MathsModuleVol II

Distance Education Programme – Sarva Shiksha Abhiyan(DEP-SSA)

(An IGNOU-MHRD, Govt. of India Project)Maidan Garhi, New Delhi – 110 068

SLM on Teaching of Mathematics at Upper Primary Level

Volume-II

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Volume II

Distance Education Programme – Sarva Shiksha Abhiyan (DEP-SSA)

(An IGNOU-MHRD, Govt. of India project) Maidan Garhi, New Delhi 110 068

© Distance Education Programme- Sarva Shiksha Abhiyan, IGNOU, 2009

Printed at: Laxmi Printindia, 556, G.T. Road, Shahadra, Delhi-110 032.

ALL RIGHTS RESERVED

• No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any other means – electronics, mechanical, photocopying, recording or otherwise without the prior permission from the publishers.

• This book is an unpriced publication and shall not be sold, hired out or otherwise

disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published.


Volume II

Guidance

Prof. K.R. Srivathsan Dr. S.S. Jena

Pro-Vice Chancellor, IGNOU Former Project Director In-charge, Director DEP-SSA DEP-SSA, IGNOU

Academic Coordination

Dr. Sarat Kumar Rout

Programme Officer DEP-SSA, IGNOU

Distance Education Programme – Sarva Shiksha Abhiyan (DEP-SSA)

(An IGNOU-MHRD, Govt. of India project) Maidan Garhi, New Delhi 110 068

Expert Committee

Prof. Provin Sinclair PVC, IGNOU

Dr. S. S. Jena Former Project Director, DEP-SSA

Prof. M.L.Koul Former In-charge Project Director, DEP-SSA

Prof. P.R Ramanujam Director, STRIDE, IGNOU

Dr. Mohit Mohan Mohanty Addl. Director (Retd.) OPEPA, Orissa

Shri Madan Mohan Mohanty Deputy Director (Retd.), BSE, Orissa

Ms. Jai Chandiram Media Advisor DEP-SSA, IGNOU

Ms. Avantika Dam Asst. Teacher, CIE Basic School, University of Delhi

Unit Writer Dr. Vijay S. Patel Lecturer, SCERT, Gujurat

Shri Thimmaraju, Lecturer, DIET, Gadog, Karnataka

Shri Manoj Kumar Shukla, Lecturer, SCERT, Uttarakhand

Shri C.P. Mantri Udaipur Rajasthan

Shri B.B.P Gupta lecturer, SCERT, M.P

Shri Sachindananda Mishra Lecturer, DIET, Cuttack, Orissa

Shri Tapas Kr. Nayak Lecturer, SCERT, Orissa

Shri P. S. Rawat Lecturer, SCERT, Haryana

Ms. Pankaj Lohani SISE, Allahabad Uttar Pradesh

Shri Avtar Singh DIET Fatehgarh Punjab

Shri Ashok Kr. Sharma DIET Fatehgarh Punjab

Shri Sanjay Kr. Gupta SCERT Solan Himachal Pradesh

Dr. C. Saroja DIET Chennai Tamil Nadu

Dr. S. Suresh Babu SCERT Andhra Pradesh

Editorial Team Content Editing Language Editing Unit Design, Format

Editing, Course Coordination

Proof Reading

Dr. Mohit Mohan Mohanty, Addl. Director (Retd.), OPEPA, Orissa

Prof. C.B. Sharma School of Education, IGNOU

Dr. Sarat Kumar Rout Programme Officer DEP-SSA, IGNOU

Dr. Sarat Kumar Rout Programme Officer DEP-SSA, IGNOU

Shri Madan Mohan Mohanty Deputy Director (Retd.), BSE, Orissa

Dr. Eisha Kannadi Sr. Lecturer, School of Education, IGNOU

Graphic Designer Cover Page Editing

Mr. S.S. Chauhan SOS, IGNOU

Mr. Mitrarun Haldar M/s Pink Chilli Communications, Dwarka, New Delhi-110078

Mrs. Kashish Thakkar Computer Programmer, DEP-SSA, IGNOU

Secretarial Support All Support Staff DEP-SSA, IGNOU

Production Sh. Deepak Israni AFO, DEP-SSA


Volume-II

Block 3 Page No. INTRODUCTION TO ALGEBRA

UNIT 8

Algebraic Expression and Operations 1

UNIT 9

Factorization 29

UNIT 10

Algebraic Equations 51

Block 4 GEOMETRICAL SHAPES AND FIGURES

UNIT 11

Introduction to Geometrical Figures and Shapes 81

UNIT 12

Construction of Geometrical Figures 111

UNIT 13

Perimeter, Area and Volume 151

UNIT 14

Symmetry 181

ACKNOWLEDGEMENTS Sarva Shiksha Abhiyan (SSA) is a flagship programme of Govement of India to provide quality elementary education to all. The Distance Education Programme (DEP) is a national component, created by the MHRD, Govt. of India on July 1, 2003 covering all the States and UTs. Indira Gandhi National Open University (IGNOU) has been entrