, or 1 hours to fill the pool to its, ., 10 3, 2, 10, , 42, , VII CLASS MATHEMATICS
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Exercise 4, 1., , Find the reciprocal of each of the following fractions., 5, 8, , (i), , 8, 7, , (iii), , 13, 7, , (ii), , 8÷, , 7, 3, , (iii), , (ii), , 7, ÷5, 8, , (iii), , (iv), , 3, 4, , Find, 18 ÷, , (i), , (i), , 4., , 2, ÷3, 5, , Deepak can paint, , 1, 3, , (iv), , 5÷3, , 4, 7, , 4 4, ÷, 9 5, , G, , Find, , 3÷2, , 2, of a house in one day. If he continues working at this rate, how many, 5, , LA, N, , 3., , 3, 4, , AN, A, , 2., , (ii), , days will he take to paint the whole house?, 2.3, , Decimal numbers or Fractional decimals, , TE, , In class VI we have learnt about decimal numbers and their addition and subtraction. Let us review, our understanding and then learn about multiplication and division., Let us write 12714 in its expanded form:, , 12714 = 1× 10000 + 2 ×1000 + 7 × ......+ 1 × ...... + 4 × 1, , T, , What will the expended form of 12714.2 be?, , ×10 10 thousands, =, 10 10,000, thousand, , ×10 10 hundreds, =, 1thousand, 1,000, , ×10, , 10 tens =, , ×10, , 100, 1 hundred, , 10 units, =, 1 10, ten, , ×10, , 1 unit, , SC, , so on, , ER, , You will find that on moving from right to left, the value increase in multiples of 10., , Now, what happens when we move from left to right? You will find that the value gets, divided by, 10. Now think, if the unit is divided by 10, what will happen? Remember you have learnt that, 1 ÷ 10 =, , 1, = 0.1, 10, , th, , 1 unit, , ÷ 10, , 10 part, 1, or, 10, or 0.1, , th, , ÷ 10, , Free distribution by T.S. Government 2018-19, , 100 part, 1, or, 100, or 0.01, , th, , ÷ 10, , 1000 part, 1, or, 1000, or 0.001, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , ÷ 10 so on, , 43
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Thus, the expanded form of 12714.2 is12714.2 = 1× 10000 + 2 ×1000 + 7 × ......+ 1 × ...... + 4 × 1 + 2 ×, , 1, 10, , AN, A, , Now find the place value of all the digits of 3.42. You might have noticed that a dot ( . ) or a, decimal point separates whole part of the number from the fractional part. The part right side of the, decimal point is called the decimal part of the number as it represents a part of 1. The part left to, the decimal point is called the integral part of the number., In the number 3.42-, , 3 × 1=3, , 4×, , Try This, , 2 is at the second, place after the, decimal point, , G, , 4 is at the first, place after the, decimal point, , 1, 4, =, or 0.4, 10 10, , LA, N, , Place value, , 3 is at units, place, , 2×, , 1, 2, =, or 0.02, 100 100, , 1. Look at the following table and fill up the blank spaces., , (100), 5, 3, 2, , Tenth, , Hundredths, , Thousandths, , Number, , (1), , ⎛ 1⎞, ⎜ ⎟, ⎝ 10 ⎠, , ⎛ 1 ⎞, ⎜, ⎟, ⎝ 100 ⎠, , ⎛ 1 ⎞, ⎜, ⎟, ⎝ 1000 ⎠, , 4, , 7, , 8, , 2, , 9, , 547.829, , 7, , 2, , 1, , 7, , 7, , _______, , 2, , __, , __, , 5, , 4, , 327.154, , __, , 4, , __, , 2, , __, , 614.326, , __, , 6, , 5, , __, , 2, , 236.512, , ER, , 6, , Units, , (10), , T, , 0, , Tens, , TE, , Hundreds, , 2. Write thefollowing numbers in their expanded form., , SC, , (i) 30.807, , (ii) 968.038, , (iii) 8370.705, , To convert money, length, weight, etc from one unit to the other we often use decimals. For e.g. 5, paise = `, , 5, 220, 5, = `0.05; 220 g =, kg = 0.220 kg ; 5 cm =, m = 0.05 m, 100, 1000, 100, , Do This, Find (i) 50 paise = ` _____, 44, , VII CLASS MATHEMATICS, , (ii) 22 g = _____ kg, , (iii) 80 cm =, , m
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2.3.1 Comparison of decimal numbers, , AN, A, , Who has more money?, Abhishek and Neha have ` 375.50 and `375.75 respectively in their kiddy bank. To find who, has more money, we first compare the digits on the left of the decimal point. Since both the children, have ` 375 we compare the digits to the right of the decimal point starting from the tenth place. We, find that Abhishek has 7 tenths and Neha has 5 tenths, 7 tenths >5 tenths, therefore, Abhishek has, more money than Neha, i.e., 375.75 > 375.50., Now compare quickly, which of the following pair of numbers is greater ?, (ii) 1.775 with 19.780, , (iii) 364.10 and 363.10, While adding or subtracting, Let us see how to add or subtract decimals., decimal numbers, the digits in, (i), 221.85 + 37.10, (ii) 39.70 - 6.85, the same places must be added, 221.85, 39.70, or subtracted, i.e., while, writing numbers one below the, +37.10, - 06.85, other, see that decimal points, 258.95, 32.85, must come one below the other., Do This, Decimal places may be made, equal by placing zeroes on the, Find (i) 0.25 + 5.30, (ii) 29.75 - 25.97., right side of the decimal, number., , TE, , LA, N, , G, , (i) 37.65 and 37.60, , The equal sides of an isosceles triangle are 3.5 cm each and the other side is 2.5, cm. What is the perimeter of the triangle?, , Solution :, , The sides of isosceles triangle are 3.5 cm, 3.5 cm and 2.5 cm. Therefore, the, perimeter of the given triangle is = sum of lengths of three sides = 3.5 cm + 3.5 cm, + 2.5 cm = 9.5 cm, , 1., , ER, , T, , Example 9 :, , Which one is greater?, , SC, , (i) 0.7 or 0.07, , (iii) 1.47 or 1.51, , 2., , 1 cm = 10 mm, (ii) 7 or 8.5, , 1 m = 100cm, , (iv), , 1 km = 1000m, , 6 or 0.66, , Express the following as rupees using decimals., (i) 9 paise, , 3., , Exercise - 5, , (ii) 77 rupees 7 paise, , 1 kg = 1000gm, (iii), , 235 paise, , (i) Express 10 cm in metre and kilometre., (ii) Express 45 mm in centimeter, meter and kilometer., , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 45
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Express the following in kilograms., (i) 190 g, , (i) 55.5, , (ii), , 5.55, , (iv) 30.303, , (v), , 1234.56, , (iii) 303.03, , Write the place value of 3 in the following decimal numbers., 32.46, , (iv) 90.30, , (v), , 794.037, , (iii) 7.43, , A, , Aruna and Radha start their journey from two different, places. A and E. Aruna chose the path from A to B then to, C, while Radha chose the path from E to D then to C., , C, , 7 km, , E, , 8.25 km, , D, , LA, N, , Find who travelled more and by how much?, , B, , 9.50 km, , m, , (ii), , 5k, , (i) 3.46, , m, 0k, 2.41, , 7., , Write the following decimal numbers in expanded form., , 3.7, , 6., , (iii) 44 kg 80 gm, , AN, A, , 5., , (ii) 247 g, , G, , 4., , Upendra went to the market to buy vegetables. He brought 2 kg 250 gm tomatoes,, 2 kg 500 gm potatoes, 750 gm lady fingers and 125 gm green chillies. How much weight, did Upendra carry back to his house?, , 2.4, , Multiplication of decimal numbers, , TE, , 8., , Rajendra of class 7 went with his mother to the bazar to buy vegetables. There they purchased 2.5, kg potatoes at the rate of ` 8.50 per kg. How much money do they need to pay?, , T, , We come across various situations in day-to-day life where we need to know how to multiply two, decimals. Let us now learn the multiplication of two decimal numbers., , ER, , Let us first multiply- 0.1 × 0.1, , 0.1 means one part of 10 parts. This is represented as, , 1, using fractions, 10, , SC, , and pictorially in Fig.1., Thus, 0.1 × 0.1 =, , 1 1, 1, 1, ×, which means, of, . So here we are, 10 10, 10, 10, , finding the 10th part of, , Figure 1, , 1, 1, . Thus, we divide, into 10 equal parts and, 10, 10, , take one part. This is represented by one square in Figure 2. How many, squares are there in Figure 2? There are 100 squares. So one square, represents one out of 100 or 0.01. So we can conclude that, Figure 1, , 46, , VII CLASS MATHEMATICS
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0.1× 0.1 =, , 1 1, 1, × =, = 0.01, 10 10 100, , Let us now find 0.4 × 0.2, 4, 2, 4, 2, ×, of, or, 10 10, 10, 10, , AN, A, , 0.4 × 0.2 =, Pictorially, , 4, of, 10, , G, , 2, 10, , =, , 8, 100, , LA, N, , =, , Since there are 8 double shaded squares out of 100, they represent 0.08., , TE, , While finding 0.1 × 0.1 and 0.4 × 0.2, you might have noticed that we first multiplied them as, Whole numbers ignoring the decimal point. In 0.1 × 0.1, we found 01 × 01 or 1 × 1. Similarly in, 0.4 × 0.2 we found 04 × 02 or 4 × 2. The products obtained are 1 and 8 respectively., , ER, , T, , We then counted the total number of digits to the right of the decimal point in the numbers being, multiplied. In both 0.1×0.1 and 0.4×0.2, the total number of digits to the right of the decimal point, in the numbers being multiplied is 2 each. Thus, in each of their products we put the decimal point, by counting two places from right to left., For any decimal number which has no, Thus, 0.1 × 0.1 = .01 or 0.01, 0.4 × 0.2 = .08 or 0.08, , integral number part, we generally place a, zero on the left side of decimal point to give, prominance to decimal point., , SC, , If we had multiplied 0.5 × 0.05 then we would have put the decimal point in the product by, counting three places from right to left i.e. 0.5 × 0.05 = 0.025., Let us now find 1.2 × 2.5, Multiply 12 and 25. We get 300. In both 1.2 and 2.5, there is 1 digit to the right of the decimal, point. So, count 1 + 1 = 2 digits. From the rightmost digit (i.e., 0) in 300, move two places, towards left. We get 3.00 or 3. Thus, 1.2 × 2.5 = 3, While multiplying 2.5 and 1.25 you will first multiply 25 and 125. For placing the decimal in the, product obtained, we will count 1 + 2 = 3 (Why?). Thus, 2.5 × 1.25 = 3.225., Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 47
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Do These, 1., , Find, , (i) 1.7 × 3, , (ii) 2.0 × 1.5, , (iii) 2.3 × 4.35, , 2., , Arrange the products obtained in (1) in descending order., , Solution :, , Length of the rectangle, , = 7.1 cm, , Breadth of the rectangle, , = 2.5 cm, , AN, A, , Example 10 : The length of a rectangle is 7.1 cm and its breadth is 2.5 cm. What is the area of, the rectangle?, , Therefore, area of the rectangle = 7.1 × 2.5 = 17.75 cm2, , 32, 235, whereas 2.35 =, . Thus, she found that depending on the, 10, 100, , LA, N, , Reshma observed that 3.2 =, , G, , 2.4.1 Multiplication of decimal number by 10, 100, 1000 etc.,, , position of the decimal point, the decimal number can be converted to a fraction with denominator, 10 or 100etc., She wondered what would happen if a decimal number is multiplied by 10, 100 or, 1000 etc.,, , TE, , Let us see if we can find a pattern in multiplying numbers by 10 or 100 or 1000., Have a look at the table given below and fill in the blanks :, 1.76 × 10 =, , 176, × 10 = 17.6, 100, , 1.76 × 100 =, , 176, × 100 = 176 or 176.0 2.35 × 100 = ........., 100, , 12.356 × 100 = ........., , 176, × 1000 = 1760 or, 100, , 12.356 × 1000 = ........., , T, , ER, , 0.5 × 10 =, , 2.35 × 1000 = ........., , 12.356 × 10 = ........., , 1760.0, , SC, , 1.76 × 1000=, , 2.35 × 10 = ........., , 5, × 10 = 5 ; 0.5 × 100 = ......... ;, 10, , 0.5 × 1000 = ........., , Look at your answers. Could you find any pattern? The decimal point in the products shifts to the, right by as many zeroes as in 10, 100, 1000 etc,., , 48, , VII CLASS MATHEMATICS
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2.4.2 Division of decimal numbers, Gopal was preparing a design to decorate his classroom. He needed a few coloured strips of, paper of length 1.6 cm each. He had a strip of coloured paper of length 9.6 cm . How many pieces, of the required length will he get out of this strip? He thought it would be, , 9.6, cm. Is he correct?, 1.6, , 2.4.2 (a) Division by numbers like 10, 100, 1000 etc.,, Let us now divide a decimal number by 10, 100 and 1000., , 315, 315 1 315, ÷ 10 =, × =, = 3.15, 10, 10 10 100, , Similarly, 31.5 ÷ 100 =, , LA, N, , 31.5 ÷ 10 =, , G, , Consider 31.5 ÷ 10., , AN, A, , Both 9.6 and 1.6 are decimal numbers. So we need to know the division of decimal numbers too!, , 315, 315 1, 315, ÷ 100 =, ×, =, = 0.315, 10, 10 100 1000, , TE, , Is there a pattern while dividing numbers by 10, 100 or 1000? This may help us in dividing, numbers by 10, 100 or 1000 in a shorter way., Observe the pattern in the table, given bellow and complete it., 132.7 ÷ 10 = .........., , 1.5 ÷ 10 = .........., , 17.36 ÷ 10 =, , 29.5 ÷ 100 = 0.295, , 132.7 ÷ 10 = .........., , 1.5 ÷ 100 = .........., , 17.36 ÷ 100 = .........., , T, , 29.5 ÷ 10 = 2.95, , .........., , ER, , 29.5 ÷ 1000 = 0.0295 132.7 ÷ 1000 = .......... 1.5 ÷ 1000 = ......... 17.36 ÷ 1000 = .........., 2.4.2 (b) Division of a decimal number by a whole number, 6.4, . Remember we also write it as 6.4 ÷ 2., 2, , SC, , Let us find, , So, 6.4 ÷ 2 =, , =, , 64, 64 1, ÷2 =, × (as learnt in fractions), 10, 10 2, , 64 × 1 1× 64 1 64 1, 32, =, = ×, = × 32 =, = 3.2, 10 × 2 10 × 2 10 2 10, 10, , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 49
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Now, let us calculate 12.96 ÷ 4 =, , 1296, 1296 1, 1 1296, 1, ÷4=, × =, ×, =, × 324 = 3.24, 100, 100 4 100, 4, 100, , Do This, (ii) 25.5 ÷ 3, , AN, A, , 1. Find (i) 35.7 ÷ 3, , Example 11 : Find the average of 4.2, 3.8 and 7.6., , 2.4.2 (c), , The average of 4.2, 3.8 and 7.6 is, , 4.2 + 3.8 + 7.6 15.6, =, = 5.2, 3, 3, , G, , Solution :, , Division of a decimal number by another decimal number, , LA, N, , Let us find how we divide a decimal number by another decimal number, For example 35.5 ÷ 0.5., 355 5 355 10, ÷ =, × = 71, 10 10 10 5, , =, , Thus 35.5 ÷ 0.5 = 71., , TE, , Example 12 : A truck covers a distance of 92.5 km in 2.5 hours. If the truck is travelling at the same, speed through out the journey what is the distance covered by it in 1 hour?, Solution : Distance travelled by the truck = 92.5 km., , T, , Time required to travel this distance, , 2., , 50, , =, , 92.5 925, =, = 37 km., 2.5, 25, , Exercise - 6, , Solve the following., , SC, , 1., , ER, , So distance travelled by it in 1 hour, , = 2.5 hours., , (i), , 0.3 × 6, , (ii), , 7 × 2.7, , (iii), , 2.71 × 5, , (iv), , 19.7 × 4, , (v), , 0.05 × 7, , (vi), , 210.01 × 5, , (vii), , 2 × 0.86, , Find the area of a rectangle whose length is 6.2 cm and breadth is 4 cm., , VII CLASS MATHEMATICS
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3., , Solve the following., (i), , 21.3 × 10, , (ii), , 36.8 × 10, , (iii), , 53.7 × 10, , (iv), , 168.07 × 10, , (v), , 131.1 × 100, , (vi), , 156.1 × 100, , (vii), , 3.62 × 100, , (viii), , 43.07 × 100, , (ix), , 0.5 × 10, , (x), , 0.08 × 10, , (xi), , 0.9 × 100, , (xii), , 0.03 × 1000, , A motor bike covers a distance of 62.5 km.consuming one litre of petrol. How much, distance does it cover for 10 litres of petrol?, , 5., , Solve the following., 1.5 × 0.3, , (ii), , 0.1 × 47.5, , (iii), , 0.2 × 210.8, , (iv), , 4.3 × 3.4, , (v), , 0.5 × 0.05, , (vi), , 11.2 × 0.10, , (vii), , 1.07 × 0.02, , (viii), , 10.05 × 1.05 (ix), , (x), , 70.01 × 1.1, (ii), , 0.45 ÷ 5, , (iii), , 44.3 ÷ 10, , 7 ÷ 3.5, , (vi), , 88.5 ÷ 0.15, , 2.3 ÷ 100, , (iv), , 127.1 ÷ 1000 (v), , (vii), , 0.4 ÷ 20, , 101.01 × 0.01, , LA, N, , Solve the following., (i), , G, , (i), , TE, , 6., , AN, A, , 4., , A side of a regular polygon is 3.5 cm in length. The perimeter of the polygon is 17.5 cm., How many sides does the polygon have?, , 8., , A rain fall of 0.896 cm. was recorded in 7 hours, what was the average amount of rain per, hour?, , 2.5, , Introduction to Rational numbers, , 2.5.1, , Positive fractional numbers:, , ER, , T, , 7., , SC, , We have learnt about integers and fractions. Let us see how the number line looks when both are, marked on it., –4, , We have, , –3, , –2, , –1, , 0, , 1, 4, , 2, 4, , 3, 4, , 1, , 4, 3, , 5, 3, , 2, , 3, , 4, , 1 2 3, , , ..... between 0 and 1 on the number line. All these are numbers that are less than, 4 4 4, , one. We call them as proper fractions and say that all proper fractions lie between 0 and 1., Similarly, we know, , 4, 5, and would lie between 1 and 2. We can recall them as improper, 3, 3, , fractions. All these are called positive fractional numbers., Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 51
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Do These, 1., , Write 5 more fractions between (i) 0 and 1 and (ii) 1 and 2., , 2., , Where does 4, , 3, lie on the number line?, 5, , On the left side of 0 we have integers –1, –2, –3 ......., , AN, A, , Do the numbers increase or decrease as we move further left on the number line?, , You know that number decreases as we move further left. The farther the number is from 0 on the, left the smaller it is., Do These, , Find the greatest and the smallest numbers among the follwing groups?, (i), , G, , 1., , 2, –2, –3, 4, 0, –5, , 2., , LA, N, , (ii) –3, –7, –8, 0, –5, –2, , Write the following numbers in ascending order., –5, –75, 3 – 2, 4,, , 3, 2, , 2 3, , , 0, –1, –2, 5, 3 2, , (ii), , TE, , (i), 2.5.2, , Negative fractional numbers, , Consider the point A shown on the line., –2, , B, , –1, , T, , –3, , A, , 0, , 1, , 2, , 3, , ER, , It lies on the number line between 0 and –1. Is it more than 0 or less than 0?, 1, 1, Neha found an easy way to represent, ? We cannot say it is as it is less than zero., 2, 2, −9, . She first wrote it in a mixed fraction, 4, 1, 1, We write A as − since it is less than zero., −9, 1, 2, 2, = −2, and then represented it, 4, 4, 3, between –2 and –3., Similarly, B the mid point of –1 and –2 is. −, 2, , SC, , Is it, , 1 3 9, You can see that negative fractional numbers like − , − , − give us points in between any two, 2 2 4, , negative integers or between zero and a negative integer., 52, , VII CLASS MATHEMATICS
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Do These, On the number line given below represent the following numbers., 7, 2, , (ii), , 3, 2, , –2, , –1, , 7, 4, , −1, 2, , −7, 4, , (iv), , (v), , 0, , (i), , 11, ,, 943, , 54, ,, 17, , –68,, , Which of these are to the left of, (a) 0, , (b) –2, , (vi), , 1, , Consider the following numbers on a number line., 7, 27, − ,, 8, , −1, 4, , –3,, , G, , 2., , (iii), , 1, 4, , 2, , AN, A, , (i) −, , 9, − ,, 6, , 7, 2, , LA, N, , 1., , (c) 4, , (d) 2, , (ii) Which of these would be to the right of, , 2.5.3, , (b) –5, , Rational Numbers, , (c) 3, , 1, 2, , (d), , −5, 2, , TE, , (a) 0, , T, , We know 0, 1, 2, 3, 4, 5 are whole numbers. We also know that –7, –6, –5, –4, –3, –2, –1, 0,, 1, 2, 3, 4, 5 is a bigger collection of numbers called integers., , ER, , Rani says “All whole numbers are integers but the converse is not true.” Do you agree with her?, Rani is right as negative numbers like –6, –5, –4, –3, –2, –1 etc are integers but not whole, numbers. Thus, all whole numbers are integers and all integers are not whole numbers., , SC, , We further know that positive fractional numbers like, , 1 1 1 5 11 8, , , , , , are ratios of whole num2 3 4 6 5 8, , w1, bers. All fractional numbers can in general be written as w with the condition w1 and w2 are, 2, , whole numbers and that w2 is not equal to zero., Try This, Write 5 fractional numbers and identify w1 and w2 in each of these., Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 53
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Rational numbers are a bigger collection of numbers, which includes all integers, all, positive fractional numbers and all negative fractional numbers., The numbers, , −7 −5 −7 −2 1 4 17 6, , , , , 0, , , , etc. are all rational numbers., 3 2 7 7, 4 4 5 1, , p, , where p, q, , AN, A, , In all these we have a ratio of two integers, thus the numbers in the form of, , and q are integers except that q is not equal to zero are called as rational numbers., The set of rational numbers is denoted by Q, Try These, , G, , (i) Take any 5 integers and make all possible rational numbers with them., , 2.5.4, , LA, N, , (ii) Consider any 5 rational numbers. Find out which integers constitute them?, Comparing rational numbers, , We know that, , 3, 9, and, are equivalent fractional numbers. We also know that when we com4, 12, , TE, , pare fractional numbers we convert each of them to equivalent fractional numbers and then compare the ones with a common denominator., For example, to compare, , 3, 5, and ., 4, 7, , We write equivalent fractional numbers for both, , ER, , T, , 3 6 9 12 15 18 21, = , , , , ,, and, 4 8 12 16 20 24 28, , SC, , 5 10 15 20, = , ,, .... ..., 7 14 21 28, , We can compare, , 21, 20, with, as they have same denominators., 28, 28, , 21, 20, is bigger than, 28, 28, , Therefore,, , 54, , 3 5, >, 4 7, , VII CLASS MATHEMATICS
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Try These, 1., , Write three more equivalent fractions of, , 3, and mark them on the number line., 4, , What do you observe?, , Now compare, , Do all equivalent fractions of, , 6, represent the same point on the number line?, 7, , AN, A, , 2., , −1, −2, and, 2, 3, , We write equivalent fractions for both, , LA, N, , G, , −1 −2 −3 −4, =, , ,, ..... ....., 2, 4 6 8, , −2 −4 − 6, =, ,, ..... ....., 3, 6 9, , ∴, , −3, −4, −4 −3, <, and, as they have same denominators., 6, 6, 6, 6, , TE, , We can compare, −2 −1, <, 3, 2, , T, , Try These, Are, , −1, −3, and, represent same point on the number line?, 2, 6, , 2., , Are, , −2, −4, and, equivalent, 3, 6, , SC, , ER, , 1., , Eg: When we place them on the number line we find that they occupy the same point. Thus We, can say that, , −1, −2, and, are equivalent rationals., 2, 4, , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 55
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Do These, 5, −7, 3, (ii), (iii) −, 2, 9, 7, , Write 5 equivalent rational numbers to (i), , 2., , Identify the equivalent rational numbers in each question:, (i), , −1 −3 −2 − 4, , , ,, 2 4 4 8, , (ii), , 1 3 5 10 2 20, , , , , ,, 4 4 3 6 4 12, , AN, A, , 1., , G, , We can say that to get equivalent rational numbers we multiply or divide the integer in the numerator and in the denominator by the same number., , LA, N, , For example,, , 1, 1× 2 2, 1× 3 3, =, = ., we would have, as one equivalent number another is, 5, 5 × 2 10, 5 × 3 15, , For, , −2, −2 × 2 −4, 2 × 3 −6, =, =, we would have, as one and −, as another., 7, 7 × 2 14, 7 × 3 21, , TE, , For, , We can go on to build more such equivalent rational numbers, just by multiplying with, , Exercise - 7, , Write any three equivalent rational numbers to each of the following, 2, 3, , SC, , 1., , ER, , T, , 1 2 3 4, = = = ..., 1 2 3 4, , (i), , (ii) −, , 3, 8, −15, with (i) denominator 12 (ii) numerator -75?, 36, , 2., , What is the equivalent rational number for, , 3., , Mark the following rational numbers on the number line., (i), , 56, , 1, 2, , (ii), , 3, 4, , VII CLASS MATHEMATICS, , (iii), , 3, 2, , (iv), , 10, 3
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3., (, , Find whether the following statements are true or false., (i) Every integer is a rational number and vice versa, ), (ii) In a rational number of the form, , (, , p, , q must be a non zero integer., q, , ), , (, , ), (iv), , (, , AN, A, , (iii) Every decimal number can be represented as a rational number., 5 6 7, , , are equivalent rational numbers., 7 7 7, , ), , (, , ), , LA, N, , G, , (v) Equivalent rational numbers of a positive rational numbers are all positive, , Looking back, , 1. We have learnt that for addition and subtraction of fractions; the fractions, should be like fractions., Pr oduct of numerators, , TE, , 2. We have also learnt how to multiply fractions i.e, Product of denominators, , T, , 3. “of” can be used to represent multiplication. For example,, , 1 2, 1, of 6 = × 6 = 2 ., 3, 3, , ER, , 4. The product of two proper fractions is less than each of the fractions that are, multiplied. The product of a proper and improper fraction is less than the improper, fraction and greater than the proper fraction.The product of two improper fractions, is greater than each of the fractions., , SC, , 5. A reciprocal of a fraction is obtained by inverting the numerator and denominator., 6. We have seen how to divide two fractions., (i) While dividing a whole number with a fraction, we multiply the whole number, with the reciprocal of that fraction., (ii) While dividing a fraction by a whole number we multiply the fraction with the, reciprocal of the whole number., , Free distribution by T.S. Government 2018-19, , FRACTIONS, DECIMALS AND RATIONAL NUMBERS, , 57
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(iii) While dividing one fraction by another fraction, we multuiply the first fraction, with the reciprocal of the second. So, , 3 5 3 7 21, ÷ = × =, ., 4 7 4 5 20, , AN, A, , 7. We also learnt how to multiply two decimal numbers. While multiplying two, decimal numbers, we first multiply them as whole numbers. We then count the, total number of digits to the right of the decimal point in both the decimal numbers, being multiplied. Lastly, we put the decimal point in the product by counting the, digits from its rightmost place., 8. To multiply a decimal number by 10, 100, 1000 ... etc., we move the decimal, point in the number to the right by as many places as there are zeros in the, numbers 10, 100, 1000 . . ., , G, , 9. We have learnt how to divide decimal numbers., , LA, N, , (i) To divide a decimal number by a whole number, we first divide them as, whole numbers. We then place the decimal point in the quotient as in the, decimal number., Note that here we are considering only those divisions in which the remainder, is zero., , TE, , (ii) To divide a decimal number by 10, 100, 1000 or any other multiple of 10,, we shift the decimal point in the decimal number to the left by as many places, as there are zeros in 10, 100, 1000 etc.,, (iii) While dividing two decimal numbers, first shift the decimal point to the right, by equal number of places in both, to convert the divisor to a whole number., , ER, , T, , 10. Rational numbers are a bigger collection of numbers, which includes all integers,, all positive fractional numbers and all negative fractional numbers. In all these we, have a ratio of two integers, thus, , i) p, q are integers and, ii) q ≠ 0, , SC, , In this, , p, represents a rational number., q, , The set of rational numbers is denoted by Q, , John Napier (Scotland), , 1550-1617 AD, Found logarithms., Introduced napier rods for multiplictions., Also introduced System of decimal fractions., , 58, , VII CLASS MATHEMATICS
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3.0, , Introduction, , LA, N, , G, , You have already come across simple equations like, 4x = 44, 2m = 10 and their solutions in Class VI. You have, seen how these equations help us in solving puzzles and daily, life problems. Let us review what we have already learnt, about simple equations and their solutions through the, following exercise., , 3, , AN, A, , SIMPLE EQUATIONS, , Exercise - 1, , 3.1, , 2x = 10, , (ii), , 2x–3 = 9, , (iii), , 4z + 1 = 8, , (iv), , 5p + 3 = 2p + 9, , (v), , 14 = 27 – y, , (vi), , 2a – 3 = 5, , (vii), , 7m = 14, , (viii), , 8=q+5, , TE, , (i), , Solve the following equations by trial and error method., (i), , 2+y=7, , (iii), , 5m = 15, , (ii), , T, , 2., , Write L.H.S and R.H.S of the following simple equations., , (iv), , a–2=6, 2n = 14, , ER, , 1., , Equation - Weighing balance, , SC, , You have seen in class VI that an equation is compared with a, weighing balance with equal weights on both sides., What will happen if the left pan of a weighing balance holds 5, kg and the right pan holds 2 kg?, What will happen if the left pan of a weighing balance holds 3, kg and the right pan holds 7 kg?, What will happen if the left pan of a weighing balance holds 3 kg and the right pan holds 3 kg?, A weighing balance needs to have equal weights on both sides to be perfectly balanced., , Free distribution by T.S. Government 2018-19, , SIMPLE EQUATIONS, , 59
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The same principle will hold in an equality., Consider the equality 12–2 = 6 + 4, HereLHS = 12 – 2 = 10, , and, , AN, A, , RHS = 6 + 4 = 10, Since both sides are equal, the equality holds., , What will happen if we add 3 on both sides of an equation? Will the values of both sides remain, equal? Will the values be equal if 10 is added? Try with some other number of your choice., , G, , What will happen if we subtract 5 from both sides of the equation? Will both sides remain equal?, Will the values be equal if 7 is subtracted? Try with some other numbers of your choice., , LA, N, , What will happen if we multiply both sides by 6? Will both sides remain equal? Will they be equal, if 8 multiplied by 8? Try with some other numbers of your choice., What will happen if we divide both sides of the equation by 5? Will both sides remain equal? Will, they be equal if both sides are divided by 2?, , 3.2, , TE, , You will find that answer is ‘yes’ in all cases. If the same number is added or subtracted on both, sides or if both sides of the equality are either multiplied or divided by same number ; the equality, remains unchanged. This principle of equality is going to help in solving equations ahead., Solving equations, , You have already learnt how to solve equations using the trial and error method. Now we will use, the above principles of equality to solve equations in a much lesser time., , ER, , T, , To solve equations we first need to separate the numerical terms from the terms containing, variables/unknowns by taking them on the two different sides of the equality and then use the, principles of equality., Let us study the examples given below., Solve x + 3 = 7, , SC, , Example 1 :, , Given equation is, , Solution :, , x + 3 = 7 ..................... (1), The L.H.S of the equation is x + 3 ., , Total value of L.H.S. is 3 more than x, To find the value of 'x' we have to remove 3 from the LHS. Thus, we need to, subtract 3 from the LHS. If 3 is subtracted from LHS it should also be subtracted, from RHS, to balance the equality., 60, , VII CLASS MATHEMATICS
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We have,, , x+3=7, x+3–3=7–3, x = 7 – 3 ........................ (2), , Thus,, , x = 4., , AN, A, , From (1) and (2) it is clear that removing ‘+3’from, LHS is equivalent to 'subtracting 3' from the RHS., That means '+3' on LHS transforms as '–3' to RHS., , x=4, , Check: let us substitute 4 for x in the given equation and find whether LHS = RHS., LHS = x + 3, = 4+3, , (substituting x = 4), , G, , =7, RHS = 7, , LA, N, , LHS = RHS., , 7, , x+3–3, , 7–3, , x, , 4, , T, , x+3, , TE, , Let us also understand the above solution with a weighing balance:, , ER, , Example 2 : Solve y – 7 = 9, Solution :, , y – 7 = 9 ..................... (1), , Here the L.H.S of the equation is y – 7, , SC, , So to get the value of ‘y’ we have to add 7 on both sides of the equation., , Therefore,, , Thus,, , y–7+7=9+7, y = 9 + 7 ..................... (2), , From (1) and (2) it is clear that '–7'on, LHS transforms to RHS as '+7'., , y = 16, y = 16., , Check : Substitute '16' for ‘y’ and check whether LHS = RHS., , Free distribution by T.S. Government 2018-19, , SIMPLE EQUATIONS, , 61
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Example 3 :, , Solve 5x = –30, , Solution :, , 5x = –30 ..................... (1), , x=, , (dividing both sides by 5), , −30, ..................... (2), 5, , ∴x = –6, , From (1) and (2) it is clear that, multiplier ‘5’on LHS transforms, to RHS as ‘divisor’., , AN, A, , 5 x −30, =, 5, 5, , Example 4 :, , Solve, , z, = −3, 6, , ⎛z⎞, 6 ⎜ ⎟ = 6 × (−3), ⎝6⎠, , LA, N, , z, = −3 ..................... (1), 6, , Solution :, , G, , Check: Substitute x = – 6 and check whether LHS = RHS., , (multiplying both sides by 6), From (1) and (2) it is clear that, divison ‘6’on LHS transforms to, RHS as ‘multipier 6’., , z = 6 × ( −3) ..................... (2), , TE, , ∴ z = –18, , Check : Substitute z = –18 and check whether LHS = RHS., Example 5 :, , Solve 3x + 5 = 5x – 11, , 3x + 5 = 5x – 11, , T, , Solution :, , ER, , 3x + 5 –5x = 5x – 11 – 5 x (subtracting 5x from both sides), –2x + 5 = –11, , –2x + 5 –5 = –11 – 5, , (subtracting '5' from both sides), , SC, , –2x = –16, , −2 x −16, =, (Dividing both sides by '–2'), −2, −2, , ∴x=8, , Check : Substituting x=8 in the given equation:, LHS = 3x + 5 = 3 (8) + 5 = 24 + 5 = 29, RHS = 5x – 11 = 5 (8) – 11 = 40 – 11 = 29, ∴ LHS = RHS, , 62, , VII CLASS MATHEMATICS
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Thus, in transforming terms from L.H.S. to R.H.S., ‘+ quantity’ becomes ‘– quantity’, ‘– quantity’ becomes ‘+ quantity’, ‘× quantity’ becomes ÷ quantity, , Example 6 :, , AN, A, , ‘÷ quantity’ becomes ‘ × quantity’, Solve 12 = x + 3, , Here if 12 is moved from LHS to RHS it becomes –12 and if x+3 is moved from, RHS to LHS it becomes –x – 3., , G, , i.e –x –3 = –12, Multiplying both sides by –1, , LA, N, , –1 (–x –3) = –1 (–12), x + 3 =12, x = 12 –3, ∴x=9, , TE, , Therefore, if both LHS and RHS of an equation are moved (transposed) from one side to, another side, the values of terms remain same., Exercise - 2, , T, , Solve the following equations without transposing and check your result., (i), , x+5=9, , (ii), , y – 12 = –5, , (iii), , 3x + 4 = 19, , (iv), , 9z = 81, , (v), , 3x + 8 = 5x + 2, , (vi), , 5y + 10 = 4y – 10, , SC, , ER, , 1., , (i), , 2+y=7, , (ii), , 2a – 3 = 5, , (iii), , 10 - q = 6, , (iv), , 2t – 5 = 3, , (v), , 14 = 27 – x, , (vi), , 5 (x+4) = 35, , (vii), , –3x = 15, , (viii), , 5x –3 = 3x – 5, , (ix), , 3y + 4 = 5y – 4, , (x), , 3 (x – 3) = 5 (2x + 1), , 2., , Solve the following equations by transposing the terms and check your result., , Free distribution by T.S. Government 2018-19, , SIMPLE EQUATIONS, , 63
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3.3, , Usage of algebraic equations in solving day to day problems., , Let us look at the following examples:, The total number of boys and girls in a class is 52. If the number of girls is 10 more than, boys, find the number of boys?, , (ii), , The present age of Ramu's father is three times that of Ramu. After five years the sum of, their ages will be 70 years. Find their present ages., , (iii), , A purse contains ` 250 in the denomination of `10 and ` 50. If the number of `10 notes, is one more than that of `50 notes find the number of notes of each denomination., , (iv), , Length of a rectangle is 8 m less than twice its breadth. The perimeter of the rectangle is 56 m., Find its length and breadth., , AN, A, , (i), , G, , Likes in all the problems given above, we can use simple equations to solve various problems of, day to day life. The following steps can be followed in doing so, Step 1: Read the problem carefully., , LA, N, , Step 2 : Denote the unknown or the quantity to be found with some letters such as x, y, z, u, v, w, p,t., Step 3: Write the problem in the form of an algebraic equation by making a relation among the, quantities., Step 4: Solve the equation., , TE, , Step 5: Check the solution., Example 7:, , Total number of the boys and girls in a class is 52. If the number of girls is 10 more, than that of boys, find the number of boys?, , Solution :, , Let us assume the number of boys to be x., , T, , The number of girls will be x + 10., The total number of boys and girls = x + (x + 10), , ER, , = x + x + 10, = 2x + 10, , SC, , According to the question, the total number of boys and girls is 52., Therefore, 2x + 10 = 52, , Solving this equation, 2x + 10, , = 52, , 2x = 52 – 10 (transposing 10 from LHS to RHS), 2x = 42, x=, , 42, (transposing 2 from LHS to RHS), 2, , ∴ x = 21, 64, , VII CLASS MATHEMATICS
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Thus, the number of boys = 21, and the number of girls = 21 + 10 = 31, Check :, , 21 + 31 = 52 i.e. the total number of boys and girls is 52., And 31 – 21=10 i.e. the number of girls is 10 more than the number of boys., The present age of Ramu's father is three times that of Ramu. After five years the, sum of their ages would be 70 years. Find their present ages., , Solution :, , Let Ramu's present age, , = x years., , Then the present age of his father, , = 3x years., , After 5 years Ramu's age, , = x+5 years., , His father's age, , = 3x + 5 years., , Sum of their ages after 5 years is, , = (x + 5) + (3x + 5) = 4x + 10 years., , LA, N, , G, , AN, A, , Example 8:, , According to the problem, Sum of their ages, , 4x + 10 = 70, , 4x = 70 – 10, , TE, , 4x = 60, x=, , 60, = 15, 4, , Thus, Ramu's present age = 15 years., , T, , So, present age of his father = 3 × 15 years = 45 years., , ER, , Check : 45 is three times of 15 i.e., at present Ramu's father is 3 times that of Ramu,, After 5 years Ramu's age = 15 + 5 = 20 years and his father's age = 45 + 5 = 50 years., Sum of their ages 20 + 50 = 70 years., A purse contains `250 in the denomination of `10 and ` 50. If the number of, `10 notes is one more than that of `50 notes, find the number of notes of each, denomination., , Solution :, , Let the number of `50 notes, , =x, , Then the total value of `50 notes, , = 50x, , Number of ` 10 notes, , =x+1, , SC, , Example 9 :, , Free distribution by T.S. Government 2018-19, , SIMPLE EQUATIONS, , 65
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Then the total value of ` 10 notes, ∴, , = 10 (x+1), , Total value of money = 50x + 10 (x+1), = 50x + 10x + 10, =60x + 10, , Therefore,, , AN, A, , Given, total value of the money that the purse contains is 250, 60x + 10 = 250, 60x = 250 – 10, 60x = 240, , G, , 240, 60, , ∴x = 4, , LA, N, , x=, , Thus, the number of `50 notes = 4, Number of `10 notes = 4 + 1 = 5, , Check :`10 notes (5) are one more than `50 notes (4)., , TE, , Value of the money = (50×4) + (10×5), = 200 + 50, = 250, , Let the breadth of the rectangle = x m., , ER, , Solution :, , T, , Example 10: Length of a rectangle is 8 m less than twice its breadth. If the perimeter of the, rectangle is 56 m, find its length and breadth., , = 2x m., , Therefore, its length, , = (2x – 8) m. (by problem), , SC, , Twice the breadth, , Perimeter of the rectangle, , = 2 (length + breadth), , Thus, perimeter, , = 2 (2x – 8 + x) m., = 2 (3x – 8) m., = (6x – 16) m., , Given, the perimeter of the rectangle is 56 m., Therefore, 6x – 16 = 56, 66, , VII CLASS MATHEMATICS
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6x = 56 + 16, 6x = 72, x=, , 72, 6, , ∴ x = 12, , AN, A, , Breadth of the rectangle = 12 m., Length of the rectangle = 2 × 12 – 8 = 16 m., Check : Perimeter = 2 (16 + 12) = 2 × 28, , = 56 m., , Write the information given in the picture in the form of an equation. Also, find 'x' in the, following figure., , LA, N, , 1, , G, , Exercise 3, , 15 cm, , x cm, , Write the information given in the picture in the form of an equation. Also, find 'y' in the, 13 cm, following figure., , TE, , 2., , 11 cm, , 8 cm, , ‘y’ cm, , If we add 7 to twice a number, we get 49. Find the number., , 4., , If we subtract 22 from three times a number, we get 68. Find the number., , 5., , Find the number when multiplied by 7 and then reduced by 3 is equal to 53., , 6., , Sum of two numbers is 95. If one exceeds the other by 3, find the numbers., , 7., , Sum of three consecutive integers is 24. Find the integers., , 8., , Find the length and breadth of the rectangle given below if its perimeter is 72m., , SC, , ER, , T, , 3., , 5x + 4, x–4, , 9., , Length of a rectangle exceeds its breadth by 4 m. If the perimeter of the rectangle is 84 m,, find its length and breadth., , Free distribution by T.S. Government 2018-19, , SIMPLE EQUATIONS, , 67
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After 15 years, Hema's age will become four times that of her present age. Find her, present age., , 11., , A sum of .3000 is to be given in the form of 63 prizes. If the prize money is either .100, or .25. Find the number of prizes of each type., , 12., , A number is divided into two parts such that one part is 10 more than the other. If the two, parts are in the ratio 5:3, find the number and the two parts., , 13 ., , Suhana said, “multiplying my number by 5 and adding 8 to it gives the, same answer as subtracting my number from 20”., Find Suhana's numbers., , 14., , The teacher tells the class that the highest marks obtained by a student in, her class is twice the lowest marks plus 7. The highest mark is 87. What is the lowest, mark?, , 15., , In adjacent figure find the magnitude of each of the three, angles formed?, , LA, N, , G, , AN, A, , 10., , (Hint: Sum of all angles at a point on a line is 180o), 16., , Solve the following riddle:, I am a number, , TE, , Tell my identity., , Take me two times over, , And add a thirty six., , T, , To reach a century, , ER, , You still need four., Looking Back, , Simple equations help in solving various problems in daily life., , •, , For balancing an equation we, , SC, , •, , (i) add the same number on both the sides or, (ii) subtract the same number form both the sides or, (iii) multiply both sides with the same number or, (iv) divides both the sides by the same number, so that the equality remains, undisturbed., , •, 68, , An equation remains same if the LHS and the RHS are interchanged., , VII CLASS MATHEMATICS
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4, , LINES AND ANGLES, Introduction, , AN, A, , 4.0, , You have learnt some geometrical ideas in previous classes. Let us have fun trying some thing we, have already done., Exercise - 1, , B, , D, , C, , (i), , (ii), , P, , (iii), , (iv), , TE, , Draw the figures for the following., , JJJG, (i) OP, 3., , X, , Y, , A, , 2., , G, , Name the figures drawn below., , LA, N, , 1., , (ii) Point X, , HJJG, (iii) RS, , (iv) CD, , Name all the possible line segments in the figure., , T, , A, , B, , C, , D, , Write any five examples of angles that you have observed arround., Example : The angle formed when a scissor is opened., , 5., , Identify the following given angles as acute, right or obtuse., , SC, , ER, , 4., , (i), , (ii), , (iv), Free distribution by T.S. Government 2018-19, , (iii), , (v), LINES AND ANGLES, , 69
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6., , Name all the possible angles you can find in the following figure. Which are acute, right,, obtuse and straight angles?, E, , F, , D, C, B, , A, , Which of the following pairs of lines are parallel? Why?, , (ii), , (iii), , (iv), , LA, N, , (i), , G, , 7., , AN, A, , O, , Which of the following lines are intersecting?, , TE, , 8., , (i), 4.1, , (ii), , (iii), , (iv), , Learning about Pairs of Angles, , ER, , T, , We have learnt how to identify some angles in the previous chapter. Now we will learn about, some more angles as well as various pairs of angles., 4.1.1, , Complementary Angles, , SC, , When the sum of two angles is equal to 90o, the angles are called complementary angles., , o, , 30, , 60o, , These are complementary angles, as their sum is 300 + 600 = 900., o, , We can also say that the complement of 300 is 600 and the complement of 600 is 30 ., , 70, , VII CLASS MATHEMATICS
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70o, o, , 40, , AN, A, , In the above figures, the sum of the two angles is 700 + 400 ≠ 900. Thus, these angles are not a pair, of complementary angles., Try This, , Draw five pairs of complementary angles of your choice., , Do This, , LA, N, , G, , Draw an angle ∠AOB = 40 0. With the same vertex ‘O’ draw, JJJG, ∠BOC = 500, taking OB as initial ray as shown in the figure., , C, , Since the sum of these angles is 90o, they together form a right angle., Take another pair 600 and 500 and join in the same way. Do they form, complementary angles? Why? Why not?, , O, , B, 50o o, 40, A, , Which of the following pairs of angles are complementary?, , T, , 1., , TE, , Exercise - 2, , 90, , 80, , o, , o, 10o, , o, , ER, , 30, , SC, , 110o, , 2., , 90o, , (i), , (ii), , (iii), , Find the complementary angles of the following., , (i) 250, , (ii) 400, , (iii) 890, , (iv) 550, , 3., , Two angles are complement to each other and are also equal. Find them., , 4., , Manasa says, “Each angle in any pair of complementary angles is always acute”. Do you, agree? Give reason., , Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 71
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4.1.2, , Supplementary Angles, , When the sum of two angles are equal to 1800, then the angles are called supplementary angles., , 120o, , AN, A, , 60o, , These are a pair of supplementary angles as their sum is 1200 + 600 = 1800 ., , We say that the supplement of 1200 is 600 and the supplement of 600 is 1200., , G, , 130o, , o, , LA, N, , 100, , 1300 and 1000 angles are not a pair of supplementary angles. Why?, Do This, , TE, , Draw an angle ∠AOB = 1000. With the same vertex O, draw∠BOC = 800, JJJG, such that OB is common to two angles., , ER, , T, , B, , 80o, C, , 100o, O, , A, , You will observe that the above angles form a straight angle that is 1800., , SC, , Thus, the angles 1000 and 800 are supplementary to each other., , Are 1300 and 700 supplementary angles? Why? Why not?, , Try This, Write any five pairs of supplementary angles of your choice., , 72, , VII CLASS MATHEMATICS
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Exercise - 3, 1., , Which of the following pairs of angles are supplementary?, o, , 50, o, , 90, , (i), 2., , 140, , o, , 90, , (ii), , (iii), , Find the supplementary angles of the given angles., (i) 1050, , (ii) 950, , o, , AN, A, , o, , 70, , o, , 110, , (iii) 1500, , (iv) 200, , Two acute angles cannot form a pair of supplementary angles. Justify., , 4., , Two angles are equal and supplementary to each other. Find them., , 4.1.3, , Adjacent Angles, , LA, N, , G, , 3., , The angles having a common arm and a common ‘vertex’ are called as adjacent angles., C, 2, O, , O, , TE, , B, , X, , Y, , 1, , A, , (i), , (ii), , Z, , ER, , T, , The angles ∠AOB and ∠BOC in Figure (i) are adjacent angles, as they have a common vertex, JJJG, ‘O’ and common arm OB ., , SC, , Are the angles in Figure (ii) adjacent angles? Which is the common vertex and which is the, common arm?, P, Q, , O, , Now, look at figure (iii)., R, S, , (iii)., , Are ∠POQ and ∠ROS adjacent angles. Why? Why not?, Which angles are adjacent to each other in the adjacent figure?, Why do you think they are adjacent angles?, Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 73
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Exercise - 4, 1., , Which of the following are adjacent angles?, , f, c, , b, , d, , (i), , (ii), , (iii), , Name all pairs of adjacent angles in the figure. How many pairs of adjacent angles are, formed? Why these angles are called adjacent angles?, B, , C, , O, , D, , LA, N, , A, , G, , 2., , e, , AN, A, , a, , 3., , Can two adjacent angles be supplementary? Draw figure., , 4., , Can two adjacent angles be complementary? Draw figure., , 5., , Give four examples of adjacent angles in daily life., , TE, , Example : Angles between the spokes at the centre of a cycle wheel., (i) ___________________, , (ii) ___________________, , (iii) __________________, , (iv) ___________________, , ER, , T, , 4.1.3 (a) Linear Pair, , P, , C, , SC, , A, , O, , (i), , O, R, , B, , Q, , (ii), , Look at Figure (i). ∠AOC and ∠BOC are adjacent angles. What is the sum of these angles?, , These angles together form a straight angle. Similarly, look at Figure (ii). Do ∠POQ and ∠ROQ, together form a straight angle?, A pair of adjacent angles whose sum is a straight angle (or 1800) is called a Linear Pair., 74, , VII CLASS MATHEMATICS
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Do This, Two adjacent angles are 40o and 140o. Do they from a linear pair?, , 140o, , AN, A, , Draw a picture and check. Renu drew the picture like this., , 40o, O, , Has she drawn correctly? Do these adjacent angles form a linear pair?, Exercise - 5, , G, , Draw the following pairs of angles as adjacent angles. Check whether they form linear pair., , 135o, , 45, , o, , (i), , o, , 55, , o, , 140, , o, , o, , 90, , 90, , (ii), , (iii), , TE, , Niharika took two angles- 1300 and 500 and tried to check whether they form a linear, pair. She made the following picture., , 50o, 130, , o, , ER, , T, , 2., , LA, N, , 1., , SC, , Can we say that these two angles do not form a linear pair? If not, what is Niharika’s mistake?, 4.1.4, , Vertically Opposite Angles, , When two lines intersect, the angles that are formed opposite to each other at the point of intersection (vertex) are called vertically opposite angles., l, 4 O, , In above figure two lines ‘l’and ‘m’ intersect each other at ‘O’., Angle ∠1 is opposite to angle ∠3 and the other pair of opposite, angles is ∠2 and ∠4. Thus, ∠1, ∠3 and ∠2, ∠4 are the two, pairs of vertically opposite angles., , m, , Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 1, , 3, , 2, , 75
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e, , d, , E, , D, , Which are the vertically opposite angles in the adjacent figure., f, , O, , F, A, , c, , B, , AN, A, , a, , C, , Do This, , b, , HJJG, HJJG, Draw two lines AB and CD such that they intersect at point ‘O’., Trace the figure given below on a tracing paper. Place the traced figure, , G, , over the figure given below and rotate it such that ∠BOD coincides ∠AOC., Observe the angles ∠AOD and ∠BOC also ∠AOC and ∠BOD., , C, , LA, N, , You will notice that ∠AOD = ∠BOC and ∠AOC = ∠BOD., , B, , 3, , 2, , O1, , TE, , 4, , A, , D, , We can conclude that vertically opposite angles are equal., , Exercise - 6, , Name two pairs of vertically opposite angles in the figure., , SC, , 1., , ER, , T, , Note : Take two straws. Fix them at a point ‘O’ with a pin. Place them such that the straw on top, covers the one below. Rotate the straws. You will find that they make vertically opposite angles., , C, , A, , D, , 76, , VII CLASS MATHEMATICS, , B, , O, , E
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2., , Find the measure of x, y and z without actually measuring them., y, z, , x, o, , 160, , Give some examples of vertically opposite angles in your surroundings., , 4.2, , Transversal, , AN, A, , 3., , LA, N, , G, , You might have seen railway track. This figure is an example when two lines are intersected by a, transversal., , A line which intersects two or more lines at distinct points is called a transversal., n, , b, , p, , q, , a, , r, , Figure (ii), , Figure (iii), , two lines ‘l’ and ‘m’ are intersected by a line ‘n’, at two distinct points., , ER, , In Fig (i), , T, , m, , TE, , l, , Figure (i), , c, , s, , Therefore, ‘n’ is a transversal to ‘l’ and m”., , In Fig (ii), , three lines ‘p’, ‘q’ and ‘r’ are intersected by a line ‘s’, at three distinct points., , SC, , So, ‘s’ is a transversal to ‘p’ ‘q’ and ‘r’., , In Fig (iii) two lines ‘a’ and ‘b’ are intersected by a line ‘c’. The point of intersection of ‘c’ is the, same as that of ‘a’ and ‘b’. The three lines are thus intersecting lines and none of them, is a transversal to the other., Try This, How many transversals can be drawn for two distinct lines?, , Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 77
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4.2.1 Angles made by a transversal, p, , When a transversal cuts two lines, 8 angles are formed. This is because at, each intersection 4 angles are formed. Observe the figure., , 1 2, , Here ‘l’ and ‘m’ are two lines intersected by the transversal ‘p’. Eight, angles ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 are formed., , 4 3, 5 6, , m, , 8 7, , AN, A, , Angles ∠3, ∠4, ∠5 and ∠6, are lying inside ‘l’ and ‘m’., , l, , They are thus called interior angles. The angles ∠1, ∠2, ∠7 and ∠8 are on the outside of the lines, ‘l’ and ‘m’. They are thus called exterior angles., Look at adjacent figure., , ∠3, ∠4, ∠5 and ∠6 are interior angles., , G, , ∠1, ∠2, ∠7 and ∠8 are exterior angles., , LA, N, , We have learnt about vertically opposite angles and noted the fact that they are equal., Renu looked at figure for vertically opposite angles, and said ∠1 = ∠3 and ∠2 = ∠4., Which are the other two pairs of vertically opposite angles?, , TE, , She said that each exterior angle is paired with an vertically opposite angle which is in the interior., The angles in these pairs are equal. Do you agree with Renu?, Do This, , Identify the transversal in Figure (i) and (ii)., , T, , 1., , ER, , Identify the exterior and interior angles and fill the table given below:, , SC, , a, d, , Figure, , l, , b, , c, , e, , h, , 1 2, , m, , f, g, , B, 8, p, , Figure (i), , 7, , 3 4, 6 5A, , r, q, , Figure (ii), , Transversal, , (i), (ii), 78, , O, , VII CLASS MATHEMATICS, , Exterior angles, , Interior angles
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2., , Consider the following lines. Which line is a transversal? Number and list all the angles, formed. Which are the exterior angles and which are the interior angles?, p, , d, a, , l, m, n, c, , AN, A, , b, , 4.2.1 (a) Corresponding Angles, Look at figures (i), (ii), (iii) and (iv) below1, , G, , 2, , 5, , 3, , 4, , 6, , 7, , (i), , (ii), , LA, N, , 8, , (iii), , (iv), , TE, , Consider the pairs of angles (∠1, ∠5), (∠2, ∠6), (∠4, ∠8), (∠3, ∠7). Is there something, common among these pairs of angles? These angles lie on different vertices. They are on the same, side of the transversal and in each pair one is an interior angle and the other is an exterior angle., Each of the above pair of angles is called corresponding angles., What happens when a line is transversal to three lines? Which are, , T, , the corrosponding angles in this case? What is the number, , ER, , of exterior and interior angles in this case?, What would happen if number of lines intersected by the transversal becomes 4,5 and more?, , SC, , Can you predict the number of exterior and the interior angles that are corrosponding to each, other., 4.2.1 (b) Interior and Exterior Alternate Angles, Look at the adjacent figure. Find the angles which have the following three properties:, (i), , Have different vertices., , (ii), , Are on the either side of the transversal, , (iii), , Lie ‘between’ the two lines (i.e are interior angles)., , 12, 4 3, 5 6, 8 7, , Such pairs of angles are called interior alternate angles., Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 79
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4, , 3, , 6, , 5, , AN, A, , The pairs of angles (∠3, ∠5) and (∠4, ∠6) are the two pairs of interior alternate angles., Similarly, you may find two pairs of exterior alternate angles., 1, , G, , 2, , 7, , LA, N, , 8, , The pairs of angles (∠2, ∠8) and (∠1, ∠7) are called alternate exterior angles., 4.2.1 (c) Interior Angles on the same side of the transversal, , TE, , Interior angles can be on the same side of the transversal too., , 4, 5, , T, , 3, 6, , ER, , Angles ∠4, ∠5 and ∠3, ∠6 are the two pairs of interior angles on the same side of the transversal., Do This, , Name the pairs of angles in each figure by their property., , SC, , 1., , 1, , 5, , 3, , 2, , 4, , (i), 80, , (ii), VII CLASS MATHEMATICS, , 6, , 7 8, , (iii), , (iv), , (v)
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4.2.2, , Transversal on parallel lines, , You know that two coplanar lines which do not intersect are called parallel lines. Let us look at, transversals on parallel lines and the properties of angles on them., , G, , These give examples for parallel lines with a transversal., , AN, A, , Look at the pictures of a window and a graph paper., , Do This, Draw a transversal ‘p’ on these lines., , LA, N, , Take a ruled sheet of paper. Draw two lines ‘l’ and ‘m’ parallel to each other., Label the pairs of corresponding angles as shown in Figures (i), (ii), (iii) and (iv)., P, 1, , TE, , l, , 2, , m, , T, , Figure (i), , ER, SC, , 6, , Figure (iii), , m, , Figure (ii), , P, , l, , m, , l, , 3, , 4, , P, , 5, , P, , l, , 7, , m, , 8, Figure (iv), , Place the tracing paper over Figure (i). Trace the lines ‘l’, ‘m’ and ‘p’. Slide the tracing paper, along ‘p’, until the line ‘l’ coincides with line ‘m’. You find that ∠1 on the traced figure coincides, with ∠2 of the original figure., Thus ∠1 = ∠2, Are the remaining pairs of corrosponding angles equal? Check by tracing and sliding., Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 81
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You will find that if a pair of parallel lines are intersected by a transversal then the angles, in each pair of corresponding angles are equal., We can use this ‘corresponding angles’ property to get another result., In the adjacent figure ‘l’ and ‘m’ are a pair of parallel lines and ‘p’ is a transversal., P, , As all pairs of corresponding angles are equal,, But ∠1 = ∠3 (vertically opposite angles), , 2, , 4, , 3, , 5, , Thus, ∠3 = ∠5, Similarly, you can show that ∠4 = ∠6 ., , l, , AN, A, , 1, , ∠1 = ∠5, , 8, , 6, , G, , Therefore, if a pair of parallel lines are intersected by a transversal then the angles in each pair of alternate interior angles, are equal., , P, 1, , LA, N, , Do you find the same result for exterior alternate angles? Try., , Now, we find one more interesting result about interior angles on, the same side of the transversal., , 2, , 4, 5, 8, , l, , 3, 6, , m, , 7, , TE, , In the adjacent figure ‘l’ and ‘m’ a pair of parallel lines intersected, by a transversal ‘p’., , m, , 7, , ∠ 3 = ∠ 5 ( alternate interior angles), But ∠ 3 + ∠ 4 = 1800 (Why?), Thus, ∠ 4 + ∠ 5 = 1800, , T, , Similarly ∠ 3 + ∠ 6 = 1800 (Give reason), , ER, , Thus, if a pair of parallel lines are intersected by a transversal then the angles in each, pair of interior angles on the same side of the transversal are supplementary., , SC, , Example 1 : In the figure given below, ‘l’ and ‘m’ are a pair of parallel lines., ‘p’ is a transversal. Find ‘x’., , Solution :, , Given l || m, p is a transversal., ∠ x and 200 are a pair of exterior, alternate angles, therefore are, equal., Thus, ∠ x = 200 ., , 82, , VII CLASS MATHEMATICS, , l, o, , 20, , x, , m
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Do This, , r, , n, 1 2, , 1 2, , l, , 3, , 5 6, 8, , 3, , 4, 5 6, , m, , q, , AN, A, , 4, , p, , 8, , 7, , Figure (i), , 7, , Figure (ii), , LA, N, , G, , Trace the copy of figures (i) and (ii) in your note book. Measure the angles using a protractor and, fill the table., , Table 1 : Fill the table with the measures of the corresponding angles., , Pairs of corresponding angles, , st, , 1 pair, , nd, , th, , 3 pair, , 4 pair, , ∠2 = ........., , ∠3 = ........., , ∠4 = ........., , ∠6 = ........., , ∠7 = ........., , ∠8 = ........., , ∠1 = ........., , ∠2 = ........., , ∠3 = ........., , ∠4 = ........., , ∠5 = ........., , ∠6 = ........., , ∠7 = ........., , ∠8 = ........., , ∠1= ........., , SC, , ER, , ∠5 = ........., (ii), , rd, , 2 pair, , T, , (i), , TE, , Figure, , Find out in which figure the pairs of corresponding angles are equal?, What can you say about the lines ‘l’ and ‘m’ ?, What can you say about the lines ‘p’ and ‘q’ ?, Which pair of lines is parallel?, Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 83
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Thus, when a transversal intersects two lines and the pair of corresponding angles are, equal then the lines are parallel., Table 2 : Fill the table with the measures of the interior alterate angles., Pairs of interior alternate angles, st, , (ii), , 2 pair, , ∠3 = ........., , ∠4 = ........, , ∠5 = ........., , ∠6 = ........, , ∠3 = ........., , ∠4 = ........, , ∠5 = ........., , ∠6 = ........, , G, , (i), , nd, , 1 pair, , AN, A, , Figure, , LA, N, , Find out in which figure the pairs of interior alternate angles are equal?, What can you say about the lines ‘l’ and ‘m’ ?, What can you say about the lines ‘p’ and ‘q’?, , TE, , Thus, if a pair of lines are intersected by a transversal and the alternate interior angles, are equal then the lines are parallel., Table 3 : Fill the table with the measures of interior angles on the same side of the transversal, Pairs of interior angles on the same side of the transversal., , T, , Figure, , st, , ER, , 1 pair, , (i), , ∠3= ........., , ∠3+∠6 = ........., , ∠6= ........., , ∠3= ........., , SC, , (ii), , nd, , 2 pair, ∠4= ........., , ∠4+∠5= ........., , ∠5= ........., ∠3+∠6 = ........., , ∠6= ........., , ∠4= ........., , ∠4+∠5= ........., , ∠5= ........., , In which figure the pairs of interior angles on the same side of the transversal are supplementary, (i.e sum is 1800 )?, What can you say about the lines ‘l’ and ‘m’ ?, What can you say about the lines ‘p’ and ‘q’ ?, 84, , VII CLASS MATHEMATICS
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Thus, if a pair of lines are intersected by a transversal and the interior angles on the, same side of the transversal are supplementary then the lines are parallel., Example 2 : In the figure given below, two angles are marked as 300 each., Is AB || CD? How?, 30, , o, , B, , o, , 30, , C, , The given angles form a pair of interior alternate angles with transversal BC ., , LA, N, , As the angles are equal, AB || CD ., , G, , Solution :, , D, , AN, A, , A, , Exercise - 7, , 2., , Fill up the blanks-, , The line which intersects two or more lines at distinct points is called _________, , (ii), , If the pair of alternate interior angles are equal then the lines are _____________, , (iii), , The sum of interior angles on the same side of the transversal are supplementary, then the lines are ___________, , (iv), , If two lines intersect each other then the number of common points they have, _____________., , T, , TE, , (i), , ER, , 1., , In the adjacent figure, the lines ‘l’ and ‘m’ are parallel and ‘n’ is a transversal., , SC, , Fill in the blanks for all the situations given below(i), , If ∠ 1 = 800, , then, , n, l, , ∠ 2 = ______________, , (ii), , If ∠ 3 = 450, , then, , ∠ 7 = ______________, , (iii), , If ∠ 2 = 900, , then, , ∠ 8 = ______________, , (iv), , If ∠ 4 = 1000 then, , ∠ 8 = ______________, , Free distribution by T.S. Government 2018-19, , 8, 7, , 1, , m, 2, , 3, 6, 4, 5, , LINES AND ANGLES, , 85
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3., , Find the measures of x,y and z in the figure, where l ll BC, A, x, , l, , 45o, , o, , o, , y, , B, , 4., , o, , z, , C, , AN, A, , 75o, , ABCD is a quadrilateral in which AB ll DC and AD ll BC. Find ∠ b, ∠ c and ∠ d., D, , B, , In a given figure, ‘l’ and ‘m’ are intersected by a transversal ‘n’. Is l || m?, , LA, N, , 5., , b, , 50o, , G, , A, , C, , c, , d, , n, , P, , 100o, , Q, , A, , B, , R, , C 80, , m, , D, , o, , l, , Find ∠a, ∠b, ∠c, ∠d and ∠e in the figure? Give reasons., , T, , 6., , TE, , S, , b, , e, d, , o, , 50, , ER, , a, , c, , o, , 50, , Note: Two arrow marks pointing in the same direction represent parallel lines., Looking Back, , SC, , 1.(i) If the sum of two angles is equal to 90o , then the angles are called, complementary angles., (ii)Each angle in a pair of complementary angles is acute., , 2.(i) If the sum of two angles is equal to 180o, then the angles are called, supplementary angles., (ii)Each angle in a pair of supplementary angles may be either acute or right or, obtuse., (iii) Two right angles always supplement to each other., , 86, , VII CLASS MATHEMATICS
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3., , The angles formed on both sides of a common arm and a common vertex are, adjacent angles., , 4., , A pair of complementary angles or a pair of supplementary angles need not be, adjacent angles., , 5., , A pair of angles that are adjacent and supplementary form a linear pair., , AN, A, , 6.(i) When two lines intersect each other at a point (vertex), the angles formed, opposite to each other are called vertically opposite angles., (ii)A pair of vertically opposite angles are always equal in measure, , 7.(i) A line which intersects two or more lines at distinct points is called a transversal, to the lines., , 2, , LA, N, , 1, , 3, , 4, , 5, 8, , Types of angles, , ∠3, ∠4, ∠5, ∠6, , —-, , ∠1, ∠2, ∠7, ∠8, , Exterior angles, , 3., , Vertically opposite angles 4 pairs, , ER, , T, , 2., , 7., , 8., , (∠1, ∠3) ; (∠4, ∠2 ); (∠5, ∠7) ; (∠8, ∠6), , Corresponding angles, , 4 pairs, , (∠1, ∠5) ; (∠2, ∠6 ); (∠4, ∠8) ; (∠5, ∠7), , Alternate interior angles, , 2 pairs, , (∠3, ∠5); (∠4, ∠6), , Alternate exterior angles, , 2 pairs, , (∠1, ∠7) ; (∠2, ∠8), , Interior angles on the, 2 pairs, same side of transversal, , (∠3, ∠6) ; (∠4, ∠5), , SC, , 6., , Angles, , —-, , Interior angles, , 5., , 7, , No.of, Pairs, , 1., , 4., , 6, , TE, , S.No., , G, , (ii) A transversal makes eight angles with two lines as shown in the adjacent figure., , When a transversal intersects a pair of parallel lines, the angles in, , (i) Each pair of corresponding angles are equal., (ii) Each pair of alternate interior angles are equal., (iii) Each pair of alternate exterior angles are equal., (iv) Each pair of interior angles on the same side of the transversal are supplementary., Free distribution by T.S. Government 2018-19, , LINES AND ANGLES, , 87
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5, , TRIANGLE AND ITS PROPERTIES, Introduction, , AN, A, , 5.0, , LA, N, , G, , You have been introduced to triangles in your previous class. Look at the figures given below., Which of these are triangles?, , (i), , (ii), , (iii), , (iv), , Discuss with your friends why you consider only some of these as triangles., A triangle is a closed figure made up of three line segments., , TE, , In ΔPQR, the:, , Three sides are PQ, QR, RP, , (ii), , Three angles are ∠PQR, ∠QRP, ∠RPQ, , (iii), , Three vertices are P, Q, R, , Q, , R, , T, , (i), , P, , ER, , The side opposite to vertex P is QR Can you name the sides which are opposite to vertices Q and R?, Likewise, the side opposite to ∠QPR is QR . Can you name the side which is opposite to ∠PQR, , SC, , Try This, , Uma felt that a triangle can be formed with three collinear points. Do you agree? Why?, Draw diagrams to justify your answer., [If three or more points lie on the same line, then they are called collinear points], , Note: LM = Length of Line segment of LM ;, JJJG, LM = Ray LM, , 88, , VII CLASS MATHEMATICS, , LM = Line segment LM, , ;, , HJJG, LM = Line LM
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5.1, , Classification of triangles, , Triangles can be classified according to properties of their sides and angles., Based on the sides, triangles are of three types:, A triangle having all three sides of equal length is called an Equilateral Triangle., , •, , A triangle having two sides of equal length is called an Isosceles Triangle., , •, , If all the three sides of a triangle are of different length, the triangle is called a Scalene Triangle., , AN, A, , •, , Based on the angles, triangles are again of three types:, , A triangle whose all angles are acute is called an acute-angled triangle., , •, , A triangle whose one angle is obtuse is called an obtuse-angled triangle., , •, , A triangle whose one angle is a right angle is called a right-angled triangle., , LA, N, , G, , •, , Triangles, , TE, , T, , 3c, m, , ER, , 40o, , o, , 50, , o, , 30, , 5cm, , 5cm, , Isosceles, triangle, , Scalene triangle, , 110o, o, , 70, , 60, , o, , o, , 30, , o, , 90 60o, , Acute-angled Right-angled Obtuse-angled, triangle, triangle, triangle, , SC, , Equilateral, triangle, , m, 3c, , m, 3c, , 3cm, , Based on angles, , m, 4c, , 3cm, , 3cm, , Based on sides, , Do This, , 1. Classify the following triangles according to their (i) sides and (ii) angles., T, L, , B, , 8 cm, , C, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , N, , 6 cm, , E, , 7 cm, , 8 cm, , 8c, m, , 10, cm, , 5c, , m, , A, , M, , 7 cm, , TRIANGLE AND ITS PROPERTIES, , N, , 89
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P, , F, , 6 cm, , T, , Q, , 6 cm, , R, , AN, A, , 5 cm, , cm, , S, , 6, , H, , 5.2 cm, , 7c, m, , m, 5c, , G, , cm, 5.2, , 5.2, cm, , R, , (2) Write the six elements (i.e. the 3 sides and 3 angles) of ΔABC., (3) Write the side opposite to vertex Q in ΔPQR., , (4) Write the angle opposite to side LM in Δ LMN., , G, , (5) Write the vertex opposite to side RT in ΔRST., , LA, N, , If we consider triangles in terms of both sides and angles we can have the following types of, triangles:, Type of Triangle, , Equilaterial, , T, , Right-angled, , Scalene, , TE, , Acute-angled, , Isosceles, , ER, , Obtuse-angled, , SC, , Try This, , 90, , 1., , Make paper-cut models of the various types of triangles discussed above., Compare your models with those of your friends., , 2., , Rashmi claims that no triangle can have more than one right angle. Do you agree, with her. Why?, , 3., , Kamal claims that no triangle can have more than two acute angles. Do you, agree with him. Why?, , VII CLASS MATHEMATICS
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5.2, , Relationship between the sides of a triangle, , 5.2.1 Sum of the lengths of two sides of a triangle, Draw any three triangles say ΔABC, ΔPQR and ΔXYZ as given below:, P, , B, , Z, , R, , Q, , C, , AN, A, , A, , X, , Y, , ΔPQR, , Is this true?, , AB =, , AB+BC =, , BC =, , BC+CA =, , BC + CA > AB, , CA=, , CA+AB =, , CA + AB > BC, , PQ =, , PQ+QR =, , PQ + QR > RP, , QR+RP =, , QR + RP > PQ, , RP+PQ =, , RP + PQ > QR, , XY=, , XY+YZ =, , XY+ YZ > ZX, , YZ=, , YZ+ZX =, , YZ + ZX > XY, , ZX, , ZX+XY =, , ZX + XY > YZ, , QR =, , ER, , T, , RP =, ΔXYZ, , Sum of two sides, , Yes/No, , AB + BC > CA, , LA, N, , ΔABC, , Sides of Δ, , TE, , Name of Δ, , G, , Use your ruler to find the lengths of their sides and tabulate your results as follows:, , SC, , We can see that in all the above examples, the sum of the lengths of any two sides of a triangle is, greater than the length of the third side., For eg., , In ΔABC,, , AB + BC > CA, BC + CA > AB, CA + AB > BC, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , TRIANGLE AND ITS PROPERTIES, , 91
Page 100 : 5.2.2 Difference between the lengths of two sides of a triangle, Take the same triangles as in the above example and tabulate your results as follows:, Name of Δ Length of sides, , Difference between two sides, , ΔABC, , AB =, , BC–CA =, , BC – AB <AC, , BC =, , CA–AB =, , CA – AB <BC, , CA=, , AB–BC =, , PQ =, , QR–RP =, , QR =, , RP–PQ =, , RP =, , PQ–QR =, , XY=, , YZ–ZX =, , YZ =, , ZX–XY =, , ZX =, , XY–YZ =, , AN, A, QR – RP <PQ, RP – PQ <QR, , PQ – QR <RP, YZ – ZX <XY, , G, , ΔXYZ, , Yes/No, , AB – BC <CA, , ZX – XY <YZ, , LA, N, , ΔPQR, , Is this true?, , XY – YZ <ZX, , From these observations we can conclude that the difference between the lengths of any two sides, of a triangle is less than the length of the third side., BC – AB < CA, , BC – CA < AB ;, , CA – BC < AB, , CA – AB < BC ;, , AB – CA < BC, , T, , Try This, , AB – BC < CA ;, , TE, , For eg. In ΔABC,, , ER, , The lengths of two sides of a triangle are 6 cm and 9 cm. Write all the possible lengths, of the third side., Can a triangle have sides with lengths 6 cm, 5 cm and 8 cm?, , Solution :, , Let the sides of the triangle be AB = 6 cm, , SC, , Example 1:, , BC = 5 cm, CA = 8 cm, , Sum of any two sides i.e, AB + BC = 6 + 5 = 11 > 8, BC + CA = 5 + 8 = 13 > 6, CA + AB = 8 + 6 = 14 > 5, , Since, the sum of the lengths of any two sides is greater than the lenght of the, third side. The triangle is possible., 92, , VII CLASS MATHEMATICS
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Exerciese - 1, Is it possible to have a triangle with the following sides?, (i), , 5.3, , 3 cm, 4 cm and 5 cm., , (ii), , 6 cm, 6 cm and 6 cm., , (iii) 4 cm, 4 cm and 8 cm., , (iv), , 3 cm, 5 cm and 7 cm., , Altitudes of a triangle, , AN, A, , 1., , In your daily life you might have come across the word 'height' in different situations. How will you, measure the height of different figures given below:, , h, , h, , G, , h, , LA, N, , h, , You will measure it from the top point of the object to its base as shown in the figures. Let us use, this creteria to measure the height for a triangle., A, In a given Δ ABC, the height is the distance from vertex A to the, , TE, , base BC . However, you can think of many line segments from A, , B, , to BC . Which among them will represent the height?, , C, , D, , The height is given by lenght of the line segment that starts from A and is perpendicular to BC ., , 1., , ER, , Try This, , T, , Thus, the line segment AD is the altitude of the triangle and its length is height. An altitude can be, drawn from each vertex., , (i) Draw altitudes from P to QR for the following triangles. Also, draw altitudes, , SC, , from the other two vertices.( you can use a set squares if needed), , Q, , P, , R, , P, , Q, , P, , R, , Obtuse-angled, Right-angled, (ii) Will an altitude always lie in the interior of a triangle?, , Q, , R, , Acute-angled, , (iii) Can you think of a triangle in which the two altitudes of a triangle are two of its, sides?, Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , TRIANGLE AND ITS PROPERTIES, , 93
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5.4, , Medians of a triangle, A, , Make a paper cut out of ΔABC., Now fold the triangle in such a way that the vertex B falls on, vertex C. The line along which the triangle has been folded will, , B, , C, , AN, A, , intersect side BC as shown in Figure 1. The point of intersection is the mid-point of side BC, which we call D. Draw a line joining vertex A to this mid-point D (as can be seen in Figure 2)., A, , A, , B, , C, , B, , Figure 2, , G, , Figure 1, , C, , D, , Similarly, fold the triangle in such a way that the vertex A falls on vertex C. The line along which the, , LA, N, , triangle has been folded will intersect side AC . The point of intersection is the mid-point of side, , AC . Draw a line joining vertex B to this mid-point, which we call E., Lastly, fold the triangle in such a way that the vertex A falls on vertex B. The line along which the, , TE, , triangle has been folded will intersect side AB . The point of intersection is the mid-point of side, A, AB . Draw a line joining vertex C to this mid-point, which we call F., Centroid, F, , Line segments AD , BE and CF join the vertices of the triangle, , G, , E, , ER, , T, , to the mid-points of the opposite sides. They are called the me- B, C, D, dians of the triangle., You will observe that the three medians intersect each other at a point in the interior of the triangle., This point of concurrent is called the Centroid., Thus, line segments which join the vertex of the triangle to the mid-point of the opposite side are, called medians of the triangle. Their point of concurrent is called the Centroid., Try This, , SC, , Take paper cut outs of right-angled triangles and obtuse-angled triangles and find, their centroid., , 1., , Exercise - 2, A, , In ΔABC, D is the midpoint of BC ., (i) AD is the ____________________, , (ii) AE is the ____________________, 94, , VII CLASS MATHEMATICS, , B, , E D, , C
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2., , Name the triangle in which the two altitudes of the triangle are two of its sides., , 3., 4., 5., , Does a median always lie in the interior of the triangle?, Does an altitude always lie in the interior of a triangle?, (i) Write the side opposite to vertex Y in ΔXYZ., , (iii) Write the vertex opposite to side AC in ΔABC., 5.5, , Properties of triangles, , 5.5.1, , Angle-sum property of a triangle, , AN, A, , (ii) Write the angle opposite to side PQ in ΔPQR., , Let us learn about this property through the following four activities, Activity 1, , LA, N, , 2. Using a scissors, cut out the three angular regions., , G, , 1. On a white sheet of paper, draw a triangle ABC. Using colour pencils mark its angles as, shown., C, 3. Draw a line XY and mark a point 'O' on it., , A, , X, , O, , B, , Y, , TE, , 4. Paste the three angular cut outs adjacent to each other to form one angle at 'O' as shown in the, figure below., X, , O, , Y, , Activity 2, , T, , You will find that three angles now constitute a straight angle. Thus, the sum of the measures of, angles of a triangle is equal to 180o., , ER, , Take a piece of paper and cut out a triangle, say ABC., Make the altitude AM by folding ΔABC., , SC, , Now, fold the three corners such that all the vertices A,B and C touch at M as shown in the, following figures., , (i), (ii), (iii), You will see that all the three angles A, B and C form a straight line and thus, ∠A + ∠B + ∠C = 180o., Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , TRIANGLE AND ITS PROPERTIES, , 95
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Activity 3, , AN, A, , Take three copies of any triangle, say ABC. Mark its angles as 1,2 and 3 as shown below:, , G, , Arrange the triangular cut-outs as shown in the above and in the following figures., , LA, N, , What do you observe about ∠1+ ∠2 + ∠3 at the point ‘O’?, , You will observe that three angles form a straight line and so measure 180o., Activity 4, , TE, , Draw any three triangles, say ΔABC, ΔPQR and ΔXYZ in your note book. Use your protractor, and measure each of the angles of these triangles., Name of the, , Measure of angles, , Triangle, , Sum of the measures of the three angles, , ∠A =....., ∠B =....., ∠C =....., ∠A + ∠B + ∠C =, , ΔPQR, , ∠P =....., ∠Q =....., ∠R =.....,, , ∠P + ∠Q + ∠R =, , ∠X =....., ∠Y =....., ∠Z =.....,, , ∠X + ∠Y + ∠Z =, , ER, , ΔXYZ, , T, , ΔABC, , Allowing marginal errors in measurements, you will find that the sum of the three angles of a triangle, is 180o., , SC, , You are now ready to give a formal justification of your assertion that the sum of the angles of a, triangle is equal to 180o through logical argumentation., , Proof of angle-sum property of a triangle, Statement, , : The sum of the angles of a triangle is 180o, , Given, , : A triangle ABC, , To prove, , : ∠A + ∠B + ∠C = 180o, , HJJG, Construction : Through A draw a line segment PQ parallel to BC., , 96, , VII CLASS MATHEMATICS, , P, , B, , 5, , 2, , A, 1, , Q, , 4, , 3, , C
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Proof :, Mark the angles as indicated in the figure:, = ∠5, , (alternate interior angles), , ∠3, , = ∠4, , (alternate interior angles), , = ∠5 + ∠4, , (adding (1) and (2)), , = ∠1 + ∠5 + ∠4, , (adding ∠1 to both sides), , = 180o, , (angles forming a straight line), , ∠2 + ∠3, ∠1 + ∠2 + ∠3, But ∠1 + ∠5 + ∠4, , AN, A, , ∠2, , Therefore, ∠1 + ∠2 + ∠3 = 180o, ∠A + ∠B + ∠C = 180o., , Thus, the sum of the angles of a triangle is 180o., , G, , ∴, , In ΔABC, ∠A = 30o, ∠B = 45o, find ∠C., , Solution :, , In ΔABC, ∠A + ∠B + ∠C = 180o, , LA, N, , Example 1:, , 30o + 45o + ∠C, , (angle-sum property of a triangle), , = 180o, , (substituting given values in question), , 75o + ∠C = 180o, , TE, , ∠C, , = 180o – 75o, , ∠C = 105o, , Therefore,, Example 2 :, , In ΔABC, if ∠A = 3 ∠B and ∠C = 2 ∠B. Find all the three angles of ΔABC., , Solution :, , ∠A + ∠B + ∠C, , [angle-sum property of a triangle], , T, , = 180o, , [∠A = 3 ∠B, ∠C = 2 ∠B], , ER, , 3 ∠B + ∠B + 2 ∠B = 180o, Therefore,, , ∠B = 30o, , Thus,, , ∠A = 3 ∠B = 3 × 30o = 90o, , SC, , 6 ∠B = 180o, , ∠C = 2 ∠B = 2 × 30o = 60o, , Example 3 :, , ΔABC is right angled at C and CD⊥AB, ∠A = 55o, Find, , Solution :, , B, , (i) ∠ACD, , (ii) ∠BCD, , In ΔACD,, ∠CAD + ∠ADC + ∠ACD, 55o + 90o + ∠ACD, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , D, , (iii) ∠ABC, A, , 55o, , C, , = 180o (angle-sum property of a triangle), = 180o (substituing values given in question), TRIANGLE AND ITS PROPERTIES, , 97
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o, , o, , o, , y = z = 110 (corresponding angles since DE || BC), , Exercise - 3, 1., , AN, A, , (iii), , Find the value of the unknown ‘x’ in the following triangles., A, , P, , o, , x, , 30o, , X, , Q, , 30o, , o, , 110, , B, , 50, , xo, , o, , 60, , R, , C, , (ii), , xo, Z, , (iii), , LA, N, , (i), , 2., , G, , o, , Y, , Find the values of the unknowns ‘x’ and ‘y’ in the following diagrams., o, , 80, R, , o, , x, , yo, , 120o, , R, , 100, , A, , (iii), x, L, , xo, , C 60o, , o, , F, , xo, , xo, , G, , xo, , (v), , E, , T, , (vi), , Find the measure of the third angle of triangles whose two angles are given below:, , (i) 38o, 102o, 4., , T, , 60o, , yo, , (iv), , 3., , o, xo y, N, , o, , y, , E, , o, , SC, , B, , 30o, , o, , 50o, , 90o, , A, , y, , x, , (ii), , ER, , (i), , M, , o, , 50, , S, , T, , Q, , 50o, , TE, , P, , (ii) 116o, 30o, , (iii) 40o, 80o, , In a right-angled triangle, one acute angle is 30o. Find the other acute angle., , VII CLASS MATHEMATICS, , xo
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6., , State true or false for each of the following statements., (i), , A triangle can have two right angles., , (ii), , A triangle can have two acute angles., , (iii), , A triangle can have two obtuse angles., , (iv), , Each angle of a triangle can be less than 60o., , AN, A, , 5., , The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles., , A, , 30o, , In the figure, DE || BC , ∠A = 30o and ∠B = 50o. Find the, values of x, y and z., , xo, , yo, , D, , 8., , LA, N, , G, , 7., , 50o, , B, , zo, , In the figure, ∠ABD = 3 ∠DAB and ∠BDC = 96o. Find ∠ABD., , E, , C, , TE, , B, , A, , o, , 96, , C, , D, , In ΔPQR ∠P= 2 ∠Q and 2 ∠R = 3 ∠Q , calculate the angles of ΔPQR., , 10., , If the angles of a triangle are in the ratio 1 : 4 : 5, find the angles., , 11., , The acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle., , 12., , In the figure, ΔPQR is right angled at Q, ML || RQ and, , ER, , T, , 9., , P, , SC, , ∠LMR = 130o. Find ∠LPM, ∠PML and ∠PRQ., 13., , M, , L, o, , 130, , In Figure ABCDE, find ∠1 + ∠2 + ∠3 + ∠4 + ∠5., A, , R, , Q, , 1, B 2, , 5, , 3, C, , E, , 4, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , D, TRIANGLE AND ITS PROPERTIES, , 101
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5.5.2, , A, , Exterior angle of a triangle, , Draw ΔABC and produce one of its sides say BC as shown, in the Figure 1. Observe the ∠ACD formed at point C. This, angle lies in the exterior of ΔABC. We call it the exterior angle, of ΔABC formed at vertex C., , B, , D, , C, , Can you say that ∠ACD = ∠A + ∠B ?, , G, , Do these two pieces together entirely cover ∠ACD?, , AN, A, , Figure 1, Clearly ∠BCA is an adjacent angle to ∠ACD. The remaining two angles of the triangle namely, A, ∠BAC or ∠A and ∠CBA or ∠B are called the two interior, opposite angles of ∠ACD. Now cut out (or make trace copies of) ∠A and ∠B and place them adjacent to each other as, shown in the Figure 2., B, , D, , C, , Do This, , LA, N, , Figure 2, From the above activity, we can say that an exterior angle of a triangle is equal to the sum of two, interior opposite angles., , Draw ΔABC and form an exterior ∠ACD. Now take a protractor, , TE, , and measure ∠ACD, ∠A and ∠B., , Find the sum ∠A + ∠B and compare it with the measure ∠ACD., Do you observe that ∠ACD is equal (or nearly equal) to ∠A + ∠B?, , T, , A logical step- by- step argument can further confirm that the exterior angle of a triangle is equal to, the sum of the interior opposite angles., : An exterior angle of triangle is equal to the sum of its interior opposite angles., , Given, , : ΔABC with ∠ACD as exterior angle, , A, , To prove, , : ∠ACD = ∠A + ∠B, , 1, , SC, , ER, , Statement, , E, , Construction : Through C draw CE parallel to BA, , Justification, , B, , x, , 2, , o, o, , y, , C, , ∠1 = ∠x ( BA & CE and AC is transversal therefore, alternate angles are equal), ∠2 = ∠y ( BA & CE and BD is transversal therefore, corresponding angles are equal), ∠1 + ∠2 = ∠x + ∠y, Therefore , ∠ACD = ∠1 + ∠2 (from the figure ∠x + ∠y = ∠ACD), 102, VII CLASS MATHEMATICS, , D
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Thus, the exterior angle of a triangle is equal to the sum of the interior opposite angles. This, property is called the exterior-angle property of a triangle., Do This, Copy each of the following triangles. In each case verify that an exterior, P, , B, , C, , R, , Q, , Solution :, , S, , O, , LA, N, , Example 8 : In the figure, find the values of x and y., , Y, , G, , A, , AN, A, , angle of a triangle is equal to the sum of the two interior opposite angles., , A, o, , x, , ∠ACD = ∠ABC + ∠BAC, ( exterior angle property), 135o, , B, , = 65o + xo, , X, , Z, , 65o, , o, y 135, , D, , C, , Therefore,, , TE, , 135o – 65o = xo, xo, , = 70o, , ∠ABC + ∠BAC + ∠BCA = 180o, = 180o, , 135o + yo, , = 180o, , ER, , T, , 65o + 70o + yo, , yo, , Therefore,, , ( angle-sum property of a triangle), , = 180o – 135o, , yo = 45o, , One of the exterior angles of a triangle is 120o and the interior opposite angles are, A, in the ratio 1 : 5. Find the angles of the triangle., , Solution :, , ∠ACD, , SC, , Example 9 :, , =, , 120o (from the question), o, , 120, B, , ∠ACD, , =, , ∠A + ∠B (exterior angle property), , ∠A + ∠B, , =, , 120o, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , TRIANGLE AND ITS PROPERTIES, , D, , C, , 103
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Therefore,, , ∠ABC + yo, , Therefore.,, , = 180o, , (linear pair), , yo, , = 180o – xo, , yo, , = 180o – 44o = 136o, , AN, A, , Also, , 88, = 44o, 2, , xo =, , Example 13 : Find the value of ∠A + ∠B + ∠C + ∠D + ∠E of the following figure., Solution :, , C, , D, , Name the angles as shown in the figure., , In ΔGHC ∠3 + ∠6 + ∠7 = 180o .......(1), , G, , (angle-sum property of triangle), , .......(2), , In ΔAGD, ∠7 = ∠1 + ∠4, , .......(3), , E, , LA, N, , In ΔEHB, ∠6 = ∠5 + ∠2, , A, , C, , (exterior angle property of a triangle), , D, , 3, , Substituting (2) and (3) in (1), , 4, , ⇒, , ∠3 + ∠5 + ∠1 + ∠2 + ∠4, , = 180, , ⇒, , ∠1 + ∠2 + ∠3 + ∠4 + ∠5, , = 180o, , TE, , B, , o, , Therefore, ∠A + ∠B + ∠C + ∠D + ∠E = 180o, , H 6, I, , E, , J, , T, , ER, SC, 2., , A, , B, , C, , Z, , X, , For ΔABC, find the measure of ∠ACD., , A, 66o, , B, , 106, , VII CLASS MATHEMATICS, , 45o, , 1, , F, A, , In ΔABC, name all the interior and exterior angles of the triangle., Y, , C, , G, 2, , 5, , Exercise - 4, , 1., , 7, , D, , B
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3., , Find the measure of angles x and y., C, 30o, , 4., , o, , 65o, , x, , yo, , D, , B, , AN, A, , A, , In the following figures, find the values of x and y., , E, , C, , o, , 107, , 57o, , B, , 5., , o, x 65, , C, , A, , G, , xo, , o, , y, , o, , o, , 107, , 40o, , B, , D, , LA, N, , A, , In the figure ∠BAD = 3∠DBA, find ∠CDB, ∠DBC and ∠ABC., A, , D, , o, , TE, , 104, , o, , 65, , C, , Find the values of x and y in the following figures., , ER, , 6., , T, , B, , x, , o, , SC, , 70, , xo, , y0, , o, , y, , o, , o, , 50, , (i), , 30, , (ii), , xo, , o, , (iii), yo, , 80o, , xo, , o, , 30, 110, , yo, , o, , o, , (iv), Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , yo, , xo, , xo, , (v), , (vi), TRIANGLE AND ITS PROPERTIES, , 107
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One of the exterior angles of a triangle is 125o and the interior opposite angles are in the, ratio 2 : 3. Find the angles of the triangle., , 8., , The exterior ∠PRS of ΔPQR is 105o. If Q = 70o, find ∠P. Is ∠PRS >∠P ?, , 9., , If an exterior angle of a triangle is 130o and one of the interior opposite angle is 60o. Find, the other interior opposite angle., , 10., , One of the exterior angle of a triangle is 105o and the interior opposite angles are in the, ratio 2 : 5. Find the angles of the triangle., , 11., , In the figure find the values of x and y., , AN, A, , 7., , C, o, , 30, , o, , LA, N, , G, , x, , A, , Looking Back, , 50o, , o, , B, , o, y 55, , D, , TE, , 1 (i) A triangle is a simple closed figure made up of three line segments., (ii) Based on the sides, triangles are of three types, A triangle having all three sides of same length is called an Equilateral, Triangle., , •, , A triangle having at least two sides of equal length is called an Isosceles, Triangle., , ER, , T, , •, , •, , If all the three sides of a triangle are of different length, the triangle is called a, Scalene Triangle., , SC, , (iii) Based on the angles, triangles are of three types, •, , A triangle whose all angles are acute is called an acute-angled triangle., , •, , A triangle whose one angle is obtuse is called an obtuse-angled triangle., , •, , A triangle whose one angle is a right angle is called a right-angled triangle., , 2. The six elements of a triangle are its three angles and the three sides., 3. Properties of the lengths of the sides of a triangle:, , 108, , VII CLASS MATHEMATICS
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(i) The sum of the lengths of any two sides of a triangle is greater than the length, of the third side., (ii) The difference between the lengths of any two sides of a triangle is smaller, than the length of the third side., , AN, A, , 4. The line segment joining a vertex of a triangle to the mid-point of its, opposite side is called a median of the triangle. A triangle has 3 medians., 5. The perpendicular line segment from a vertex of a triangle to its opposite, side is called the altitude of the triangle., 6. The total measure of the three angles of a triangle is 180o. This is called the, angle sum property of a triangle., , LA, N, , G, , 7. The measure of any exterior angle of a triangle is equal to the sum of its, interior opposite angles. This is called the exterior angle property of the, triangle., 8. Representation of line, line segment and ray, , TE, , LM = Lenght of Line segment of LM ; LM = Line segment LM, JJJG, HJJG, ; LM = Line LM, LM = Ray LM, , T, , Fun with Card board shapes, , SC, , ER, , Take square card board sheet., Mark the mid points of sides and, draw lines as shown in the figure., Cut the square into four parts and, rearrange them to get a triangle, , Triangle, and Its, Properties, Free distribution, by T.S., Government 2018-19, , TRIANGLE AND ITS PROPERTIES, , 109
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RATIO - APPLICATIONS, , Introduction, , AN, A, , 6.0, , 6, , In your previous class, you have learnt how to use ratio and proportion to compare quantities. In, this class, we will first review our understanding of the same and then learn about ratios expressed, in the form of percentages., 6.1, , Ratio, , Madhuri's weight is 50 kg and her daughter's weight is 10 kg. We say that Madhuri's weight is, 5 times her daughter's weight. We can also say that the daughter's weight is one-fifth of her, mother's weight. Thus, the ratio of Madhuri's weight and her daughter's weight is 50 : 10 or, 5 : 1. Inversely, the ratio of the daughter's weight and the mother's weight is 1 : 5., , •, , In a class there are 60 boys and 40 girls. The number of boys is, , LA, N, , G, , •, , 3, times the number of girls., 2, , TE, , we can also say that the number of girls is two-thirds of the boys. Thus, the ratio of the number, of boys and the number of girls is 60 : 40 or 3 : 2. Inversely, the ratio of number of girls and, number of boys is 2 : 3., , ER, , T, , Anand has a wire of length 100 cm and Rashmi has a wire of length 5 m. Anand said to Rashmi,, “the wire with me is 20 times longer than yours.” You know that this is not true as 5 m is much, longer than 100 cm. The length of Rashmi's wire has been expressed in meters and that of Anand, has been expressed in centimeters. Both have to be expressed in the same units before they are, compared., , SC, , We know that 1 m = 100 cm. So the length of the wire with Rashmi is 5 m = 5 × 100 = 500 cm., Thus, the ratio of Rashmi and Anand's wire is 500 : 100 or 5 : 1. We can also say that the length, of Rashmi's wire is 5 times that of Anand., In all the above examples quantities have been comapred in the form, of ratios. Thus, a ratio is an ordered comparison of quantities of the, same units. We use the symbol ':' to represent a ratio. The ratio of, two quantities a and b is a: b and we read this as "a is to b". The two, quantities 'a' and 'b' are called terms of the ratio. The first quantity 'a', is called first term or antecedent and the second quantity 'b' is called, second term or consequent., 110, , VII CLASS MATHEMATICS
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Try This, Think of some real life situations in which you have to compare quantities in the form, of a ratio., Exercise 1, What is the ratio of ` 100 and `10 ? Express your answer in the simplest form., , 2., , Sudha has ` 5. Money with Radha is 3 times the money with Sudha. How much money, does Radha have?, , AN, A, , 1., , (i) What is the ratio of Radha's money and Sudha's money?, , (ii) What is the ratio of Sudha's money and Radha's money?, , Divide 96 chocolates between Raju and Ravi in the ratio 5 : 7, , 4., , The length of a line segment AB is 38 cm. A point X on it divides it in the ratio 9 : 10. Find, the lengths of the line segments AX and XB., , LA, N, , G, , 3., , X, , A, , B, , A sum of ` 1,60,000 is divided in the ratio of 3 : 5. What is the smaller share?, , 6., , To make green paint, a painter mixes yellow paint and blue paint in the ratio of 3 : 2. If he, used twelve liters of yellow paint, how much blue paint did he use?, , 7., , A rectangle measures 40 cm at its length and 20 cm at its width. Find the ratio of the length, to the width., , 8., , The speed of a Garden-Snail is 50 meters per hour and that of the Cheetah is 120 kilometers per hour. Find the ratio of the speeds., , 9., , Find, , The ratio of boys and girls in your class., , ER, , (i), , T, , TE, , 5., , (ii), , The ratio of number of doors and number of windows of your classroom., , (iii), , The ratio of number of text books and number of note books with you, , SC, , Classroom Project, 1. Take a tape and with the help of your friend measure the length and breadth, of your classroom Find the ratio of length and breadth., 2. Take a ` 10 note. Find its length and breadth. Roundoff the answers to the, nearest whole number, with the help of your teacher, find the ratio of the, length and breadth., Repeat this activity with ` 20 and ` 50 notes and record the lengths in your, note book., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 111
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6.2, , Proportion, , Srilekha’s mother prepares tea by using 2 spoons of tea powder for 1 cup of tea. One day 3 guests, visited their home. How many spoons of tea powder must she use to prepare 3 cups of tea? Yes,, you are right. She uses 6 spoons of tea powder to prepare 3 cups of tea. Here, Srilekha's mother, 4 cm, used the ‘law of proportion’ to solve the problem., Let us see one more example:, , AN, A, , Ravi took a photo. He got the picture developed in a photo lab, in a size 4 cm × 6 cm ., 4 cm, , 6 cm, , G, , He wanted to get the photo enlarged so he went to the photo lab again., The lab-man gave him this photo. In turn Ravi said, “there seems to be, something wrong with this picture”., Do you think, is Ravi right?, Can you say what is wrong with this picture?, , LA, N, , 12 cm, , Ravi decided to measure the length and breadth of the photo. He knew, that the ratio of length and breadth of the original photo should be equal to, the ratio of length and breadth of the enlarged photo., , TE, , Ratio of length and breadth of the original photo = 4 : 6 = 2 : 3, Ratio of length and breadth of the enlarged photo = 4 : 12 = 1 : 3, , T, , Are the two ratios equal? Ravi also realised that the ratio of length and breadth of the enlarged, photo was not equal to that of the original photo. He understood that the second picture was not, 8 cm, proportionate to the first., , ER, , He asked the lab-man to develop another enlarged photo., , 12 cm, , SC, , This time the photo was good. He again measured the length, and breadth and calculated the ratio., Ratio of length and breadth = 8 : 12 = 2 : 3, Now, Ravi understood that the original photo and the new, enlarged photo looked fine to him because the ratios of their, length and breadth were equal i.e., they were in proportion., , Thus, two ratios are said to be in proportion when they are equal. The symbol we use for propotion, is ‘: :’ (is as). If two ratios a : b and c : d are equal, we write a : b = c : d or a : b : : c : d. We read, this as ‘a is to b is proportionate to c is to d’. This can also be read as ‘a is to b is as c is to d’., 112, , VII CLASS MATHEMATICS
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The four quantities a, b, c and d are called first, second, third and fourth terms respectively. The, first and fourth terms are known as extreme terms or extremes. The second and third terms are, known as middle terms or means., a:b=c:d, i.e., Therefore,, , a c, =, b d, , AN, A, , In a proportion,, , ad = bc, , i.e.,, , Means, a : b=c : d, Extremes, , Let us consider some examples, , Solution:, , Find, , to complete the proportion., , (i), , 2:5=6:, , TE, , Example 1 :, , b.c, a, , LA, N, , Here ‘d’ is called the fourth proportional and d =, , G, , Thus, The product of the means is equal to the product of the extremes., , The product of the means is equal to the product of the extremes,, i.e., , 2 : 5 = 6 :, , =5×6, , 2×, , ER, , T, , Therefore,, , SC, , (ii), , =, , 16 : 20 =, , 30, = 15, 2, , : 35, , The product of the means is equal to the product of the extremes,, , i.e., , 16 : 20 =, , Therefore,, , 20 ×, , = 16 × 35, =, , ∴, , : 35, , 560, = 28, 20, , 6 : 20 = 28 : 35, , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 113
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Exercise 2, Find the missing numbers in the following proportions in the table given below ., , (i), , 1:2::4:8, , (ii), , 5 : 6 : : 75 : 90, , (iii), , 3 : 4 : : 24 : 32, , (iv), , 2:5::, , (v), , 3 : 6 : : 12 :, , Product of extremes, , : 15, , 30, , 72, , Write true or false., (i) 15 : 30 : : 30 : 40, (ii) 22 : 11 : : 12 : 6, (iii) 90 : 30 : : 36 : 12, , TE, , (iv) 32 : 64 : : 6 : 12, , Product of means, , AN, A, , Proportion, , LA, N, , 2., , S.No., , G, , 1., , (v) 25 : 1 : : 40 : 160, , Madhu buys 5 kg of potatoes at the market. If the cost of 2 kg is . 36, how much will, Madhu pay?, , 4., , A man whose weight is 90 kgs weighs 15 kg on the moon, what will be the weight of a man, whose weight is 60 kg?, , ER, , 5., , T, , 3, , A disaster relief team consists of engineers and doctors in the ratio of 2 : 5., , SC, , (i) If there are 18 engineers, find the number of doctors., (ii) If there are 65 doctors, find the number of engineers., , 6., , The ratio of two angles is 3 : 1. Find the, (i) larger angle if the smaller is 180o, , 114, , VII CLASS MATHEMATICS, , (ii) smaller angle if the larger is 63o.
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Do This, , Rate, , LA, N, , 6.3, , G, , AN, A, , Enlarge the square and rectangle such that the enlarged square and rectangle, remain proportional to the original square and rectangle., , Sometimes ratios appear as rates. Some examples are given below :, i), , My father drives the vehicle with a speed of 60 km per hour., , ii), , I bought apples at the rate of, , iii), , My heart rate is 72 beats per minute., , iv), , The cost of eggs is 60 per dozen., , v), , The birth rate of India is 21 (approximately). (Birth rate is the number of live births, per thousand people in a given time - Refer: http://www.indexmundi.com/g/, g.aspx?c=in&v=25), , T, , TE, , 120 per kg., , SC, , ER, , In the first example the distance travelled by the vehicle is compared with the time taken. In the, second example cost of apples is compared to the quantity of apples. In the third example the, number of heart beats is compared to the time taken. In the fourth example, the cost of eggs is, compared to the quantity of eggs. In the fifth example, the number of live births is compared to, 1000 poeple., The word per can be replaced with the symbol '/' and the above examples can be written as 60km/, hour, 120/kg, 72 beats/ minute, 60/dozen and 21 births per1000 people., 6.4, , Unitary Method, , The method in which we first find the value of one unit and then the value of the required number of, units is known as unitary method., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 115
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Example 2 :, Solution, , A shopkeeper sells 5 tumblers for 30. What would be the cost of 10 such, tumblers?, Cost of 5 tumblers = 30, 30, Therefore, Cost of 1 tumbler =, = 6, 5, Thus, cost of 10 tumblers = 6 × 10 = 60., , Therefore, cost of 1 banana =, , 20, × 9 = 15, 12, , G, , Thus, cost of 9 bananas =, , 20, 12, , AN, A, , Example 3 : What is the cost of 9 bananas, if the cost of a dozen bananas is ` 20?, Solution, 1 dozen = 12 units., Cost of 12 bananas = 20, , Do This, , 40 benches are required to seat 160 students. How many benches, will be required to seat 240 students at the same rate?, , 2., , When a Robin bird flies, it flaps wings 23 times in ten seconds., How many times will it flap its wings in two minutes?", , 3., , The average human heart beats at 72 times per minute. How many times does it beat in, , LA, N, , 1., , TE, , 15 seconds? How many in an hour? How many in a day?, 6.5, , Direct Proportion, , For example:, , T, , There are various situations in day-to-day life, when a change in one quantity leads to a change in, the other quantity., If the number of things purchased increases, the cost incurred also increases. Alternately, if the, number of things purchased decreases, the cost incurred also decreases., , •, , If the money deposited with a bank increases, the interest earned on that sum also increases., Alternately, if the money in the bank decreases, the interest also decreases., , SC, , ER, , •, , •, , At a constant speed, if the distance travelled increases, the time taken for it also increases., Alternately, if the distance travelled decreases, time also decreases., , In the above examples, when one quantity increases the other also increases and vice-versa. Let, us understand such situations with the help of an example., A tap takes 1 hour to fill 300 litres of a tank. How many litres will be filled up in 2 hours?, The tank will filled up by 600 litres. How many in 4 hours, how many in 8 hours? How do you, make this calculation?, 116, , VII CLASS MATHEMATICS
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Look at the table given below :, ×8, , ×4, ×2, , Capacity filled (lts), , 1, , 2, , 4, , 300, , 600, , 1200, , ×2, , 8, 2400, ×8, , ×4, , AN, A, , Time taken to fill tank (hours), , LA, N, , G, , You will find that in each case above, if the time taken increases the quantity of water filled also, increases such that the ratio of the time taken and the ratio of the quantity filled is same. Thus,, when the time taken doubles, the quantity filled will also doubled; when the time taken is 4 times,, the quantity filled is also four times the original. And when the time taken is 8 times, the quantity, filled is also 8 times. The ratio of the time taken is 1 : 2 and the ratio of quantity filled is also 1 : 2., Thus, we can say that time taken to fill the tank and quantity filled are in direct proportion., Example 4 :, , A shopkeeper sells 6 eggs for, , Solution, , Let the cost of 10 eggs be x., We know that as the number of eggs increases, the cost will also increases such, that the ratio of the number of eggs and the ratio of their costs will remain the, same. In other words, the ratio of the number of eggs and the ratio of the cost of, eggs is in proportion., , TE, , :, , 30. What would be the cost of 10 eggs?, , Thus,, , 6 : 10 = 30 : x, , Since the product of the means is equal to the product of the extremes :, , T, , 6 × x = 10 × 30, , ER, , 6x = 10 × 30, x=, , x=, , 10 × 30, = 50, 6, , 50, , SC, , Thus, the cost of 10 eggs is equal to 50., This problem can be solved by using unitary method too i.e. finding the cost of, one egg and then multiplying the unit cost with the number of eggs required., Cost of 6 eggs is 30, Therefore, cost of 1 egg =, , 30, 6, , =, , 5, , Cost of 10 eggs = 5 × 10 = 50, Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 117
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20 kgs of rice is needed for a family of 4 members. How many kgs of rice will be, required if the number of members in the house increases to 10?, , Method 1 :, , Girija said as the number of members increases, the amount of rice required will, also increase such that the ratio of number of members and the ratio of the amount, of rice is the same. Thus, the number of members and amount of rice are in direct, proportion., , Solution :, , Let x be the amount of rice required for 10 members, Then, , AN, A, , Example 5:, , x : 20 = 10 : 4, , Since the product of the means is equal to the product of the extremes:, , x=, , 20 ×10, = 50, 4, , LA, N, , x = 50 kgs, , G, , 4x = 20 × 10, , ∴ Amount of Rice required for 10 members = 50 Kgs., , Method 2 :, , Sarala used the unitary method to solve this problem :, Amount of Rice required for 4 members =, , 20 kgs., , TE, , Thus, amount of Rice required for one member =, , 20, = 5 kgs., 4, , ∴ Amount of Rice required for 10 members = 10 × 5 = 50 kgs., , A jeep travels 90 km in 3 hours at a constant speed. In how many hours will the, jeep covers 150 kms?, , T, , Example 6 :, , ER, , We know that as the distance travelled increases the time taken will also increases, such that the ratio of the number of km and the ratio of the times taken is the same., Thus, the number of kms and the time taken is directly proportional., , Solution :, , Let x be the number of hours for the jeep to cover 150 kms., , SC, , Thus, x : 3 = 150 : 90, Since the product of the means is equal to the product of the extremes, 90 x = 150 × 3, x=, , 150 × 3, =5, 90, , x=5, ∴ Time taken to cover 150 Km = 5 hours., , 118, , VII CLASS MATHEMATICS
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Example 7 :, , The scale of a map is given as 1:30000. Two cities are 4 cm apart on the map., Find the actual distance between them., , Solution :, , Let the actual distance be x cm. Since the distance on the map is directly proportional to the actual distance,, 1:30000 = 4 : x, , x = 4 × 30, 000, =1,20,000 cm, =1.2 kms, , (1 km = 1,00,000 cm), , AN, A, , Since the product of the means is equal to the product of the extremes, , G, , Thus, two cities, which are 4 cm apart on the map, are actually 1.2 kms away, from each other., , LA, N, , Try This, , Place a 1litre empty bottle under a tap from which water is falling drop by drop, due to leakage. How much time did it take to fill the bottle? Calculate how much, water would be wasted in a year?, , 2., , Take a clock and fix its minutes hand at 12., , Time Passed, , TE, , 1., , (T1), , (T2), , (T3), , (T4), , (in minutes), , 15, , 30, , 45, , 60, , Angle turned, , (A1), , (A2), , (A3), , (A4), , (in degree), , 90, , ...., , ...., , ...., , ER, , T, , Note the angles made by minutes hand in the given intervals of time :, , SC, , Is the angle turned through by the minute hand directly proportional to the time that, has passed? Yes!, From the above table, you can also see, T1 : T2 = A1 : A2, because, T1 : T2 = 15 : 30 = 1 : 2, , A1 : A2 = 90 : 180 = 1 : 2, Check if T2 : T3 = A2 : A3 and T3 : T4 = A3 : A4, You can repeat this activity by choosing your own time interval., Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 119
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Exercise - 3, A length of a bacteria enlarged 50,000 times attains a length of, 5 cm. What is the actual length of the bacteria? If the length is, enlarged 20,000 times only, what would be its enlarged length?, , 2., , Observe the following tables and find if x is directly proportional ., , (iii), , 20, , 17, , 14, , 11, , 8, , y, , 40, , 34, , 28, , 22, , 16, , x, , 6, , 10, , 14, , 18, , 22, , y, , 4, , 8, , 12, , 16, , 20, , x, , 5, , 8, , 12, , 15, , y, , 15, , 24, , 36, , 60, , 5, , 2, , 10, , 4, , 26, , 30, , 24, , 28, , G, , (ii), , x, , 18, , 20, , 25, , 72, , 100, , 125, , LA, N, , (i), , AN, A, , 1., , Sushma has a road map with a scale of 1 cm representing 18 km. She drives on a road for, 72 km. What would be her distance covered in the map?, , 4., , On a Grid paper, draw five squares of different sizes. Write the following information in a, tabular form., , TE, , 3., , Square 1, , Square 2, , Square 3, , Square 4, , Square 5, , Length of a side (L), , Area (A), , T, , Perimeter (P), , ER, , Find whether the length of a side is in direct proportion to:, the perimeter of the square., , (ii), , the area of the square., , SC, , (i), , Ratios also appear in the form of percentages. We will learn about, percentages and the various ways in which we use them in day-to-day life., , 6.6, , Percentages, , •, , Soumya got 65% marks in Mathematics and Ranjeet got 59% marks., , •, , A cloth seller in whole-sale market makes a profit of 25% on silk sarees in the retail-market, makes a profit of 10%., , 120, , VII CLASS MATHEMATICS
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•, , Anita borrowed a loan of ` 10000 from the bank for one year. She has to pay a 10% interest, at the end of the year., , •, , During festival season a T.V. seller was offering a discount of 10% and another was offening a, discount of 15%., , 1, or 0.01, 100, 27, or 0.27, 100, , 93% can also be written as, , 93, or 0.93, 100, , LA, N, , 27% can also be written as, , Do This, 1., , G, , 1% can also be written as, , AN, A, , The word ‘percent’ means ‘per every hundred’ or ‘for a hundred’. The symbol ‘%’ is used to, represent percentage. Thus, 1% (one percent) means 1 out of a 100; 27% (27 percent) means 27, out of 100 and 93% (ninty three percent) means 93 out of a 100., , Given below are various grids of 100 squares., , TE, , Each has a different number of squares coloured., , In each case, write the coloured and white part in the, , SC, , ER, , T, , form of a (1) Percentage, (2) Fraction and (3) Decimal., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 121
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2., , Look at the grid paper given below. It is shaded in various designs., Find the percentage of each design, ?, , What percent represents, , ?, , What percent represents, , ?, , What percent represents, , ?, , AN, A, , What percent represents, , No. of children, , VI, , 17, , VII, , 15, , VIII, , 20, , IX, , 30, , X, , 18, , Total, , 100, , As a fraction, , As a percentage, , LA, N, , Class, , G, , The strength particular of a school are given below. Express the strength of each class as, a fraction, percentage of total strength of the school., , TE, , 3., , In all the above examples the total number is 100. How do we find percentages when the total is, not hundred?, , ER, , T, , Example. 8 : In a class there are 35 girls and 15 boys. What is the percentages, of boys and what is the percentage of girls?, Sudhir solved it like this;, Table - 1, , Student Number Fraction Converting denominator into As a percentage, hundred, , SC, , Solution:, , 122, , Girls, , 35, , 35, 50, , 35 100 70, ×, =, 50 100 100, , 70%, , Boys, , 15, , 15, 50, , 15 100 30, ×, =, 50 100 100, , 30%, , Total, , 50, , VII CLASS MATHEMATICS
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Table - 2, , Table - 3, , Anwar found the percentage of girls and boys like this., , Reena solved it like this, , Total students 35 + 15 = 50, , 35 2, 70, × =, = 70%, 50 2 100, , Thus, out of 100 students there will be, , AN, A, , Out of 50 students there are 35 girls, 35, ×100 = 70 girls, 50, , We see that there are three methods that can be used to find percentage when the, , G, , total does not add up to 100. In the first table, we multiply the fraction by, , 100, ., 100, , This does not change the value of the fraction. Subsequently, only 100 remains in, 2, to get 100 in the denominator., 2, , LA, N, , the denominator. Reena has multiplied by it, , Anwar has used the unitary method. You can choose any of the methods or you, can also find your own method., , TE, , Does Anwar's method work for all ratios? Does the method be used by Reena, also work for all ratios?, , Shirt "A" has, , 3, 3, cotton where as shirt "B" has, cotton., 5, 4, , ER, , Example 9 :, , T, , Anwar says Reena's method can be used only if you can find a natural number, which on multiplication with the denominator gives 100. Since denominator was, 50, she could multiply it by 2 to get 100. If the denominator was 60, she would not, have been able to use this method. Do you agree?, , (i) What is the percentage of cotton in each shirt?, , SC, , (ii) Which shirt has more percentage of cotton?, , Solution :, , The percentage of cotton in shirt "A" =, The percentage of cotton in shirt "B" =, , 3, × 100 = 60%, 5, , 3, × 100 = 75%, 4, , shirt “B ”has more percentage of cotton., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 123
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Example 10 : Ganga went to a tailor with 1 mt. cloth. She asked, him to make a blouse to her. The tailor used 0.75, mts of cloth to make the blouse and returned the, remaining cloth to Ganga., , Solution :, , The tailor used 0.75 mts of cloth., , AN, A, , What percentage of the cloth (i) is used in making, the blouse (ii) is given back to Ganga?, , The percentage of cloth used = 0.75 × 100%, , 75, ×100%, 100, , LA, N, , = 75%, , G, , =, , The tailor returned 1 – 0.75 = 0.25 mts of cloth., , TE, , The percentage of cloth returned = 0.25 × 100%, , =, , 25, ×100 %, 100, , = 25 %, , ER, , T, , Example 11 : Last year the cost of a commodity was ` 40. This year, the cost of the comodity, increased to ` 50. What is the percentage change in its price?, , Percentage increase in price, , SC, , Solution :, , amount changed, = original amount ×100 %, , =, , 50 − 40, ×100 %, 40, , =, , 10, 1000, × 100% =, % = 25 %, 40, 40, , Example 12 : Shyam’s monthly income is ` 10,000. He spends 60% of it on family expenses,, 10% on medical expenses, 5% on donations and saves by 25%. Find the amount, he spends on each item?, 124, VII CLASS MATHEMATICS
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Similarly, amount spent on medical expenses, , =, , 60% of total income, , =, , 60% of .10000, , =, , 60, × 10000 = .6000, 100, , =, , 10, × 10000 = .1000, 100, , Amount spent on donations, , =, , Amount saved, , =, , 5, × 10000 = .500, 100, , 25, × 10000 = . 2500, 100, , LA, N, , Exercise 4, , AN, A, , Amount spent on family expenses, , G, , Solution :, , In a school X, 48 students appeared for 10th class exam out of which 36 students passed., In another school Y, 30 students appeared and 24 students passed. If the District, Educational Officer wants to give an award on the basis of pass percentage. To which, school will he give the award?, , 2., , Last year the cost of 1000 articles was 5000. This year it goes down to, What is the percentage of decrease in price?, , 3., , Sri Jyothi has a basket full of bananas, oranges and mangoes. If 50% are bananas,, 15% are oranges, then what percent are mangoes?, , 4., , 64% + 20% + .....?....... = 100%, , 5., , On a rainy day, out of 150 students in a school 25 were absent. Find the percentage of, students absent from the school? What percentage of students are present?, , 6., , Out of 12000 voters in a constituency, 60% voted. Find the number of people voted in the, constituency?, , 4000., , SC, , ER, , T, , TE, , 1., , 7., , A local cricket team played 20 matches in one season. If it won 25% of them and lost rest., How many matches did it loose?, , 8., , In every gram of gold, a goldsmith mixes 0.25 grams of silver and 0.05 grams of copper., What is the percentage of gold, silver and copper in every gram of gold?, , 9., , 40% of a number is 800 then find the number?, , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 125
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Try This, 1., , Population of our country as per 2011 census is about 12 × 108 (120,00,00,000), If the population of our country increases by 3% every year what will be the, population by 2012?, , 2., , (i), , Can you eat 75% of a dosa?, , AN, A, , (ii) Can the price of an item go up by 90%?, , (iii) Can the price of an item go up by 100%?, , G, , Project Work, , LA, N, , Fill up the following table showing the amount of time you spend on various activities, in a day and calculate the percentage of time on each activity., Activity, , No. of hours, , % of a day, , For brushing bathing and getting ready for school, In school, , For reading and doing home work, , For sleeping, 6.7, , TE, , For playing / watching TV/helping parents, , Some situations in which we use percentages, , ER, , T, , We use percentages to express profit and loss, discount and interest. Expressing these in percentages makes comparisons easy., 6.7.1, , A potter makes pots on the wheel, then bakes them in a kiln, and decorates them with paint. He spends 3 on material,, 2 on baking and 1 on painting the each pot. He sells each, pot for ` 10. Does the potter make profit or loss?, , SC, , •, , Profit and Loss, , •, , A toy maker makes a toy for, , •, , A trader buys shirts at 540 each. The shirts remain unsold till the end of the year. The trader, sells them at 500 each at year end. Did the trader make a profit or a loss?, , 126, , VII CLASS MATHEMATICS, , 50 and sells it for, , 75. Does he make profit or loss?
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•, , Amar is a gold merchant. He bought 10 gms of gold worth 15000 in the last year. Now its, rate has gone up to 20000. Will Amar make a profit or a loss on selling the gold?, , For each of the above situations you can calculate the amount of profit or loss. However, many a, times percentages are used in expressing the profit or loss made in a transaction., , Solution :, , Ramayya’s Profit = 240 – 200 =, , 40, , Somayya’s Profit = 575 – 500 =, , 75, , AN, A, , Example 13 : Ramayya bought pens for 200 and he sold them for 240 whereas Somayya, bought pens for 500 and he sold them for 575. Who made more profit?, , 40 when he invested an amount of 200 where as, 75 when he invested an amount of 500., , LA, N, , Ramayya made a profit of, Somayya made a profit of, , 75 where as, , G, , It appears like Somayya made more profit as he made a profit of, Ramayya made a profit of 40 only. Is this correct?, , Thus, Ramayya’s ratio of profit and cost =, , 75, 500, , TE, , Somayya’s ratio of profit and cost is =, , 40, and, 200, , To compare profit, cost ratios we convert them in to percentages., Thus,, , Ramayya’s profit percentage, , T, , 40, × 100 %, 200, , ER, , =, , = 20 %, , Somayya’s profit percentage, 70, × 100 %, 500, , SC, =, , = 15%, , Ramayya earn a profit of 20% or 20 on investment of 100 and Somayya, earns a profit of 15% or 15 on investment of 100. Thus, Ramaya earns more, profit percent than Somayya., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 127
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Example 14 : A shop keeper bought a TV for ` 9000 and he sold it for ` 10,000. Find the, profit or loss? calculate percentage., Solution :, , Gopal solved the problem in the following way:, Cost price (CP) of the TV = ` 9000, Selling price (SP) of the TV = ` 10,000, , Profit (P) = 10000 – 9000 =, , 9000, the shopkeeper makes a profit of, , The ratio of profit and cost is, , 1000, 9000, , 1000, , G, , Thus, when the CP is, , 1000, , AN, A, , As SP is greater than CP, the shopkeeper makes a profit:, , To find the profit percentage we multiply this ratio with 100%, , 1000, 100, 1, ×100% =, % = 11 %, 9000, 9, 9, , LA, N, , i.e., , Madhu solved this problem using proportion., When the CP is, , 9000, the profit is, , 100, let the profit be, , x., , TE, , Now, when CP is, , 1000., , We know that the CP and profit are directly proportional thus, ratio of profit, and the ratio of cost price (CP) will be same in both cases., Therefore, x : 1000 = 100 : 9000, , ER, , T, , x, 100, =, 1000 9000, , 9000 × x = 1000 × 100, , SC, , x=, , 1000 × 100, 1, = 11 ., 9000, 9, , 1, Thus, the profit % = 11 %, 9, , Try This, The cost price of 12 mangoes is equal to the selling price of 15 mangoes. Find the, loss percent?, 128, , VII CLASS MATHEMATICS
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Example 15 : Suppose a person buys an article for 650/- and gains 6% on selling it. Find the, selling price?, Solution :, , Ravi solved it like this:, CP =, , 650, , So, if the CP is 100 then gain is, , AN, A, , Gain % = 6%, 6 and SP is 100 + 6 = 106, , Now, when the CP is 650 let the SP be, The CP and SP are directly proportional, , G, , Therefore, The ratio of CP = ratio of SP, , x., , LA, N, , 100 : 650 = 106 : x, 100 106, =, x, 650, , Therefore, 100 x = 106 × 650, x=, , 106 × 650, = 689, 100, , TE, , Therefore,, , Thus, the SP = 689, , 650, , ER, , CP =, , T, , Arun solved it like this:, , Profit % = 6%, , Thus, profit = 6% of 650, , SC, , 6, × 650 = 39, 100, , We know that SP = CP + Profit, = 650 + 39 = 689, , Thus, the SP = 689, , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 129
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Example 16 : Ramesh sold a D.V.D player for ` 2800 at a gain of 12%. For how much did he, buy it?, Solution :, , Naik uses proportion:, Gain % = 12%, , So, If CP is 100, then SP is, , 112, , When SP = 2800, let its CP be x., CP and SP are directly proportional, , G, , Thus, ratio of CP = ratio of SP, , LA, N, , x : 100 = 2800 : 112, x, 2800, =, 100 112, , AN, A, , SP = 2800, , Therefore, 112 × x = 100 × 2800, , 100 × 2800, = 2500, 112, , TE, , Therefore, x =, , Thus, CP = 2500, , Meena uses the unitary method:, , T, , S.P = 2800, , ER, , Gain = 12%, , If CP is 100, then profit is 12, SP= 100+12=112, , SC, , So, when SP in 112 then CP is 100, Therefore, when SP is 1 then CP is, , Thus, when SP is 2800 then CP is, CP = 2500, 130, , VII CLASS MATHEMATICS, , 100, 112, 100, × 2800 =, 112, , 2500
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Example 17 : A man sold two cycles for 3000 each, gaining 20% on one and losing 20% on, the other. Find his gain or loss percentage on the whole transaction?, Solution:, , SP = 3000, Gain% on first cycle = 20%, Loss% on second cycle = 20%, , Thus, if SP is, , AN, A, , Using the unitary method:, For first cycle, If CP is 100, then the profit is 20 and SP = 100 + 20 = 120, 120 then CP is 100, , Now, if SP is 1 then CP is =, , 100, 120, , 100, × 3000 = 2500, 120, , LA, N, , Now, if SP is 3000 then CP =, For second cycle, , G, , Method (i) :, , If CP is 100 then the loss is 20 and the SP is 100 – 20 = 80, Thus, if SP is 80 then CP is = 100, 100, 80, , TE, , Now, if SP is Rs. 1 then CP is =, , Now, if SP is 3000 then CP is =, , 100, × 3000 = 3750, 80, , T, , Total CP = 2500 + 3750 = 6250, , ER, , Total SP = 3000 + 3000 = 6000, Since SP is less than CP, loss = 6250 – 6000 = 250, loss, × 100 =, CP, , 250, × 100 = 4%, 6250, , SC, , Loss % =, , Method (ii): Using proportion:, On the first cycle:, When CP increases SP will increase, thus CP and SP are in direct proportion., CP, , SP, , 100, , 120, , x, , 3000, , Thus, the ratio of CP = ratio of SP, Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 131
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100 : x = 120 : 3000, 100 120, =, 3000, x, , 100 × 3000 = 120 x, , AN, A, , 100 × 3000, =x, 120, , x = 2500, Thus, CP of first cycle =, , 2500, , SP, , 100, , 80, , x, , 3000, , 100 : x = 80 : 3000, 100 80, =, x 3000, , 100 × 3000, = 3750, 80, , TE, , x=, , LA, N, , CP, , G, , On the second cycle:, , Therefore, total CP of two cycles =, 6250, , T, , =, , 2500 + 3750, , 6000, , ER, , Total SP of cycles =, , Since SP is less than CP, he has a loss, Loss =, , 6250 – 6000 =, , SC, , Therefore, loss percentage =, , Method (iii):, , SP of first cycle =, , 250, Loss, 250, ×100 = 4%, × 100 =, C .P, 6250, , 3000, , Gain% = 20%, Let the CP be, , x, , Then, the profit =, 132, , VII CLASS MATHEMATICS, , 20, 20, ×x =, x, 100, 100
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We know that SP = CP + profit, Thus, x +, , 20, x = 3000, 100, , x=, , 3000 × 100, = ` 2500, 120, , Thus, CP of the first cycle = ` 2500, SP of second cycle = ` 3000, , LA, N, , Loss % = 20%, Let the CP be ` x, Then, the loss, , G, , 120 x, = 3000, 100, , AN, A, , 100 x + 20 x, = 3000, 100, , 20, 20, ×x =, x, 100, 100, , Thus, x −, , TE, , We know that SP = CP – loss, 20, x = 3000, 100, , T, , 80, x = 3000, 100, , ER, , 80 x = 3000 ×100, x=, , 3000 ×100, = ` 3750, 80, , SC, , Thus, CP of the second cycle = ` 3750, , Therefore, total CP of two cycles = ` 2500 + ` 3750 = ` 6250, Total SP of cycles = ` 6000, Since SP is less than CP, he has a loss, Loss = ` 6250 – ` 6000 = ` 250, Therefore, loss =, , Loss, 250, ×100 = 4%, × 100 =, C .P, 6250, , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 133
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Example 18 : The cost of an article goes down every year by 20% of its previous value. Find its, original cost if the cost of it after 2 years is 19,200?, Solution :, , Cost of an article at the end of 2nd year =, , 19,200, , The cost decreases every year by 20%, , AN, A, , Let cost at the beginning of 1st year be 100. At the begining of 2nd year it will be, ` 80 (i.e. 100–20% of 100), At the begning of the 3rd year = 64 (80 – 20% of 80), , Thus, an article that costs 100 will cost 64 at the begining of third year., 19200 after 2 years, , Let the original cost be, , x., , G, , The cost of an article is, , Thus, ratio of the original cost = ratio of cost after 2 years, , x, 19200, =, 100, 64, , LA, N, , x : 100 = 19200 : 64, , x=, , TE, , 64 x = 19200 × 100, , 19200 ×100, 64, , T, , = 30000, , ER, , Thus, the original cost of an article was ` 30000., 6.7.2, , Discount, , SC, , Situation 1 : Vijay opened a new cloth shop. To, attract people, he advertised in the, following way., , 134, , VII CLASS MATHEMATICS
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On some special occassions such as dussera, deepawali, sankranthi business men, offer discounts on marked price., , Situation 3 :, , Some times to clear their old stock or out, dated stock, businessmen offer clearance, sales in the form of discounts in the, following way., , LA, N, , G, , AN, A, , Situation 2 :, , Example 19 : A shopkeeper marks his goods 25% above the cost price and allows a discount, of 12% on them. What percent does he gain?, , TE, , Solution: Let the cost price be 100., , Then marked price (MP) = 100 + 25 = 125., Discount percent on marked price = 12%, 12, × 125 = 15, 100, , T, , Discount =, , ER, , SP = MP – Discount, = 125 – 15 = 110, , SC, , Gain = SP – CP, =110 – 100, =, , 10, , Gain% =, , 10, × 100 = 10%, 100, , Thus, the shopkeeper gains 10% after discount., , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 135
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Exercise - 5, A shopkeeper bought a suit case for, , 480 and sold it for, , 540. Find his gain percent?, , 2., , Ajay bought a TV for 15000 and sold it for 14100. Find the loss percent?, , 3., , Ramu sold a plot of land for 2,40,000 gaining 20%. For how much did he purchase the, plot?, , 4., , On selling a mobile for 750, a shop keeper looses 10%. For what amount should he sell, it to gain 5%?, , 5., , A farmer sold 2 bullocks for 24000 each. On one bullock he gained 25% and on the, other he lost 20%. Find his total profit or loss percent?, , 6., , Sravya bought a watch for, , AN, A, , 1., , G, , 480. She sold it to Ridhi at a gain of 6, , 1, %. Ridhi sold, 4, , LA, N, , it to Divya at a gain of 10%. How much did Divya pay for it?, The marked price of a book is, Find the selling price of it?, , 225.The publisher allows a discount of, , 10% on it., , 8., , A carpenter allows 15% discount on his goods. Find the marked price of a chair, which is sold by him for 680?, , 9., , A dealer allows a discount of 10% and still gains by 10%. What should be the, marked price if the cost price is 900?, , 6.7.3, , Simple Interest, , TE, , 7., , Sir, I need some money for agricultural purposes., , ER, , Ramayya:, , T, , Ramayya has 10,000. He requires 15,000 for agriculture. He approaches an agricultural bank, manager. The conversation with the bank manager is as follows:, , Bank manager :How much money do you require?, Ramayya :, , 5000, , SC, , Bank manager :How long will you take to repay?, Ramayya :, , One year., , Bank manager :You have to pay an interest of 6% on the loan along with the lent amount after one year., Ramayya :, , Yes sir, I will repay after one year the whole amount., , Bank manager :Do you know how much you have to pay after one year., Ramayya :, 136, , Yes, On, , 100 I have to pay, , VII CLASS MATHEMATICS, , 6.
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6, and on, 100, , So, on, , 1, I have to pay, , 5000, I have to pay, , that is, , 300.Thus, I have to pay a total amount of 5300., , 6, × 5000, 100, , AN, A, , The money borrowed or lent out for a certain period is called the Principle. This money would be, used by the borrower for some time before it is returned. For keeping this money for some time the, borrower has to pay some extra money to the bank. This is known as Interest., The amount that is to be repayed back is equal to the sum of the borrowed principle and the, interest. That is, Amount = Principle + Interest., Interest is generally expressed as percent of the principle for a period of one year. It is written as, say 10% per year or per anmum or in short as 10% p.a., , G, , 10% p.a. means on every 100 borrowed, 10 is the interest you have to pay for one year., Let us take an example and see how this works., , Solution :, , LA, N, , Example 20 : Sunita takes a loan of 5000 at 12% rate of interest. Find the interest she has to, pay at the end of one year., Principle = 5000, Rate of interest = 12 % per year, , =, , TE, , If 100 is borrowed, sunita has to pay 20 interest for one year. Since the, amount borrowed is 5000 the interest she has to pay for one year, 12, × 5000 =, 100, , 600, , 5000 + 600 = 5600, , T, , So, at the end of the year she has to pay an amount of, , ER, , In general, when P is principle, R% is rate of interest per annum and I is the interest, the, amount to be received at the end of the year is:, P× R, 100, , SC, , A = P+, , If Ramayya, due to unavoidable circumstances, can not pay the total amount as requested by the, manager in one year then the loan can be extended for one more year, The interest for next year, will also be 300. Thus, Ramayya will pay 600 interest for 2 years., , For 100 borrowed for 3 years at 18%, the interest be paid at the end of 3 years will be 18 + 18, + 18 = 3 ×18 = 54, As the number of year increase the interest also increases. This interest being charged unformly for, each year is called simple interest., Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 137
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In general, for Principle = P, Rate of Interest = R and Time = T years., Interest to be paid (I) = P×R%×T or P ×, , R, PRT PTR, ×T =, =, 100, 100 100, , Do This, , 2., , AN, A, , 1. Find the interest on a sum of 8250 for 3 years at the rate of 8% per annum., 3000 is lent out at 9% rate of interest. Find the interest which will be recieved, at the end of 2½ years., , Example 21 : In what time will 6880 amount to 7224, if simple interest is calculated at 10%, per annum?, Amount = 7224, , G, , Solution :, , Principle = 6880, , LA, N, , S.I = Amount - Principle = 7224 – 6880 = 344, R% = 10%, Now I = P ×, , R, ×T, 100, , 10, ×T, 100, , TE, , 344 = 6880 ×, , 344 × 100 = 6880 × 10 × T, , 344 ×100, 1, = year = 6 months, 6880 × 10, 2, , ER, , T, , Therefore, T =, , Example 22 : What sum will yield an interest of, annum?, S.I = 3927,, , SC, , Solution :, , R% = 8 %, T = 2 year + 4 months, , 138, , 4⎞, ⎛, ⎜ 2 + ⎟ Yrs., 12 ⎠, ⎝, 1⎞, 7, ⎛, ⎜ 2 + ⎟ Yrs. = Yrs., 3⎠, 3, ⎝, , VII CLASS MATHEMATICS, , 3927 in 2 years and 4 months at 8% per
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Substituting in I = P ×, 3927 = P ×, , R, ×T, 100, , 8 7, ×, 100 3, , 3927 ×100 × 3, 8× 7, , Therefore, P =, Thus, P =, , 21037.50, , G, , Therefore, Principle = 21037.50, , AN, A, , 3927 × 100 × 3 = P × 8 × 7, , Solution :, , 6360, , Principle (P) =, Time (T) = 2, , 1, years?, 2, , LA, N, , Example 23 : At what rate per annum will 6360 yield an interest of 1378 in 2, , 1, 5, years = years, 2, 2, , Simple interest(S.I) = 1378, R, ×T, 100, , TE, , Substituting in I = P×, , R 5, ×, 100 2, , T, , 1378 = 6360 ×, , ER, , 1378 ×100 × 2 = 6360 × 5 × R, Therefore, R =, , 1378 × 100 × 2, 26, 2, =, =8 %, 6360 × 5, 3, 3, , SC, , Example 24 : At what rate per annum will the principle triples in 16 years?, Solution :, , Let the principle be, , x, , Amount after 16 years = 3x, Amount – Principle = Interest, Therefore, 3x – x = 2 x, For P = x, T = 16, I = 2x, , Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 139
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I = P×, 2x = x ×, , R, ×T, 100, , R, × 16, 100, , 2 x × 100= x × 16 × R, 2 x × 100, 25, 1, =, = 12 %, x × 16, 2, 2, , Exercise - 6, , AN, A, , Therefore, R =, , 1., , How long will it take for a sum of, 15624?, , 2., , At what rate a sum doubles itself in 8 year 4 months?, , 3., , A child friendly bank announces a savings scheme for school children. They will give kiddy, banks to children. Children have to keep their savings in it and the bank collects all the, money once in a year. To encourage children savings, they give 6% interest if the amount, exceeds by 10000, and other wise 5%. Find the interest received by a school if they, deposit is 9000 for one year., , 4., , A sum of money invested at 8% per annum for simple interest amounts to 12122 in, 2 years. What will it amounts to in 2 year 8 months at 9% rate of interest?, , 5., , In 4 years, 6500 amounts to 8840 at a certain rate of interest. In what time will 1600, amounts to 1816 at the same rate?, , T, , TE, , LA, N, , G, , 12600 invested at 9% per annum become to, , ER, , Lets earn Interest, , Children! Let us play a game on simple interest., 5 members can play this game., Take 3 bowls each labelled as P, R and T. Drop, 5 pieces of paper in each bowl such that every, paper is marked with a number., , SC, , 1., , (Hint: All the numbers in bowl P must be multiples, of 100 or 1000., , 2., , Pick out 3 pieces of papers, one from each of the bowls, one after another., , 3., , The number on the paper picked from bowl P relates to principle, number on the paper, picked from bowl T relates to time, number on the paper picked from bowl R relates to, rate of interest., , 140, , VII CLASS MATHEMATICS
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4., , Now calculate interest and tell I, P, T and R to every one., , 5., , If you say the right answer enter the interest amount in your account other wise put a 0 in, your account., , Interest amount, Name, , 1st round, , 2nd round, , Looking Back, , 3rd round, , AN, A, , Note : Repeat two or three rounds as per your wish and note down the values in the table given, below:, , Total, , Many times in day-to-day life we compare quantities using ratios. For e.g., my, income is 10000 and my friend’s is 20000. Thus, my income is half of, my friend’s income or we can say that my friend’s income is twice my, income. The ratio of my income and my friends income is 1 : 2. and the ratio, of my friend’s income and my income is 2 : 1., , •, , When two ratio’s are equal they are said to be in a proportion. The idea of, proportion helps us solve various problems in our daily life., , •, , If some increase (decrease) in one quantity leads increase (decrease) in other, quantity, the quantities are said to be in direct proportion., , •, , Ratio’s can be expressed in the form of percentages. The word ‘percent’ means, per hundred or out of every hundred. The symbol for percentage is ‘%’. 13%, means 13 out of 100., , T, , TE, , LA, N, , G, , •, , 13, = 0.13, 100, , ER, 13% =, , Percentages are used in various situations like profit and loss, discount and, simple interest etc.,, , SC, , •, , Fun with Fascinating Ratios, The digits 1,2,3,...9 can be arranged to form two numbers whose, ratio is 1:2, as, , 7329 1, = = 1 : 2 . This is interesting itself., 14658 2, , But even more fascinating is the fact that the nine digits can also, be arranged to form numbers whose ratio is 1:3, 1:4, 1:5, 1:6, 1:7,, 1:8 and 1:9. Enjoy by finding them., Free distribution by T.S. Government 2018-19, , RATIO - APPLICATIONS, , 141
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7.0, , Introduction, , 7, , AN, A, , DATA HANDLING, , Ravi is reading the sports section of a newspaper. There are two tables on the sports page of the, newspaper., Top 5 Batsmen in World Cup 2011, , Top 5 Bowlers in World Cup 2011, , Runs, scored, , T Dilshan (Sri Lanka), , 500, , Shahid Affridi (Pakistan), , 21, , Sachin Tendulkar (India), , 482, , Zahir Khan (India), , 21, , K. Sangakkara (Sri Lanka), , 465, , TG Southee (New Zealand), , 18, , Jonadhan Trott (England), , 422, , Robin Peterson ( South Africa), , 15, , U Tharanga (Sri Lanka), , 395, , M. Muralitharan (Sri Lanka), , 15, , Wickets, Taken, , Table - 2, , TE, , What do the two tables tell us?, , LA, N, , Table - 1, , Name of the Bowler, , G, , Name of the Batsman, , T, , Table 1 tells us the names of batsmen who scored the most runs in the World, Cup, 2011 as well as the number of runs they scored. This information can, help in taking decisions or in drawing conclusions. For e.g. it can help the, organisers of the World Cup in deciding whom to award the prize for the best, batsman., , SC, , ER, , Table-2 tells us the names of bowlers who took the most wickets in the World, Cup, 2011 as well as the number of wickets they took. This information can, also help in taking decisions or in drawing conclusions. For e.g. it can help the, organisers of the World Cup in deciding whom to award the prize for the best, bowler., Information which is in the form of numbers or words and helps in taking decisions, or drawing conclusions is called data. The names of batsmen and the runs they scored as, well as the names of bowlers and the number of wickets they took is data. Tables and, graphs are the ways in which data is presented., The numerical entries in the data are called ‘Observations’., Try This, Look at your school information board. Do you find any data tables there? Find out, who uses this data., 142, , VII CLASS MATHEMATICS
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7.1, , Organising data, , Details of seven students of class VIII in a school are collected under the Javahar Bala Arogya, Raksha Scheme., Krishna noted the heights of the following students in his notebook as, , AN, A, , Amala-125cm, Lekhya-133cm, Thabasum-121cm, Sudha-140cm, Vanaja-117cm, Lenin-129cm, and Rajesh-132cm., Another student Kumar wrote the same data in the form of a table and arranged the heights in, ascending order., , Sudha, , G, , Height (in cms), 117, 121, 125, 129, 132, 133, , LA, N, , Name of the Student, Vanaja, Thabassum, Amala, Lenin, Rajesh, Lekhya, , 140, , Now, let us answer these questions., , (i) Who is the tallest amongst the students?, , TE, , (ii) Who is the shortest amongst the students?, , (iii) Whose height is between that of Amala and Rajesh?, , T, , Did you use the data written by Krishna? or by Kumar? to answer the question. You must have, used Kumar’s data as it is organised and thus easier to read and understand., Do This, , SC, , ER, , In a unit test Amar secured 20, 18, 23, 21, 24 and 22 marks in Telugu, Hindi,, English, Mathematics, Science and Social Science respectively. Peter got 23, 21,, 20, 19, 24 and 17 marks in the above subjects respectively. Interpret the data in, an organized manner., Classroom Project, Use the weighing machine to find the weights of all your classmates. Organise, this data in the form a table. Make sure to arrange the weights in either ascending, or descending order. Then answer the following questions:, a., , Who is the lightest student in your class?, , b., , How many students weigh more than 25 kg?, , c., , How many students weigh between 20 and 30 kg?, , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 143
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7.2 Representative Values, In a hostel, Average consumption of rice per child per day is 150 g., , •, , Average age of children is 13 years., , •, , Average height of children is 135 cm., , AN, A, , •, , G, , On studying this data, can we say that every child consumes, exactly 150 gms of rice per day? Can we say that the age of, each child in the class is 13 years? Can we say that the height, of each child in class is 135 cm? Obviously not, we know that, some children may take more than 150 gms of rice some may take less and some may take exactly, 150 gms. A similar situation will hold for children’s weight and height., , TE, , 7.3.1 Arithmetic Mean, , LA, N, , At the same time, 150 gms gives us an idea of the amount of rice consumed by each child in the, hostel. It is a representative value of the amount of rice consumed by each child. Similarly, 13 years, gives us an idea of the age of each child in the hostel. It is a representative value of the age of each, child. The same holds for the height. All the above examples are of a particular representative value, called arithmetic mean. In the section ahead, we shall learn about ‘arithmetic mean’ and also two, other types of representative values called ‘median’ and ‘mode’., , The physical education teacher in a school instructed his students to practice regularly Rajender, had his practice sessions for a week as follows., Tue, , Wed, , Thu, , Fri, , Sat, , Sun, , 20, , 35, , 40, , 30, , 25, , 45, , 15, , ER, , Minutes, , Mon, , T, , Day, , Can we compute the time spent by Rajender for practice in terms of time spent per day? Let us, observe., , SC, , What is the total time Rajender spent during the week on practice?, Total time = 20 + 35 + 40 + 30 + 25 + 45 + 15 = 210 minutes, Now to find the time spent on practice, per day, we divide the total time spent by the number of, days., i.e., , 20 + 35 + 40 + 30 + 25 + 45 + 15 210, =, = 30 minutes, 7, 7, , This is the average time spent on practice per day or the average practice session per day., 144, , VII CLASS MATHEMATICS
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Example 1 : Earnings (in rupees) of a vegetable vendor in a week are ` 200, `150, `180,, `300, `160, `170 and `170. Find his average earning, per day., Solution :, , Total earnings (in rupees), , = 200+150+180+300+160+170+170, = `1330, =7, , Average earning or mean earning =, , AN, A, , Number of days, , 1330, = ` 190, 7, , Sum of all observations, Number of observations, , LA, N, , Average or Mean or Arithmetic Mean (A.M) =, , G, , The average of a data is also called Arithmetic Mean or Mean., , Try This, , TE, , 1. The ages (in years) of players are in a team of 16, 16, 16, 14, 17, 18. Then find, the following:, (i), , Age of the youngest and the oldest player., , (ii), , Mean age of the players., , ER, , T, , What is the average number of glasses of water that you drink per day? in a, week. How did you find the average?, 7.3.2 Where does the mean lie?, , SC, , The marks obtained by Anil, Amar, Anthony and Inder in Telugu, Hindi and English are given, below., Telugu, , Hindi, , English, , Anil, , 15, , 8, , 10, , Amar, , 10, , 10, , 12, , Antony, , 11, , 6, , 11, , Inder, , 12, , 12, , 13, , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 145
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Now let us calculate the average marks obtained by the students in each subject., , Telugu, AM =, , Hindi, , 15 + 10 + 11 + 12, 4, , AM =, , English, , 8 + 10 + 6 + 12, 4, , AM =………, , =, , 36, 4, , = ………, , = 12, , = …….., , = ………, , 48, , Highest marks = 15, , Highest marks = ……., , Least marks = 10, , Least marks, , = ……., , Mean, , Mean, , = ……., , = 12, , AN, A, , = 4, , Highest marks = ……., Least marks = ……., Mean, , = ……., , G, , Does the mean lie between the minimum and maximum value in each case?, , LA, N, , You will find this to be true., , The arithmetic mean always lies between the highest and lowest observations of the, data., 7.3.3, , A property of mean, , TE, , Example 2 : In a family, ages (in years) of members; Krishna, Radhika, Niharika and Nikhil, are 44, 39, 17 and 12. (i) Find the arithmetic mean of their ages. (ii) What were, their ages 5 years before? Find their mean age. (iii) Can you see a relationship, between the change in mean and the number of years., = 44, 39, 17, 12 years, , Number of family members, , =4, , T, , Present ages of family members are, , ER, , Solution :, , Therefore, Arithmetic Mean of their ages =, , SC, , Ages of family members, 5 years ago, , ∴ mean of their ages 5 years ago =, , 44 + 39 + 17 + 12 112, =, = 28 years, 4, 4, , = 44 – 5, 39 – 5, 17 – 5, 12 – 5, = 39, 34, 12, 7, , 39 + 34 + 12 + 7 92, =, = 23 years, 4, 4, , Thus, on reducing the age of each family member by 5 years, we find that the mean age of the, family also decreases by 5 years from the present mean age., Now calculate the mean age of the family, 3 years from now. What do you think will be the mean, age of the family 10 years from now?, 146, , VII CLASS MATHEMATICS
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You will find that when all the values of data set are increased or decreased by a certain, number, the mean also increases or decreases by the same number., Try This, 1. A data of 10 observations has a minimum value 15 and maximum value 25., What is the mean of the data?, (ii) 15, , (iii) 21, , (iv) 27, , AN, A, , (i) 12, , 2. Observations of a data are 23, 45, 33, 21, 48, 30, 34, 36 and 35. Without actual, calculation choose the mean of the data., (ii) 35, , (iii) 48, , (iv) 50, , 1., , LA, N, , Exercise - 1, , G, , (i) 20, , Maximum day time temperatures of Hyderabad in a week (from 26th February to 4th, March, 2011) are recorded as 26 oC, 27 oC, 30 oC, 30 oC, 32 oC, 33 oC and 32 oC., What is the maximum temperature of the week?, , (ii), , What is the average temperatures of the week?, , TE, , (i), , Rice consumed in a school under the mid-day meal program for 5, consecutive days is 15.750 kg, 14.850 kg, 16.500 kg, 14.700, kg, and 17.700 kg. Find the average rice consumption for the 5, days., , 3., , In a village three different crops are cultivated in four successive years. The profit, (in rupees) on the crops, per acre is shown in the table below-, , ER, , T, , 2., , year, , 2005, , 2006, , 2007, , 2008, , Ground nuts, , 7000, , 8000, , 7500, , 7500, , Jawar, , 6000, , 1000, , 8000, , 1000, , Millets, , 9000, , 5000, , 3000, , 4000, , SC, , Crop, , (i), , Calculate the mean profit for each crop over the 4 years., , (ii), , Based on your answers, which crop should be cultivated in the next year?, , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 147
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The number of passengers who travelled in APSRTC, bus from Adilabad to Nirmal in 4 trips in a day are 39,, 30, 45 and 54. What is the occupancy ratio (average, number of passengers travelling per trip) of the bus, for the day?, , 5., , The following table shows the marks scored by Anju,, Neelesh and Lekhya in four unit tests of English., Name of the Student, , Unit Test I, , Unit Test II, , Unit Test III Unit Test IV, , Anju, , Absent, , 19, , 23, , 21, , Neelesh, , 0, , 20, , 22, , 24, , Lekhya, , 20, , 24, , 24, , 24, , Find the average marks obtained by Lekhya., , (ii), , Find the average marks secured by Anju. Will you divide the total marks by 3 or, 4? Why?, , (iii), , Neelesh has given all four tests. Find the average marks secured by him. Will you, divide the total marks by 3 or 4? Why?, , (iv), , Who performed best in the English?, , LA, N, , G, , (i), , Three friends went to a hotel and had breakfast to their taste, paying ` 16, ` 17 and ` 21, , TE, , 6., , AN, A, , 4., , 8., , Find the mean of the first five prime numbers., , In a set of four integers, the average of the two smallest integers is 102, the average of the, three smallest integers is 103, the average of all four is 104. Which is the greatest of these, integers?, , SC, , 9., , Find the mean of the first ten natural numbers., , ER, , 7., , T, , respectively (i) Find their mean expenditure.(ii) If they have spent 3 times the amount that, they have already spent, what would their mean expenditure be? (ii) If the hotel manager, offers 50% discount, what would their mean expenditure be? (iii) Do you notice any, relationship between the change in expenditure and the change in mean expenditure., , 10., , Write at least two questions to find the mean, giving suitable data., Project Work, Find out the number of family members in the houses on your street. Calculate the, average family size of your street., , 148, , VII CLASS MATHEMATICS
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7.4 Mode, The second type of representative value that we will learn about is mode., Let us read the example given below-, , Mon, , GGGSSSSPP, , Tue, , GGGSSSSSPP, , Wed, , GGSSSSSP, , Thu, , GGGSSSP, , Fri, , GGGSSPP, , Sat, , GSSSSSSSS, , Sun, , GGGSSSP, , G, , Packets of oil sold, , LA, N, , Day, , AN, A, , Example 3 : A shop keeper wants to find out which cooking oil he should stock in more number., For this, he maintains a record of cooking oil sale for the week in the form of the, table given below., , G = Ground nut oil packet, S = Sunflower oil packet, and P= Palmolien oil packet., , Solution :, , TE, , In such a situation will calculating the mean number of oil packets sold help the shopkeeper to take, a decision?, The shopkeeper first calculates the averrage number of packets that he can order., 18 + 30 + 9 57, =, = 19., 3, 3, , T, , Average number of packets =, , SC, , ER, , Should the shopkeeper stock 19 packets for each type of oil? The shopkeeper looked at his sales, figures again. He finds sunflower oil to be the most frequently demanded oil and palmolien oil to be, the least demanded oil. If he was to order 19 packets of each he would fall short of sunflower oil, and palmolien oil would be in surplus. The shopkeeper decides to stock more packets of sunflower, oil and lesser number of packets of palmolien oil. Thus, the number of packets of sunflower oil i.e., 30 is the representative value for the shopkeeper’s data as it tells him the most frequently purchased, oil., , The most frequently occurring value for a set of observations is called the mode., The longest bar in a bar graph represents the mode, as can be seen in the bar graph given in the, next page., , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 149
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y, , 35, , 25, 20, 15, 10, 5, Groundnut, Sunflower, , Sunflower, Groundnut, , G, , 0, , AN, A, , Number of Packets, , 30, , Palmolien, , x, , Type of Oil, , Solution :, , LA, N, , Example 4 : Find the mode of the given set of numbers- 2,3,5,3,4,7,3,2,1,7,3, Arranging the numbers with same value together, we get 1,2,2,3,3,3,3,4,5,7,7, 3 occurs more frequently than the other observations. Thus, Mode = 3, Find the mode of the data 3, 5, 9, 6, 5, 9, 2, 9, 3, 5., , Solution :, , Arranging the numbers with the same value together we get 2, 3, 3, 5, 5, 5, 6, 9,, 9, 9., , TE, , Example 5 :, , T, , Here both 5 and 9 occurs more and equal number of times i.e., 3 times., , ER, , Thus, the given data contains two modes, i.e., 5 and 9, This kind of data is called Bimodal Data., , SC, , Note : If each observation in a data set is repeated an equal number of times then the data set has, no mode., , Try This, 1. Find the modes of the following data., (i) 5, 6, 3, 5, 4, 9, 5, 6, 4, 9, 5, (ii) 25, 14, 18, 15, 17, 16, 19, 13, 12, 24, (iii) 10, 15, 20, 15, 20, 10, 15, 20, 10, , 150, , VII CLASS MATHEMATICS
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Example 6 :, , Following are the marks obtained by 50 students in a unit test, which is administered, for 10 marks. Find the mode of the data., , LA, N, , G, , AN, A, , Marks obtained, No of students, 00, 2, 1, 1, 2, 2, 3, 1, 4, 5, 4, 6, 10, 7, 15, 8, 9, 9, 5, 10, 1, Total, 50, Solution: In the data marks are observations. From the data table it is clear that 7 marks are, obtained by many students., Mode of the data is 7, , TE, , Note: The observation 7 that repeats fifteen times is the mode and number of times i.e.15 should, not be confused as the mode., Example 7 : In which of the following situations, is the mode an appropriate representative value?, A shopkeeper selling shirts, needs to decide which size of shirts to order more., , (b), , For purchasing rice for a party of 20 people., , (c), , For finding the height of the door in your house., , ER, , T, , (a), , SC, , Solution : (a) Let us look at the first situation. Supposing the shopkeeper is selling 4 sizes of shirts, and his sale for the month of February isShirt Size, , Number, , M, L, XL, XXL, Total, , 15, 18, 40, 22, 92, , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 151
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The average number of shirts sold by the shopkeeper is, , 12 + 18 + 40 + 22, = 23 shirts., 4, , AN, A, , In such a situation does it make sense for the shopkeeper to order 23 shirts of each size? The, shopkeeper looks at his data again. He finds that the most frequently purchased size is XL, If he, orders 23 shirts of each size, he will fall short of size XL shirts. He thus finds it more sensible to, order more shirts of this size and lesser of the rest., Thus, the shopkeeper uses mode or the most frequently occurring value to take his decision., , G, , (b) Neither we know how many take maximum and how much nor how many take minimum and, how much. If we purchase 20 times of maximum, it would be waste, or if we purchase 20 times of, munimum, it is not sufficient. So mode cann’t be suggested here., , LA, N, , (c) If there are 5 members in the house, and whose heights are 134cm, 125cm, 100cm, 125cm, and 144cm, as mode of the data is 125cm, we may suggest the height of the door must be 125cm., But it is difficult for the person of height 144cm. Even if we take mean of their heights, it is difficult, for tall persons. So neither the mode nor the mean can be used here., Try This, , TE, , 1. One situation where mean would be an appropriate representative value., 2. One situation where mode would be an appropriate representative value., , T, , Exercise - 2, , Ages of players in a cricket team are 25, 26, 25, 27, 28, 30, 31, 27, 33, 27, 29., (i) Find the mean and mode of the data.(ii) Find the minimum number of players to be, added to the above team so that mode of the data changes and what must be their ages., , SC, , 2., , Long jumps by 7 students of a team are 98cm, 125cm, 140cm, 155cm, 174cm, 140cm, and 155cm. Find the mode of the data., , ER, , 1., , 3., , Find the mode of the following data. 12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46,, 25, 12, 24, 46, 25, 72, 12, 24, 36,25, 38 and 36., , 4., , Decide whether mean or mode is a better representative value, in the following situations., (i), , 152, , A shop keeper, who sells tooth paste tubes of different, sizes, wants to decide which size is to be ordered more., VII CLASS MATHEMATICS
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An invigilator wants to bring sufficient number of additional papers to the examination, hall., , (iii), , Preparation of the number of laddus for a marriage., , (iv), , For finding the favorite cricketer in a class., , Median, , AN, A, , 7.5, , (ii), , -, , ` 40,000, , Worker 1, , -, , ` 3,300, , Worker 2, , -, , ` 5,000, , Worker 3, , -, , Worker 4, , -, , Worker 5, , -, , Worker 6, , -, , Worker 7, , -, , Worker 9, , ` 4,200, ` 3,500, ` 4,500, ` 4,200, , -, , ` 4,300, , -, , ` 3,500, , -, , ` 3,500, , T, , Worker 10, , ` 4,000, , TE, , Worker 8, , G, , Manager, , LA, N, , We have looked at situations where mean and mode are representative values of the data. Now let, us look at another situation. The following are the salaries (in rupees) earned by the manager and, the workers in a production unit., , ER, , Will the mean salary or the mode of salaries be a representative value for this data?, Let us calculate the mean salary in the production unit., , SC, , Total salary, Mean salary = Number of employees, , =, , 3300 + 5000 + 4000 + 4200 + 3500 + 4500 + 4200 + 4300 + 3500 + 3500 + 40000, 11, , = `.7272.72, Is this salary a representative of the salaries of either the manager or the workers? No it is not. It, is much lesser than the manager’s salary and more than the salary of all the workers., , Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 153
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Now let us consider the mode. 3500 is the most frequently occurring value in the data. However,, it occurs only thrice thus, cannot be a representative of the data., Now, let us use another way of calculating the representative value., Let us arrange the numbers in ascending order-, , AN, A, , 3300, 3500, 3500, 3500, 4000, 4200, 4200, 4300, 4500, 5000, 40000, The middle value of this data is 4200 as it divides employees into 2 equal groups – 5 are earning, more than 4200 and 5 are earning less. This value is called Median and as you can see it provides, a representative picture for all., , LA, N, , Now what if the number of observations were even?, , G, , In the above example, the number of observations is 11 i.e. an odd number, thus the median, divides the data into 2 equal groups., , Let us the take the example of the production unit again. What if a new worker earning ` 4000, joined the production unit?, Arranging the number in ascending order we get-, , TE, , 3300, 3500, 3500, 3500, 4000, 4000, 4200, 4200, 4300, 4500, 5000, 40000, Here both 4000 and 4200 lie in the middle of the data. Here the median will be calculated by, finding the average of these two values. Thus, the median salary =, , 4000 + 4200, = `.4100., 2, , ER, , T, , Example 8 : The monthly incomes of 7 graduates is ` 8000, ` 9000, ` 8200, ` 7900, ` 8500,, ` 8600 and ` 60000. Find the median income., Solution :, , Arranging the incomes in ascending order we get : 7900, 8000, 8200, 8500,, 8600, 9000, 60000, , SC, , Number of observations, , =7, , Middle term, i.e., 4th term in the data = 8500, Thus, the median income, , = ` 8500, , Example 9 : Find the median of 49, 48, 15, 20, 28, 17, 14 and 110., Solution :, , Ascending order of observations, , = 14, 15, 17, 20, 28, 48, 49, 110, , Number of observations, , =8, , Middle terms, i.e. the 4th and 5th values are 20 and 28., 154, , VII CLASS MATHEMATICS
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Median = average of 4th and 5th values =, , 20 + 28, = 24, 2, , Thus, median of the given data is 24, Exercise - 3, Say true or false and why?, , AN, A, , 1., , (i) The difference between the largest and smallest observations in a data set is called the, mean., (ii) In a bar graph, the bar which has greater length may contains mode., , G, , (iii)Value of every observation in the data set is taken into account when median is calculated., (iv) The median of a set of numbers is always one of the numbers, , The monthly income (in rupees) of 7 households in a village are 1200, 1500, 1400, 1000,, 1000, 1600, 10000. (i) Find the median income of the house holds. (ii) If one more, household with monthly income of `1500 is added, what will the median income be?, , 3., , Observations of a data are16, 72, 0, 55, 65, 55, 10, and 41. Chaitanya calculated the, mode and median without taking the zero into consideration. Did Chaitanya do the right, thing?, , 4., , How many distinct sets of three positive integers have a mean of 6, a median of 7, and no, mode?, , 5., , Four integers are added to a group of integers 3, 4, 5, 5 and 8 and the mean, median, and, mode of the data increases by 1 each. What is the greatest integer in the new group of, integers?, , ER, , T, , TE, , LA, N, , 2., , Play the Game, , SC, , Take a dice numbered 1, 2, 3, 4, 5 and 6 on its faces. Make a group of three, students. Ask each student to roll the dice and record the number, turn by turn., Repeat the process for 10 rounds. Now each student will have 10 numbers each., Find the mean, median and mode of data of each student., 7.6, , Presentation of data, , We have already learnt how to present data in bar graphs and pictographs in class 6. Pictographs, represent data using pictures of objects. However, presenting data by a pictograph is often time, consuming and difficult. Bar graphs help in presenting data with much more ease., Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 155
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7.6.1, , Bar Graph, , In this section we will learn a little more about bar graphs. We know that bar graphs are made up, of bars of uniform width which can be drawn horizontally or vertically with equal spacing between, them. The length of each bar tells us the frequency of the particular item. We also know that the, length of the bar graph is as per scale., , (i) What are taken on x-axis and y axis?, , x, , (ii) What is the scale selected on the y-, , 30, , AN, A, , Example 10 : The bar graph shows the one day sales of various items in a shop., Sale of Items, , 25, 20, , G, , (iii) Which of these provisions has most, , Weight (kg), , axis?, , 15, 10, , sale? How much?, , LA, N, , 5, , (iv) Is the sale of onions more than red, gram?, , 0, , Wheat, , Rice, , TE, , Sugar, , o, , 1 cm = 50 C, , o, , Boilling points ( C), , Alchohol, , Example 11 : Observe another bar graph., , Bromine, , (i) What information does the graph give us?, , Chlorofrm, , ER, , T, , (ii) What are taken on x-axis and y-axis?, Ether, , (iii) Which of these liquids has highest boiling point?, Mercury, , (iv) Which of these liquids has the lowest boiling, , SC, , point?, , (v) What is the approximate ratio between the, boiling point of mercury and the boiling point, of ether?, , 156, , VII CLASS MATHEMATICS, , Water, , 0, , 50, , y, , Onions, , Item, , (v) What is the ratio between the sale of jowar, and the sale of red gram?, , Red gram, , Jowar, , 1 cm = 5 kgs, , 100, , 150, , 200, , Temperature, , 250, , 300, , 350, , 400
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7.6.2, , Double Bar Graph, , Now let us learn about another type of bar graph, Example 12 : Study the following graph representing the total enrolment of boys and girls in ZPP, High School and answer the following questions., Enrollment in ZPP High School, , AN, A, , y, 180, 160, 140, , G, , 100, 80, 60, 40, , 2006-07, , 2008-09, , 2009-10, , Academic Year, , TE, , Boys, , 2007-08, , LA, N, , Enrolment, , 120, , Girls, , x, , 2010-11, 1 cm = 20 students, , T, , Did you notice that there are two bars for each year? What does the first bar tell you? What does, the second bar tell you? This kind of bar graph is called Double bar graph. It presents two, observations side by side., , ER, , (i) In which year is the enrolment of girls more than the boys?, (ii) In which year is the enrolment of boys and girls the same?, (iii) In which year is the enrolment of girls minimum?, (iv) What is the total enrolment in the year 2007-08?, , SC, , Example 13 : The following are the marks in Maths and Science of five students in class VII., Present this data in the form of a double bar graph., Name of Student, Saravan, Raman, Mani, Renuka, Girija, Sharmila, , Free distribution by T.S. Government 2018-19, , Maths, 70, 35, 65, 90, 22, 50, , Science, 75, 30, 75, 100, 35, 50, DATA HANDLING, , 157
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Solution : Steps in drawing a double bar graph., 1. Draw x-axis (horizontal line) and y-axis (vertical line) on the graph paper and mark their, intersection as O., 2. Take names of students on x-axis., , AN, A, , 3. Take Maths and Science marks on y-axis,, 4. Take an appropriate scale on y-axis so that maximum marks of both the subjects fit on the, graph sheet. Here the maximum value to be plotted on Y- axis is 100, so the scale 1 cm = 10, marks, is appropriate., , G, , 5. Find the length of each bar by dividing the value by 10 (Scale is 1 cm = 10 marks)., , LA, N, , 6. Draw bars representing ‘Maths marks ’ and ‘Science marks’ side by side of every student., Marks in Maths and Science, , y, , 90, 80, , TE, , 70, , Marks, , 60, 50, 40, , T, , 30, , ER, , 20, 10, 0, , Saravan, , SC, , Maths Marks, , 7.6.3, , Raman, , Science Marks, , Mani, , Renuka, , Name of Students, , Girija, , Sharmila, , x, , 1 cm = 10 marks, , Pie Charts, , Another way in which data can be presented is through pie charts., The monthly budget of a family is given in the table on the left. This data has been presented in a pie, chart on the right. The higher the share of expenditure of perticular item of the total income, the, more the area occupied by the item in the pie chart., 158, , VII CLASS MATHEMATICS
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Amount (`), , Food, , 1500, , Education, , 750, , Others, , 2250, , Savings, , 4500, , Total income, , 9000, , AN, A, , Budget head, , Looking at the pie chart answer the following questions., (i) What is the shape of the pie chart?, , (ii) What is the name of each shape used to present different items in the pie chart?, , G, , (iii) Say true or false (a) The largest part of the income is saved., , 7.6.4, , LA, N, , (b) Least amount of money is spent on education., Drawing a pie chart, , Now, let us learn about how data is presented on a pie chart., , The pie chart represents each item as a portion of the circle, as how much part of the total incomeis, is shared by the particular item., , TE, , We know that the total angle at the centre of a circle is 360º. We can assume that it represents the, total of all observations i.e. ` 9000., Each item of expenditure is a part of the total income thus, the angle of the sector or the area of the, sector will depend on the ratio between the item of expenditure and total income., Amount of Expenditure, × 360, Total Income, We make a table to find the angle of the sectors. The table is shown below., , ER, , T, , Thus, the angle of each sector =, , Amount of, expenditure, , Food, , 1500, , SC, , Budget, head, , Education, , 750, , Others, , 2250, , Savings, , 4500, , Ratio between, expenditure and total income, , Angle of sector or, area of the sector, , 1500 1, =, 9000 6, 750, 1, =, 9000 12, 2250 1, =, 9000 4, , 1, × 360o = 60o, 6, 1, × 360o = 30o, 12, 1, × 360o = 90 o, 4, , 4500 1, =, 9000 2, , 1, × 360o = 180o, 2, , Note: Check whether the sum of all the angles of the sectors equal to 360 º ?, Free distribution by T.S. Government 2018-19, , DATA HANDLING, , 159
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Steps of construction, 1. Draw a circle with any convenient radius and mark its centre ‘O’., 2. Mark a point A, somewhere on the circumference and join OA., 3. Construct angle of the sector for food = 60º. Draw ∠AOB = 60º., , AN, A, , 4. Construct angle of the sector for education = 30º. Draw ∠BOC = 30º., 5. Construct angle of the sector for other = 90º. Draw ∠COD = 90º., , 2, , 3, , 4, , 5, , TE, , 1, , LA, N, , G, , 6. Now ∠DOA = 180º represents the angle sector for savings., , Exercise - 4, , 1., , Draw a bar graph for the following data., , ER, , Year, , T, , Population of India in successive census years1941, , 1951, , 1961, , 1971, , 1981, , 1991, , 2001, , 320, , 360, , 440, , 550, , 680, , 850, , 1000, , Population, (in millions) (approx), , SC, , Source : Data from census of India 1991 and 2001., , 2., , 160, , Draw a pie chart for the following data., Item of expenditure, , Food, , Health, , Clothing, , Education, , Savings, , Amount spent in rupees, , 3750, , 1875, , 1875, , 1200, , 7500, , VII CLASS MATHEMATICS
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3., , Draw a double bar graph for the following data., Birth and Death rates of different states in 1999., Death Rate (Per 1000), , Andhra Pradesh, , 22, , 8, , Karnataka, , 22, , 8, , Tamil Nadu, , 19, , Kerala, , 18, , Maharashtra, , 21, , Orissa, , 24, , AN, A, , Birth Rate (Per 1000), , 8, , 6, , 8, , 11, , G, , State, , Source : The table is taken from vittal statistics SRS 1999., , Draw a pie chart for the following data., , LA, N, , 4., , Time spent by a child during a day-, , Time spent, , 8 hrs, , School, , Play, , Others, , 6 hrs, , 2 hrs, , 8 hrs, , TE, , Sleep, , The adjoining pie chart gives the expenditure on various items during a month for a family., (The numbers written around the pie chart tell us the angles made by each sector at the, centre.), , T, , 5., , Time spent for, , ER, , Answer the following-, , On which item is the expenditure minimum?, , (ii), , On which item is the expenditure maximum?, , (iii), , If the monthly income of the family is ` 9000,, what is the expenditure on rent?, , (iv), , If the expenditure on food is ` 3000, what is, the expenditure on education of children?, , SC, , (i), , Free distribution by T.S. Government 2018-19, , 90o, , 120o, , 60o, o, , 90, Savings, , Education, , Rent, , Food, , DATA HANDLING, , 161
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Project Work, 1. Gather information of the number of different kinds of houses in your locality, (ward / colony / village). Then find mode., 2. Collect the item-wise expenditure of your family in a month and represent it as a, pie chart., , AN, A, , 3. Collect different data presented in the form of bar graphs and pie charts in, magazines, newspapers etc. and present them on your school bulletin board., Looking back, , Mean, mode and median are representative values for a data set., , •, , Arithmetic mean or mean is equal to sum of all the observations of a data set, divided by the number of observations. It lies between the lowest and highest, values of the data., , •, , An observation of data that occurs most frequently is called the mode of the, data. A data set may have one or more modes and sometimes none., , •, , Median is simply the middle observation, when all observations are arranged, in ascending or descending order. (In case of even number of observations, meadian is the average of middle two observations.), , •, , A pie chart is a circular chart /graph divided into sectors, and is used to, present data., , LA, N, , TE, , T, , The central angle of each sector (and consequently its area) in a pie chart, is, proportional to the quantity that it represents., , ER, , •, , G, , •, , SC, , Dr.C.R.Rao (India), 1920 AD, , A well known Statistician, famous for his, “Theory of Estimation” (1945)., He worked on Cramer-Rao Inequality and, Fisher-Rao theorm., , 162, , VII CLASS MATHEMATICS
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CONGRUENCY OF TRIANGLES, Introduction, , AN, A, , 8.0, , 8, , G, , If we take a pile of one rupee coins and place them one on top of the, other, they would match perfectly. Do you know why this happens?, This is because all the coins have the same size and shape. In the same, way papers of a blank note book have the same size and shape., , LA, N, , Look around you and find some examples of objects that share this kind of similarity i.e. they are, identical in shape and size. Think of at least 5 such examples., When we talk about objects of the same size and shape we say that the objects are congruent. A, practical test of congruence is to place one object over the other and see if they superimpose, exactly., , TE, , Activity:, , ER, , T, , Are all ten rupee notes congruent? How will you check?, , SC, , Similarly, check if the five rupee note you find are congruent., , We see many examples of congruent objects all around us. Now, think of some shapes that are, congruent., 163, Free distribution by T.S. Government 2018-19, CONGRUENCY OF TRIANGLES
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Do This, 1., , Here are some shapes. See whether all the shapes given in a row, are congruent to each other or not. You can trace the figures and check., , AN, A, , (i), , LA, N, , G, , (ii), , (iii), , Which of the following pairs of figures are congruent?, , T, , TE, , 2., , (ii), , (iii), , (iv), , (v), , (vi), , SC, , ER, , (i), , 164, , VII CLASS MATHEMATICS
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8.1, , Congruency of line segments, , Observe the two pairs of line segments given below., , C, , Q, , P, , B, D, , R, , Figure 1, , S, , Figure 2, , AN, A, , A, , Copy the line segment AB on a tracing paper. Place it on CD. You will find that AB covers CD., Hence the line segments are congruent. We write AB ≅ CD ., , Repeat this activity for the pair of line segments in Figure 2. What do you find? Are they, congruent?, , G, , You will notice that the pair of line segments in Figure 1 match with each other because they have, same length and this is not the case in Figure 2., , LA, N, , The line segment has only one dimension i.e., length. So if two line segments have the same length,, they are congruent. Conversely, if two line segments are congruent, they have the same length., When we write AB = CD, what we actually mean is AB ≅ CD ., , TE, , 8.2 Congruency of triangles, , We learnt that two line segments are congruent when their lengths are equal. We extend this idea to, triangles. Two triangles are congruent if they are copies of one another and when superimposed,, they cover each other exactly., B, , ER, , T, , F, , A, , C, , E, , G, , SC, , ΔABC and ΔEFG cover each other exactly i.e. they are of the same size and shape. They are, congruent triangles. We express congruency of the two triangles as ΔABC ≅ ΔEFG ., , If two triangles are congruent then all the corresponding six elements of the two triangles i.e. the, three angles and three sides are congruent. We also say that if the corresponding parts of two, triangles are congruent, then the triangles are congruent. This means that, when you place ΔABC, on ΔEFG, their corresponding corners coincides with each other. A lies on E, B lies on F and C, lies on G. Also ∠A coincides with ∠E, ∠B coincides with ∠F and ∠C coincides with ∠G and, lastly AB coincides with EF, BC coincides with FG and AC coincides with EG., Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 165
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Thus, for two triangles that are congruent, their corresponding parts i.e. vertices, angles and sides, match one another or are equal., In ΔABC and ΔEFG, B→F, , C→G, , (corresponding vertices), , ∠A ≅ ∠E, , ∠B ≅ ∠F, , ∠C ≅ ∠G, , (corresponding angles), , AB ≅ EF, , BC ≅ FG, , AC ≅ EG, , (corresponding sides), , AN, A, , A→E, , Do This, ΔEFG ≅ ΔLMN, , N, , LA, N, , 1., , G, , So, when we say that ΔABC ≅ ΔEFG. The order of the alphabet in the names of congruent, triangles displays the corresponding relationships., , E, , G, , F, , M, , L, , TE, , Write the corresponding vertices, angles and sides of the two triangles., If ΔABC ≅ ΔDEF write the parts of ΔABC that correspond to-, , 2., , (ii) ∠E, , (i) DE, , (iii) DF, , (iv) EF, , (v) ∠F, , Name the congruent triangles in each of the following pairs. Write the statement using ‘ ≅’., , T, , 3., , E, , ER, , R, , SC, , S, , U, , 4., , K, , J, , I, , TA, , Name the congruent angles and sides for each pair of congruent triangles., 1. ΔTUV ≅ ΔXYZ, , 166, , B, , VII CLASS MATHEMATICS, , 2. ΔCDG ≅ ΔRSW, , C
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8.3, , Criterion for congruency of triangles, , Is it necessary for congruency to check whether all the corresponding parts of two triangles are, congruent? How can we check if the given triangles are congruent using a minimum number of, measurements. Let us explore and find out., 8.3.1, , Side-Side-Side congruency (SSS), , LA, N, , G, , AN, A, , Will all of you draw the same triangle if you only knew that the measure of one side of the triangle, is 5 cm? Kamal, Namrita and Susana have drawn them like this., , Kamal, , Namrita, , Susana, , As you can see all the triangles are different. Kamal drew an equilateral triangle, Namrita drew a, right-angled triangle and Susana drew an obtuse-angled triangle., , ER, , T, , TE, , Now can all of you draw the same triangle, if you knew the measures of only two sides of a triangle, say, 4 cm and 5 cm. Again Kamal, Namrita and Susana drew different triangles., , SC, , Kamal, , Namrita, , Susana, , What if you know all the sides of the triangle? Kamal, Namrita and Susana all drew the same, triangle with the three sides- 4 cm, 5cm and 6 cm., , Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 167
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Thus, if we want to make a copy of ABC or a triangle congruent to ABC, we need the lengths of, the three sides. This is referred to as the Side-Side-Side(SSS) criterion for congruency of, triangles., If two triangles are congruent because the lengths of their corresponding sides are equal, then will, their angles also be equal?, , AN, A, , Side-Side-Side (SSS) criterion for congruence of triangles: If three sides of a triangle are, equal to the corresponding three sides of another triangle, then the triangles are, congruent., Try This, , L, , G, , Measure the lengths of ΔLMN. Now, construct a, triangle with these measurements on a sheet of, paper. Place this triangle over ΔLMN. Are the, triangles congruent? What criterion of congruency, applies over here?, , LA, N, , 6 cm, , cm, , Y, , m, , 6 cm, , Z, , T, , P, , 5.3, c, , 2., 9, , cm, , 5.3, , X, , cm, , Q, , 2., 9, , R, , According to the given figure of ΔPQR and ΔXYZ, we have, , ER, , Solution :, , N, , Is ΔPQR ≅ ΔXYZ? Also, write the corresponding angles of the two triangles., , TE, , Example 1 :, , M, , PQ = XY = 2.9 cm, QR = YZ = 6 cm, , SC, , RP = ZX = 5.3 cm, Therefore, by Side-Side-Side congruence criterion, Δ PQR ≅ ΔXYZ, Clearly, the point P corresponds to point X, point Q corresponds to point Y and, the point R corresponds to point Z., So, ∠P, ∠X ; ∠Q, ∠Y ; ∠R, ∠Z are pairs of corresponding angles., , 168, , VII CLASS MATHEMATICS
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Exercise - 1, Decide whether the SSS congruence is true with the following figures. Give reasons, (i), , S, , (ii), , C, , 4, , 2.5, , A, N, , R, , AN, A, 4, , E, , B, , D, , L, 3, , 1., , A, , For the following congruent triangles, find the pairs of corresponding angles., T, , (ii), , S, , 5 cm, , P, , TE, , (iii), , ΔPQR ≅ ΔSQP, , (iv), , ΔPQR ≅ ΔSPQ, , P, , ER, , T, , ΔPQR ≅ ΔQPS, , Q, m, 5c, , S, , In the figure given below, AB = DC and AC = DB. Is ΔABC ≅ ΔDCB., , SC, 8.3.2, , m, 5c, , cm, , (ii), , S, , 3.5, , ΔPQR ≅ ΔPQS, , cm, , (i), , 5 cm, , R, , 3.5, , 4., , In adjacent figure, choose the correct answer!, , m, , R, , 3., , O, , m, 4c, , R, , Q, , Q, , 34 c, , P, , LA, N, , m, 4c, , G, , (i), , 4c, m, , 2., , A, , D, , B, , C, , Side-Angle-Side Congruence, , We have seen that it is not possible to draw congruent triangles, if we are given only the, measurements of one side. Now, what if you were given one angle and one side? Kamal, Namrita, and Susana were told to draw triangles with one side equal to 5 cm and one angle equal to 65o., They drew the following dissimilar triangles., Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 169
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65o, , 5 cm, , B, , A, , Kamal, , 65o, , 5 cm, , AN, A, , A, , B, , A, , Namrita, , 65o, , B, , 5 cm, , Susana, , G, , Now, what if the three of them knew the two sides of the triangle and the angle included between, these sides. The three children decided to draw triangles with sides 5 cm and 4.5 cm and the, included angle of ∠65., , 5 cm, , C, , B, , 5 cm, , 65 o, , C, , A, , cm, 4.5, , B, , TE, , LA, N, , Kamal drew ΔABC. He drew BC as the base= 5 cm. He then made ∠C= 65o using a protractor, and then marked point A at a length of 4.5 cm on the angular line. He then joined points A and B., , 65 o, , B, , 5 cm, , C, , ER, , T, , Can you draw the 65o angle at point B with side AB = 4.5 cm. Will the triangle that is formed be, congruent to Kamal's triangle? Can you take the base to be 4.5 cm, side = 5 cm and included, angle = 65o? Will the triangle that is formed be congruent to Kamal's triangle? You will find that the, triangles formed in all these situations are congruent triangles., Therefore, if we want to make a copy of ΔABC or a triangle congruent to ΔABC, we need the, lengths of the two sides and the measure of the angle between the two sides. This is referred to as, the Side-Angle-Side(SAS) criterion for congruence of triangles., , SC, , Side-Angle-Side(SAS) criterion for congruence of triangles: If two sides and the angle included between the two sides of a triangle are congruent to the corresponding two sides and the, included angle of another triangle, then the triangles are congruent., Try This, In ΔPQR measure the lengths PQ and QR as well as, ∠Q. Now, construct a triangle with these three, measurements on a sheet of paper. Place this triangle over, ΔPQR. Are the triangles congruent? What criterion of, congruency applies over here?, , 170, , VII CLASS MATHEMATICS, , P, , Q, , R
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Example 2 : See the measurements of the triangles given below. Are the triangles congruent?, Which are the corresponding vertices and angles?, C, , R, , o, , 40, , Q, , P, , B, , Solution :, , m, , AN, A, , A, , 5c, , 4 cm, , 4 cm, , m, 5c, , o, , 40, , In ΔABC and ΔPQR AC = QR and BC = PR and included angle ∠C ≅ ∠R, So, ΔABC ≅ ΔQPR., The correspondence is as follows:, B↔P, , and, , ∠B ≅ ∠P, , ∠C ≅ ∠R, , and, , LA, N, , Therefore, ∠A ≅ ∠Q,, , C↔R, , G, , A ↔ Q,, , Example 3 : In ΔPQR, PQ = PR and PS is angle bisector of ∠P., P, , Are ΔPQS and ΔPRS congruent? If yes, give reason., , In ΔPQS and ΔPRS, , TE, , Solution :, , Q, , PQ = PR ( given), , S, , R, , PS = PS (common side in both the triangles), , T, , and included angle ∠QPS ≅ ∠RPS (PS is the angle bisector), , Exercise - 2, , What additional information do you need to conclude that the two triangles given here, under are congruent using SAS rule?, , SC, , 1., , ER, , Therefore, ΔPQS ≅ ΔPRS (by SAS rule), , R, , H, , G, , T, J, , S, Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 171
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2., , The map given below shows five different villages. Village M lies exactly halfway between, the two pairs of villages A and B as well as and P and Q. What is the distance between, village A and village P. (Hint: check if ΔPAM ≅ ΔQBM), Q, , B, , G, , 3 km, , P, , Look at the pairs of triangles given below. Are they congruent? If congruent write the, corresponding parts., B, , R, , o, , 4c, m, , 6 cm, , T, , ER, , 3.7 cm, , SC, , (iv), , 30o, , 4 cm, , B, , o, , D, , 4.5 cm, , O, , 4., , A, , 3.7 cm, , S, , S, , R, , D, , 4.5 cm, , 4 cm, , 3 cm, , (iii), , T, , O, , 4 cm, , 130o S, , C, , W, , Q, , TE, , 130, , 3 cm, , 6 cm, , A, , P, , (ii), , 4c, m, , (i), , LA, N, , 3., , AN, A, , M, , A, , 4k, m, , m, 6k, , 7 cm, , 30, , C, , R, , Which corresponding sides do we need to know to prove that the triangles are congruent, using the SAS criterion?, (i), , A, , 5 cm, , P, , o, , 40, , D, , (ii), , Q, , o, , A, , 35, o, 35, , C, , o, , B, , 172, , 40, , 5 cm, , C, , VII CLASS MATHEMATICS, , R, , B
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8.3.3, , Angle-Side-Angle congruency (ASA), , Can the children construct a triangle if they know only one angle of the triangle? What if they know, two angles? Will children be able to draw congruent triangles if they know all the angles of the, triangle?, Kamal, Namrita and Susana drew the following triangles of angles- 40o, 60o and 80o., o, , 80o, o, , 80, , o, , 40, , 60, , 40o, , o, , Kamal, , 60o, , Namrita, , AN, A, , 80, , o, , 60, , o, , 40, , Susana, , G, , Therefore, though the angles of all the triangles are congruent, the lengths of their sides is not and, they are not congruent., , LA, N, , Thus, we need to know the length of, the sides, to draw congruent triangles., What if we have two angles and one, 60, 60, 40, side? Kamal and Namrita drew the, 5 cm, 5 cm, following triangles with angles 60o and, Kamal, Susana, 40o and side 5 cm. When both the, children constructed their triangles they made the given side, the included side., o, , o, , 40o, , TE, , o, , We can conclude that if we want to make a copy of a triangle or a triangle congruent to another, triangle, then we need to know two angles and the length of the side included between the two, angles. This is referred to as the Angle-Side-Angle criterion of congruence., , ER, , T, , Angle-Side-Angle criterion of congruence: If two angles and the included side of a triangle are, congruent to the two corresponding angles and included side of another triangle then the triangles, are congruent., Try This, , SC, , Teacher has asked the children to construct a triangle with angles 60°, 40° and with a, side 5 cm. Sushma calculated the third angle of the triangle as 80° using angle - sum, property of triangle., Then Kamal, Sushma and Namratha, constructed triangles differently using the following measurements., Kamal: 60°, 40° and 5cm side (as teacher said), Sushma: 80°, 40°, and 5 cm side, Namratha: 60°, 80° and 5 cm side., They cut these triangles and place them one upon the other. Are all of them, congruent? You also try this., , Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 173
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Example 4 :, , Two triangles ΔCAB and ΔRPQ are given below. Check whether the two are, congruent? If they are congruent, what can you say about the measures of the, remaining elements of the triangles., C, R, , Solution :, , In ΔCAB and ΔRPQ,, , 120o, , AB = PQ = 3 cm A, , 120o, , B, , 3 cm, , AN, A, , ∠B = ∠Q = 120°, , 4c, m, , 4c, m, , BC = QR = 4 cm, , P, , Q, , 3 cm, , Thus, two sides and included angle of ΔCAB are equal to the corresponding sides, and included angle of ΔRPQ., , G, , Therefore, by Side-Angle-side criterion of congruency ΔCAB ≅ ΔRPQ, Thus, in the two triangles, , LA, N, , AC ≅ PR, , ∠C ≅ ∠R and ∠A ≅ ∠P, , Example-5 : In the following picture, the equal angles in the two triangles are shown. Are the, triangles congruent?, A, , In ΔABD and ΔACD, , TE, , Solution :, , ∠BAD ≅ ∠CAD (given in the question), ∠ADB ≅ ∠ADC (given in the question), AD ≅ AD (common side, seen in the figure), , T, , B, , D, , ER, , Thus, by Angle-side-Angle congruence criterion ΔABD ≅ ΔACD, , Try This, , SC, , Is the following pair of triangles congruent? Give reason to support your answer., T, , M, 70o, o, , 70, , 5, o, , 30, , cm, , V, , K, 5 cm, , 30o, , L, W, , 174, , VII CLASS MATHEMATICS, , C
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Exerciese - 3, In following pairs of triangles, find the pairs which are congruent? Also, write the criterion, of congruence., A, , (i), , P, , 60o, , Q, , m, 6.5, c, , 70o, , 60o, , D, , C, , R, , B, , (iv), , cm, , (iii), 5.5, , O, , LA, N, , 5., 5, , cm, , 60o D, , C, , B, , 50o, , 50o, , C E, , 70o, , F, , D, , A, 30o, , 30o, , TE, , In the adjacent figure., , 30o o, 20, , G, , 70o, , C, , B, , 20o o, 30, , D, , A, , o, A 60, , 2., , A, , (ii), , m, 6 .5 c, , B, , 70o, , AN, A, , 1., , (i) Are ΔABC and ΔDCB congruent?, , O, , (ii) Are ΔAOB congruent to ΔDOC?, , B, , 50o, , o, , 50, , C, , Right-Angle Hypotenuse Side congruence, , ER, , 8.3.4, , T, , Also identify the relation between corresponding elements and, give reason for your answer., , In right-angled triangles we already know that one of the angles is a right angle. So what else do we, need to prove that the two triangles are congruent?, , SC, , Let us take the example of ΔABC with ∠B = 90. Can we draw a triangle congruent to this triangle, if,, (i), , only BC is known, , (ii), , only ∠C is known, , (iii), , ∠A and ∠C are known, , (iv), , AB and BC are known, , (v), , ∠C and BC are known, , (vi), , BC and the hypotenuse AC are known, , Free distribution by T.S. Government 2018-19, , A, , B, , C, , CONGRUENCY OF TRIANGLES, , 175
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When you try to draw the rough sketches of these triangles, you will find it is possible only, in cases (iv), (v) and (vi)., The last of the situations is new to us and it is called the Right-Angle Hypotenuse Congruence, Criterion., , Given below are measurements of some parts of two triangles. Examine whether, the two triangles are congruent or not, using RHS congurence rule. In case of, congruent triangles, write the result in symbolic form :, , G, , Example 6 :, , AN, A, , Right-Angle Hypotenuse Congruence Criterion: If the hypotenuse and one side of a, right angled triangle are equal to the corresponding hypotenuse and side of the other, right angled triangle, then the triangles are congruent., , ΔABC, , ∠P = 90o, PR = 3 cm, QR = 8 cm, , (ii) ∠A = 90o, AC = 5 cm, BC= 9 cm, , ∠Q = 90o, PR = 8 cm, PQ = 5 cm, , LA, N, , (i) ∠B = 90o, AC = 8 cm, AB = 3 cm, , TE, , Solution :, , Here, ∠B = ∠P = 90o, , T, , hypotenuse, AC = hypotenuse, RQ (= 8 cm) and, , R, , A, 8 cm, , 3 cm, , (i), , ΔPQR, , side AB = side RP (= 3 cm), , 8 cm, Q B, , P, , C, , Here, ∠A = ∠Q = 90o and, , SC, , (ii), , side AC = side PQ (= 5 cm)., hypotenuse, BC ≠ hypotenuse, PR, So, the triangles are not congruent., , 176, , VII CLASS MATHEMATICS, , R, , C, , 5 cm, , ER, , So, ΔABC ≅ ΔRPQ (By RHS Congruence rule)., , A, , 8 cm, , 9 cm, , B, , Q, , 5 cm, , P
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Example 7 :, , In the adjacent figure, DA ⊥ AB , CB ⊥ AB and, AC = BD., State the three pairs of equal parts in ΔABC and, ΔDAB., , D, , C, , A, , B, , Which of the following statements is meaningful?, Solution :, , (ii) ΔABC ≅ ΔABD, , The three pairs of equal parts are :, ∠ABC = ∠BAD (= 90o), AC = BD (Given), AB = BA (Common side), , G, , ΔABC ≅ ΔBAD (By RHS congruence rule)., , AN, A, , (i) ΔABC ≅ ΔBAD, , From the above,, , LA, N, , statement (i) is true;, , statement (ii) is not meaningful, in the sense that the correspondence among the, vertices is not staisfied., Try This, , TE, , In the figures given below, measures of some parts of triangles are given. By, applying RHS congruence rule, state which pairs of triangles are congruent. In, caseP of congruent triangles, write the result in symbolic form., , (i), , o, , 90, , 3.6, , 3c, m, , A, cm, , D, , (iii), , m, , 90o, , 90o, , 3.5 cm, , B, , (ii), , P, , cm, , C, , A, , E, , D, , m, 3c, , SC, , 3.6, , 2c, , T, , ER, o, , 90, , o, , 90, , R F, , B, , 2., , 6 cm, , C, , m, 2c, , 90o, , Q, , D, , 2.5 cm, , 6 cm, , 3 cm, , 1., , o, , Q, , 90, , R, , (iv), , It is to be established by RHS congruence rule that ΔABC ≅ ΔRPQ. What, additional information is needed, if it is given that ∠B = ∠P = 90o and AB = RP?, , Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 177
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3., , In the adjacent figure, BD and CE are altitudes of ΔABC, such that BD = CE., , A, , (i) State the three pairs of equal parts in ΔCBD and, ΔBCE., , AN, A, , 4., , C, , B, , (ii) Is ΔCBD ≅ ΔBCE? Why or why not?, (iii) Is ∠DBC = ∠EBC? Why or why not?, , D, , E, , ABC is an isosceles triangle with AB = AC and AD is one, of its altitudes (fig ...)., , A, , (i) State the three pairs of equal parts in ΔADB and, ΔADC., (iii) Is ∠B ≅ ∠C? Why or why not?, , C, , D, , LA, N, , (iv) Is BD ≅ CD? Why or why not?, , B, , G, , (ii) Is ΔADB ≅ ΔADC? Why or why not?, , Exercise - 4, , Which congruence criterion do you use in the following?, (i), , TE, , 1., , Given : AC = DF, , D, , A, , AB = DE, BC = EF, , Given : ZX = RP, , ER, , (ii), , T, , So, ΔABC ≅ ΔDEF, , C, , B, , R, , E, , F, , X, , Y, , Z, , RQ = ZY, , SC, , ∠PRQ ≅ ∠XZY, So, ΔPQR ≅ ΔXYZ, , (iii), , Given : ∠MLN ≅ ∠FGH, L, , ∠NML ≅ ∠GFH, , So, ΔLMN ≅ ΔGFH, , VII CLASS MATHEMATICS, , H, N, , ML = FG, , 178, , Q, , P, , F, M, , G
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(iv), , Given : EB = DB, , D, , AE = BC, , E, , ∠A = ∠C = 90o, So, ΔABE ≅ ΔCDB, , C, , B, , You want to show that ΔART ≅ ΔPEN,, , A, , AN, A, , 2., , A, , (i) If you have to use SSS criterion, then you need to show, (a) AR =, , (b) RT =, , (c) AT =, , R, , T, , (ii) If it is given that ∠T = ∠N and you are to use SAS, criterion, you need to have, and, , (ii) PN =, , G, , (a) RT =, , P, , E, , LA, N, , N, , (iii) If it is given that AT = PN and you are to use ASA criterion, you need to have, (a) ?, 3., , (b) ?, , You have to show that ΔAMP ≅ ΔAMQ., , Steps, , Reasons, , PM = QM, , (i), , ........., , (ii), , ∠PMA ≅ ∠QMA, , (ii), , ........., , (iii), , AM = AM, , (iii) ........., , (iv) ΔAMP ≅ ΔAMQ, , (iv) ........., , T, , (i), , ER, , 4., , TE, , In the following proof, supply the missing reasons., , A, , M, , P, , Q, , In ΔABC, ∠A = 30o, ∠B = 40o and ∠C = 110o, , SC, , In ΔPQR, ∠P = 30o, ∠Q = 40o and ∠R = 110o, A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or, why not?, R, , 5., , In the figure, the two triangles are congruent. The, corresponding parts are marked. We can write, ΔRAT ≅ ?, , T, , A, N, , O, , W, Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 179
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6., , Complete the congruence statement., , B, , Q, , P, , T, , A, , T, , AN, A, , C, , S, , ΔABC ≅ ?, , ΔQRS ≅?, , In a squared sheet, draw two triangles of equal areas such that, (i), , the triangles are congruent., , (ii), , the triangles are not congruent., , A, , LA, N, , What can you say about their perimeters?, , G, , 7., , R, , A, , 9., , Explain, why, , B, , TE, , If ΔABC and ΔPQR are to be congruent,, name one additional pair of corresponding, parts. What criterion did you use?, , 8., , D, , R, , C, , P, , E, , T, , ΔABC ≅ ΔFED., , C, , F, , ER, , B, , Looking Back, , 1. Congruent objects are objects having the same shape and size., , SC, , 2. The method of superimposition examines the congruence of plane figures., 3. Two line segments say, AB and CD are congruent if they have equal lengths. We, write this as AB ≅ CD. However, it is common to write it as AB = CD., , 4. If all the parts of one triangle are equal to the corresponding parts of other, triangle, then the triangles are congruent., , 180, , VII CLASS MATHEMATICS, , Q
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5. The necessary and sufficient conditions for two triangles to be congruent are as, follows:, , AN, A, , (i) Side-Side-Side (SSS) criterion for congruence: If three sides of a triangle, are equal to the corresponding three sides of another triangle, then the, triangles are congruent., (ii) Side-Angle-Side(SAS) criterion for congruence: If two sides and the angle, included between the two sides of a triangle are equal to the corresponding, two sides and the included angle of another triangle, then the triangles are, congruent., , LA, N, , G, , (iii) Angle-Side-Angle criterion of congruence: If two angles and the included, side of a triangle are equal to the corresponding two angles and included, side of another triangle then the triangles are congruent., , SC, , ER, , T, , TE, , (iv) Right-Angle Hypotenuse criterion of congruence: If the hypotenuse and, one side of a right-angled triangle are equal to the corresponding, hypotenuse and side of the other right-angled triangle, then the triangles are, congruent., , Free distribution by T.S. Government 2018-19, , CONGRUENCY OF TRIANGLES, , 181
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CONSTRUCTION OF TRIANGLES, Introduction, , AN, A, , 9.0, , 9, , You will learn how to construct triangles in this chapter. You do not need all the six elements i.e. the, three angles and three sides of a triangle to construct a triangle. A triangle can be drawn if you, know the elements that are required for two triangles to be congurent. Thus, a triangle can be, drawn in any of the situations given below i.e., if we know theThree sides of the triangle. (S.S.S), , (ii), , Two sides and the angle included between them. (S.A.S), , (iii), , Two angles and the side included between them (A.S.A), , (iv), , Hypotenuse and one adjacent side of a right-angled triangle. (R.H.S), , LA, N, , G, , (i), , A triangle can also be drawn if two of its sides and a non-included angle are given. However, it is, important to remember that this condition is not suffiecient to make two triangles, congruent., , TE, , Let us learn to construct triangles in each of the above cases., 9.1, , Construction of a triangle when measurements of the three sides are given (S.S.S), , ER, , T, , In the construction of any geometrical figure, drawing a rough sketch first, helps in indentifying the, sides. So we should first draw a rough sketch of the triangle we want to construct and label it with, the given measurements., Example 1: Construct a Δ PQR with sides PQ = 4 cm, QR = 5 cm and RP = 7 cm., , SC, , STEP 1 : Draw a rough sketch of the triangle and, label it with the given measureents., , STEP 2 : Draw a line segment QR of length 5 cm., 5 cm, , Q, 1, , 182, , 2, , R, 3, , VII CLASS MATHEMATICS, , 4, , 5
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G, , LA, N, , STEP 4 : Since P is at a distance of 7 cm from, R, draw another arc from R with, radius 7 cm such that it intersects first, arc at P., , AN, A, , STEP 3 : With centre Q, draw an arc of, radius 4 cm., , ER, , T, , TE, , STEP 5 : Join Q,P and P,R. The required Δ PQR, is constructed., , Try This, , Construct a triangle with the same measurements given in above example, taking, PQ as base. Are the triangles congurent?, , 2., , Construct a Δ PET, PE = 4.5 cm, ET = 5.4 cm and TP = 6.5 cm in your, , SC, , 1., , notebook. Now construct Δ ABC, AB = 5.4 cm, BC = 4.5 cm and CA = 6.5 cm, on a piece of paper. Cut it out and place it on the figure you have constructed, in your note book. Are the triangles congruent? Write your answer using, mathematical notation., , Free distribution by T.S. Government 2018-19, , CONSTRUCTION OF TRIANGLES, , 183
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Exercise - 1, Construct Δ ABC in which AB = 5.5 cm, BC = 6.5 cm and CA = 7.5 cm., , 2., , Construct Δ NIB in which NI = 5.6 cm, IB = 6 cm and BN = 6 cm. What type of triangle, is this?, , 3., , Construct an equilateral Δ APE with side 6.5 cm., , 4., , Construct a Δ XYZ in which XY = 6 cm, YZ = 8 cm and ZX = 10 cm. Using protractor, find the angle at X. What type of triangle is this?, , 5., , Construct Δ ABC in which AB = 4 cm, BC = 7 cm and CA= 3 cm. Which type of, triangle is this?, , 6., , Construct Δ PEN with PE = 4 cm, EN = 5 cm and NP = 3 cm. If you draw circles instead, of arcs how many points of intersection do you get? How many triangles with given, measurements are possible? Is this true in case of every triangle?, , LA, N, , G, , AN, A, , 1., , Try This, , TE, , Sushanth prepared a problem: Construct Δ XYZ in which XY = 2 cm, YZ = 8 cm, and XZ = 4 cm., , T, , He also drew the rough sketch as shown in, Figure 1., , ER, , Reading the problem, Srija told Sushanth that it would not be possible to draw a, triangle with the given measurements., , SC, , However, Sushanth started to draw the diagram as shown in Figure 2., , Check whether Sushanth can draw the triangle. If not why? Discuss with your friends., What property of triangles supports Srija’s idea?, , 184, , VII CLASS MATHEMATICS
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9.2, , Construction of a triangle with two given sides and the included angle (S.A.S), , Example 2 : Construct Δ ABC in which AB = 4 cm, BC = 5cm and ∠ B = 50°., , AN, A, , STEP 1 : Draw a rough sketch of a triangle and label it with, the given measurements., , STEP 2 : Draw a line segment AB of length 4 cm., , G, , X, , TE, , LA, N, , JJJG, STEP 3 : Draw a ray BX making an angle 50° with AB., , SC, , ER, , T, , JJJG, STEP 4 : Draw an arc of radius 5 cm from B, which cuts BX at C., , STEP 5 : Join C, A to get the required Δ ABC., , Free distribution by T.S. Government 2018-19, , CONSTRUCTION OF TRIANGLES, , 185
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Exercise - 2, Draw Δ CAR in which CA = 8 cm, ∠ A = 60° and AR = 8 cm. Measure CR, ∠ R and, ∠ C. What kind of triangle is this?, , 2., , Construct Δ ABC in which AB = 5 cm, ∠ B = 45° and BC = 6 cm., , 3., , Construct Δ PQR such that ∠ R=100°, QR = RP = 5.4 cm., , 4., , Construct Δ TEN such that TE = 3 cm, ∠ E = 90° and NE = 4 cm., , 9.3, , Construction of a triangle when two angles and the side between the angles is given, , AN, A, , 1., , LA, N, , G, , Example 3 : Construct Δ MAN with MA = 4 cm, ∠ M = 45° and ∠ A = 100°., , ER, , T, , TE, , STEP 1 : Draw rough sketch of a triangle and label it with, the given measurements., , STEP 2 : Draw line segment MA of length 4 cm., , SC, , 4 cm., , X, , JJJJG, STEP 3 : Draw a ray MX , making an angle 45° at M., , M, , 186, , VII CLASS MATHEMATICS, , 45o, , 4 cm., , A
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JJJG, STEP 4 : Draw a ray AY , making an angle 100°, at A., , AN, A, , JJJJG, Extend the ray MX if necessary to, JJJG, intersect ray AY ., , LA, N, , G, , 4 cm, , STEP 5 : Mark the intersecting point of the two rays, , 4 cm, , ER, , T, , TE, , as N. You have the required Δ MAN, , Try This, , SC, , Construct a triangle with angles 105° and 95° and a side of length of your choice., Could you construct the triangle? Discuss and justify., Exercise - 3, , 1., , Construct Δ NET with measurement NE = 6.4 cm, ∠ N = 50° and ∠ E = 100°., , 2., , Construct Δ PQR such that QR = 6 cm, ∠ Q = ∠ R = 60°. Measure the other two sides, of the triangle and name the triangle., , 3., , Construct Δ RUN in which RN = 5cm, ∠ R = ∠ N = 45°. Measure the other angle and, other sides. Name the triangle., , Free distribution by T.S. Government 2018-19, , CONSTRUCTION OF TRIANGLES, , 187
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9.4, , Construction of right-angled triangle when the hypotenuse and a side are given., , Example 4 : Construct Δ ABC, right-angled at A, and BC = 6 cm; AB = 5 cm., STEP 1 : Draw a rough sketch of right-angled triangle and, label it with given information., Note: side opposite to the right angle is called, , AN, A, , hypotenuse., , 5cm, , G, , STEP 2 : Draw a line segment AB of length 5 cm., , TE, , LA, N, , JJJG, STEP 3 : Construct a ray AX perpendicular to AB at A., , SC, , ER, , T, , STEP 4 : Draw an arc from B with radius 6cm to intersect, JJJG, AX at ‘C’., , STEP 5 : Join B,C to get the required Δ ABC., , 188, , VII CLASS MATHEMATICS, , C
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Exercise - 4, Construct a right-angled Δ ABC such that ∠ B = 90°, AB = 8 cm and AC = 10 cm., , 2., , Construct a Δ PQR, right-angled at R, hypotenuse is 5 cm and one of its adjacent sides is, 4 cm., , 3., , Construct an isosceles right-angled Δ XYZ in which ∠ Y = 90° and the two sides are, 5 cm each., , 9.5, , Construction of triangle when two sides and the non-included angle are given, , AN, A, , 1., , G, , Example 5 : Construct Δ ABC such that AB = 5 cm, AC = 4 cm, ∠ B = 40°., , TE, , LA, N, , STEP 1 : Draw rough sketch of Δ ABC and label it with, the given measurements., , STEP 2 : Draw a line segment AB of length 5 cm., , B, , 5 cm, , X, , SC, , ER, , T, , A, , JJJG, STEP 3 : Draw a ray BX making an angle 40° at B., 400, A, , Free distribution by T.S. Government 2018-19, , 5 cm, , CONSTRUCTION OF TRIANGLES, , B, , 189
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X, , STEP 4 : With A as centre and radius 4 cm, draw an arc to, JJJG, cut ray BX ., C, , AN, A, , 400, 5 cm, , A, , B, , X, , STEP 5 : Mark the intersecting point as C and join, , LA, N, , G, , C, A to get the required Δ ABC., , C, 400, , A, , TE, , JJJG, Can you cut the ray BX at any other point?, , B, , 5 cm, , X, C’, , You will see that as ∠ B is acute, the arc from, , T, , JJJG, A of radius 4 cm cuts the ray BX twice., , C, , ER, , 400, B, , 5 cm, , A, , SC, , So we may have two triangles as given below:, X, , X, , C’, , 4 cm, , C’, , C, , 4 cm, , 400, , 400, A, , 190, , 5 cm, , VII CLASS MATHEMATICS, , C, , B, , A, , 5 cm, , B
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Try This, Construct a triangle with two sides of length of your choice and the non-included, angle as an obtuse angle. Can you draw two triangles in this solution?, Exercise - 5, Construct Δ ABC in which AB = 4.5 cm, AC = 4.5 cm and ∠ B = 50°. Check whether, you get two triangles., , 2., , Construct Δ XYZ such that XY = 4.5 cm, XZ = 3.5 cm and ∠ Y = 70°. Check whether, you get two triangles., , 3., , Construct Δ ANR with the sides AN and AR of lengths 5 cm and 6 cm respectively and, ∠ N is 100°. Check whether you get two triangles., , 4., , Construct Δ PQR in which QR = 5.5 cm, QP = 5.5 cm and ∠ Q = 60°. Measure RP., What kind of triangle is this?, , 5., , Construct the triangles with the measurement given in the following table., , LA, N, , G, , AN, A, , 1., , Measurements, , Δ ABC, , BC = 6.5 cm, CA = 6.3 cm, AB = 4.8 cm., , Δ PQR, Δ XYZ, , PQ = 8 cm, QR = 7.5 cm, ∠ PQR = 85°, , XY = 6.2 cm, ∠ Y = 130°, ∠ Z = 70°, AB = 4.8 cm, AC = 4.8 cm, ∠ B = 35°, , T, , Δ ABC, , TE, , Triangle, , ∠ N = 90°, MP = 11.4 cm., MN = 7.3 cm., , Δ RKS, , RK = KS = SR = 6.6 cm., , Δ PTR, , ∠ P = 65°, PT = PR = 5.7 cm., , ER, , Δ MNP, , SC, , Looking Back, , •, , A triangle can be constructed when., (i) The three sides of the triangle are given., (ii) Two sides and the angle included between them is given., (iii) Two angles and their included side is given., (iv) The hypotenuse and one adjacent side of a right angle triangle are given., (v) Two sides and the not included angle are given., , Free distribution by T.S. Government 2018-19, , CONSTRUCTION OF TRIANGLES, , 191
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10, , ALGEBRAIC EXPRESSIONS, Introduction, , AN, A, , 10.0, , In class VI you had already learnt that variables can take on different values and the value of, constants is fixed. You had also learnt how to represent variables and constants using letters like x,, y, z, a, b , p, m etc. You also came across simple algebraic expressions like 2x –3 and so on. You, had also seen how these expressions are usefull in formulating and solving problems., , LA, N, , G, , In this chapter, you will learn more about algebraic expressions and their addition and subtraction., However, before doing this we will get acquainted to words like ‘terms’, ‘like terms’,‘unlike, terms’ and ‘coefficients’., Let us first review what you had learnt in class VI., , Exercise - 1, , Find the rule which gives the number of matchsticks required to make the following, patterns(i), , A pattern of letter 'H', , (ii), , A pattern of letter 'V', , Given below is a pattern made from coloured tiles and white tiles., 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 12345, 1234, 1234, 1234, 1234, 12345, 1234, 1234, 1234, 12345, 1234, 1234, 1234, 12345, 1234, 1234, 1234, 1234, 1234, 12345, 1234, 1234, 1234, 1234, 1234, , 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 12345, 1234, 1234, 1234, 12345, 1234, 12341234, 1234, 1234, 1234, 1234, 12341234, 12341234, 12341234, 12341234, 1234, 1234, 123412341234, 1234, 123412341234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, , Figure 2, , Figure 3, , T, , 2., , TE, , 1., , SC, , ER, , 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, 1234, , Figure 1, , 192, , (i), , Draw the next two figures in the pattern above., , (ii), , Fill the table given below and express the pattern in the form of an algebraic, expression., Figure Number, , 1, , Number of coloured tiles, , 4, , VII CLASS MATHEMATICS, , 2, , 3, , 4, , 5
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(iii), , 5., , 1, , Number of total tiles, , 5, , 2, , 3, , 4, , 5, , AN, A, , Write the following statements using variables, constants and arithmetic operations., 6 more than p, , (ii), , 'x' is reduced by 4, , (iii), , 8 subtracted from y, , (iv), , q multiplied by '-5', , (v), , y divided by 4, , (vi), , One-fourth of the product of 'p' and 'q', , (vii), , 5 added to the three times of 'z', , (viii), , x multiplied by 5 and added to '10', , (ix), , 5 subtracted from two times of 'y', , (x), , y multiplied by 10 and added to 13, , TE, , LA, N, , G, , (i), , Write the following expressions in statements., (i), , x+3, , (iv), , x, 5, , (ii), , y–7, , T, , 4., , Figure Number, , (v), , ER, , 3., , Fill the table given below and express the pattern in the form of an algebraic, expression., , 3m + 11, , (iii), , 10l, , (vi), , 2y – 5, , Some situations are given below. State the number in situations is a variable or constant?, , SC, , Example : Our age - its value keeps on changing so it is an example of a variable quantity., (i), , The number of days in the month of January, , (ii), , The temperature of a day, , (iii), , Length of your classroom, , (iv), , Height of the growing plant, , Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 193
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10.1, , Algebraic Term and Numeric term, , Consider the expression 2x + 9., Here 'x' is multiplied by 2 and then 9 is added to it. Both ‘2x’ and ‘9’ are terms in the expression 2x + 9. Moreover 2x is called algebraic term and 9 is called numeric term., Consider another expression 3x2 – 11y., , G, , AN, A, , 3x2 is formed by multiplying 3, x and x. 11y is the product of 11 and y. 11y is then subtracted from, 3x2 to get the expression 3x2 – 11y. In the expression 3x2–11y, 3x2 is one term and 11y is the, other term., When we multiply x with x we can write this as x2. This is, similar to writing 4 multiplied by 4 as 42., Similarly when we multiply x three times i.e., x ×x ×x we, ×6 ×6 as 63., can write this as x3. This is similar to writing 6×, , LA, N, , Do This, In the expressions given below identify all the terms., (i), , 5x2 + 3y + 7, , (iv) 5x – 7, , (ii) 5x2y + 3, , (iii) 3x2y, , (v) 5x + 8 –2(– y), , (vi) 7x2 – 2x, , TE, , 10.1.1 Like and unlike terms, , Let us observe the following examples., 5x and 8x, , (iii), , 3xy and 4xy, , (ii), , 7a2 and 14a2, , (iv), , 3xy2 and 4x2y, , T, , (i), , ER, , In the first example, both terms contain the same variable i.e. x and the exponent of the variable is, also the same i.e. 1, , SC, , In the second example, both terms contain the same variable i.e. a and the exponent of the variable, is also the same i.e. 2, In the third example, both terms contain the same variables i.e. x and y and the exponent of, variable x is 1 and the exponent of variable y is 1., In the fourth example, both terms contain the same variables x and y. However, their exponents, are not the same. In the first term, the exponent of x is 1 and in the second it is 2. Similarly, in the, first term the exponent of y is 2 and in the second term it is 1., The first three pairs of terms are examples of ‘like terms’ while the fourth is a pair of ‘unlike terms’., Like terms are terms which contain the same variables with the same exponents., 194, , VII CLASS MATHEMATICS
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Do This, 1., , Group the like terms together., 12x, 12, 25x, –25, 25y, 1, x, 12y, y, 25xy, 5x2y, 7xy2, 2xy, 3xy2, 4x2y, , 10.2, , State true or false and give reasons for your answer., (i), , 7x2 and 2x are unlike terms, , (ii), , pq2 and –4pq2 are like terms, , (iii), , xy, –12x2y and 5xy2 are like terms, , Co-efficient, , In 9 xy ;, , '9' is the co-efficient of 'xy' as 9 (xy) = 9xy, , G, , 'x' is the co-efficient of '9y' as x (9y) = 9xy, , AN, A, , 2., , ‘y’ is the co-efficient of ‘9x’ as y (9x) = 9xy, , LA, N, , '9x' is the co-efficient of 'y' as 9x (y) = 9xy, 9y is the co-efficient of ‘x’ as, , 9y (x) = 9xy, , xy is the co-efficient of ‘9’ as, , xy (9) = 9xy, , TE, , Since 9 has a numerical value it is called a numerical coefficient x, y and xy are literal coefficients, because they are variables., Similarly in '-5x', ‘–5’ is the numerical coefficient and ‘x’ is the literal coefficient., Try This, , (i) What is the numerical coefficient of 'x' ?, , T, , (ii) What is the numerical coefficient of '–y’ ?, , ER, , (iii) What is the literal coefficient of ‘–3z’ ?, (iv)Is a numerical coefficient a constant?, (v) Is a literal coefficient always a variable?, , Expressions, , SC, , 10.3, , An expression is a single term or a combination of terms connected by the symbols ‘+’, (plus) or ‘–’ (minus)., For example : 6x + 3y, 3x2+ 2x+y, 10y3 + 7y + 3, 9a + 5, 5a + 7b, 9xy, 5 + 7 – 2x, 9+ 3– 2, Note: multiplication '×' and division ‘÷’ do not separate terms. For example 2 x × 3 y and, , 2x, 3y, , are single terms., Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 195
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Do This, How many terms are there in each of the following expressions?, , (i), , x+y, , (iv) x2z + 3, , x−, , (vii), , 11, 3, , (ii), , 11x – 3y – 5, , (iii), , 6x2 + 5x – 4, , (v), , 5x2y, , (vi), , x+3+y, , (viii), , 3x, 7y, , (ix), , AN, A, , 1., , 2z – y, , (x) 3x + 5, , 10.3.1 Numerical expressions and algebraic expressions, , G, , Consider the following examples., , 11, 3, , 1 + 2 –9, , (ii), , –3 – 5, , (iii), , x−, , (v), , 9 + (6–5), , (vi), , 3x + 5, , (vii), , (17–5) +4, , LA, N, , (i), , (iv), , 4y, , (viii) 2x – y, , TE, , Do you find any algebraic terms in the examples (i), (ii), (v) and (vii)?, If every term of an expression is a constant term, then the expression is called numerical, expression. If an expression has at least one algebraic term, then the expression is called, an algebraic expression., , T, , Which are the algebraic expressions in the above examples?, , ER, , Try This, , SC, , Write 3 algebraic expressions with 3 terms each., , Aryabhata (India), 475 - 550 AD, He wrote an astronomical treatise, Aryabhatiyam (499AD)., He was the first Indian mathematician who used algebraic, expressions. India’s first satellite was named Aryabhata., , 196, , VII CLASS MATHEMATICS
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10.3.2 Types of algebraic expressions, Algebraic expressions are named according to the number of terms present in them., No. of terms, , Name of the Expression, , Examples, , One term, , Monomial, , (a) x (b) 7xyz, , Two unlike terms, , (d) qz2, , AN, A, , (c) 3x2y, Binomial, , (a) a + 4x, , (b) x2 + 2y, , (c) 3x2 – y2, , Trinomial, , (a) ax2 + 4x + 2, , G, , Three unlike terms, , (b) 7x2 + 9y2 + 10z3, , Multinomial, , (a) 4x2 + 2xy + cx + d, , LA, N, , More than one, unlike terms, , (b) 9p2–11q + 19r + t, , Note: Binomial, trinomials are also multinomial algebraic expressions., Do This, , Give two examples for each type of algebraic expression., , 2., , Identify the expressions given below as monomial, binomial, trinomial, and multinomial., (iii), , 5x2y + 6x, , (ii), , 3xy, , T, , 5x2 + y + 6, , (iv), , a + 4x – xy + xyz, , Degree of algebraic expressions, , ER, , 10.4, , (i), , TE, , 1., , Before discussing the degree of algebraic expressions let us understand what we mean by the, degree of a monomial., , SC, , 10.4.1 Degree of a monomial, Consider the term 9x2 y2, , 1. What is the exponent of 'x' in the above term?, 2. What is the exponent of 'y' in the above term?, 3. What is the sum of these two exponents?, The sum of all exponents of the variables present in a monomial is called the degree of, the term or degree of the monomial., Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 197
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Study the following table., S. No., , Monomial, , Exponents, , Degree of the monomial, , x, , y, , z, , x, , 1, , -, , -, , 1, , 2, , 7x 2, , 2, , -, , -, , 2, , 3, , -3xyz, , 1, , 1, , 1, , 4, , 8y2z2, , -, , 2, , 2, , Let us discuss the degree of the constant term 5., , 1+1+1=3, 2+2=4, , G, , 10.4.2 Degree of constant terms, , AN, A, , 1, , Degree of constant term is zero., , LA, N, , Since xo = 1, we can write 5 as 5xo. Therefore, the degree of 5 is '0'., , 10.4.3 Degree of algebraic expressions, Study the following table., , Degree of each term, , TE, , S. No. Algebraic Expression, , Highest Degree, , First, term, , Second, term, , Third, term, , Fourth, term, , 3, , -, , -, , -, , 3, , 7xy2, , 2, , 3y – x2y2, , 1, , 4, , -, , -, , 4, , 3, , 4x2 + 3xyz + y, , 2, , 3, , 1, , -, , 3, , pq – 6p2q2 – p2q + 9, , 2, , 4, , 3, , 0, , 4, , ER, , 4, , T, , 1., , SC, , In the second example the highest degree of one of the terms is 4. Therefore, the degree of the, expression is 4. Similarly, the degree of the third expression is 3 and the degree of the fourth, expression is 4., The highest of the degrees of all the terms of an expression is called the degree of the, expression., , 198, , VII CLASS MATHEMATICS
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learn how., 1., , Number of pencils with Siddhu is not given in the problem, we shall take the number as 'x'., Vinay has 4 times of Siddu i.e., 4 × x = 4x, To find the total number of pencils, we have to add x and 4x, Therefore, the total number of pencils = x + 4x = (1 + 4) x = 5x (distributive law), Since the cost of each book is not given, we shall take it as 'y'., , AN, A, , 2., , Therefore, Tony spends 7 × y = 7y, Basha spends 2 × y = 2y, , To find how much more Tony spends,we have to subtract 2y from 7y, , Therefore, the amount spent more = 7y– 2y = (7–2)y = 5y (distributive law), , G, , Thus, we can conclude that., , LA, N, , The sum of two or more like terms is a like term with a numerical coefficient equal to the, sum of the numerical coefficients of all the like terms in addition., The difference between two like terms is a like term with a numerical coefficient equal to, the difference between the numerical coefficients of the two like terms., , 5x,7x, , (iv), , 18ab,5ab, 12ab, , (vii), , 18pq, –15pq, 3pq, , (ii), , 7x2y, –6x2y, , (iii), , 2m, 11m, , (v), , 3x2, –7x2, 8x2, , (vi), , 4m2, 3m2, –6m2, m2, , (iii), , 12y, 3y, , T, , (i), , Subtract the first term from the second term., (i), , 2xy, 7xy, , (ii), , 5a2, 10a2, , (iv), , 6x2y, 4x2y, , (v), , 6xy, –12xy, , SC, , 2., , Find the sum of the like terms., , ER, , 1., , TE, , Do This, , 10.5.1 Addition and subtraction of unlike terms, 3x and 4y are unlike terms. Their sum can be wirtten as 3x + 4y., However, 'x' and 'y' are different variables so we can not apply distributive law and thus cannot, add them., , 200, , VII CLASS MATHEMATICS
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10.6, , Simplification of an algebraic expression, , Consider the expression 9x2 – 4xy + 5y2 + 2xy – y2 – 3x2 + 6xy, , S.No. Steps, , Process, , AN, A, , We can see that there are some like terms in the expression. These are 9x2 and –3x2 ; 5y2 and y2, and 2xy and +6xy. On adding the like terms we get an algebraic expression in its simplified form., Let us see how the expression given above is simplified., , 9 x 2 − 4 xy + 5 y 2 + 2 xy − y 2 − 3 x 2 + 6 xy, , 1., , Write down the expression, , 2., , Group the like terms together (9 x 2 − 3 x 2 ) + (2 xy − 4 xy + 6 xy ) + (5 y 2 − y 2 ), , 3., , Addding the like terms, , (9 − 3) x 2 + (2 − 4 + 6) xy + (5 − 1) y 2 = 6 x 2 + 4 xy + 4 y 2, , G, , Note : If no two terms of an expression are alike then it is said to be in the simplified, form., , LA, N, , Let us study another example: 5 x 2 y + 2 x 2 y + 4 + 5 xy 2 − 4 x 2 y − xy 2 − 9, Step 1: 5 x 2 y + 2 x 2 y + 4 + 5 xy 2 − 4 x 2 y − xy 2 − 9, , Step 3: 3x 2 y + 4 xy 2 − 5, Do This, , TE, , Step 2: (5 x 2 y + 2 x 2 y − 4 x 2 y ) + (5 xy 2 − xy 2 ) + (4 − 9) (bringing the like terms together), , Simplify the following., , (i), , 3m + 12m − 5m, , (ii), , 25 yz − 8 yz − 6 yz, , (iii), , 10m 2 − 9m + 7 m − 3m 2 − 5m − 8, , (iv), , 9 x 2 − 6 + 4 x + 11 − 6 x 2 − 2 x + 3 x 2 − 2, , (v), , 3a 2 − 4a 2b + 7 a 2 − b 2 − ab, , (vi), , 5 x 2 + 10 + 6 x + 4 + 5 x + 3 x 2 + 8, , 10.7, , Standard form of an expression, , ER, , T, , 1., , SC, , Consider the expression 3 x + 5 x 2 − 9 . The degrees of first, second and third terms are 1, 2, and 0, respectively. Thus, the degrees of terms are not in the descending order., By re-arranging the terms in such a way that their degrees are in descending order; we get the, expression 5 x 2 + 3 x − 9 . Now the expression is said to be in standard form., Let us consider 3c + 6a –2b. Degrees of all the terms in the expression are same. Thus the, expression is said to be already in standard form. If we write it in alphabetical order as, 6a –2b + 3c it looks more beautiful., , Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 201
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In an expression, if the terms are arranged in such a way that the degrees of the terms, are in descending order then the expression is said to be in standard form., Examples of expressions in standard form (i) 7 x 2 + 2 x + 11 (ii), , 5 y2 − 6 y − 9, , Do This, Write the following expressions in standard form., (i), , 3 x + 18 + 4 x 2, , (ii), , 8 − 3x 2 + 4 x, , (iii), , −2m + 6 − 3m 2, , (iv), , y3 + 1 + y + 3 y 2, , Identify the expressions that are in standard form?, 9x2 + 6x + 8, , (ii), , 9x2 + 15 + 7x, , (iii), , 9x2 + 7, , (iv), , 9x3 + 15x + 3, , (v), , 15x2 + x3 + 3x, , (vi), , (vii), , x3 + x2y2 + 6xy, , G, , (i), , 3., , Write 5 different expressions in standard form., , 10.8, , Finding the value of an expression, , Find the value of 3x2 if x = –1, Step 1: 3x 2, (write the expression), 2, Step 2 : 3(–1) (substitute the value of variable), Step 3 : 3(1) = 3, , TE, , Example 1:, Solution :, , x2y + xy + 3, , LA, N, , 2., , AN, A, , 1., , Step 1 : x 2 − y + 2, , ER, , Solution :, , T, , Example 2 : Find the value of x 2 − y + 2 if x = 0 and y = −1, , x= –1, , 3x 2, 3, , (write the expression), , Step 2 : 02 − (−1) + 2 (substitute the value of variable), , Step 3 : 1 + 2 = 3, , SC, , 1, Example 3 : Area of a triangle is given by A = bh . If b = 12 cm and h = 7cm find the area, 2, of the triangle., , Solution :, , 202, , Step 1:, , 1, A = bh, 2, , Step 2:, , 1, A = ×12 × 7, 2, , Step 3:, , A = 42 sq. cm., , VII CLASS MATHEMATICS
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Try This, 1. Find the value of the expression '–9x' if x = –3., 2. Write an expression whose value is equal to -9, when x = –3., , 1., , 2., , AN, A, , Exercise - 3, , Find the length of the line segment PR in the following figure in terms of 'a'., , (i), , Find the perimeter of the following triangle., , G, , 2x, 5x, , LA, N, , 6x, , (ii) Find the perimeter of the following rectangle., 2x, , 4. ., , 8x, 5x, , TE, , (i), , (ii), , 5p, 11p, , (iii), , 13m2, 2m2, , (iii), , –2x2, , Find the value of following monomials, if x = 1., –x, , (ii), , 4x, , ER, , (i), , T, , 3., , 3x, Subtract the second term from first term., , Simplify and find the value of 4 x + x − 2 x 2 + x − 1 when x = –1., , 6., , Write the expression 5x2 – 4 – 3x2 + 6x + 8 + 5x – 13 in its simplified form. Find its value, when x = –2, , SC, , 5., , 7., , If x = 1 ; y = 2 find the values of the following expressions, , (i), , 4x–3y + 5, , (ii), , x2 + y2, , (iii), , xy + 3y – 9, , 8., , Area of a rectangle is given by A = l × b . If l = 9cm, b = 6cm, find its area?, , 9., , Simple interest is given by I =, , PTR, . If P = . 900, T = 2 years; and R = 5%, find the, 100, , simple interest., Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 203
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The relationship between speed (s), distance (d) and time (t) is given by s =, , 10., , d, . Find the, t, , value of s, if d = 135 meters and t = 10 seconds., 10.9, , Addition of algebraic expressions, , AN, A, , Consider the following problems., 1., , Sameera has some mangoes. Padma has 9 more than Sameera., Mary says that she has 4 more mangoes than the number of, mangoes Sameera and Padma have together. How many, mangoes does Mary have?, , G, , Since we do not know the number of mangoes that Sameera has, we shall take them to be x, mangoes., , LA, N, , Padma has 9 more mangoes than Sameera., , Therefore, the number of mangoes Padma has = x + 9 mangoes, , Mary has 4 more mangoes than those Sameera and Padma have together., , TE, , Therefore, the number of mangoes Mary has = x + (x + 9) + 4 mangoes, = 2x + 13 mangoes, , In a Mathematics test Raju got 11 marks more than Imran. Rahul got 4 marks less than, what Raju and Imran got together. How much did Rahul score?, , T, , 2., , ER, , Since we do not know Imran’s marks, we shall take them to be x marks., Hint : Why are we taking Imran's marks as x?, , SC, , Raju got 11 more marks than Imran = x + 11 marks, Rahul got 4 marks less than the marks Raju and Imran scored together = x + x + 11–4 marks, = 2 x + 7 marks, , In both the situations given above, we have to add and subtract algebraic expressions. There are, number of real life situations in which we need to do this. Let us now learn how to add or subtract, algebraic expressions., , 204, , VII CLASS MATHEMATICS
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10.9.1 Addition of Expressions, The sum of expressions can be obtained by adding like terms.This can be done in two ways., Column or Vertical Method, , (ii), , Row or Horizontal Method, , Column or Vertical Method, , AN, A, , (i), , (i), , Example 4 : Add 3 x 2 + 5 x − 4 and 6 + 6x 2, Solution:, S. No., , Steps, Write the expressions in standard form, , (i) 3x2 + 5x – 4 = 3x2 + 5x – 4, , G, , 1, , Process, , (ii) 6 + 6x2 = 6x2 + 6, , LA, N, , if necessary, , 3., , Write one expression below the other such that, , 3x2 + 5x – 4, , the like terms come in the same column, , 6x2, , Add the like terms column wise and write the, , 3x2 + 5x – 4, , result just below the concerned column, , 6x2, , +6, +6, , 9x2 + 5x + 2, , TE, , 2, , Example 5 : Add 5 x 2 + 9 x + 6, 4 x + 3 x 2 − 8 and 5 − 6x, Step 1 :, , 5 x 2 + 9 x + 6 = 5x 2 + 9 x + 6, , T, , Solution:, , ER, , 4 x + 3x 2 − 8 = 3x 2 + 4 x − 8, , 5 − 6x, , SC, , Step 2 :, , Step 3 :, , =, , − 6x + 5, , 5x 2 + 9 x + 6, 3x 2 + 4 x − 8, − 6x + 5, 5x 2 + 9 x + 6, 3x 2 + 4 x − 8, − 6x + 5, , 8x2 + 7x + 3, Free distribution by T.S. Government 2018-19, , ALGEBRAIC EXPRESSIONS, , 205
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(ii), , Row or Horizontal Method, , Example 6 :, , Add 3x2 + 5x – 4 and 6 + 6x2, , S. No., , Steps, , Process, , Write all expressions with addition, symbol in between them., , 3x2 + 5x – 4 + 6 + 6x2, , 2, , Re-arrange the term by grouping, the like terms together., , (3x2 + 6x2) + (5x) + (–4 + 6), , 3, , Simplify the coefficients, , (3+6) x2 + 5x + 2, , 4, , Write the resultant expression in, , 9x2 + 5x + 2, , AN, A, , 1, , Do This, Add the following expressions., (i), , x − 2 y , 3x + 4 y, , LA, N, , 1., , G, , standard form., , (ii) 4m2 – 7n2 + 5mn, 3n2 + 5m2 – 2mn, , TE, , (iii) 3a – 4b, 5c – 7a + 2b, , 10.9.2 Subtraction of algebraic expressions, 10.9.2(a)Additive inverse of an expression, , T, , If we take a positive number '9' then there exists '–9' such that 9 + (–9) = 0., Here we say that ‘–9’ is the additive inverse of ‘9’ and ‘9’ is the additive inverse of ‘–9’., , ER, , Thus, for every positive number, there exists a negative number such that their sum is, zero. These two numbers are called the additive inverse of the each other., , SC, , Is this true for algebraic expressions also? Does every algebraic expression have an additive, inverse?, If so, what is the additive inverse of ‘3x’?, For ‘3x’ there also exists ‘–3x’ such that 3x + (–3x ) = 0, Therefore, ‘–3x’ is the additive inverse of ‘3x’ and ‘3x’ is the additive inverse of ‘–3x’., Thus, for every algebraic expression there exists another algebraic expression such that, their sum is zero. These two expressions are called the additive inverse of the each other., , 206, , VII CLASS MATHEMATICS
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8., , What should be taken away from 3x2– 4y2 + 5xy + 20 to get –x2– y2 + 6xy + 20., , 9., , The sum of 3 expressions is 8 + 13a + 7a2. Two of them are 2a2 + 3a + 2 and 3a2 –4a + 1., Find the third expression., , 10., , If, , A = 4x2 + y2 – 6xy;, , C = 6x2 + 8y2 + 6xy, Find (i) A + B + C (ii) (A – B) –C, , (iii) 2A+B, , Looking Back, , AN, A, , B = 3y2 + 12x2 + 8xy;, , (iv) A-3B, , G, , • An algebraic expression is a single term or a combination of terms connected by, the symbols ‘+’ (plus) or ‘–’ (minus)., , LA, N, , • If every term of an expression is a constant term, then the expression is called a, numerical expression. If an expression has at least one algebraic term, then the, expression is called an algebraic expression., , TE, , • An algebraic expression contaning one term is called a monomial. An algebraic, expression contaning two unlike terms is called a binomial. An algebraic, expression contaning three unlike terms is called a trinomial. An algebraic, expression contaning more than three unlike terms is called a multinomial., • The sum of all the exponents of the variables in a monomial is called the degree of, the term or degree of monomial., , T, , • The degree of any constant term is zero., , ER, , • The highest of the degrees of all the terms of the expression is called the degree of, the expression., , SC, , • If no two terms of an expression are alike then the expression is said to be in its, simplified form., • In an expression, if the terms are arranged in a manner such that the degrees of the, terms are in descending order then the expression is said to be in standard form., • The sum of two or more like terms is a like term with a numerical coefficient equal, to the sum of the numerical coefficients of all the like terms., • The difference between two like terms is a like term with a numerical coefficient, equal to the difference between the numerical coefficients of the two like terms., , 210, , VII CLASS MATHEMATICS
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POWERS AND EXPONENTS, Introduction, , AN, A, , 11.0, , 11, , The population of India according to 2011 census is about 120,00,00,000, , The approximate distance between the sun and the earth is 15,00,00,000 km., , The speed of the light in vacuum is about 30,00,00,000 m/sec. Light travell a distance, of 30,00,00,000 mts., approximately in 1 sec.,, , G, , The population of Andhra Pradesh according to 2011 census is about 8,50,00,000., , LA, N, , These are all very large numbers. Do you find it easy to read, write and understand such large, numbers? No, certainly not., Thus, we need a way in which we can represent such larger numbers in a simpler manner. Exponents help us in doing so. In this chapter you will learn more about exponents and the laws of, exponents., 11.1, , Exponential Form, 4+4+4+4+4, , TE, , Let us consider the following repeated additions:, , 5+5+5+5+5+5, , T, , 7+7+7+7+7+7+7+7, , ER, , We use multiplication to shorten the representation of repeated additions by writing 5×4, 6×5 and, 8×7 respectively., Now can we express repeated multiplication of a number by itself in a simpler way?, , SC, , Let us consider the following illustrations., The population of Bihar as per the 2011 Census is about 10,00,00,000., Here 10 is multiplied by itself for 8 times i.e. 10×10×10×10×10×10×10×10., So we can write the population of Bihar as 108. Here 10 is called the base and 8 is called the, exponent. 108 is said to be in exponential form and it is read as 10 raised to the power of 8., The speed of light in vacuum is 30,00,00,000 m/sec. This is expressed as 3×108 m/sec in, exponential form. In 3×108, 108 is read as ‘10 raised to the power of 8’. 10 is the base and 8 is, the exponent., Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 211
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The approximate distance between the sun and the earth is 15,00,00,000 km. This is expressed as, 15×107 km in exponential form. In 107, 10 is the base and 7 is the exponent., The population of Andhra Pradesh according to 2011 census is about 8,50,00,000. This is, expressed as 85×106 in exponential form. 106 is read as ‘10 raised to the power of 6’. Here 10, is the base and 6 is the exponent., , AN, A, , We can also use exponents in writing the expanded form of a given number for example the, expanded form of 36584 = (3×10000) + (6×1000) + (5×100) + (8×10) + (4×1), = (3×104) + (6×103) + (5×102) + (8×101) + (4×1), Do This, , Write the following in exponential form. (values are rounded off), , G, , 1., , LA, N, , (i) Total surface area of the Earth is 510,000,000 square kilometers., (ii) Population of Rajasthan is approximately 7,00,00,000, (iii) The approximate age of the Earth is 4550 million years., (iv) 1000 km in meters, , Express (i) 48951 (ii) 89325 in expanded form using exponents., , TE, , 2., , 11.1.1 Exponents with other bases, , So far we have seen numbers whose base is 10. However, the base can be any number., 81 = 3 × 3 × 3 × 3, , T, , For example, , = 34, , ER, , Here 3 is the base and 4 is the exponent., Similarly,, , 125 = 5 × 5 × 5, , = 53, , SC, , Here 5 is the base and 3 is the exponent., Example 1:, , Which is greater 34 or 43 ?, 34 = 3 × 3 × 3 × 3 = 81, 43 = 4 × 4 × 4 = 64, 81 > 64, , Therefore, 34 > 43, , 212, , VII CLASS MATHEMATICS
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Do This, 1., , Is 32 equal to 23 ? Justity., , 2., , Write the following numbers in exponential form. Also state the, (a) base, , (b)exponent and (c) how it is read., , (i) 32, , (ii) 64, , (iv) 243, , (v) 49, , AN, A, , (iii) 256, , Squared and cubed, , When any base is raised to the power 2 or 3, it has a special name., , 10 2 = 10 × 10 and is read as '10 raised to the power 2' or '10 squared'., Similarly, 42 = 4 × 4 and can be read as '4 raised to the power of 2' or '4 squared'., , G, , 10 × 10 × 10 = 103 is read as '10 raised to the power 3' or '10 cubed '., , LA, N, , Similarly, 6 × 6 × 6 = 63 and can be read as '6 raised to the power 3' or '6 cubed'., In general, we can take any positive number 'a' as the base and write., a×a, , = a2, , (this is read as 'a raised to the power of 2' or 'a squared'), , a×a×a, , = a3, , (this is read as 'a raised to the power of 3' or 'a cubed'), (this is read as 'a raised to the power of 4'), , TE, , a × a × a × a = a4, , ___________ = a5 ( ____________________________________ ), ___________ = a6 ( _____________________________________ ), , and so on., , ER, , T, , Thus, we can say that a × a × a × a × a × a ×..................... 'm' times = am where 'a' is the, base and 'm' is the exponent., Do This, 1., , Write the expanded form of the following., , SC, , (i) p7, , 2., , (ii) l4, , (iii) s9, , (iv) d6, , (v) z5, , Write the following in exponential form., (i) a × a × a × ................. 'l' times, (ii) 5 ×5 ×5 × 5.................... 'n' times, (iii) q × q × q × q × q .................. 15 times, (iv) r × r × r × .................. 'b' times, , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 213
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11.2, , Writing a number in exponential form through prime factorization., , Let us express the following numbers in the exponential form using prime factorization., (ii) 450, , Solution (i):, , 432 = 2 × 216, , 2, , 432, , = 2 × 2 × 108, , 2, , 216, , = 2 × 2 × 2 × 54, , 2, , 108, , 2, , 54, , 3, , 27, , 3, , 9, , 3, , 3, , AN, A, , (i) 432, , = 2 × 2 × 2 × 2 × 27, =2×2×2×2×3×9, =2×2×2×2×3×3×3, = (2 × 2 × 2 × 2) × (3 × 3 × 3), , 450, , = 2 × 225, = 2 × 3 × 75, = 2 × 3 × 3 × 25, , TE, , =2×3×3×5×5, , LA, N, , Therefore, 432 = 24 × 33, (ii), , 1, , G, , = 24 × 33, , 2, , 450, , 3, , 225, , 3, , 75, , 5, , 25, , 5, , 5, 1, , = 2 × 3 2 × 52, Therefore, 450 = 2 × 32 × 52, , T, , Do This, , ER, , Write the following in exponential form using prime factorization., (i) 2500, , (i) 34, , 2., , (iii) 8000, , (iv) 6300, Exercise - 1, , Write the base and the exponent in each case. Also, write the term in the expanded form., , SC, , 1., , (ii) 1296, , (ii) (7x)2, , (iii) (5ab)3, , Write the exponential form of each expression., (i) 7 × 7 × 7 × 7 × 7, (ii) 3 × 3 × 3 × 5 × 5 × 5 × 5, (iii) 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5, , 214, , VII CLASS MATHEMATICS, , (iv) (4y)5
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Express the following as the product of exponents through prime factorization., (i) 288, , 4., , (ii)1250, , (iii) 2250, , (iv) 3600, , (v) 2400, , Identify the greater number in each of the following pairs., (i) 23 or 32, , (ii) 53 or 35, , (iii) 28 or 82, , AN, A, , 3., , 5., , If a = 3, b = 2 find the value of (i) ab + ba (ii) aa + bb (iii) (a + b)b (iv) (a–b)a, , 11.3, , Laws of exponents, , Multiplying terms with the same base, , LA, N, , 11.3.1, , G, , When we multiply terms with exponents we use some rules to find the product easily. These rules, have been discussed here., , Example 2 :, , 24 × 2 3, , Solution :, , 24 × 23 = (2 × 2 × 2 × 2) × (2 ×2 × 2), 4 times, , 3 times, , TE, , =2×2×2×2×2×2×2, , = 27 and this is same as 24+3, , (as 4 + 3 = 7), , Therefore, 24 × 23 = 24+3, , = (5 × 5) × (5× 5 × 5), , ER, , Solution : 52 × 53, , T, , Example 3:52 × 53, , 2 times, , 3 times, , =5×5×5×5×5, (as 2 + 3 = 5), , SC, , = 55 and this is same as 52+3, , Therefore, 52 × 53 =52+3, , Do This, , Find the values of 24, 23 and 27, verify whether 24 × 23 = 27, Find the values of 52, 53 and 55 and verify whether 52 × 53 = 55, , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 215
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Example 4 :, Solution:, , a4 × a5, a4 × a5 = (a × a × a × a) × (a × a × a × a × a), = (a × a × a × a × a × a × a × a × a), = a9 and this is same as a4+5, Therefore, a4×a5 = a4+5, , Based on the above observations we can say that., , AN, A, , (as 4 + 5 = 9), , am × an = ( a × a × a …………..'m' times) × ( a × a × a × ………'n' times)= am+n, , G, , For any non-zero integer 'a', and integers 'm' and 'n', , Do This, 1., , LA, N, , am × an = am+n, , Simplify the following using the formula am × an = am+n, (i), , (ii), , p5 × p8, , Find the appropriate number in place of the symbol '?' in the following., , TE, , 2., , 311 × 39, , Let 'k' be any non zero integer, (i) k3 × k4 = k?, , (ii) k15 × k? = k31, , T, , 11.3.2 Exponent of exponent, , ER, , Example 5 : Consider (32)3, , Solution : Here ' 32 ' is the base and' 3' is the exponent, (32)3, , = 32 × 32 × 32, , = 32+2+2, , (multplying terms with the same base), , SC, , = 36 and this is the same as 32×3, , (as 2 × 3 = 6), , Therefore, (32)3 = 32×3, , Do This, , Compute 36 , cube of 32 and verify whether (32)3 = 36 ?, , 216, , VII CLASS MATHEMATICS
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Example 6 : Let us consider (45)3, Solution : (45)3 = 45 × 45 × 45, =, , 45+5+5, , (multplying terms with the same base), , = 415 and this is same as 45×3, , (as 5 × 3 = 15), , AN, A, , Therefore, (45)3 = 45 × 3, Example 7: (am)4, Solution :, , (am)4, , = am × am × am × am, , = am+m+m+m, , (multplying terms with the same base), , = a4m and this is same as am×4, , (as 4 × m = 4m), , G, , Therefore, (am)4 = am×4, , LA, N, , Based on all the above we can say that (am)n = am × am × am...... n times = am+m+m+... n times, = amn, , For any non-zero integer 'a' and integers 'm' and 'n', (am)n = amn, , TE, , 11.3.3 Exponent of a product, Example 8 :, , Consider 35 × 45, , Solution :, , Here 35 and 45 have the same exponent 5 but different bases., , T, , 35 × 45 = (3×3×3×3×3) × (4×4×4×4×4), , ER, , = (3×4) × (3×4) × (3×4) × (3×4) × (3×4), = (3×4)5, , Therefore, 35 × 45 = (3 × 4)5, , SC, , Example 9: Consider 44 × 54, Solution :, , Here 44 and 54 have the same exponent 4 but have different bases., 44 × 54, , = (4 × 4 × 4 × 4) × (5 × 5 × 5 × 5), , = (4 × 4 × 4 × 4 × 5 × 5 × 5 × 5), = (4 × 5)× (4 × 5) × (4 × 5) × (4 × 5), = (4 × 5)4, Therefore, 44 × 54 = (4 × 5)4, , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 217
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Example 10 : Consider p7 × q7, Solution :, , Here p7 and q7 have the same exponent 7 but different bases., , p7 × q7 = (p × p × p × p × p × p × p) × (q × q × q × q × q × q × q), = (p × p × p × p × p × p × p × q × q × q × q × q × q × q), , AN, A, , = (p × q ) × (p × q ) × (p × q ) × (p × q ) × (p × q ) × (p × q ) × (p × q), = (p × q )7, Therefore, p7 × q7 = (p × q)7, , Based on all the above we can conclude that am × bm = (a × b)m = (ab)m, , G, , For any two non-zero integers 'a', 'b' and any positive integer 'm', , LA, N, , am × bm = (ab)m, Do This, , Simplify the following using the law am × bm = (a b)m, , TE, , (i) (2 × 3)4(ii) xp × yp (iii) a8 × b8 (iv) (5×4)11, 11.3.4 Division of exponential forms, , Before discussing division of exponents we will now discuss about negative exponents., , T, , 11.3.4(a) Negative exponents, , ER, , Observe the following pattern., =, , 32, , 35, , =, , 243, , 24, , =, , 16, , 34, , =, , 81, , 23, , =, , 8, , 33, , =, , 27, , SC, , 25, , 218, , 22, , =, , 4, , 32, , =, , 9, , 21, , =, , 2, , 31, , =, , 3, , 20, , =, , 1, , 30, , =, , 1, , 2-1, , =, , 3-1, , =, , (Hint: half of 1), , (Hint: one-third of 1), , 2-2, , 3-2, , =, VII CLASS MATHEMATICS, , =
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What part of 32 is 16?, What is the difference between 26 and 24?, You will find that each time the exponent decreases by 1,the value becomes half of the previous., From the above patterns we can say., , 3–1 =, , 1, 1, and 3–2 =, 3, 9, , and, , Furthermore, we can see that 2–2 =, , 1, 1, = 2, 4, 2, , 1, 1, 1, and 3–2 =, = 2, 3, 9, 3, , LA, N, , similarly, 3–1 =, , AN, A, , 1, 1, and 2–2 =, 2, 4, , G, , 2–1 =, , For any non -zero integer 'a' and any integer 'n', , Do This, , Write the following, by using a, (i) x–7, , (ii) a–5, , −n, , =, , 1, , with positive exponants., an, , (iii) 7–5, , T, , 1., , 1, an, , TE, , a −n =, , (iv) 9–6, , ER, , 11.3.4(b) Zero exponents, , In the earlier discussion we have seen that, 20 = 1, , SC, , 30 = 1, , Similarly we can say, 40 = 1, , 50 = 1 and so on, Thus for non zero integer ‘a’, a0 = 1, , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 219
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11.3.4(c) Division of exponents having the same base, 77, Example 11 : Consider 3, 7, Solution :, , 77 7 × 7 × 7 × 7 × 7 × 7 × 7, =, = 7×7× 7× 7, 73, 7×7×7, , 77, = 7 7 −3, 3, 7, , Therefore,, , 38 3×3×3×3×3× 3× 3× 3, =, =3×3×3×3×3, 33, 3× 3× 3, , LA, N, , Solution :, , 38, 33, , G, , Example 12: Consider, , (as 7 – 3 = 4), , AN, A, , = 74 which is same as 77–3, , = 35 which is same as 38–3, Therefore,, , 38, = 38−3, 33, , 55, 58, , TE, , Example 13: Consider, , 55, 5×5×5×5×5, 1, 1, =, =, = 3, 8, 5 5× 5× 5× 5 × 5 × 5 × 5 × 5, 5× 5× 5, 5, , T, , Solution :, , (as 8 – 3 = 5), , ER, , 1, 1, 3 which is same as 8 − 5, 5, 5, , (as 8 – 5 = 3), , SC, , 1, 55, Therefore, 8 = 8−5, 5, 5, , Example 14: Consider, , Solution :, , a2, a/ × a/, 1, =, =, 7, a a × a × a × a × a × a/ × a/ a × a × a × a × a, =, , 220, , a2, a7, , 1, 1, 5 which is the same as, a, a 7−2, , VII CLASS MATHEMATICS, , (as 7 – 2 = 5)
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Therefore,, , 1, a2, 7 =, a 7−2, a, , Based on all the above examples we can say thatam, am, 1, m− n, =, a, = n − m if m<n, if, m, >, n, and, n, n, a, a, a, , AN, A, , For 'd' a non-zero integer ‘a’ and integers 'm' and 'n', 1, am, am, m− n, =, = n− m if n > m, a, if, m, >, n, and, n, n, a, a, a, , Solution :, , 43, 43, , LA, N, , Example 15 : Consider, , G, , What happens when m = n? Give your answer., , 43, 4×4×4, 1, =, = = 1 . . . . . (1), 3, 4, 4×4×4, 1, , ∴, , TE, , am, m −n, Also we know that n = a, a, , 43, = 43−3 = 40 =1 from (1), 3, 4, , T, , 74, Similarly find 4 = ?, 7, , ER, , What do you observe from above?, Also consider, , a4 a × a × a × a, =, =1, a4 a × a × a × a, , am, = a m− n, an, , We have, , a4, = a 4−4 = a 0 =1, a4, , SC, , But from, , For any non zero number ‘a’ we have a0 = 1., Observe here m, n (m = n), Thus if m = n, , am, =1, an, , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 221
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Do This, 1. Simplify and write in the form of am-n or, , (ii), , a, , ., , 34, 314, , 2. Fill the appropriate number in the box., 88, = 8 8 −3 = 8 5, 83, , (i), , 1212, = 12, 127, , = 12, , (ii), , a18, a, , =a, , =a, , 10, , G, , Ex :, , AN, A, , 138, 135, , (i), , 1, n−m, , ⎛7⎞, Example 16: Consider ⎜ ⎟, ⎝4⎠, , LA, N, , 11.3.4(c) Dividing terms with the same exponents, 5, , 5, , 7×7×7×7×7, 4× 4× 4× 4× 4, , 75, 45, , T, , =, , TE, , Solution :, , ⎛7⎞ 7 7 7 7 7, ⎜ ⎟ = × × × ×, ⎝4⎠ 4 4 4 4 4, , ER, , =, , (by the definition of exponent), , 5, , SC, , 75, ⎛7⎞, Therefore, ⎜ ⎟ = 5, 4, ⎝4⎠, , ⎛ p ⎞, Example 17: Consider ⎜, ⎟, ⎝ q ⎠, Solution :, , 6, , ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞, ⎜ ⎟ = ⎜ ⎟×⎜ ⎟×⎜ ⎟×⎜ ⎟×⎜ ⎟×⎜ ⎟, ⎝q⎠ ⎝q⎠ ⎝q⎠ ⎝q⎠ ⎝q⎠ ⎝q⎠ ⎝q⎠, =, , 222, , 6, , p× p× p× p× p× p, q×q×q×q×q×q, , VII CLASS MATHEMATICS
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=, , p6, q6, , (By the definition of exponent), 6, , ⎛ p⎞, p6, =, Therefore, ⎜ ⎟, q6, ⎝q⎠, , AN, A, , Based on the above observations we can say that., m, a × a × a × a × ................ × a ' m ' times, am, ⎛a⎞, =, =, ⎜ ⎟, b × b × b × b × ................ × b ' m ' times, bm, ⎝b⎠, , m, , G, , am, ⎛a⎞, For any non-zero integers a, b and integer 'm' ⎜ ⎟ = m, b, ⎝b⎠, , 1., , LA, N, , Do This, Complete the following, 3, , (ii), , 35, ⎛3⎞, ⎜ ⎟ = 5, 2, ⎝2⎠, , TE, , (i), , 53, ⎛5⎞, =, ⎜ ⎟, ⎝7⎠, , 11, , 4, , ⎛8⎞, (iii) ⎜⎝ 3 ⎟⎠ =, , T, , (iv), , ⎛ x⎞, ⎜ ⎟ = 11, y, ⎝ y⎠, , ER, , 11.3.5 Terms with negative base, Example 18 :, , Evaluate (1)4 , (1)5 ,(1)7, (–1)2 , (–1)3 , (–1)4 , (–1)5, , Solution : (1)4 = 1 × 1 × 1 × 1= 1, , SC, , (1)5 = 1 × 1 × 1 × 1 × 1= 1, (1)7 = 1 × 1 ×1 × 1 × 1 × 1 × 1 = 1, (–1)2 = (–1) × (–1) = 1, (–1)3 = (–1) × (–1) × (–1) = –1, (–1)4 = (–1) × (–1) × (–1) × (–1) = 1, (–1)5 = (–1) × (–1) × (–1) × (–1) × (–1) = –1, , Free distribution, distributionby, byT.S., A.P.Government, Government2018-19, , POWERS AND EXPONENTS, , 223
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From the above illustrations we observe that:, (i) 1 raised to any power is 1., (ii) (–1) raised to odd power is (–1) and (–1) raised to even power is (+1)., Thus (–a)m = –am If ‘m’ is odd, , AN, A, , (–a)m = am If ‘m’ is even, Now, let us look at some more examples., (–3)4 = (–3) (–3) (–3) (–3) = 81, , 1, 1, 1, 1, −1, ×, ×, = 3= 3, (−a) (−a) (− a) − a, a, , LA, N, , (–a)–3 =, , G, , (–a)4 = (–a) (–a) (–a) (–a) = a4, , −27, in exponential form, 125, , Express, , Solution :, , –27 = (–3) (–3) (–3) = (–3)3, , TE, , Example 19 :, , 125 = 5 × 5 × 5 = (5)3, , T, , − 27 ( − 3) 3, =, Therefore,, 125, (5) 3, , ER, , − 27 ⎛ − 3 ⎞, =⎜, Thus,, ⎟, 125 ⎝ 5 ⎠, , am ⎛ a ⎞, as m = ⎜ ⎟, b, ⎝b⎠, , m, , 3, , SC, , Do This, 1., , Write in expanded form., , (i), , 2., , 224, , (a)–5, , (ii), , (–a)4, , (iii), , (–7)–5 (iv), , (–a)m, , (ii), , (–b) × (–b) × (–b) × (–b), , Write in exponential form, (i), , (–3) × (–3) × (–3), , (iii), , 1, 1, 1, ×, ×, ...... 'm' times, (−2) (−2) (−2), VII CLASS MATHEMATICS
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Exercise 2, Simplify the following using laws of exponents., 2 ×2, , 4, , (ii), , (3 ) × (3 ), , (v), , ⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞, ⎜ ⎟ ×⎜ ⎟ ×⎜ ⎟, ⎝5⎠ ⎝5⎠ ⎝5⎠, , 2, , 4, , (iv), , 92 × 918 × 910, , (vii) (32)2, , (viii) 24 × 34, , (102)3, , ⎛2⎞, (xiii) ⎜ ⎟, ⎝3⎠, , 3, , ⎡⎛ −5 ⎞ 2 ⎤, ⎢⎜ ⎟ ⎥, ⎢⎣⎝ 6 ⎠ ⎥⎦, , (xi), , 8, , (vi) (–3)3 × (–3)10 × (–3)7, (ix), , 5, , 24a × 25a, , (xii) 23a+7 × 27a+3, , 5, , (xiv) (–3) × (–5), , 57, 52, , (−4)6, (xv), (−4)3, , LA, N, , (x), , (iii), , 2 4, , AN, A, , (i), , 10, , G, , 1., , 3, , (−6)5, (xvii), (−6)9, , (xix) (–64)4, , (xx) ax × ay × az, , (xiii) (–7)7 × (–7)8, , TE, , 97, (xvi) 15, 9, , 3, , By what number should 3–4 be multiplied so that the product is 729?, , 3., , If 56 × 52x = 510, then find x., , 4., , Evaluate 20 + 30, , 5., , ⎛ x a ⎞ ⎛ xb ⎞ ⎛ x a ⎞, Simplify ⎜ b ⎟ × ⎜ a ⎟ × ⎜ a ⎟, ⎝x ⎠ ⎝x ⎠ ⎝x ⎠, , 6., , State true or false and justify your answer., , T, , 2., , a, , ER, , a, , b, , (ii) 32 × 43 = 125, , (iii)) 50 = (100000)0, , (iv) 43 = 82, , (v) 23 > 32, , (vi) (–2)4 > (–3)4, , SC, , (i) 100×1011 = 1013, , (vii) (–2)5 > (–3)5, Classroom Project, Collect the annual income particulars of any ten families in your locality and, round it to the nearest thousands / lakhs and express the income of each family in, the exponential form., , Free distribution by T.S. Government 2018-19, , POWERS AND EXPONENTS, , 225
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11.3.6 Expressing large numbers in standard form, The mass of the Earth is about 5976 x 1021 kg., The width of the Milky Way Galaxy from one edge to the other edge is about 946 ×105 km., These numbers are still not very easy to comprehend. Thus, they are often expressed in standard, form. In standard form the:, , AN, A, , Mass of the Earth is about 5.976 × 1024 kg, , Width of the Milky Way Galaxy from one edge to the other edge is about 9.46 ×107 km., , Thus, in standard form (Scientific notation) a number is expressed as the product of, largest integer exponent of 10 and a decimal number between 1 and 10., , G, , Exercise 3, , LA, N, , Express the number appearing in the following statements in standard form., The distance between the Earth and the Moon is approximately 384,000,000m., , (ii), , The universe is estimated to be about 12,000,000,000 years old., , (iii), , The distance of the sun from the center of the Milky Way Galaxy is estimated to be, 300,000,000,000,000,000,000 m., , (iv), , The earth has approximately 1,353,000,000 cubic km of sea water., , TE, , (i), , Looking Back, , Very large numbers are easier to read, write and understand when expressed in, exponential form., , •, , 10,000 = 104 (10 raised to the power of 4); 243 = 35 (3 raised to the power of 5);, 64 = 26 (2 raised to the power of 6). In these examples 10, 3, 2 are the respective, bases and 4, 5, 6 are the respective exponents., , •, , Laws of Exponents: For any non-zero integers 'a' and 'b' and integers 'm' and 'n', , SC, , ER, , T, , •, , 226, , (i), , am × an = am+n (ii), , (iv), , a −n =, , (vi), , am, 1, = n − m if n > m, n, b, a, , (viii), , a0 = 1 (where a ≠ 0), , 1, an, , VII CLASS MATHEMATICS, , (v), , (am)n=amn, , (iii), , am, = a m − n if m > n, an, (vii), , am ⎛ a ⎞, =⎜ ⎟, bm ⎝ b ⎠, , m, , am × bm = (ab)m
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12, , QUADRILATERALS, , Quadrilateral, , LA, N, , What is common among all these pictures?, , G, , 12.0, , AN, A, , In Class VI, we have been introduced to quadrilaterals. In this unit you will learn about the different, types of quadrilaterals and their properties in detail., , (Hints: Number of sides, angles, vertices. Is it an open or closed figure?), Thus, a quadrilateral is a closed figure with four sides, four angles and four vertices., Quadrilateral ABCD has, , Four sides, namely AB, BC , CD and DA, , (ii), , Four vertices, namely A, B, C and D., , (iii), , Four angles, namely ∠ABC, ∠BCD, ∠CDA and ∠DAC., , (iv), , The line segments joining the opposite vertices of a quadri-, , T, , TE, , (i), , lateral are called the diagonals of the quadrilateral. AC and, , C, B, , A, D, , (v), , ER, , BD are the diagonals of quadrilateral ABCD., , The two sides of a quadrilateral which have a common vertex are called the 'adjacent, , SC, , sides' of the quadrilateral. In quadrilateral ABCD, AB is adjacent to BC and B is their, common vertex., , (vi), , The two angles of a quadrilateral having a common side are called the pair of 'adjacent, angles' of the quadrilateral. Thus, ∠ABC and ∠BCD are a pair of adjacent angles and, , BC is the common side., , Do This :, 1.Find the other adjacent sides and common vertices., 2. Find the other pairs of adjacent angles and sides., Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 227
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(vii), , The two sides of a quadrilateral, which do not have a common vertex, are called a pair of, 'opposite sides' of the quadrilateral. Thus AB , CD and AD , BC are the two pairs of, 'opposite sides' of the quadrilateral., The two angles of a quadrilateral which do not have a common side are known as a pair, of 'opposite angles' of the quadrilateral. Thus ∠BAD, ∠DCB and ∠ADC, ∠CBA are, the two pairs of opposite angles of the quadrilateral., , AN, A, , (viii), , Try This, , D, , How many different quadrilaterals can be obtained, from the adjacent figure? Name them., , E, , F, , H, , G, , G, , LA, N, , A, , 12.1, , Interior-Exterior of a quadrilateral, , In quadrilateral ABCD which points lie inside the quadrilateral?, , B, , L, , D, , C, , N, , O, P, , Which points lie outside the quadrilateral?, , TE, , Which points lie on the quadrilateral?, , C, , Q, , M, , B, , A, , Points P and M lie in the interior of the quadrilateral. Points L, O and Q lie in the exterior of the, quadrilateral. Points N, A, B, C and D lie on the quadrilateral., , T, , Mark as many points as you can in the interior of the quadrilateral., , ER, , Mark as many points as you can in the exterior of the quadrilateral., How many points, do you think will be there in the interior of the quadrilateral?, 12.2, , Convex and Concave quadrilateral, , SC, , Mark any two points L and M in the interior of quadrilateral ABCD and join them with a line, segment., Does the line segment or a part of it joining these points lie in the exterior, of the quadrilateral? Can you find any two points in the interior of the, quadrilateral ABCD for which the line segment joining them falls in the, exterior of the quadrilateral?, You will see that this is not possible., Now let us do similar work in quadrilateral PQRS., 228, , VII CLASS MATHEMATICS, , D, C, L, M, B, , A, Figure1
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Mark any two points U and V in the interior of quadrilateral PQRS and join, them. Does the line segment joining these two points fall in the exterior of the, quadrilateral? Can you make more line segments like these in quadrilateral, PQRS., Can you also make line segments, joining two points, which lie in the interior, of the quadrilateral. You will find that this is possible too., , S, , Q, U, , V, R, , P, , AN, A, , Quadrilateral ABCD is said to be a convex quadrilateral if all line segments joining points, in the interior of the quadrilateral also lie in interior of the quadrilateral., Quadrilateral PQRS is said to be a concave quadrilateral if all line segment joining points, in the interior of the quadrilateral do not necessarily lie in the interior of the, quadrilateral., , E, , F, , LA, N, , 1., , G, , Try This, , W, , V, , T, , G, , H, , TE, , (i) Is quadrilateral EFGH, a convex quadrilateral?, , U, , (ii) Is quadrilateral TUVW, a concave quadrilateral?, , (iii) Draw both the diagonals for quadrilateral EFGH. Do they intersect each other?, (iv) Draw both the diagonals for quadrilateral TUVW. Do they intersect each other?, , 12.3, , ER, , T, , You will find that the digonals of a convex quadrilateral intersect each other in the, interior of the quadrilateral and the diagonals of a concave quadrialteral intersect each, other in the exterior of the quadrilateral., , Angle-sum property of a quadrilateral, , SC, , Activity 1, Take a piece of cardboard. Draw a quadrilateral ABCD on it. Make a cut of it. Then cut, quadrilateral into four pieces (Figure 1) and arrange them as shown in the Figure 2, so that all, angles ∠1, ∠2, ∠3, ∠4 meet at a point., , Figure 1, Free distribution by T.S. Government 2018-19, , Figure 2, QUADRILATERALS, , 229
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Is the sum of the angles ∠1, ∠2, ∠3 and ∠4 equal to 360o ?, , (sum of angles at a point), , The sum of the four angles of a quadrilateral is 360°., [Note: We can denote the angles by ∠1, ∠2, ∠3, etc., as their respective measures i.e. m∠1,, m∠2, m∠3, etc.], You may arrive at this result in several other ways also., Let P be any point in the interior of quadrilateral ABCD. Join P to vertices A, B, C and D., In the figure, consider ΔPAD., m∠2 + m∠3= 180° – x, , ................ (1), , AN, A, , 1., , D, , 4, , 3, , Similarly, in ΔPDC, m∠4 + m∠5 = 180° – y ....... (2), , P, , ................. (4), , A, , x, , 2, , y, , w, , z, , 8, , 1, , 7, B, , LA, N, , in ΔPBA, m∠8 + m∠1 = 180º – w., , G, , in ΔPCB, m∠6 + m∠7 = 180º – z and ............ (3), , C, , 5, 6, , (angle-sum property of a triangle), , Adding (1), (2), (3) and (4) we get, , m∠1 + m∠2 + m∠3 + m∠4 + m∠5 + m∠6 + m∠7 + m∠8, , TE, , = 180° – x + 180° – y + 180° – z + 180° – w, , = 720° – (x + y + z + w), , (x + y + z + w = 360o ; sum of angles at a point), = 360°, , T, , =720° – 360°, , B, , Thus, the sum of the angles of the quadrilateral is 360o., Take any quadrilateral, say ABCD. Divide it into two triangles, by drawing a diagonal. You get six angles 1, 2, 3, 4,, 5 and 6., , ER, , 2., , 2, , A, , 1, , C, , 4, , 3, , 6, , 5, D, , SC, , Using the angle-sum property of a triangle and you can easily find how the sum of the, measures of ∠A, ∠B, ∠C and ∠D amounts to 360°., B, , Try This, What would happen if the quadrilateral is not convex? Consider, quadrilateral ABCD. Split it into two triangles and find the sum of, the interior angles. What is the sum of interior angles of a concave, quadrilateral?, , 230, , VII CLASS MATHEMATICS, , A, , C, D
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Example 1 : The three angles of a quadrilateral are 55°, 65° and 105°. What is the fourth, angle?, Solution :, , The sum of the four angles of a quadrilateral = 360°., The sum of the given three angles, , = 55° + 65° + 105° = 225°, , Therefore, the fourth angle, , = 360° – 225° = 135°, , Solution :, , AN, A, , Example 2 : In a quadrilateral, two angles are 80° and 120°. The remaining two angles are, equal. What is the measure of each of these angles?, The sum of the four angles of the quadrilateral = 360°., Sum of the given two angles = 80° + 120° = 200°, , Both these angles are equal., , LA, N, , Therefore, each angle = 160° ÷ 2 = 80°, , G, , Therefore, the sum of the remaining two angles = 360° – 200° = 160°, , Example 3 : The angles of a quadrilateral are x°, (x – 10)°, (x + 30)° and 2x°. Find the angles., The sum of the four angles of a quadrilateral = 360°, Therefore, x + (x – 10) + (x + 30) + 2x, , = 360°, , Solving,, , = 360°, , 5x + 20, , TE, , Solution:, , x, , = 68°, , T, , Thus, the four angles are = 68° ; (68–10)° ; (68+30)° ; (2×68)°, = 68°, 58°, 98° and 136°., , ER, , Example 4 : The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Find the angles., Solution :, , The sum of four angles of a quadrilateral = 360°, The ratio of the angles is 3 : 4 : 5 : 6, , SC, , Thus, the angles are 3x, 4x, 5x and 6x., 3x + 4x + 5x + 6x = 360, 18x = 360, x=, , 360, = 20, 18, , Thus, the angles are = 3 × 20; 4 × 20; 5 × 20; 6 × 20, = 60°, 80°, 100° and 120°, Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 231
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Exercise - 1, Q, , 1., , R, , In quadrilateral PQRS, (i) Name the sides, angles, vertices and diagonals., P, S, , AN, A, , (ii) Also name all the pairs of adjacent sides, adjacent angles,, opposite sides and opposite angles., , o, , 80, , 2., , The three angles of a quadrilateral are 60°, 80° and 120°. Find the, fourth angle?, , o, , 60, , o, , 120, , The angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find the measure of each of the, four angles., , 4., , The four angles of a quadrilateral are equal. Draw this quadrilateral in your notebook., Find each of them., , 5., , In a quadrilateral, the angles are x°, (x + 10)°, (x + 20)°, (x + 30)°. Find the angles., , 6., , The angles of a quadrilateral cannot be in the ratio 1 : 2 : 3 : 6. Why? Give reasons., , LA, N, , G, , 3., , TE, , (Hint: Try to draw a rough diagram of this quadrilateral), 12.4 Types of quadrilaterals, , Based on the nature of the sides and angles, quadrilaterals have different names., , T, , 12.4.1 Trapezium, , SC, , ER, , Trapezium is a quadrilateral with one pair of parallel sides., , These are trapeziums, (Note: The arrow marks indicate parallel lines)., Why the second set of figures not trapeziums?, 232, , VII CLASS MATHEMATICS, , These are not trapeziums
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12.4.2 Kite, A Kite is a special type of quadrilateral. The sides with the same markings in each figure are equal, in length. For example AB = AD and BC = CD., A, , AN, A, , D, , C, , These are kites, , These are not kites, , LA, N, , Why the second set of figures are not kites?, , G, , B, , Observe that:, (i), , A kite has 4 sides (It is a convex quadrilateral)., , (ii), , There are exactly two distinct, consecutive pairs of sides of equal length., , TE, , Activity 2, , Take a thick sheet of paper. Fold the paper at the, centre. Draw two line segments of different lengths, , T, , as shown in Figure 1. Cut along the line segments, , ER, , and open up the piece of paper as shown in Figure 2., You have the shape of a kite., , Figure1, , Figure2, , Does the kite have line symmetry?, , SC, , Fold both the diagonals of the kite. Use the set-square to check if they cut at right angles., Are the diagonals of the kite equal in length? Verify (by paper-folding or measurement) if the, diagonals bisect each other., Try This, , Prove that in a kite ABCD, ΔABC and ΔADC are congruent., , Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 233
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12.4.3 Parallelogram, Activity 3, , 5cm, , Arrange them as shown in the figure given below:, , AN, A, , 3c, m, , 4c, m, , m, , m, , 5cm, , 5cm, 3c, , 3c, , m, , 4c, m 4, cm, , 3c, , Take two identical cut-outs of a triangle of sides 3 cm, 4 cm, 5 cm., , 5cm, , You get a parallelogram. Which are the parallel sides here? Are the parallel sides equal? You can, get two more parallelograms using the same set of triangles. Find them out., , G, , A parallelogram is a quadrilateral with two pairs of opposite sides arallel., Activity 4, , LA, N, , Take a ruler. Place it on a paper and draw two lines along its two sides as shown in Figure1. Then, place the ruler over the lines as shown in Figure2 and draw two more lines along its edges again., 0, , 1, , 2, , 3, , 3, , 4, , 5, , 6, , 7, , TE, , 2, , 4, , 0 1, , C, , D, , 5, , A, , B, , 6, , 7, , Figure 1, , Figure 2, , Figure 3, , ER, , T, , These four lines enclose a quadrilateral which is made up of two pairs of parallel lines. It is a, parallelogram., 12.4.3(a) Properties of a parallelogram, Sides of parallelogram, , SC, , Activity 5, , Take cut-outs of two identical parallelograms, say ABCD and A1B1C1D1., , C, , D, , 1, , C, , 1, , D, , O, A, , 234, , O, B, , VII CLASS MATHEMATICS, , 1, , A, , 1, , B
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Here AB is same as A'B' except for the name. Similarly, the other corresponding sides are equal, too. Place A'B' over DC . Do they coincide? Are the lengths A'B' and DC equal?, Similarly examine the lengths AD and B'C' . What do you find?, You will find that the sides are equal in both cases. Thus, the opposite sides of a parallelogram, are of equal length., , AN, A, , You will also find the same results by measuring the side of the parallelogram with a scale., , 2 3 4 5 6 7 8 9 10 11 12 13, , 9, 8, 7, 6, 5, 4, 3, 2, , 2, 3, 4, 5, 6, 7, 8, 2 3 4 5 6 7 8 9 10 11 12 13 9, , G, , Take two identical set squares with angles 30° –, 60° – 90° and place them adjacently as shown in, the adjacent figure. Does this help you to verify the, above property? Can we say every rectangle is a, parallelogram?, , 1, , Try This, , LA, N, , 1, , Example 5 :, , Find the perimeter of the parallelogram PQRS., , Solution :, , In a parallelogram, the opposite sides have same length., , According to the question, PQ= SR = 12 cm and QR = PS = 7 cm, , TE, , Thus, Perimeter = PQ + QR + RS + SP, , = 12 cm + 7 cm + 12 cm + 7 cm = 38 cm, , Activity 6, , T, , Angles of a parallelogram, , SC, , ER, , Let ABCD be a parallelogram. Copy it on a tracing sheet. Name this copy as A′, B′, C′, D′. Place, A′, B′, C′, D′ on ABCD as shown in Figure 1. Pin them together at the point where the diagonals, meet. Rotate the transparent sheet by 90o as shown in Figure 2. Then rotate the parallelogram, again by 90o in the same direction. You will find that the parallelograms coincide as shown in Figure, 3. You now find A′ lying exactly on C and C′ lying on A. Similarly B′ lies on D and D′ lies on B as, shown in Figure 3., A, ', , D, , ', , C, , D, , ', , A, A, , B, , B, , ', , D', , D, , C, , ', , A, , Figure 1, Free distribution by T.S. Government 2018-19, , B, , B, , ', , Figure 2, , ', , C, , D, , C ', A, , B', , C, , A, C', , B, , D', , Figure 3, QUADRILATERALS, , 235
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Does this tell you anything about the measures of the angles A and C? Examine the same for, angles B and D. State your findings., You will conclude that the opposite angles of a parallelogram are of equal measure., Try This, , You can justify this idea through logical arguments-, , AN, A, , Take two identical 30° – 60° – 90° set squares and form a parallelogram as, before. Does the figure obtained help you confirm the above property?, , If AC and BD are the diagonals of the parallelogram, ABCD you find that ∠1 =∠2 and ∠3 = ∠4 (alternate angles, property), , D, , A, , 4, , 2, , 1, , C, 3, , B, , Therefore, m∠B = m ∠D (c.p.c.t.)., , G, , ΔABC and ΔCDA are congruent ΔABC ≅ ΔCDA (ASA congruency)., , LA, N, , Similarly, ΔABD ≅ ΔCDB, therefore, m ∠A = m ∠C. (c.p.c.t.)., , Thus, the opposite angles of a parallelogram are of equal measure., We now turn our attention to adjacent angles of a parallelogram., , TE, , In parallelogram ABCD, DC || AB and DA is the transversal., Therefore, ∠A and ∠D are the interior angles on the same side of, the transversal.thus are supplementary., ∠A and ∠B are also supplementary. Can you say 'why'?, , D, , A, , C, , B, , T, , AD & BC and BA is a transversal, making ∠A and ∠B interior angles., , ER, , Do This, , Identify two more pairs of supplementary angles from the, parallelogram ABCD given above., BEST is a parallelogram. Find the values x, y and z. T, , Solution :, , ∠S is opposite to ∠B., , SC, , Example 6 :, , So,, , x, , S, , x = 100° (opposite angles property), y = 100° (corresponding angles), z = 80° (since ∠y, ∠z is a linear pair), , B, , 100o, , z, E, y, , The adjacent angles in a parallelogram are supplementary. You have observed the same, result in the previous example., 236, , VII CLASS MATHEMATICS
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Example 7 :, , In parallelogram RING if m ∠R = 70°, find all the other angles., , Solution :, , According to the question,, Then, , m∠R = 70°, G, , m∠N = 70°, , N, , (opposite angles of a parallelogram), o, , Since ∠R and ∠I are supplementary angles,, , 70, , I, , AN, A, , R, , m∠I = 180° – 70° = 110°, , m∠G = 110° since ∠G and ∠I are opposite angles of a parallelogram., , Also,, , Try this, , G, , Thus, m∠R = m∠N = 70° and m∠I = m∠G = 110°, , For the above example, can you find m ∠I and m ∠G by any other method?, , LA, N, , Hint : angle-sum property of a quadrilateral, 12.4.3 (b) Diagonals of parallelogarm, , C, , D, , Activity 7, , O, , TE, , Take a cut-out of a parallelogram, say, ABCD. Let its, diagonals AC and DB meet at O., , B, , A, , T, , Find the mid-point of AC by folding and placing C on A. Is the mid-point same as O?, , ER, , Find the mid-point of DB by folding and placing D on B. Is the mid-point same as O?, , SC, , Does this show that diagonal DB bisects the diagonal AC at the point O? Discuss it with your, friends. Repeat the activity to find where the mid point of, C, D, DB could lie., 2, 4, O, , The diagonals of a parallelogram bisect each other, It is not very difficult to justify this property using, ASA congurency:, ΔAOB ≅ ΔCOD, , A, , 1, , 3, , B, , (How is ASA used here?), , This gives AO = CO and BO = DO, Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 237
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Example 8 :, , HELP is a parallelogram. Given that OE = 4 cm, where O is the point of, intersection of the diagonals and HL is 5 cm more than PE? Find OH., , Solution :, , If OE = 4 cm then OP also is 4 cm (Why?), (Why?), , 4 cm, , HL is 5 cm more than PE, HL = 8 + 5 = 13 cm, , Thus,, , OH =, , E, , 1, × 13 = 6.5 cms, 2, , 12.4.4 Rhombus, , H, , AN, A, , Therefore,, , P, O, , So PE = 8 cm, , L, , A, , LA, N, , G, , Recall the paper-cut kite you made earlier. When you cut along ABC and, opened up, you got a kite. Here lengths AB and BC were different. If you, draw AB = BC, then the kite you obtain is called a rhombus., Note that all the sides of rhombus are of same length; this is not the case with, the kite., , B, , C, , ER, , T, , TE, , Rhombus-cut, Since the opposite sides of a rhombus are parallel, it is also a parallelogram., So, a rhombus has all the properties of a parallelogram and also that of a kite. Try to list them out., You can then verify your list with the check list at the end of the chapter., , Kite, , Rhombus, , SC, , The diagonals of a rhombus are perpendicular bisectors of one another, Activity 8, , Take a copy of a rhombus. By paper-folding verify if the point of intersection is the mid-point of, each diagonal. You may also check if they intersect at right angles, using the corner of a set-square., Now let us justify this property using logical steps., ABCD is a rhombus. It is a parallelogram too, so diagonals bisect each other., Therefore, OA = OC and OB = OD., 238, , VII CLASS MATHEMATICS
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We now have to show that m∠AOD = m∠COD = 90°., , D, , C, , It can be seen that by SSS congruency criterion., ΔAOD ≅ ΔCOD, Therefore,, , O, , m ∠AOD = m∠COD, A, , B, , AN, A, , Since ∠AOD and ∠COD are a linear pair,, m ∠AOD = m ∠COD = 90°, , We conclude, the diagonals of a rhombus are perpendicular bisectors of each other., , A rectangle is a parallelogram with equal angles., , G, , 12.4.5 Rectangle, , LA, N, , What is the full meaning of this definition? Discuss with your friends., , If the rectangle is to be equiangular, what could be the measure of each angle?, Let the measure of each angle be x°., 4x° = 360°, , Therefore,, , x° = 90°, , (Why)?, , TE, , Then, , Thus, each angle of a rectangle is a right angle., , T, , N, , x, , o, , x, , o, , x, , x, , o, , K, , o, , E, , So, a rectangle is a parallelogram in which every angle is a right angle., , T, , Being a parallelogram, the rectangle has opposite sides of equal length and its diagonals, bisect each other., , ER, , In a parallelogram, the diagonals can be of different lengths. (Check this); but surprisingly, the rectangle (being a special case) has diagonals of equal length., This is easy to justify:, , SC, , If ABCD is a rectangle,, , ΔABC ≅ ΔABD, , This is because AB = AB, BC = AD, , D, , C, , A, , B, , (Common), (Why?), , m ∠A = m ∠B = 90° (Why?), Thus, by SAS criterion ΔABC ≅ ΔABD and AC = BD (c.p.c.t.), Thus, in a rectangle the diagonals are of equal length., Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 239
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Solution :, , OT is half of the diagonal TE and OR is half of the, diagonal RN ., , T, , O, , Diagonals are equal here. (Why?), , 3x + 1 = 2x + 4, , or, , x=3, , 12.4.6 Square, , G, , A square is a rectangle with equal adjacent sides., , E, , AN, A, , R, , So, their halves are also equal., Therefore, , N, , 2x, +4, , RENT is a rectangle. Its diagonals intersect at O. Find x,, if OR = 2x + 4 and OT = 3x +1., , +1, 3x, , Example 9 :, , LA, N, , This means a square has all the properties of a rectangle with an additional property that all the, sides have equal length., The square, like the rectangle, has diagonals of equal length., , In a rectangle, there is no requirement for the diagonals to be perpendicular to one another (Check, this). However, this is not true for a square., , TE, , Let us justify this-, , L, , T, , BELT is a square, therefore, BE = EL = LT = TB, Now, let us consider ΔBOE and ΔLOE, , OE is common, , T, , OB=OL (why?), , O, , B, , So, , ∠BOE = ∠LOE, , ∠BOE + ∠LOE = 180o (why?), , SC, , but, , ER, , Thus, by SSS congruency ΔBOE ≅ LOE, , ∠BOE = ∠LOE =, , 180, = 90o, 2, , Thus, the diagonals of a square are perpendicular bisectors of each other., In a square the diagonals., (i), , bisect one another, , (square being a parallelogram), , (ii), , are of equal length, , (square being a rectangle) and, , (iii), , are perpendicular to one another., , 240, , VII CLASS MATHEMATICS, , E
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12.5, , Making figures with a tangram., , 2, , 4, 3, , AN, A, , 5, , 1, , 6, , G, , 7, , LA, N, , Use all the pieces of tangarm to form a trapezium, a parallelogram, a rectangle and a square., Also make as many different kinds of figures as you, can by using all the pieces. Two examples have been, given for you., , 5, , 7, , 6, 2, , 5, , 2, , TE, , 3, , 1, , 1, , 4, , 3, , 6, , 4, 7, , D, , C, , T, , Example 10 : In trapezium ABCD, AB is parallel to CD ., , ER, , If ∠ A = 50°, ∠B = 70°. Find ∠C and ∠D., Solution :, , Since AB is parallel to CD, ∠A + ∠D = 180°, , A, , 50o, , 70o B, , (interior angles on the same side of the transversal), , SC, , So ∠D = 180° – 50° = 130°, , Similarly,, , ∠B + ∠C = 180°, So ∠C =180° – 70° = 110°, , Example 11 : The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find, the angles of the parallelogram., Solution :, , The adjacent angles of a parallelogram are supplementary., , Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 241
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i.e. their sum = 180°, Ratio of adjacent angles = 3:2, 3, o, So, each of the angles is 180 × = 108 and, 5, 180 ×, , 2, = 72o, 5, , AN, A, , C, , 12, , Example 12 : RICE is a rhombus. Find OE and OR. Justify your findings., Solution :, , Diagonals of a rhombus bisect each other, , G, , i.e., OE = OI and OR = OC, , E, , LA, N, , Therefore, OE = 5 and OR = 12, , R, , Exercise - 2, , All rectangles are squares, , (, , ), , (ii), , All rhombuses are parallelogram, , (, , ), , (iii), , All squares are rhombuses and also rectangles, , ), , (iv), , All squares are not parallelograms, , (, , ), , (v), , TE, , (, , All kites are rhombuses, , (, , ), , (vi), , All rhombuses are kites, , (, , ), , (vii), , All parallelograms are trapeziums, , (, , ), , (viii), , All squares are trapeziums, , (, , ), , T, , (i), , Explain how a square is a-, , SC, , 2., , State whether true or false-, , ER, , 1., , 3., , (i), , quadrilateral, , (ii), , parallelogram, , (iii), , rhombus, , (iv), , rectangle., A, , In a rhombus ABCD, ∠CBA = 40°., Find the other angles., , D, , 40o, C, , 242, , VII CLASS MATHEMATICS, , O, , B, , 5, , I
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4., , The adjacent angels of a parallelogram are x° and (2x + 30)°., Find all the angles of the parallelogram., , 5., , Explain how DEAR is a trapezium. Which of its two sides are, parallel?, , R, , 8 o, 70, , o, , E, , 100, , A, , BASE is a rectangle. Its diagonals intersect at O. Find x, if OB = 5x+1 and OE = 2x + 4., A, , S, 1, 35x+, , O, , 2x+, 4, , B, , E, , AN, A, , 6., , D, , Is quadrilateral ABCD a parallelogram, if ∠A = 70° and ∠C = 65°? Give reason., , 8., , Two adjacent sides of a parallelogram are in the ratio 5:3 the perimeter of the, parallelogram is 48cm. Find the length of each of its sides., , 9., , The diagonals of the quadrilateral are perpendicular to each other. Is such a quadrilateral, always a rhombus? Draw a rough figure to justify your answer., , 10., , ABCD is a trapezium in which AB & DC . If ∠A = ∠B = 30o , what are the measures of, the other two angles?, , 11., , Fill in the blanks., , TE, , LA, N, , G, , 7., , A parallelogram in which two adjacent sides are equal is a ____________., , (ii), , A parallelogram in which one angle is 90° and two adjacent sides are equal is a, _____________., , (iii), , In trapezium ABCD, AB & DC . If ∠D = x° then ∠A = __________., , ER, , T, , (i), , Every diagonal in a parallelogram divides it in to ________ triangles., , (v), , In parallelogram ABCD, its diagonals AC and BD intersect at O. If AO = 5cm, then AC = ________cm., , SC, , (iv), , (vi), , In a rhombus ABCD, its diagonals intersect at 'O'. Then ∠AOB = ________, degrees., , (vii), , ABCD is a parallelogram then ∠A – ∠ C = ________ degrees., , (viii), , In a rectangle ABCD, the diagonal AC = 10cm then the diagonal BD = _______cm., , (ix), , In a square ABCD, the diagonal AC is drawn. Then ∠BAC = __________, degrees., , Free distribution by T.S. Government 2018-19, , QUADRILATERALS, , 243
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Looking back, 1. A simple closed figure bounded by four line segments is called a quadrilateral., 2. Every quadrilateral divides a plane into three parts interior, exterior and, the quadrilateral., 3. Every quadrilateral has a pair of diagonals., , AN, A, , 4. If the diagonals lie in the interior of the quadrilateral it is called convex, quadrilateral. If any one of the diagonals is not in the interior of the quadrilateral, it is called a concave Quadrilateral., , 6. Properties of Quadrilateral, Quadrilateral, , G, , 5. The sum of interior angles of a quadrilateral is equal to 360°., , Properties, , (1) Opposite sides are equal., (2) Opposite angles are equal., (3) Diagonals bisect one another., , Rhombus : A parallelogram with, all sides of equal length., , (1) All the properties of a parallelogram., (2) Diagonals are perpendicular to each, other., , TE, , LA, N, , Parallelogram : A quadrilateral, with both pair, of opposite sides, parallel, , (1) All the properties of a parallelogram., (2) Each of the angles is a right angle., (3) Diagonals are equal., , Square : A rectangle with sides, of equal length., , All the properties of a parallelogram,, rhombus and a rectangle, , Kite : A quadrilateral with exactly, two pairs of equal consecutive, sides., , (1) The diagonals are perpendicular to one, another., (2) The diagonals are not of equal length., (3) One of the diagonals bisects the other., , SC, , ER, , T, , Rectangle : A parallelogram with, all right angles., , Trapezium: A quadrilateral with, one pair sides parallel., , 244, , VII CLASS MATHEMATICS, , 1) One pair of opposite sides are parallel
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13, , AREA AND PERIMETER, Introduction, , AN, A, , 13.0, , TE, , LA, N, , G, , Ira wants to find the area of her agricultural land, which is irregular in shape (Figure 1). So she, divided her land into some regular shapes- triangles, rectangle, parallelogram, rhombus and square, (Figure 2). She thought, ‘if I know the area of all these parts, I will know the area of my land.’, , Figure 1, , Figure 2, , 1., , ER, , T, , We have learnt how to find the area of a rectangle and square in earlier classes. In this chapter we, will learn how to find the area of a parallelogram, triangle, rhombus. First let us review what we, have learnt about the area and perimeter of a square and rectangle in earlier classes., Exercise - 1, , Complete the table given below., , SC, , Diagram, , b, , Shape, , Area, , Perimeter, , Rectangle, , l×b = lb, , _____________, , Square, , _____________, , 4a, , l, a, , a, , a, a, , Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 245
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2., , The measurements of some squares are given in the table below. However, they are, incomplete. Find the missing information., Side of a square, , Area, , 15 cm, , 225 cm2, , Perimeter, , 88 cm, The measurements of some rectangles are given in the table below. However, they are, incomplete. Find the missing information., Length, , Breadth, , 20 cm, , 14 cm, , Perimeter, , 60 cm, , 150 cm, , Area of a parallelogram, , LA, N, , 15 cm, 13.1, , Area, , G, , 12 cm, , AN, A, , 3., , TE, , Look at the shape in Figure 1. It is a parallelogram. Now let us learn, how to find its area-, , Activity 1, , Draw a parallelogram on a sheet of paper., , •, , Cut out the parallelogram., , •, , Now cut the parallelogram along the dotted line as shown in, Figure 2 and separate the triangular shaped piece of paper., , SC, , ER, , T, , •, , Figure 1, , •, , Figure 2, , Keep the triangle on the other side as shown in Figure 3 and see if, both the pieces together form a rectangle., Figure 3, , Can we say that the area of the parallelogram in Figure 2 equal to the area of the rectangle in, Figure 3? You will find this to be true., , 246, , VII CLASS MATHEMATICS
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As you can see from the above activity the area of the parallelogram is equal to the area of the, rectangle., We know that the area of the rectangle is equal to length × breadth., We also know that the length of the rectangle is equal to the base of, the parallelogram and the breadth of the rectangle is equal to its height., , breadth, (height), , AN, A, , length (base), , Therefore, Area of parallelogram = Area of rectangle, = length × breadth, , Example 1 :, , height (h), , LA, N, , G, , = base × height, (length = base ; breadth = height), Thus, the area of the parallelogram is equal to the product of its, base (b) and corresponding height (h) i.e., A = bh, , base (b), , Find the area of each parallelogram., , (i), , Solution :, , TE, , Base (b) of a parallelogram = 4 units, Height (h) of a parallelogram = 3 units, , 3, , Area (A) of a parallelogram = bh, Therefore, A =4 × 3 = 12 sq. units, Thus, area of the parallelogram is 12 sq. units., , A, , B, , D, , Solution :, Base of a parallelogram (b) = 6 m., Height of a parallelogram (h) = 13 m., , SC, , (ii), , ER, , T, , 4, , Area of a parallelogram (A) = bh, Therefore, A = 6×13 = 78 m2, Thus, area of parallelogram ABCD is 78 m2, C, , Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 247
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Try This, ABCD is a parallelogram with sides 8 cm and 6 cm. In Figure 1, what is the base of, the parallelogram? What is the height? What is the area of the parallelogram?, In Figure 2, what is the base of the parallelogram? What is the height? What is the, area of the parallelogram? Is the area of Figure 1 and Figure 2 the same?, D, , AN, A, , E, , 8 cm, Figure 1, , C, , 6 cm, F, , 6 cm, , 3 cm, A, , D, , C, , B, , A, , 4 cm, , B, , 8 cm, Figure 2, , Do This, 1., , TE, , LA, N, , G, , Any side of a parallelogram can be chosen as base of the parallelogram. In Figure 1,, DE is the perpendicular falling on AB. Hence AB is the, D, C, base and DE is the height of the parallelogram. In Figure 2, BF is the perpendicular falling on side AD. Hence,, F, AD is the base and BF is the height., , (ii) The length of BF, if AD = 6 cm, , 4 cm, , A, , E, , 10 cm, , B, , ER, , Carefully study the following parallelograms., , SC, , 2., , C, , (i) The area of ABCD., , T, , Find, , F, , B, , E, , D, , In parallelogram ABCD, AB = 10 cm, and DE = 4 cm, , A, , (i), , (ii), , (v), , 248, , VII CLASS MATHEMATICS, , (iii), , (vi), , (iv), , (vii)
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(i), , Find the area of each parallelogram by counting the squares enclosed in it. For counting, incomplete squares check whether two incomplete squares make a complete square in, each parallelogram., , Complete the following table accordinglyArea by counting squares, , (i), , Base, , Height, , Area, , No. of full, No. of, Total, squares incomplete squares (full), squares, , 5 units 3 units 5 × 3 = 15 sq. units, , 12, , (ii), (iv), , 15, , LA, N, , (v), (vi), (vii), , Do parallelograms with equal bases and equal heights have the same area?, , TE, , (ii), , 6, , G, , (iii), , AN, A, , Parallelogram, , Try This, , (i) Why is the formula for finding the area of a rectangle related to the formula for, finding the area of a parallelogram?, , Exercise - 2, , Find the area of the each of the following parallelograms., , 6 cm, , 4 cm, , 3 cm, , SC, , 1., , ER, , T, , (ii) Explain why a rectangle is a parallelogram but a parallelogram may not be a, rectangle., , 7 cm, (i), , cm, 5.1, , 5 cm, (ii), , 7.6, , cm, , 4 cm, , (iii), Free distribution by T.S. Government 2018-19, , (iv), AREA AND PERIMETER, , 249
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2., , PQRS is a parallelogram. PM is the, , M, , S, , 12 cm, , R, , height from P to SR and PN is the, 7.6 cm, , height from P to QR . If SR = 12 cm, (i), (ii), , P, , Find PN, if QR = 8 cm., , Q, , G, , DF and BE are the height on sides AB and AD respectively in parallelogram ABCD. If the, area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BE, and DF., D, C, , 49 c, , E, , F, , 35 cm, , B, , TE, , A, , LA, N, , m, , 3., , Find the area of the parallelogram PQRS, , AN, A, , and PM = 7.6 cm., , N, , The height of a parallelogram is one third of its base. If the area of the parallelogram is, 192cm2, find its height and base., , 5., , In a parallelogram the base and height are in the ratio of 5:2. If the area of the, parallelogram is 360m2, find its base and height., , 6., , A square and a parallelogram have the same area. If a side of the square is 40m and the, height of the parallelogram is 20m, find the base of the parallelogram., , ER, Area of triangle, , SC, , 13.2, , T, , 4., , 13.2.1 Triangles are parts of rectangles, , A, , B, , D, , C, , Draw a rectangle Cut the rectangle along its, diagonal to get two triangles., Superimpose one triangle over the other. Are they, exactly the same in area? Can we say that the, triangles are congruent?, , 250, , VII CLASS MATHEMATICS
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You will find that both the triangles are congruent. Thus, the area of the rectangle is equal to the, sum of the area of the two triangles., Therefore, the area of each triangle =, , 1, × (area of rectangle), 2, , 1, 1, × (l × b) =, lb, 2, 2, 13.2.2 Triangles are parts of parallelograms, , AN, A, , =, , You will find that the area of the parallelogram is equal, to the area of both the triangles., , LA, N, , We know that area of parallelogram is equal to product of its base and height. Therefore,, , 1, × (area of parallelogram), 2, , Area of triangle, , 1, × (base × height), 2, , TE, , Area of each triangle =, =, , G, , Make a parallelogram as shown in the Figure. Cut the, parallelogram along its diagonal. You will get two, triangles. Place the triangles one on top of each other., Are they exactly the same size (area)?, , 1, 1, ×b×h, =, bh, 2, 2, Thus, the area of a triangle is equal to half the product of its base (b) and height (h) i.e.,, , 1, bh, 2, , ER, , A=, , T, , =, , Solution :, Base of triangle (b) = 13 cm, Height of triangle (h) = 6 cm, , 6 cm, , SC, , Example 2 : Find the area of the triangle., , Area of a triangle (A) =, , 13 cm, , Therefore,, , A=, , 1, 1, (base × height) or, bh, 2, 2, 1, × 13 × 6, 2, , = 13 × 3 = 39 cm2, Thus the area of the triangle is 39 cm2, Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 251
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Example 3 : Find the area of ΔABC., Base of the triangle (b) = 8 cm, , 1, bh, 2, , Area of the triangle (A) =, Therefore,, , 6 cm, , Height of the triangle (h) = 6 cm, , B, , 1, × 8 × 6 = 24 cm2, 2, , A=, , Thus, the area of ΔABC = 24 cm2, , C, , 8 cm, , AN, A, , Solution:, , A, , Try This, , G, , Notice that in a right angle triangle two of its sides can be the height., D, , C, , F, , LA, N, , In Figure all the triangles, are on the base AB = 24, cm. Is the height of each, of the triangles drawn on, base AB, the same?, , E, , TE, , Will all the triangles have, A, equal area? Give reasons, to support your answer., Are the triangles congruent also?, , T, , Exercise - 3, , Find the area of each of the following triangles., , ER, , ( ii ), , ( iii ), , 6 cm, , 7.5, , 5.4, 4 cm, , ( iv ), , 5 cm, , 6 cm, , 252, , cm, , cm, , 8 cm, , SC, , (i), , 4 cm, , 1., , B, , 24 cm, , VII CLASS MATHEMATICS
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2., , In ΔPQR, PQ = 4 cm, PR = 8 cm and RT = 6 cm. Find (i) the area of ΔPQR (ii) the length, of QS., P, , S, 4 cm, , T, , 87 c, , m, , AN, A, , 6 cm, , R, , Q, , 3., , ΔABC is right-angled at A. AD is perpendicular to BC, AB = 5 cm, BC = 13 cm and, AC = 12 cm. Find the area of ΔABC. Also, find the length of AD., , B, , C, , 13 cm, , D, , ΔPQR is isosceles with PQ = PR = 7.5 cm and QR = 9 cm. The height PS from P to QR,, is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?, , TE, , 4., , LA, N, , 5 cm, , 12 c, m, , G, , A, , P, , T, ER, , cm, , R, , ABCD rectangle with AB = 8 cm, BC = 16 cm and AE=4 cm. Find the area of ΔBCE., Is the area of ΔBEC equal to the sum of the area of ΔBAE and ΔCDE. Why?, , SC, , 5., , S, 9 cm, , 7.5, , Q, , 6 cm, , T, , A, , 4 cm, , E, , D, , 8 cm, B, Free distribution by T.S. Government 2018-19, , 16 cm, , C, AREA AND PERIMETER, , 253
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Ramu says that the area of ΔPQR is, A =, , 6., , Gopi says that it is, A =, , 1, × 7 × 5 cm2, 2, , 1, × 8 × 5 cm2. Who is correct? Why?, 2, , m, 7c, , 6c, m, , S, , m, 5c, , Q, , AN, A, , P, , R, , 8 cm, , Find the base of a triangle whose area is 220 cm2 and height is 11cm., , 8., , In a triangle the height is double the base and the area is 400 cm2. Find the length of the, base and height., , 9., , The area of triangle is equal to the area of a rectangle whose length and breadth are 20 cm, and 15 cm respectively. Calculate the height of the triangle if its base measures 30 cm., , 10., , In Figure ABCD find the area of the shaded region., , LA, N, , G, , 7., , C, , TE, , D, , A, , B, , 40 cm, , In Figure ABCD, find the area of the shaded region., , SC, , 11., , ER, , T, , 40 cm, , D, , C, , 8cm, E, 20 cm, , A, , 254, , VII CLASS MATHEMATICS, , 10 cm F, , 10 cm, , B
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12., , Find the area of a parallelogram PQRS, if PR = 24 cm and QU = ST = 8 cm., S, , R, U, , 8 cm, , 8 cm, , P, , 24 cm, Q, , 9 cm, , AN, A, , T, , The base and height of the triangle are in the ratio 3:2 and its area is 108 cm2. Find its base, and height., , 13.3, , Area of a rhombus, , ER, , T, , TE, , LA, N, , Santosh and Akhila are good friends. They are fond, of playing with paper cut-outs. One day, Santosh gave, different triangle shapes to Akhila. From these she made, different shapes of parallelograms. These, parellelograms are given below-, , G, , 13., , SC, , Santosh asked Akhila, 'which parallelograms has 4 equal sides?", Akhila said, 'the last two have equal sides.", Santhosh said, ‘If all the sides of a parallelogram are equal, it is called a Rhombus.’, Now let us learn how to calculate the area of a Rhombus., Like in the case of a parallelogram and triangle, we can use the method of splitting into congruent, triangles to find the area of a rhombus., , Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 255
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A, , ABCD is a rhombus., Area of rhombus ABCD = (area of ΔACD) + (area of ΔACB), ⎛1, ⎞ ⎛1, ⎞, = ⎜ × AC × OD ⎟ + ⎜ × AC × OB ⎟, ⎝2, ⎠ ⎝2, ⎠, , d1, B, , diagonals bisect perpendicularly, 1, AC × (OD + OB), 2, , C, , 1, = AC × BD, 2, , 1, d1 × d 2 (as AC = d1 and BD = d2), 2, , G, , =, , D, , d2, , AN, A, , =, , O, , 1, dd, 2 1 2, , Solution :, , Length of the diagonal (d1), , TE, , = 7.5 cm, , c, 7.5, , C, m, , 3 cm, , Find the area of rhombus ABCD, , cm, 5.6, , Example 4 :, , 3 cm, , D, , 3 cm, , A=, , LA, N, , In other words, the area of a rhombus is equal to half the product of its diagonals i.e.,, , Length of the other diagonal (d2) = 5.6 cm, Area of the rhombus (A), , =, , 1, d d, 2 1 2, , A, , 3 cm, , B, , T, , 1, × 7.5 × 5.6 = 21 cm2, 2, Thus, area of rhombus ABCD = 21 cm2, , A, , ER, , Therefore,, , =, , The area of a rhombus is 60 cm2 and one of its diagonals is 8 cm. Find the other, diagonal., , Solution :, , Length of one diagonal (d1), , = 8 cm, , Length of the other diagonal, , = d2, , Area of rhombus, , =, , SC, , Example 5 :, , 1, × d1 × d2, 2, , 1, × 8 × d2, 2, d2 = 15 cm., Thus, length of the other diagonal is 15 cm., , Therefore,, , 256, , VII CLASS MATHEMATICS, , 60 =
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Exercise 4, Find the area of the following rhombuses., , Diagonal-2 (d2), , 12 cm, , 16 cm, , Area of rhombus, , LA, N, , Diagonal-1 (d1), , 5c, m, , 4 cm, , AN, A, , Find the missing values., , G, , m, 5c, , 2., , 3 cm, , 3 cm, , m, 2c, , m, 5c, , 4 cm, , m, 5c, , m, , m, , 5c, , 2c, , 5c, m, , 1., , 27 mm, , 2025 mm2, , 24 m, , 57.6 m, , If length of diagonal of a rhombus whose area 216 sq. cm. is 24 cm. Then find the length, of second diagonal., , 4., , The floor of a building consists of 3000 tiles which are rhombus shaped. The diagonals of, each of the tiles are 45 cm and 30 cm. Find the total cost of polishing the floor, if cost per, m2 is ` 2.50., , 13.4, , Circumference of a circle, , ER, , T, , TE, , 3., , Nazia is playing with a cycle tyre. She is rotating the tyre with a stick, and running along with it., , SC, , What is the distance covered by tyre in one rotation?, The distance covered by the tyre in one rotation is equal to the length, around the wheel. The length around the tyre is also called the, circumference of the tyre ., What is the relation between the total distance covered by the tyre, and number of rotations?, Total distance covered by the tyre = number of rotations × length around the tyre., Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 257
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Activity 2, Jaya cut a circular shape from a cardboard. She wants to stick lace around, the card to decorate it. Thus, the length of the lace required by her is equal to, the circumference of the card. Can she measure the circumference of the card, with the help of a ruler?, , AN, A, , Let us see what Jaya did?, , LA, N, , G, , Jaya drew a line on the table and marked its starting point A. She then made a point on the edge of, the card. She placed the circular card on the line,, A, such that the point on the card coincided with point, B, A. She then rolled the card along the line, till the, point on the card touched the line again. She marked this point B. The length of line AB is the, circumference of the circular card. The length of the lace required around the circular card is the, distance AB., Try This, , TE, , Take a bottle cap, a bangle or any other circular object and find its circumference, using a string., It is not easy to find the circumference of every circular shape using the above method. So we need, another way for doing this. Let us see if there is any relationship between the diameter and the, circumference of circles., , T, , A man made six circles of different radii with cardboard and found their circumference using a, string. He also found the ratio between the circumference and diameter of each circle., , ER, , He recorded his observations in the following tableRadius, , Diameter, , Circumference, , Ratio of circumference and diameter, , 1., , 3.5 cm, , 7.0 cm, , 22.0 cm, , 22, = 3.14, 7, , 2., , 7.0 cm, , 14.0 cm, , 44.0 cm, , 44, = 3.14, 14, , 3., , 10.5 cm, , 21.0 cm, , 66.0 cm, , 4., , 21.0 cm, , 42.0 cm, , 132.0 cm, , 5., , 5.0 cm, , 10.0 cm, , 32.0 cm, , 6., 258, , 15.0cm, , 30.0 cm, , 94.0 cm, , SC, , Circle, , VII CLASS MATHEMATICS
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What can you infer from the above table? Is the ratio between the circumference and the diameter, of each circle approximately the same? Can we say that the circumference of a circle is always, about three times its diameter?, The approximate value of the ratio of the circumference to the diameter of a circle is, , 22, or 3.14., 7, , AN, A, , Thus it is a constant and is denoted by π (pi)., c, = π where 'c' is the circumference of the circle and 'd' its diameter., d, , Since,, , c, =π, d, , c=πd, , c=π×2r, , or, , LA, N, , Since, diameter of a circle is twice the radius i.e. d = 2r, , G, , Therefore,, , c=2πr, , Thus, circumference of a circle = πd or 2π, πr, , Find the circumference of a circle with diameter 10 cm. (Take π = 3.14), , Solution :, , Diameter of the circle (d), , TE, , Example 6 :, , = 10 cm., , Circumference of circle (c) = π d, , T, , = 3.14 × 10, , c = 31.4 cm, , ER, , Thus, the circumference of the circle is 31.4 cm., Find the circumference of a circle with radius 14 cm. (Take π =, , SC, , Example 7 :, , Radius of the circle (r), , = 14 cm, , Circumference of a circle (c), , =2πr, , Therefore,, , 22, ), 7, , 22, × 14, 7, , c, , =2×, , c, , = 88 cm, , Thus, the circumference of the circle is 88 cm., Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 259
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Exercise - 5, 1., , Find the circumference of a circle whose radius is(i) 35 cm, , (iii) 15.4 cm, , Find the circumference of circle whose diameter is(i) 17.5 cm, Note : take π =, , 3., , (i), , (ii) 5.6 cm, , 22, in the above two questions., 7, , Taking π = 3.14, find the circumference of a circle whose radius is, (b) 15 cm, , (c) 20 cm, , G, , (a) 8 cm, (ii), , (iii) 4.9 cm, , AN, A, , 2., , (ii) 4.2 cm, , Calculate the radius of a circle whose circumference is 44cm?, 22, ., 7, , If the circumference of a circle is 264 cm, find its radius. Take π =, , 5., , If the circumference of a circle is 33 cm, find its diameter., , 6., , How many times will a wheel of radius 35cm be rotated to travel 660 cm?, ( Take π =, , 22, )., 7, , TE, , LA, N, , 4., , The ratio of the diameters of two circles is 3 : 4. Find the ratio of their circumferences., , 8., , A road roller makes 200 rotations in covering 2200 m., Find the radius of the roller., , ER, , The minute hand of a circular clock is 15 cm., How far does the tip of the minute hand move in 1 hour?, (Take π = 3.14), , SC, , 9., , T, , 7., , 10., , 260, , A wire is bent in the form of a circle with radius 25 cm. It is, straightened and made into a square. What is the length of the side, of the square?, , VII CLASS MATHEMATICS
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Rectangular Paths, , AN, A, , 13.5, , A plot is 60m long and 40m wide. A path 3m wide is to be constructed around the, plot. Find the area of the path., , LA, N, , Example 8 :, , G, , We often come across such walking paths in garden areas. Now we shall learn how to measure the, areas of such paths as this often useful in calculating their costs of construction., , E, , F, , 3m, , B, , A, 3m, , 3m, , C, , TE, , D, , 3m, , H, , Let ABCD be the given plot. A 3m wide path is running all around it. To find the, area of this path we have to subtract the area of the smaller rectangle ABCD from, the area of the bigger rectangle EFGH., , ER, , T, , Solution :, , G, , =, , 60m, , Breadth of inner rectangle, , =, , 40m, , Area of the plot ABCD, , =, , (60 x 40) m2, , =, , 2400 m2, , Width of the path, , =, , 3m, , Length of outer rectangle, , =, , 60 m + (3+3) m, , =, , 66 m, , =, , 40 m + (3+3) m, , =, , 46 m, , SC, , Length of inner rectangle, , Breadth of outer rectangle, , Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 261
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Area of the outer rectangle, , =, , 66 x 46 m2, , =, , 3036 m2, , Therefore, area of the path =, , (3036 - 2400) m2, , =, , The dimensions of a rectangular field are 90 m and 60 m. Two roads are, constructed such that they cut each other at the centre of the field and are parallel, to its sides. If the width of each road is 3 m, find-, , AN, A, , Example 9 :, , 636 m2, , (i) The area covered by the roads., , (ii) The cost of constructing the roads at the rate of 110 per m2., , Let ABCD be the rectangular field. PQRS and EFGH are the 3m roads., P, , A, , Q, , LA, N, , 3m, , G, , Solution :, , K, , E, 3m, H, , L, , F, 60 m, G, , M, , TE, , N, , D, , B, , S 90 m R, , C, , ER, , T, , (i) Area of the crossroads is the area of the rectangle PQRS and the area of the, rectangle EFGH. As is clear from the picture, the area of the square KLMN, will be taken twice in this calculation thus needs to be subtracted once., , SC, , From the question we know that,, PQ = 3 m, and, , PS = 60 m, , EH = 3 m, and, , EF = 90 m, , KL = 3 m, , KN = 3 m, , Area of the roads = Area of the rectangle PQRS + area of the rectangle EFGH, – Area of the square KLMN, = (PS × PQ) + (EF × EH) – (KL × KN), = (60 × 3) + (90 × 3) – (3 × 3), = (180 + 270 – 9) m2, = 441 m2, , 262, , VII CLASS MATHEMATICS
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(ii), , Cost of construction = 110 × m2, Cost of constructing the roads = 110 × 441, =, , 48,510, , A, , B, P, , Q, , SR, , SR, , D, , G, , Let PQRS be the square park of the side 100 m. The shaded region represents, the 5m wide path., Length of AB, , LA, N, , Solution :, , C, , AN, A, , Example 10 : A path of 5m wide runs around a square park of side 100m. Find the area of the, path. Also find the cost of cementing it at the rate of 250 per 10m2, , = 100 + (5 + 5) = 110 m, , = (side)2 = (100 m)2 = 10000 m2, , Area of the square ABCD, , = (side)2 = (110 m)2 = 12100 m2, , Therefore, area of the path, , = (12100 - 10000) = 2100 m2, , Cost of the cementing per 10 m2, , =, , 250, , Therefore, cost of the cementing 1 m2 =, , 250, 10, , T, , TE, , Area of the square PQRS, , SC, , ER, , Thus, cost of cementing 2100 m2, , =, , 250, × 2100, 10, , =, , 52,500, , Exercise - 6, , 1., , A path 2.5 m wide is running around a square field whose side is 45 m. Determine the area, of the path., , 2., , The central hall of a school is 18m long and 12.5 m wide. A carpet is to be laid on the floor, leaving a strip 50 cm wide near the walls, uncovered. Find the area of the carpet and also, the uncovered portion?, , Free distribution by T.S. Government 2018-19, , AREA AND PERIMETER, , 263
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The length of the side of a grassy square plot is 80 m. Two walking paths each 4 m wide, are constructed parallel to the sides of the plot such that they cut each other at the centre, of the plot. Determine the area of the paths., , 4., , A verandah 2 m wide is constructed all around a room of dimensions 8 m × 5 m. Find the, area of the verandah, , 5., , The length of a rectangular park is 700 m and its breadth is 300 m. Two crossroads, each, of width 10 m, cut the centre of a rectangular park and are parallel to its sides. Find the, area of the roads. Also, find the area of the park excluding the area of the crossroads., , AN, A, , 3., , Looking Back, , •, , G, , The area of the parallelogram is equal to the product of its base (b) and, corresponding height (h) i.e., A = bh. Any side of the parallelogram can be taken, as the base., The area of a triangle is equal to half the product of its base (b) and height (h), 1, bh., 2, , i.e., A =, •, , The area of a rhombus is equal to half the product of its diagonals, 1, dd., 2 1 2, , TE, , i.e., A =, •, , LA, N, , •, , The circumference of a circle = 2 π r where r is the radius of the circle and, 22, or 3.14., 7, , ER, , T, , π=, , Archimedes (Greece), , SC, , 287 - 212 BC, He calculated the value of π first time., He also evolved the mathematical formulae for, finding out the circumference and area of a circle., , 264, , VII CLASS MATHEMATICS
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UNDERSTANDING 3D, AND 2D SHAPES, Introduction, , AN, A, , 14.0, , 14, , We have been introduced to various three-dimensional shapes in class VI. We have also identified, their faces, edges and vertices. Let us first review what we have learnt in class VI., Exerciese - 1, , G, , Given below are the pictures of some objects. Categorise and fill write their names according, to their shape and fill the table with name of it., , SC, , ER, , T, , TE, , LA, N, , 1., , Sphere, , Cylinder, , Pyramid, , Free distribution by T.S. Government 2018-19, , Cuboid, , UNDERSTANDING 3D, , Cone, , AND, , 2D SHAPES, , Cube, , 265
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Write names of at least 2 objects from day-to-day life, which are in the shape of the basic, 3D shapes given below:, (i), , -----------, , -----------, , -----------, , -----------, , (ii) Cube, , -----------, , -----------, , -----------, , -----------, , (iii) Cuboid, , -----------, , -----------, , -----------, , -----------, , (iv) Sphere, , -----------, , -----------, , -----------, , -----------, , (v) Cylinder, , -----------, , -----------, , -----------, , -----------, , Identify and state the number of faces, edges and vertices of the figures given below:, Edge, , Face, , G, , 3., , Cone, , AN, A, , 2., , Vertex, , Vertex, , Cube, , Faces, Edges, , 14.1, , Cuboid, , Edge, , Pyramid, , TE, , Vertices, , Vertex, , Edge, , LA, N, , Face, , Face, , Nets of 3-D shapes, , T, , We now visualise 3-D shapes on 2-D surfaces, that is on a plain paper. This can be done by, drawing the ‘nets’ of the various 3-D figures., , SC, , ER, , Take a cardboard box (cartoon of tooth paste or shoes etc.,). Cut the edges to lay the box flat., You have now a net for that box. A net is a sort of skeleton-outline in 2-D (Figure 1), which, when, folded (Figure 2), results in a 3-D shape (Figure 3)., , Figure 1, , 266, , VII CLASS MATHEMATICS, , Figure 2, , Figure 3
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8, , Here is a net pattern for a box. Trace it and paste it on, a thick paper and try to make the box by suitably folding, and gluing together. What is the shape of the box?, , 3, 6, , 6, , 6, , Similarly, take a cover, 3, 8, 3, of an ice-cream cone (, or any like shape). Cut it along it’s slant surface, as shown in Figure 1. You will get the net for the, cone (Figure 2)., , 8, , 3, , AN, A, , Figure 2, , 3, , 3, 8, , 6, , Figure, , 8, , Try This, , 8, , G, , Take objects having different shapes (cylinder, cube, cuboid and cone) and cut them, to get their nets with help of your teachers or friends., , LA, N, , You will come to know by the above activity that you have different nets for different, shapes. Also, each shape can also have more than one net according to the way we, cut it., Exercise - 2, , Some nets are given below. Trace them and paste them on a thick paper. Try to make 3-D, shapes by suitably folding them and gluing together. Match the net with it’s 3-D shape., 3D shapes, , SC, , ER, , T, , Nets, , TE, , 1., , Free distribution by T.S. Government 2018-19, , UNDERSTANDING 3D, , AND, , 2D SHAPES, , 267
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2., , Three nets for each shape are given here. Match the net with its 3D-shape., , (b), , LA, N, , (a), , G, , AN, A, , (i), , ER, , T, , TE, , (ii), , (c), , (a), , Here are two nets to make dice. Insert the suitable number of dots in blanks., , 268, , (c), , A dice is a cube with dots on each face. The opposite faces of a dice always, have a total of seven dots on them., , SC, , 3., , (b), , VII CLASS MATHEMATICS
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Play This, You and your friend sit back to back. One of you read out a net to make a 3-D shape, while the, other copies it and sketches or builds the described object., 14.2, , Drawing solids on a flat surface, , 14.2.1, , AN, A, , Our drawing surface is a paper, which is a flat surface. When you draw a solid shape, the images, are somewhat distorted. It is a visual illusion. You will find here two techniques to help you to draw, the 3-D shapes on a plane surface., Oblique Sketches, , LA, N, , G, , Here is a picture of a cube. It gives a clear idea of how the cube looks, when seen, from the front. You do not see all the faces as we see in reality. In the picture, all the, lengths are not equal, as they are in a real cube. Still, you are able to recognise it as, a cube. Such a sketch of a solid is called an oblique sketch., , T, , TE, , How can you draw such sketches? Let us attempt to learn the technique. You need a squared, (lines or dots) paper. Initially practice to draw on these sheets and later on a plain sheet (without, the aid of squared lines or dots!) Let us attempt to draw an oblique sketch of a 3 × 3 × 3 cube, (each edge is 3 units)., , SC, , ER, , Step 1, Draw the front face., , Step 3, Join the corresponding corners, , Step 2, Draw the opposite face. Sizes of the, faces have to be same, but the sketch is, somewhat off-set from step 1., , Step 4, Redraw using doted lines for hidden edges., (It is a convention) The sketch is ready now., , In the above oblique sketch did you notice the following?, , Free distribution by T.S. Government 2018-19, , UNDERSTANDING 3D, , AND, , 2D SHAPES, , 269
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(i) The sizes of the front face and its opposite face are same., (ii) The edges, which are all equal in a cube, appear so in the sketch, though the actual measures, of edges are not taken so., You could now try to make an oblique sketch of a cuboid (remember the faces in this case are, rectangles)., , AN, A, , You can draw sketches in which measurements also agree with those of a given solid. To do this, we need what is known as an isometric sheet., Let us try to make a cuboid with dimensions 7 cm length, 3 cm breadth and 4 cm height on an, isometric sheet., 14.2.2 Isometric Sketches, , G, , To draw sketches in which measurements also agree with those of the given solid, we can use, isometric dot sheets. In such a sheet the paper is divided into small equilateral triangles made up of, dots or lines., , T, , TE, , LA, N, , Let us attempt to draw an isometric sketch of a cuboid of dimensions 7 × 3 × 4 (which means the, edges forming length, breadth and height are 7, 3, 4 units respectively)., , Step 2, Draw four parallel line segments of length 3 units, starting from the four corners of the rectangle., , SC, , ER, , Step 1, Draw a rectangle to show the front face, , Step 3, Connect the matching corners, with appropriate line segments., 270, , VII CLASS MATHEMATICS, , Step 4, This is an isometric sketch, of a cuboid.
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Note that the measurements of the solid are of exact size in an isometric sketch; this is not so, in the case of an oblique sketch., Here is an oblique sketch of a cuboid. Draw an isometric, sketch that matches this drawing., , Solution :, , The length, breadth and height are 3, 3 and 6 units respectively, , LA, N, , G, , AN, A, , Example 1 :, , Exercise - 3, , Use an isometric dot paper and make an isometric sketch for each one of the given shapes., , TE, , 1., , (ii), , (i), , 1, , 3, 2, , T, , 1, , 6, , ER, , 6, , 1, , SC, , 7, , 7, , 1, , 1, , (iii), , (iv), , 4, , 4, , 4, , 4, , 4, 4, , 3, , 3, 3, , 3, 3, , 4, 4, , 4, , Free distribution by T.S. Government 2018-19, , UNDERSTANDING 3D, , AND, , 2D SHAPES, , 271
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2., , The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric, sketches of this cuboid., , 3., , Three cubes each with 2 cm edge are placed side by side to form a cuboid. Draw an, oblique or isometric sketch of this cuboid., , 4., , Make an oblique sketch for each of the given isometric shapes., (i), , LA, N, , G, , AN, A, , (ii), , (iv), , 5., , ER, , T, , TE, , (iii), , Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:, , (a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?), , SC, , (b) A cube with an edge 4 cm long., , 14.3, , Visualising solid objects, , Sometimes when you look at combined shapes, some of, them may be hidden from your view., , 272, , VII CLASS MATHEMATICS
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AN, A, , Here are some activities to help you visualise some solid objects and how they look. Take some, cubes and arrange them as shown below., , G, , Now ask your friend to guess the total number of cubes in the following arrangements., Try This, , TE, , LA, N, , Estimate the number of cubes in the following arrangements., , Such visualisations are very helpful., , ER, , T, , Suppose you form a cuboid by joining cubes. You will be able to estimate what the length, breadth, and height of the cuboid would be., Example 2 : If two cubes of dimensions 2 cm by 2cm by 2cm are placed side by side, what, would the dimensions of the resulting cuboid be?, As you can see when kept side by side, the, length is the only measurement which increases,, it becomes 2 + 2 = 4 cm., , SC, , Solution :, , The breadth = 2 cm and the height = 2 cm., , Free distribution by T.S. Government 2018-19, , 2 cm = H, , 2 cm L 2 cm, , UNDERSTANDING 3D, , AND, , 2D SHAPES, , 2 cm = B, , 273
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Try This, 1., , Two dice are placed side by side as shown. Can you, say what the total would be on the faces opposite to, them? (i) 5 + 6, (ii) 4 + 3, (Remember that in a dice the sum of numbers on opposite faces is 7), Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try, to make an oblique sketch and say what could be its length, breadth and height., , AN, A, , 2., , 14.3.1 Viewing different sections of a solid, , G, , Now let us see how an object which is in 3-D can be viewed in different ways., 14.3.1a) One way to view an object is by cutting or slicing the object, , LA, N, , Slicing game, , Here is a loaf of bread. It is like a cuboid with a square faces. You ‘slice’ it with a knife., , TE, , When you give a ‘horizontal’ cut, you get several pieces,, as shown in the figure. Each face of the piece is a square!, We call this face a ‘cross-section’ of the whole bread., The cross section is nearly a square in this case., , A kitchen play, , T, , Beware! If your cut is ‘vertical’ you may get a different cross section! Think about it. The boundary, of the cross-section you obtain is a plane curve. Do you notice it?, , ER, , Have you noticed cross-sections of some vegetables when they are cut for the purposes of cooking, in the kitchen? Observe the various slices and get aware of the shapes that results as, cross-sections., Do This, , Make clay (or plasticine) models of the following solids and make, , SC, , 1., , vertical or horizontal cuts. Draw rough sketches of the cross-sections, you obtain. Name them if possible., , 274, , VII CLASS MATHEMATICS
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2., , What cross-sections do you get when you give a (i) vertical cut (ii) horizontal cut to the, following solids?, (a) A brick (b) A round apple (c) A die (d) A circular pipe (e) An ice cream cone, , 14.3.1b) Another Way is by Shadow Play, , G, Figure 1, , LA, N, , Shadows are a good way to illustrate, how three-dimensional objects can be, viewed in two dimensions. Have you seen, a shadow play? It is a form of, entertainment using solid articulated, figures in front of an illuminated, backdrop to create the illusion of moving, images. It makes some indirect use of, ideas of Mathematics., , AN, A, , A shadow play, , You will need a source of light and a few solid shapes for this activity. If you have an overhead, projector, place the solid under the lamp and do these investigations., Keep a torchlight, right in front of a cone. What type of shadow does it cast on the screen? (Figure 1)., , TE, , The solid is three-dimensional; what about the shadow?, , If, instead of a cone, you place a cube in the above game, what type of shadow will you get?, , T, , Experiment with different positions of the source of light and with different positions of the solid, object. Study their effects on the shapes and sizes of the shadows you get., , ER, , Here is another funny experiment that you might have tried already:, Place a circular tumbler in the open when the sun at the noon time is just right above it as shown in, the figure below. What is the shadow that you obtain?, (b) evening?, , SC, , Will it be same during (a) afternoon?, , Study the shadows in relation to the position of the sun and the time of observation., , Free distribution by T.S. Government 2018-19, , UNDERSTANDING 3D, , AND, , 2D SHAPES, , 275
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Exerciese - 4, A bulb is kept burning just right above the following solids. Name the shape of the shadows, obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to, experiment first and then answer these questions)., , A book, , Here are the shadows of some 3D objects, when seen under the lamp of an overhead, projector. Identify the solid(s) that match each shadow. (There may be many answers for, these!), , LA, N, , 2., , A cylindrical pipe, , G, , A ball, , AN, A, , 1., , A square, , TE, , A circle, , (ii), , A rectangle, , (iii), , (iv), , T, , (i), , A triangle, , ER, , Looking Back, , 3D shapes can be visualised on 2D surfaces, that is on paper by drawing their nets., , SC, , Oblique sketches and isometric sketches help in visualising 3D shapes on a, plane surface., , Fun with a cube, A unit cube can fitted together with 7 other, identical cubes to make a larger cube with an, edge of 2 units as shown in figure., How many unit cubes are needed to make a, cube with an edge of 3 units?, , 276, , VII CLASS MATHEMATICS
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15, , SYMMETRY, , Introduction, , AN, A, , 15.0, , LA, N, , G, , Look around you. You will find that many objects around you are symmetrical. So are the objects, that are drawn below., , All these objects are symmetrical as they can be divided in such a way that their two parts coincide, with each other., 15.1, , Line Symmetry, , ER, , T, , TE, , Let us take some more examples and understand what we mean. Trace the following figures on a, tracing paper., , Figure 1, , Figure 2, , Figure 3, , Figure 4, , SC, , Fold Figure 1 along the dotted line. What do you observe?, You will find that the two parts coincide with each other. Is this true in Figure 2 and 3?, You will observe that in Figure 2, this is true along two lines and in Figure 3 along many lines. Can, Figure 4 be divided in the same manner?, Figure 1, 2 and 3 have line symmetry as they can be divided in such a manner that two parts of the, figure coincide with each other when they are folded along the line of symmetry. The dotted line, which divides the figures into two equal parts is the line of symmetry or axis of symmetry. As you, have seen, an object can have one or more than one lines of symmetry or axes of symmetry., 277, Free distribution by T.S. Government 2018-19, SYMMETRY
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Try This, 1. Name a few things in nature, that are symmetric., 2. Name 5 man-made things that are symmetric., Exercise - 1, Given below are some fiugres. Which of them are symmetric? Draw the axes of symmetry, for the symmetric figures., , (iv), , (v), , (vii), , (viii), , (ix), , (x), , (xi), , (xii), , (xiii), , (xiv), , SC, , ER, , T, , (vi), , (iii), , LA, N, , (ii), , TE, , (i), , G, , AN, A, , 1., , 278, , VII CLASS MATHEMATICS
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(xvi), , (xviii), , (xix), , (xvii), , LA, N, , G, , AN, A, , (xv), , (xxii), , TE, , (xxi), , (xx), , (xxiii), , 15.1.1 Lines of symmetry for regular polygons, , SC, , ER, , T, , Look at the following closed figures., , A closed figure made from several line segments is called a 'Polygon'. Which of the above figures, are polygons?, Try This, 1. Can we make a polygon with less than three line segments?, 2. What is the minimum number of sides of a polygon?, Free distribution by T.S. Government 2018-19, , SYMMETRY, , 279
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Figure 1, , Figure 2, , AN, A, , Observe the different triangles below., , Figure 3, , In Figure 3, the triangle has equal sides and congruent angles. It is thus called a regular polygon., A polygon, with all sides and all angles equal is called a 'Regular Polygon'., , Parallelogram, , LA, N, , G, , Which of the following polygons are regular polygons?, , Square, , Trapezium, , Equilateral, triangle, , Rectangle, , T, , TE, , Now draw axes of symmetry for the following regular polygons., , ER, , Equilateral Triangle, , Square, , Regular Pentagon, , Regular Hexagon, , SC, , Write down your conclusions in the table below., Regular Polygon, Equilateral Traingle, Square, , Pentagon, Hexagon, 280, , VII CLASS MATHEMATICS, , No. of sides, , No. of axes of symmetry, 3, , 3
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Did you find any relationship between the number of sides of a regular polygon and number of axes, symmetry? You will find that the number of sides is equal to number of axes of symmetry., You can verify your results by tracing out all the four figures on a paper, cuting them out and, actually folding each figure to find the axes of symmetry., Try This, , G, , AN, A, , 1. Given below are three types of triangles. Do all the triangles have the same, number of lines of symmetry? Which triangle has more?, , Square, , TE, , Rhombus, , LA, N, , 2. Given below are different types of quadrilaterals. Do all of them have the same, number of lines of symmetry? Which quadrilateral has the most?, , Rectangle, , Hint: You can trace the triangles and quadrilaterals on a tracing paper and actually fold each figure to find the axes of symmetry., , Exercise - 2, , In the figures given below find the axes of symmetry such that on folding along the axis the, two dots fall on each other., , SC, , 1., , ER, , T, , On the basis of (i) and (ii) can we say that a regular polygon has the maximum, number of axes of symmetry., , (i), Free distribution by T.S. Government 2018-19, , (ii), , (iii), SYMMETRY, , 281
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(vii), , (viii), , (iv), , AN, A, , (v), , (ix), , Given the line of symmetry, find the other dot., , LA, N, , G, , 2., , (iv), , (i), , (iii), , (iv), , In the following incomplete figures, the mirror line (i.e. the line of symmetry) is given as a, dotted line. Complete each figure, performing reflection on the dotted (mirror) line. (You, might perhaps place a mirror along the dotted line and look into the mirror for the image)., Are you able to recall the name of the figure you complete?, , ER, , T, , TE, , 3., , (ii), , (i), , 5., , 282, , (iii), , (iv), , (v), , (vi), , State whether the following statements are true or false., , SC, , 4., , (ii), , (i), , Every closed figure has an axis of symmetry., , (, , ), , (ii), , A figure with at least one axis of symmetry is called a symmetric figure. (, , ), , (iii), , A regular polygon of 10 sides will have 12 axes of symmetry., , ), , (, , Draw a square and construct all its axes of symmetry. Measure the angles between each, pair of successive axes of symmetry. What do you notice? Does the same rule apply for, other regular polygons?, VII CLASS MATHEMATICS
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15.2, , Rotational Symmetry, , Activity 1 : Trace the following diagram onto a tracing paper., , AN, A, , Try to fold the diagram so that its two parts coincide. Is this diagram symmetric?, , LA, N, , G, , Now, let us try to match the different positions of the diagram in another way. Draw the above, diagram on a piece of paper. Mark a point 'o'at the centre and name the four edges of the paper, A,B,C,D as shown in Figure 1., , Figure 1, , T, , TE, , Rotate the paper around the marked point for 180o., , Figure 2, , ER, , What do you notice? Does this diagram look different from the previous one?, Due to the rotation, the points A,B,C,D have changed position however the diagram seems to be, unchanged. This is because the diagram has rotational symmetry., , SC, , Activity 2 : Lets make a wind wheel, •, , Take a paper and cut it into the shape of a square., , •, , Fold it along the diagonals., , •, , Starting from one corner, cut the paper along the diagonals towards the centre, up to one, fourth of the length of the diagonal. Do the same from the remaining corners., , •, , Fold the alternate corners towards the centre., , •, , Fix the mid point to a stick with a pin so that the paper rotates freely., , Free distribution by T.S. Government 2018-19, , SYMMETRY, , 283
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Face it in the opposite direction of the wind. You will find it rotates, , AN, A, , •, , Now, let us rotate the wind-wheel by 90º. After each rotation you will see that the wind-wheel, looks exactly the same. The wind-wheel has rotational symmetry., Thus, if we rotate a figure, about a fixed point by a certain angle and the figure, looks exactly the same as before, we say that the figure has rotational symmetry., , G, , 15.2.1 Angle of Rotational Symmetry, , LA, N, , We know that the square has line symmetry and 4 axes of symmetry. Now, let us see if the square, has rotational symmetry., , TE, , Consider a square as in Figure (i) with P as one of its corners., , Figure 2, , Figure 3, , Figure 4, , Figure 5, , T, , Figure 1, , ER, , Figure 1 represent the initial position of square., , SC, , Rotate the square by 90 degrees about the centre. This quarter turn will lead to Figure 2. Note the, position of P. In this way, rotate the square again through 90 degrees and you get Figure 3. When, we complete four quarter turns, the square reaches its original position. After each turn, the square, looks exactly like it did in its original position. This can also be seen with the help of the position, taken by P., , In the above activity all the positions in figure 2, figure 3, figure 4 and figure 5 obtained by the, rotation of the first figure through 900 , 1800, 2700 and 3600 look exactly like the original, figure 1. Minimum of these i.e., 900 is called the angle of rotational symmetry., The minimum angle rotation of a figure to get exactly the same figure as original is, called the “angle of rotational symmetry” or “angle of rotation”., 284, , VII CLASS MATHEMATICS
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Do This, 1. What is the angle of rotational symmetry of a square?, 2. What is the angle of rotational symmetry of a parallelogram?, 3. What is the angle of rotational symmetry of a circle?, 15.2.2 Order of rotational symmetry, , AN, A, , In the above activity, the angle of rotational symmetry of square is 900 and the figure is, turned through the angle of rotational symmetry for 4 times before it comes to original position., Now we say that the square has rotational symmetry of order 4., Consider an equilateral triangle. Its angle of rotational symmetry is 1200. That means it, has to be rotated about its centre for 3 times to get exactly the same position as the original one., So the order of rotational symmetry of a equilateral triangle is 3., , LA, N, , G, , By these examples we conclude that the number of times a figure, rotated through its, angle of rotational symmetry before it comes to original postion is called order of rotational, symmetry., Let us conclude from the above examples, , The centre of rotational symmetry of a square is its intersection point of its diagonals., , •, , The angle of rotational symmetry for a square is 900., , •, , The order of rotational symmetry for a sqaure is 4., , TE, , •, , Try This, , Can you now tell the order of rotational symmetry for an equilateral triangle., , ER, , T, , 1. (i), , (i), , (ii), , (iii), , (iv), , SC, , (ii) How many lines of semetry?, (iii) What is the angle between every adjacent axes?, 2. Look around you. Which objects have rotational symmetry (i.e. rotational, symmetry of order more than 1)., , Note: It is important to understand that all figures have rotational symmetry of order 1,, as can be rotated completely through 360º to come back to its original position. So, we say that an object has rotational symmetry, only when the order of symmetry, is more than 1., Free distribution by T.S. Government 2018-19, , SYMMETRY, , 285
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Exercise - 3, Which of the following figures have rotational symmetry of order more than 1?, , (i), , (iii), , (iv), , (v), , Give the order of rotational symmetry for each figure., , (v), 3., , (iii), , (iv), , (vi), , (vii), , (viii), , (ii), , TE, , (i), , LA, N, , G, , 2., , (ii), , AN, A, , 1., , Draw each of the shapes given below and fill in the blanks., Centre of Rotation, Angle of Rotation Order of Rota, (intersection of diagonals/, -tion, Intersection of axes, of symmetry), , ER, , T, , Shape, , Square, , SC, , Rectangle, Rhombus, , Equilateral Triangle, Regular Hexagon, Circle, Semi-circle, , 286, , VII CLASS MATHEMATICS
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15.3, , Line symmetry and rotational symmetry, , By now you must have realised that some shapes only have line symmetry and some have only, rotational symmetry (of order more than 1) and some have both. Squares and equilateral triangles, have both line and rotational symmetry. The circle is the most perfect symmetrical figure, because, it can be rotated about its centre through any angle and it will look the same. A circle also has, unlimited lines of symmetry., , (i), , (ii), , G, , AN, A, , Which of the following shapes have line symmetry? Which have rotational, symmetry?, , (iii), , LA, N, , Example 1 :, , Figure, , Line symmetry, , 1., , Yes, , 2., , No, , 3., , Yes, , Yes, , 4., , No, , Yes, , Rotational symmetry, No, , Yes, , TE, , Activity 3 :, , (iv), , Take a square shaped paper., , •, , Fold it vertically first, then horizontally., , •, , Then fold along a diagnal such that the paper takes a triangular shape (Figure 4)., , •, , Cut the folded edges as shown in the figure or as you wish (Figure 5)., , •, , Now open the piece of paper., , SC, , ER, , T, , •, , Figure 1, , Figure 2, , Free distribution by T.S. Government 2018-19, , Figure 3, , Figure 4, SYMMETRY, , Figure 5, , 287
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(ii), , Does the paper have rotational symmetry?, , AN, A, , Does the paper have line symmetry?, , G, , (i), , LA, N, , Exercise - 4, 1., , Some english alphabets have fascinating symmetrical structures. Which capital letters have, only one line of symmetry (like E))? Which capital letters have rotational symmetry of, order 2 (like I)?, , TE, , Fill the following table, thinking along such lines., Line, symmetry, , Number of lines, symmetry, , Rotational, symmetry, , Order of, rotational, symmetry, , Z, , No, , 0, , Yes, , 2, , 1, , No, , -, , ER, , S, , T, , Alphabets, , H, , O, , SC, , E, , Yes, , N, C, , Home Project, Collect pictures of symmetrical figures from newspapers, magazines and advertisement pamphlets. Draw the axes of symmetry over them. Classify them., 288, , VII CLASS MATHEMATICS
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Looking Back, The line which divides a figure into two identical parts is called the line of, symmetry or axis of symmetry., , •, , An object can have one or more than one lines of symmetry or axes of, symmetry., , •, , If we rotate a figure, about a fixed point by a certain angle and the figure looks, exactly the same as before, we say that the figure has rotational symmetry., , •, , The angle of turning during rotation is called the angle of rotation., , •, , All figures have rotational symmetry of order 1, as can be rotated completely, through 360º to come back to their original position. So we say that an object, has rotational symmetry only when the order of symmetry is more than 1., , •, , Some shapes only have line symmetry and some have only rotational symmetry, and some have both. Squares, equilateral triangles and circles have both line, and rotational symmetry., , SC, , ER, , T, , TE, , LA, N, , G, , AN, A, , •, , Free distribution by T.S. Government 2018-19, , SYMMETRY, , 289
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ANSWERS, 01- Integers, Exercise - 1 (page - 2), , (i) −2, −3, −5, , (ii) −1, −2, −8, , (5), , +2, , LA, N, , −6 −5, −4, −3, −2, −1, 0, +1, −8, −7, −6, −4, −3, −2, −1, 1, 2, 3, 5, 6, 7, 9, , 6., , +3, , +4, , +5, , +6, , i) No., Name of the City, Temperature, 1, Bangalore, 20°C, 2, Ooty, 15°C, 3, Nainital, −3°C, 4, Manali, −7°C, 5, Kasauli, −9°C, (ii) Bangalore (20°C) (iii) Kasauli (−9°C), (iv) Nainital (−3°C) Manali (−7°C) Kasauli (−9°C) (v) Ooty (15°C) Bangalore (20°C), , T, , TE, , (7), , (iii) 8, 5, −2, , G, , (4), , AN, A, , (1)Biggest number =2, smallest number = -3, (2) (i) −9, −8, −7, −6, ;, biggest number = −6 ; smallest number = −9, (ii) −1, 0 +1, +2,, ;, biggest number = +2 ; smallest number = −1, (iii) −7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, biggest number = +4 ; smallest number = −7, (3) (i) −8, −5, 1, 2, (ii) −5, −4, −3, 2, (iii) −15, −10, −7, , (iv) 5+(−7), , SC, , (1), , ER, , Exercise - 2 (page - 4), , (2), , (i) 11, (vii) −2, , (ii) 5, (viii) 0, , −7, −2 −1 0 1 2 3 4 5, (iii) 14, (ix) 8, , (iv) 8, (x) 20, , (v) 2, (xi) 80, , (vi) 4, , Exercise - 3 (page - 6), , (1), , (i) 5, , (ii) 15, , (iii) -4, , (iv) 1, , (2), , (i) 31, (v) −8, , (ii) 21, (vi) 130, , (iii) 24, (vii) 75, , (iv) −13, (viii) 50, , 290, , v) 13, , vi) -1
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(3) Sl.No Negative integer, , +, , Whole No., , =, , −6, , (−6), , +, , 0, , =, , −6, , 2, , (−7), , +, , 1, , =, , −6, , 3, , (−8), , +, , 2, , =, , −6, , 4, , (−9), , +, , 3, , =, , −6 etc.,, , Exercise - 4 (page - 11), , AN, A, , 1, , (1), , (i) +600, , (ii) −1, , (iii) −600, , (2), , (i) −3, , (ii) −225, , (iii) 630, , (iv) 316, , (vi) 1320, , (vii) 162, , (viii) −360, , (ix) −24, , −10°, (i) −9, , (v) 0, , (x) 36, , (4) (i) 10, (ii) 18 (iii) 5 (5) (i) ` 5,000 profit (ii) 3200, (ii) −7, (iii) +7, (iv) −11, , G, , (3), (6), , (v) −45, , (iv) +200, , LA, N, , Exercise - 5 (page - 19), (1), , (i) True (72 = 126 − 54 = 72) (ii) True (210 = 84 + 126 = 210) (2) (i) −a (ii) −5, , (3), , (i) 480, , (ii) −53,000, , (v) −62500 (vi) 336, , (iii) −15000, , (vii) 493, , (1), , (viii) 1140, , TE, , Exercise - 6 (page - 22), , (iv) −4182, , (i) −1, , (ii) −49, , (iii) not defined, , (iv) 0, , Exercise - 7 (page - 23 & 24), (i) 24, , (ii) 20, , (2) (i) Profit 33,000, , (3), , 9 PM ; Temperature at Midnight = −14°C, , (4), , (i) 8 questions, , T, , (1), , ER, , (ii) 13 question, , (ii) 3000, , (5) 1 hour, , 02- Fractions, Decimals and Rational Numbers, , Exercise - 1 (page - 29), , SC, , 3, 4, , (1), , (i) 2, , (2), , (i), , (3), , Sum in row =, , (4), , 17, , Key, , 1 5 5, , ,, 2 8 6, , 11, cm, 15, , (ii) 1, (ii), , 1, 9, , (iii), , 3, 7, , (iv) 3, , 1, 6, , (v), , 11, 24, , (vi) 6, , 1, 6, , 3 1 2, , ,, 10 3 5, , 21, 21, 21, , Sum in column =, , Sum in diagonal =, All the sums are equal., 13, 13, 13, , (5), , 1, , 7, 8, , (6), , 7, 12, , 291
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1, Perimeter of Δ ABE = 10 cm;, 5, , (7), , BCDE perimeter = 7, , difference = 2, , Δ ABE is greater ;, , 11, cm ;, 15, , 7, 15, , Exercise - 2 (page - 34), , (2), (3), , (i) 6, (i) 4, , 0, or 5, 6, , (ii) 1, , 1, 3, , (iii), , (ii) 6, (ii) 6, , Exercise - 3, 35, 66, , (3), , (i), , 3 3 1, = of, (ii) Both are euqal, 8 4 2, , (7), , (i), , 10, 7, , 1, 5, , (iii) 7, , (i) 3, , 1, (4) 17 hrs., 2, , TE, , 2, 15, , 7, 13, , (iii), , 7, 40, , (iv), , T, , (i), , 7, 8, , (ii), , (iii), , ER, , (3), , (ii), , (2), , 7, 15, , (ii), , 2, 21, , 1, (5) 85 km., 3, , (iii) 3, (6) 1350 m., , 3, , 35 or 3, 7, 5, , (ii), , Exercise - 4 (page - 43), 8, 5, , 7, 15, , 0, or 6, 5, , LA, N, , (ii) 1, , (v) 6, , (iv) 15, , (page - 37), , (i), , (i), , (iv) 1, , (iii) 9, , (1), , (1), , 5, 7, , 1, , 1, 9, , AN, A, , (i) 5, , G, , (1), , 4, 3, , 5, 9, , (2) (i) 24 (ii) 3, (4) 2, , 3, 7, , (iii) 1, , 2, 7, , (iv), , 7, 5, , 1, days, 2, , Exercise - 5 (pages- 45 & 46), (1), , (ii) 8.5, , (4), , (i) 0.19 kg, , (5), , (i) 50 + 5 +, (iv) 30 +, , (6), , (iii) 1.51, , (i) 0.1 m, 0.0001 km, , SC, , (3), , (i) 0.7, , (i) 3, , (iv) 6 (2), , (ii) 4.5 cm, 0.045 m, 0.000045 km., (ii) 0.247 kg, , 5, 10, , 3, 3, +, 10 1000, , (ii) 30 (iii), , (i) . 0-09 (ii) . 77-07 (iii), , (ii) 5 +, , (iii) 44.08 kg, , 5, 5, +, 10 100, , (iii) 300 + 3 +, , (v) 1000 + 200 + 30 + 4 +, 3, 3, (iv), 100, 10, , (v), , 5, 6, +, 10 100, , 3, 100, , (7) Radha walked 100 m. more than Aruna (8) 5.625 kg., 292, , 3, 100, , . 2-35
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Exercise -6 (page- 50 & 51), (i) 1.8, (vi) 1050.05, , (ii) 18.9, (vii) 1.72, , (iii) 13.55, (iv) 78.8, (2), 24.8 cm2, , (v) 0.35, , (3), , (i) 213, , (ii) 368, , (iii) 537, , (iv) 1680.7, , (v) 13110, , (vi) 15610, , (vii) 362, , (viii) 4307, , (ix) 5, , (x) 0.8, , (xi) 90, (iii) 42.16, , (xii) 30, (iv) 14.62, , (4) 625 Km, (v) 0.025, , (5) (i) 0.45, (vi) 1.12, , (ii) 0.75, (vii) 0.0214, , (x) 77.011, (vi) 590, , (6) (i) 0.023, (vii) 0.02, , (ii) 0.09 (iii) 4.43, (7) 5 (8) 0.128 cm, , (viii) 10.5525 (ix) 1.0101, (iv) 0.1271, (v) 2, , AN, A, , (1), , (i), , −5, 12, , −75, (ii), 180, , 1 3, , (3), (4), , 2 4, , -3, (i) false, , -2, , -1, 0, (ii) true, , 3, , 10, , LA, N, , (2), , G, , Exercise -7 (page- 56), , 2, , 3, , 1, 2, (iii) false, , 3, , 4, (iv) false, , (v) true, , 03 - Simple Equations, (1), , (i) L.H.S = 2x, R.H.S = 10, , TE, , Exercise - 1 (page - 59 ), , (ii) L.H.S = 2x−3, R.H.S = 9, , (i) y = 5, , (ii) a = 8, , (iv) L.H.S = 5p+3, R.H.S = 2p+9, , (vii) L.H.S = 7m, R.H.S = 14, , (viii) L.H.S = 8, R.H.S = q+5, , (iii) m = 3, , (iv) n = 7, , (ii) y = 7, , (iii) x = 5, , (iv) z = 9, , (v) x = 3, , (i) y = 5, , (ii) a = 4, , (iii) q = 4, , (iv) t = 4, , (v) x = 13, , (vi) x = 3, , (vii) x = −5, , (viii) x = −1, , (ix) y = 4, , (x) x = −2, , ER, , (2), , T, , (v) L.H.S = 14 (vi) L.H.S = 2a−3, R.H.S = 27−y, R.H.S = 5, , (iii) L.H.S = 4z+1, R.H.S = 8, , Exercise - 2 (page - 63 ), (i) x = 4, , SC, , (1), (2), , (vi) y = −20, , Exercise - 3 (page - 67 ), , (1) 4 cm, (2) 5 cm, (3) 21 (4) 30 (5) 8, (8) l = 34m, b = 2m (9) l=23m, b=19m (10) 5 years, , (6) 46, 49, (11) 19, 44, , (12) 40; 25, 15, , (15) 30°, 60°, 90°, , (13) 2, , (14) 40, , (7) 7, 8, 9, , (16) 30, Key, , 293
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04 - Lines and Angles, Exercise - 1 (page - 69 ), (1), , (i) Line segment AB, , (ii) Ray CD, , (iii) Line XY, , (2), , (i) O, , (ii), , (iii) R S, , (3), , AB, AC, AD, BC, BD, CD, , (5), , (i) acute, , (6), , ∠AOF, ∠FOE, ∠EOD, ∠DOC, ∠COB, ∠FOD, ∠EOC, ∠DOB - Acute angles., , X, , (ii) obtuse, , (iv) C, , D, , AN, A, , P, , (iv) Point ‘P’, , (iii) Right, , (iv) acute, , (v) obtuse, , ∠AOE, ∠EOB, ∠FOC - Right angles ; ∠AOD, ∠AOC, ∠FOB - Obtuse angles., , (i),( ii) and (iv) are intersecting lines and (iii) non-intersecting lines., , Exercise - 2 (page - 71 ), (2) (i) 65°, , LA, N, , (8), , G, , ∠AOB - straight angle (7) (i) and (iv) are parallel ; (ii) and (iii) non parallel, , (1), , iii, , (ii) 50°, , (4), , Yes. Because the sum of the angles must be 90°, , Exercise - 3 (page - 73 ), , (iv) 35°, , (1), , (i), (ii), , (ii) 85°, , (3), , The sum of two acute angles is always less than 180°, , TE, , (2) (i) 75°, , (iii) 1°, , (3) 45°, 45°, , (iii) 30°, , (iv) 160°, (4) 90°, 90°, , Exercise - 4 (page - 74 ), (1), , (2) (i) ∠AOD, ∠DOB, , (i) a, b (ii) c, d, , (ii) ∠DOB, ∠BOC, , SC, , (4), , Yes., , ER, , (3), , T, , (iii) ∠BOC, ∠COA (iv) ∠COA, ∠AOD, , Yes ., , ., , because ∠AOC + ∠COB = 180°, , because ∠AOB + ∠BOC = 90°, , Exercise - 5 (page - 75 ), , (1), , i, ii, , (2) No. Because there is no common arm., , Exercise - 6 (page - 76 ), (1), , (i) ∠AOD, ∠BOC, , (2), , y = 160° (Vertically opposite angles) x + 160° = 180°, ∠x = ∠z Vertically opposite angles, ∴ z = 20°, , 294, , (ii) ∠AOC, ∠BOD, ∴ x = 20°
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Exercise - 7 (page - 85 ), (1), (2), , (i) Transversal (ii) Parallel, (iii) Parallel, (i) 100°, (ii) 45° (iii) 90°, , (iv) one, (iv) 100°, , (3) ∠x = 180 − (75+45) = 60° ; ∠y = 75° ; z = 45°, ∴ b = 130°, , b + 50° = 180°, , (5), , l || m, , (6), , ∠a = 50° (Alternate angles), , G, , b + c = 180° ⇒ 130°+ c = 180° ⇒ c = 50°, d + 50° = 180° ⇒ d = 130°, , AN, A, , (4), , LA, N, , ∠b = 50° (Alternate angles), ∠c = ∠d = ∠e = 50°, (all are Alternate angles), , Exercise - 1 (page - 93 ), (1) (i) Possible, , (ii) Possible, , (iii) Not possible, , T, , Exercise - 2 (page - 94 ), , (ii) Altitude (Height), , (iv) Possible, , (v) Not possible, , (2) Right angle triangle, , ER, , (1) (i) Median, , TE, , 05 - Triangle and its Properties, , (4) No, in some cases it lies in the exterior of the triangel (5) (i) XZ, , (3) Yes, (ii) ∠Ρ, , (iii) B, , Exercise - 3 (page - 100), , SC, , (1) (i) 70° (ii) 60° (iii) 40°, (2) (i) x = 70° ; y = 60°, (ii) x = 80° ; y = 50°, (iii) x = 110° ; y = 70° (iv) x = 60° ; y = 90° (v) x = 45° ; y = 90° (iv) x = 60°, (3) (i) 40° (ii) 34°, , (iii) 60° (4) 60° (5) (i) False (ii) True, , (6) (i) 30° ; 60° ; 90°, (9), , (7) x = 100° ; y = 50° ; z = 100° (8) 72°, , ∠P = 80°; ∠Q = 40° ; ∠R = 60° (10) 18° ; 72° ; 90°, , (12) ∠LPM = 40° ; ∠PML = 50° ; ∠PRQ = 50°, , Key, , (iii) False (iv) False, , (11) 36°, 54°, (13) 540°, , 295
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Exercise - 4 (page - 107), (1) Interior angles : ∠ABC, ∠ACB, ∠BAC ; Exterior angles : ∠CBX , ∠ACZ , ∠BAY, (2) ∠ACD = 111°, , (3) x = 115° ; y = 35°, , (4) (i) x = 50° (ii) x = 33° ; y = 82°, , (5) ∠CDB = 76° ; ∠DBC = 39° ; ∠ABC = 58°, , (vi) x = 50° ; y = 130°, , AN, A, , (6) (i) x = 55° (ii) x = 100° (iii) x = 120°, y =30° (iv) y = 70° (v) x = 60° ; y = 150° ;, (7) 50° ; 75° ; 55°, (11), , (10) 30° ; 75° ; 75°, , (8) ∠P=35°,yes (9) 70°, , x = 135° ; y = 80°, , 06 - Ratio - Applications, Exercise - 1 (page - 111), (2), , 15, , (4) AX = 18 cm ; XB = 20 cm., (7) 40 : 20 or 2: 1, , (3) Raju’s Share = 40 ; Ravi’s Share = 56, , LA, N, , (ii) 5 : 15 or 1 : 3 (Sudha : Radha), , (i) 15 : 5 or 3 : 1 (Radha : Sudha), , G, , (1) 100 : 10 , 10:1, , (5) .60,000, , (6) 8 liters, , (8) 1:2400 or 0.05 : 120, , (9) (i) count No. of boys and girls in your class and write inthe form of ratio. If boys or girls will be, zero, can you write it in the form of ratio? We can not compare such ratios., , TE, , (ii) Count of doors and number of windows of your classroom and number write in the form, of ratios., (iii) Count all textbooks and note books with you and write in ratio form., Exercise - 2 (page - 114), , (3), , (ii) True, , ER, , (2) (i) False, , (ii) 450, 450, , T, , (1) (i) 8, 8, .90, , (4) 10 kg, , (iii) 96, 96, , (iv) 6, 30, , (iii) True, , (iv) True, , (5) a) 45, , b) 26, , (v) 24, 72, , (v) False, (6) i) 540°, , ii) 21°, , Exercise - 3 (page - 120), 0.0001 cm ; 2cm, , (2) (i) Yes, , (ii) No, , (iii) No., , SC, , (1), , • Draw 5 different squares, measure their lengths and fill the table., , (4), , • 4 times of side will be perimeter of square find and fill the table., • Square the side of each and fill the table., , (i) Yes, lenght of side is in direct proportion to perimeters of the squares., , (ii) No, length of side is not indirect proportion to area of the square., , 296, , (3) 4 cm
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Exercise - 4 (page - 125), (1) School Y, (2) 20% decrease, (5), , 2, Abscent = 16 % or 16.66%, 3, , (7), , 15, , (3) Mangoes = 35%, , 1, Present = 83 % or 83.33%, 3, , (4) 16%, (6) 7200, , (8) gold 70% ; silver 25% ; Copper 5%, , (9) 2000, , (1), , 1, 12 % or 12.5% (2) 6%, 2, , (3) . 2,00,000, , (4) . 875, , (5), , loss = 1200 (2.44%), , (7) 202.5, , (8) 800, , (9) 1100, , G, , AN, A, , Exercise - 5 (page - 136), , (2) 12%, , (6) 561, , Exercise - 6 (page - 140), 2 years 8 months or, , (3) . 450, , 8, 2, years or 2 years, 3, 3, , 1, years, 2, , LA, N, , (1), , (5) 1, , (4) . 12958, , 07 - Data Handling, Exercise 1 (page - 147 ), (ii)30 0C, , (2) 15.9 kg, , TE, , (1) (i) 33 0C, , (3) (i) Ground nuts `:7500 ; Jawar `:4000 ;, (4) 42, , Millets `:5250 (ii) Ground nuts, , (5) (i) 23 (ii) 21 (iii) 16.5 (iv) Lekhya, , T, , (6) (i) ` 18 (ii) ` 54 (iii) Proportional, , (7) 5.5 (8) 5.6, , (9) 107, , ER, , Exercise 2 (page - 152 ), (1) 155 cm, 140cm, (3) 25, , (2) (i) Mean = 28, Mode = 27 (ii) 2 players of age 25 years each., , (4) (i) Mode, , (ii) Mean, , (iii) Mean, , (iv) Mode, , SC, , Exercise 3 (page - 155 ), (1) (i) F, , (ii) T, , (iii) F, , (iv) F, , (3) Mode is correct, but median is wrong., , (2) (i) ` 1400 (ii) ` 1450, (4), , three 1, 7, 10 ; 2, 7, 9 ; 3, 7, 8 (5) 11, , Exercise 4 (page - 160 ), (5) (i) Education, , Key, , (ii) Food, , (iii)` 2250, , (iv) ` 1500, , 297
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Exercise - 4 (Page. 178), (1), , (i) S.S.S. (ii) S.A.S. (iii) A.S.A. (iv) R.H.S. (2) (i) a) AR = PE, b) RT = EN, c) AT = PN (ii) a) RT = EN b) PN = AT (iii) a) ∠ A = ∠ P b) ∠ T = ∠ N, , (3), , (i) Side, , (iii) Common side, , (iv) S.A.S., , We can’t say ΔABC ≅ ΔPQR when the corresponding angles are equal, but can say that, the triangles are simlar., , (5), (7), (8), (9), , AN, A, , (4), , (ii) Angle, , ΔRAT ≅ ΔWON, , (6) ΔABC ≅ ΔABT and ΔQRS ≅ ΔTPQ, , (i) Draw two triangles with same measures. (ii) Draw two triangles of different measures., BC = QR (A.S.A.) or AB = PQ (A.A.S.) or AC = PR (A.A.S.), ∠ B = ∠ E ; ∠ A = ∠ F by A.A.S. ΔABC ≅ ΔFED are congruent ; BC = ED, , G, , 10- Algebraic Expressions, , LA, N, , Exercise - 1 (page - 192), (1), , (i) 3n, , (2), , (i) • In fig. 4 number of coloured tiles will be 4 on each side., •, , (ii) 2n, , In fig. 5 number of coloured tiles will be 5 on each side., , (ii) Algebraic expression for the pattern = 4n ; 4, 8, 12, 16, 20 . . . . . expression = 4n, , (3), , (ii) x −4, , (i) p + 6, , (iii) y −8, , (iv) −5q, , (v) y ÷ 4 or, , 1, pq, of pq or, (vii) 3z+5, (viii) 10+5x, (ix) 2y −5, 4, 4, (i) ‘ 3 more than x’ or 3 is added to x (ii) 7 is substracted from‘y’, , ER, , T, , (vi), (4), , TE, , (iii) Algebraic expression for the pattern = 4n + 1 ; 9, 13, 17, 21 . . .. expression = 4n + 1, , (iii) l is multiplied by 10., , y, 4, , (x) 10 y + 13, , (iv) x is divided by 5, , (v) m is multiplied by 3 and added to 11, , SC, , (vi) y is multiplied by 2 and subtracted 5 or 5 is subtracted from 2 times of y., (5), , (i) Constant, , (ii) Variable, , (iii) Constant, , (iv) Variable, , Exercise - 2 (page -199), , (i) (a2, − 2a2) (ii) (− yz , 2zy) (iii) (−2xy2 , 5y2x) (iv) (7p, −2p, 3p) and (8pq, −5pq), Algebaric expression : Problem Numbers : i, ii, iv, vi, vii,ix, xi, Numerical expression : Problem Numbers : iii, v, viii, x, (3), Monomial i, iv, vi ; binomial : ii, v, vii ; trinomial : iii, viii, ix multinomial : x, (4) (i) 1, (ii) 3 (iii) 5 (iv) 4 (v) 2 (vi) 3 (5) (i) 1, (ii) 2 (iii) 4 (iv) 3, 2, (v) 4, (vi) 2 (6) xy + yz 2x + 3x + 5, (1), (2), , Key, , 299
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12 - Quadrilaterlals, Exercise - 1 (page - 232 ), (1), , (i), , Sides:, , PQ , QR , RS , SP Angles:, , diagnals: PR , QS, , Vertices: P, Q, R, S, , Pairs of adjacent sides PQ, QR; QR, RS; RS,SP and SP, PQ, , AN, A, , (ii), , ∠ QPS, ∠ PSR, ∠ SRQ, ∠ RQP, , Pairs of adjacent angles: ∠ QPS, ∠ PSR; ∠ PSR, ∠ SRQ; ∠ SRQ, ∠ RQP, and ∠ RQP, ∠ QPS, Pairs of opposite sides : PS, QR and QP, RS, , Pairs of opposite angles: ∠ QPS, ∠ SRQ and ∠ PSR, ∠ RQP, (3) 48°, 72°, 96°, 144°, , (4) 90°, 90°, 90°, 90°, , LA, N, , (2) 100°, , G, , A, , (5) 75°, 85°, 95°, 105°, , B, , 90°, , 90°, , 90°, , 90°, , C, , D, , (6) Angle of the quadrilateral cannot be 180°, Exercise - 2 (page - 242 ), , (ii) true (iii) true (iv) false (v) false (vi) true (vii) true (viii) true, , (2) (i) Since it has 4 sides, , TE, , (1) (i) false, , (ii) Since opposite sides in a square are parallel, , (iii) Since diagonals of a square are perpendicular bisectors, , T, , (iv) Since opposite sides of a square are of equal length., (4) 50°, 130°, 50°, 130°, , (5) It has 4 sides and one pair of parallel sides; EA, DR, , (6) 1, , (7) Opposite angles are not equal., , (8) 15 cm, 9cm, 15cm, 9cm, , (9) No, Rhomsun should have equal length of sides, , (10) ∠C = 150o , ∠D = 150o, , (11) (i) Rhombus, , (ii) Squrare, , (iii) 180° - x°, , (iv) equal/congruent, , (v) 10, , (vi) 90°, , (vii) 0, , (viii) 10, , SC, , ER, , (3) ∠ BAD = 140°, ∠ DCB = 140°, ∠ CDA = 40°, , Key, , (ix) 45, , 301
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13 - Area and Perimeter, Exercise - 1 (page - 245), (1) 2(l+b); a2 (2) 60 cm; 22cm; 484 cm2 (3) 280cm2; 68cm; 18cm; 216cm2; 10cm; 50cm, Exercise - 2 (page - 249), , Exercise - 3 (page - 252), (1) (i) 20cm2, , (ii) 12cm2, , (iii) 20.25cm2 (iv) 12cm2, , (3) 30cm2 ; 4.62 cm (4) 27cm2 ; 7.2 cm, , AN, A, , (1) (i) 28cm2, (ii) 15cm2 (iii) 38.76cm2 (iv) 24cm2 (2) (i) 91.2cm2 (ii) 11.4cm, (3) 42cm ; 30cm (4) 8cm ; 24cm (5) 30m, 12m, (6) 80m, (2) (i) 12cm2 (ii) 3cm, , G, , (5) 64cm2 ; Yes ; ΔBEC, ΔBAE and ΔCDE are three triangles drawn between the two parallel, lines BC and AD, BC = AE+ED, , LA, N, , (6) Ramu in ΔPQR, PR is the base, because QS⊥PR. (7) 40 cm, (8) 20 cm ; 40cm, 2, 2, 2, (9) 20 cm, (10) 800cm (11) 160cm, (12) 192cm (13) 18 cm ; 12cm, Exercise - 4 (page - 257), , (1) (i) 20cm2 (ii) 24cm2 (2) 96cm2 ; 150 mm : 691.2m2, Exercise - 5 (page - 260), , (3) (i) (a) 50.24 cm, (5) 10.5 cm, , (iii) 96.8 cm, , TE, , (1) (i) 220cm (ii) 26.4cm, , (b) 94.2 cm, , (6) 3 times, , (4) `506.25, , (ii) 17.6 m, , (iii) 15.4m, , (2) (i) 55m, , (c) 125.6 cm, , (7) 3 : 4, , (3) 18cm, , (ii) 7 cm, , (8) 1.75cm, , (4) 42 cm, (9) 94.20 cm, , (10) 39.25 cm, , (2) 195.5m2; 29.5m2 (3) 624 m2, , ER, , (1) 475m2, , T, , Exercise - 6 (page - 263), , (4) 68 m2 (5) 9900 m2 ; 200100m2, , 14 - Understanding 3D and 2D Shapes, , Exercise - 1 (Page - 265), , Sphere: Foot ball, Cricket ball,, , SC, , (1), , Laddu, , Cylinder: Drum, Biscuit pack, Log, Candle, Pyramid: Pyramid, Cone :, , (2), , (i) Cone:, , ;, , Cuboid: Match box, Brick, Biscuit pack, , Ice-cream, Flower pot ; Cube: Dice, Carton, Ice-cream, upper part of a funnel, , (ii) Cube: Dice, Carton, , (iii) Cuboid: Duster, Brick (iv) Sphere: Ball, Marble (v) Cylinder: Pencil, Pype., , 302
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(3), , Cube, , Cuboid, , Faces, , 6, , 6, , 5, , Edges, , 12, , 12, , 8, , 8, , 5, , Do activity, , (2) i) C ii) a (3), , Exercise - 4 (Page - 276), , •, •• •, • •, • •, , •, •, , •, , A ball : a circle., A Cylindrical pipe : a rectangle., A book : a rectangle., , •• •, • • • • • •, •• •, • • •, • • •, , G, , (1), , • • •, • • •, • ••, •, • ••, , AN, A, , Vertices, 8, Exercise - 2 (Page - 267), (1), , Pyramid, , (ii) Cube / Square sheets, , LA, N, , (2) (i) Spherical / Circular objects, , (iii) Triangular shapes or Right prism with triangular base., (iv) Cylinder / Rectangle sheets., , Exercise 1 (page - 278 ), , TE, , 15 - Symmetry, , - - - - - - - - - - -- - - - -, , (ii), , (iii), , ER, , T, , (i), , (ix), , SC, , (viii), , (xvi), , Key, , (xvii), , (x), , (xviii), , (iv), , (xii), , (xix), , (v), , (vi), , (vii), , (xiii), , (xiv), , (xv), , (xx), , (xxi), , (xxiii), , 303
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Exercise 2 (page - 281), , (ii), , (iii), , (viii), , (iv), , (v), , (ix), , (3), , (iii), , (ix), , LA, N, , (ii), , (i), (ii), (iii), (iv), (v), (i) False, (ii) True, (iii) False, Angle between successive axes = 360/2n =360/2x4 = 360/8 = 450, This is true for all regular polygons, , TE, , (4), (5), , T, , Exercise 3 (page - 286), 1., , Figures i, ii, iv and v have rotational symmetry., , 2., , (i) 2, , ER, , (iii) 3, , (iv) 4, , Square, Rectangle, Rhombus, Equilateral Triangle, Regular Hexagon, Circle, Semi-circle, Exercise 4 (page - 288 ), , yes, yes, yes, yes, yes, yes, No, , 1., , Yes, Yes, Yes, Yes, No, , SC, , 3., , (ii) 4, , 304, , S, H, O, N, C, , No, Yes, Yes, No, Yes, , 0, 2, 2, 0, 1, , (v) 4, , (vi) 5, 900, 1800, 1800, 1200, 600, infinity, -, , 2, 2, 2, 2, 1, , (vii), , G, , (2), , (i), , (vi), , AN, A, , (i), , (vii) 6 (viii) 3, 4, 2, 2, 3, 6, infinity, -, , (vi)
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INSTRUCTIONS TO TEACHERS, Dear Teachers!!, Greetings and a hearty welcome to the newly developed textbook Mathematics, for class VII., , AN, A, , • The present textbook is developed as per the syllabus and Academic standards, conceived by the mathematics position paper prepared based on SCF – 2011 and, RTE – 2009 for Upper Primary stage of education., , G, , • The new textbook constitutes 15 chapters with concepts from the main branches, of mathematics like Arithemetics, Algebra, Geometry, Mensuration and, Statistics., , TE, , LA, N, , • These chapters emphasize the prescribed academic standards in achieving the, skills like Problem Solving, Reasoning-proof, Communication, Connectivity, and representation. The staratagies in building a chapter are observation of, patterns, making generalization through deductive, inductive and logical thinking,, exploring different methods for problem solving, questioning, interaction and, the utilization of the same in daily life., • The situations, examples and activities given in the textbook are based on the, competencies acquired by the child at Primary Stage. So the child participates, actively in all the classroom interactions and enjoys learning of Mathematics., , ER, , T, , • Primary objective of a teacher is to achieve the “Academic standards” by involving, students in the discussions and activities suggested in the textbook and making, them to learn the concepts., , SC, , • Mere completion of a chapter by the teacher doesn’t make any sense. The, exhibition of prescribed academic standards by the student only ensures the, completion of the chapter., • Students are to be encouraged to answer the questions given in the chapters., These questions help to improve logical, inductive and deductive thinking of the, child., • Understanding and generalization of properties are essential. Student first finds, the need and then proceeds to understand, followed by solving similar problems, on his own and then generalises the facts. The strategy in the presentation of, concepts followed., 305
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• Clear illustrations and suitable pictures are given wherever it was found, connection and corrects the misconnection necessary., , AN, A, , • Exercises of ‘Do This’ and ‘Try This’ are given extensively after completion, of each concept. Exercises given under ‘Do This’ are based on the concept, taught. After teaching of two or three concepts some exercises are given, based on them. Questions given under ‘Try This’ are intended to test the skills, of generalization of facts, ensuring correctness of statements, questioning etc.,, ‘Do This’ exercise and other exercises given are supposed to be done by students, on their own. This process helps the teacher to know how far the students can, fare with the concepts they have learnt. Teacher may assist in solving problem, given in ‘Try This’ sections., , LA, N, , G, , • Students should be made to digest the concepts given in “looking back”, completely. The next chapter is to be taken up by the teacher only after, satisfactory performance by the students in accordance with the academic, standards designated for them (given at the end)., • Teacher may prepare his own problems related to the concepts besides solving, the problems given in the exercises. Moreover students should be encouraged, to identify problems from day- to-day life or create their own., , TE, , • Above all the teacher should first study the textbook completely thoroughly, and critically. All the given problems should be solved by the teacher well, before the classroom teaching., , SC, , ER, , T, , • Teaching learning strategies and the expected learning outcomes, have been, developed class wise and subject-wise based on the syllabus and compiled in, the form of a Hand book to guide the teachers and were supplied to all the, schools. With the help of this Hand book the teachers are expected to conduct, effective teaching learning processes and ensure that all the students attain the, expected learning outcomes., , Happy Teaching., , 306
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Syllabus, , LA, N, , G, , AN, A, , Number System: (i) Integers, (50 hrs), • Multiplication and division of integers (through patterns)., 1. Integers, • Properties of integers (including identities for addition & multiplication,, 2. Fractions,, (closure, commutative, associative, inverse, distributive) (through, Decimals &, patterns). (examples from whole numbers as well). Expressing properties, Rational, in a general form. Construction of counter examples, (eg. subtraction is, Numbers, not commutative)., • Word problems involvingintegers (all operations), (ii) Fractions, Decimals and rationalnumbers:, • Multiplication of fractions, • Fraction as an operator “of”, • Reciprocal of a fraction and its use, • Division of fractions, • Word problems involving mixed fractions ( related to daily life), • Introduction to rational numbers (with representation on number line), • Difference between fraction and rational numbers., • Representation of rationalnumber as a decimal., • Word problems on rationalnumbers (all operations), • Multiplication and division of decimal fractions, • Conversion of units (length & mass), • Word problems (including all operations), , Expressions, , SC, , ER, , T, , 3. Simple, , Equations, , Exponents and powersIntroduction Meaning of x in ax where a Ý Z, • Laws of exponents (throughobserving patterns to arrive at5, generalization.)whereM, n ∈N(i) am an = am?+n(ii)? (am)?n = amn(iii) am/an =, a m-n, where (m-n) ∈ N(iv) a m.b m = (ab) m(v) number with exponent, zerovi)Decimal number in exponential notation vii) Expressing large, number in standard form (Scientific Notation), ALGEBRAIC EXPRESSIONSIntroduction Generate algebraic, expressions(simple) involving one or two variables, • Identifying constants, coefficient, powers, • Like and unlike terms, degree of expressions e.g., x2y etc.(exponentd”?3,, number of variables d”?2), • Addition, subtraction of algebraic expressions (coefficients should be, integers)., Simple equations, • Simple linear equations in one variable (in contextual problems) with two, operations (integers as coefficients), , TE, , Algebra, (20 hrs), 11. Exponents, 10. Algebraic, , 6. Ratio Applications, (20 hrs), , • Ratio and proportion (revision), • Unitary method continued,consolidation, generalexpression., • Compound ratio : simple word problems, • Percentage- an introduction, • Understanding percentage as a fraction with denominator 100, • Converting fractions anddecimals into percentage andvice-versa., • Application to profit and loss (single transaction only), • Application to simple interest (time period in complete years)., , 307
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T, , TE, , LA, N, , G, , AN, A, , Understanding (i) Lines and Angles:, shapes /, • Pairs of angles (linear,supplementary, complementary,adjacent, vertically, Geometry, opposite)(verification and simple proofof vertically opposite angles), • Properties of parallel lines withtransversal (alternate,corresponding,, 4. Lines and, interior, exteriorangles), Angles, (ii) Triangles:, 5. Triangle and, • Definition of triangle., Its Properties, • Types of triangles acc. To sides and angles, 8.Congurencey, • Properties of triangles, of Triangles, • Sum of the sides, difference of two sides., 9.Construction, • Angle sum property (with notion of proof and verification through paper, of Triangles, folding, proofs , using property of parallel lines , difference between, 12.Quadrilaterals, proof and verification, 15. Symmetry, • Exterior angle property of triangle, 14.Understand- (iii) Congruence:, ing 3D and, • congruence through superposition ex. Blades, stamps etc.., 2D Shapes, • Extend congruence to simple geometrical shapes ex. Triange , circles,, • criteria of congruence (by verification only), • property of congruencies of triangles SAS, SSS, ASA, RHS, Properties with figures•, (iv) Construction of triangles (all models), • Constructing a triangles when the lengths of its 3 sides are known, (SSS criterion), • Constructing a triangle when the lengths of 2 sides and the measure of, the angle between them are known (SAS criterion), • Constructing a triangle when the measures of 2 of its angles and length, of the side included between them is given (ASA criterion), • Constructing a right angled triangle when the length of one leg and its, hypotenuse are given (RHS criterion), , SC, , ER, , (v) QuadrilateralsQuadrilateral-definition., • Quadrilateral, sides, angles, diagonals., • Interior, exterior of quadrilateral, • Convex, concave quadrilateral differences with diagrams, • Sum angles property (By verification) , problems, • Types of quadrilaterals, • Properties of parallelogram, trapezium, rhombus, rectangle, square, and kite., (vi) Symmetry, • Recalling reflection symmetry, • Idea of rotational symmetry,observations of rotationalsymmetry of 2-D, objects. (900,1200, 1800), • Operation of rotation through900 and 1800 of simple figures., • Examples of figures with bothrotation and reflection symmetry(both operations), • Examples of figures that havereflection and rotation symmetryand viceversa, 308
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(vii) Understanding 3-D and 2-D Shapes:, • Drawing 3-D figures in 2-Dshowing hidden faces., • Identification and counting ofvertices, edges, faces, nets (forcubes, cuboids, and cylinders,cones)., • Matching pictures with objects(Identifying names), Mensuration, , Area and Perimeter, • Revision of perimeter and Area of Rectangle, Square., • Idea of Circumference of Circle., • Area of a triangle, parallelogram, rhombus and rectangular paths., , AN, A, , (15 hrs), , 13. Area and, Perimeter, , Data Handling, • Collection and organisation ofdata, • Mean, median and mode ofungrouped data – understandingwhat they, represent.Reading bar-graphs, • Constructing double bar graphs, • Simple pie charts with reasonable data numbers, , SC, , ER, , T, , TE, , LA, N, , G, , 7. Data, Handling, (15 hrs), , 309
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Academic Standards, CONTENT, , • Solves the problems involving four fundamental operations, of integers, • Solves the word problems involving the integers., • Used brackets for solving problems to simplify numerical, statements., Reasoning, • Explains why the division by zero is meaning less., Proof:, • Differentiates and compares the set of Natural numbers, with integers., • Gives examples and counter examples to the number, properties such as closure, Commutative, Associative etc., Communication:• Expressing the number properties of integers in general, form., • Uses the negative symbol in different contexts., Connections: • Finds the usage of integers from their daily, life situations, • Understands the relation among N, W and Z., Representation:• Represents the integers on number line., • Performs the operations of integers on the number line., Problem, Solving:, , LA, N, , G, , AN, A, , Number, system, 1. Integers, , ACADEMIC STANDARDS, , SC, , ER, , T, , TE, , • Solves the problems in all operation of fractions., • Solves the word problems of all operations of rational, numbers., • Solves the problems of all operations of decimal fractions, • Converts the small units into large units and vice versa., Reasoning :, • Differentiates rational numbers with fractions., and Proof, • Justifies density property in rational numbers, Communication:• Expresses the need of set of rational numbers, • Expresses the properties of rational numbers in general, form, Connections: • Finds the usage of / inter relation among fractions,, rational numbers, and decimal numbers., Representation:• Represents rational numbers on the number line., • Represents the rational numbers in decimal form., , 2. Fractions, Problem, Decimals, Solving:, and Rational, numbers, , Algebra:, Problem, 11. Exponents Solving:, and powers Reasoning :, and Proof, , • Writes the large numbers in exponential form by using, prime factorization, • Generalizes the exponential laws through the, observation of patterns, , Communication:• Understands the meaning of x in ax where a € z., • Uses of exponential form when using large numbers, 310
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• Uses prime factorization in expression of large numbers, in exponential form, Representation:• Expresses the large numbers in standard form, Algebra:, Problem, • Finds the degree of algebraic expressions, 10. Algebraic Solving, • Doing addition, subtraction of algebraic expressions, Expression, (Co-efficient should be integers), 3. Simple, • Solves the word problems involving two operations (Which, Equations, can be expressed as simple equation and single variable), Reasoning, • Generates algebraic expressions involving one or two, and Proof, variables by using the patters, Communication:• Writes the standard form of first, second, third order, expressions in one or two variables, • Converts the daily life problems into simple equations., (Contains one variable only), Connections: • Uses closure, commutative etc. properties in addition, and subtraction of algebraic expressions., • Uses solving simple equations in daily life situations., Representation:• Represents algebraic expressions in standard forms, Finds the compound, inverse ratio of 2 ratios, Solves word problems involving unitary methods, Solves word problems involving percentage concept, Solves word problems to find simple interest (Time, period in complete years), Compares the decimals, converting into percentages and, vice versa., Formulates the general principles of ratios and, proportions, Expresses the fractions into percentages and decimal, forms and their usage., Uses profit and loss concepts in daily life situations, (Single transactions only), Understands and uses the solutions for percentage, problems in daily life., Converts fractions and decimals into percentage form, and vice versa., , TE, , •, •, •, •, •, •, , T, , Reasoning, and Proof, , ER, , 6. Ratio Problem, Applications Solving, , LA, N, , G, , AN, A, , Connections:, , Communication:•, , SC, , Connections:, , •, •, , Representation:•, , 311
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• Solves problems on angles made by transversal intersecting, parallel line, , Understanding Problem, Solving, Shapes /, Geometry, , • Differentiates the types of pair of angles from given angles, • Verifies the parallel ness of the given lines with the use of, properties of parallel lines., • Proofs and verifies the angle sum property through paper, folding and using property of parallel lines., , Reasoning, and proof, , AN, A, , 4. Lines, and, Angles, , Communication: • Gives examples of pairs of angles., Connections:, , • Observes the parallelness in surroundings., , G, , Representation:• Represents the notation of angle., , • Determines whether the given lengths of sides are shapes, suitable to make triangle., • Finds the angle which is not given from exterior and, other angles of triangle., , LA, N, , 5. Triangle Problem, and Its, Solving, Properties, , • Makes relationship between exterior angle to its opposite., • Classifies the given triangles on the basis of sides and angles., • Estimates the kind of triangle by observing the given triangle., , TE, , Reasoning, and proof, , Communication: • Explains the different types of triangles according to sides, and angles., • Explains the property of exterior angle of triangle., • Uses the concept of triangle., , ER, , T, , Connections:, , Representation:• ______________________, , 8.Conguren- Problem, , SC, , cey of, Solving, Triangles, , Reasoning, and proof, , • Identifies the congruent triangles from given triangles, suitable to make triangle., • _________________________, , Communication: • Appreciates the congruency in 2-D figures., Connections:, , • ______________________, , Representation:• Represents the congruent triangles using symbols, notation., , 312
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99.. Construc- Problem, tion of, Solving, Triangles, Reasoning, and proof, , • Construct triangles using given measurements., • _________________________________, , Connections:, , AN, A, , Communication: • _________________________________, • _________________________________, , Representation:• _________________________________, 12.Quardila- Problem, teral, Solving, , G, , • _________________________________, , • Differentiates the convex, concave quadrilaterals., • Verifies and justifies the sum angle property of quadrilaterals., , LA, N, , Reasoning, and proof, , TE, , Communication: • Explains the inter relationship between triangle and, quadrilateral., • Explains the different types quadrilaterals based on their, properties., Connections:, , • Tries to define the quadrilateral., • Classifies the given quadrilaterals using their properties and, their inter relationship., , T, , Representation:• _________________________________, , ER, , 15.Symmetry Problem, Solving, , SC, , Reasoning, and proof, , • Rotate the figure and find its angular symmetry., • Can differentiate linear and reflection symmetry using, objectives or figures., , Communication: • Gives examples that have reflection symmetry., , Connections:, , • _________________________________, , Representation:• _________________________________, , 313
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14.Understanding, 3-D and, 2-D, shapes, , Problem, Solving, , • Identifying and counting of faces, Edges, Vertices, nets, for 3D Fig (Cube, Cuboid, Cone, Cylender)., , Reasoning, and proof, , • Matches picture with 3-D objects and visualize fells the Faces,, Edges, Vertices etc., , Communication: • _________________________________, • _________________________________, , AN, A, , Connections:, , Representation:• Can draw simple 3-D shapes in to 2-D figures., , • Solves the problem of Area and perimeter for square,, rectangle, parallelogram, triangle and Rhombus shapes of, things., , G, , Mensuration Problem, 13. Area and Solving, Perimeter, , LA, N, , • Understands the relationship between square, Rectangle,, Parallelogram with triangle shapes for finding the area of, triangle., , Reasoning, and Proof, , • Understands the Area of Rhombus by using area of triangles., Communication:• Explains the concept of Measurement using a basic unit., • Applies the concept of Area perimeter to find the daily life, situation problems (Square, Rectangle, Parallelogram,, Triangle, Rhombus and Circle), • Applies the concept of area of Rectangle, Circle., • Finds the area of the rectangular paths, Circular paths., Representation:• Represent word problems as figures., , TE, , Connections:, , • Organization of raw data into classified data., • Solves the problems for finding the Mean, Medium, Mode, of ungrouped data, , T, , ER, , 7. Data, Problem, Handling Solving, , • Understands the Mean, Mode and Medium of ungrouped, data and what they represent., Communication:• Explains the Mean, Mode and Medium for ungrouped data., , SC, , Reasoning, , • Understands the usage of Mean, Mode and Medium in daily, life situation problems., • Understands the usage of double graphs and pie graphs in, daily life situation (Year wise population, Budget, Production, of crops etc.), Representation:• Representation of Mean, Medium and Mode for ungrouped, data., • Representation of the data in to double bar graphs and pie, graphs., Connections:, , 314
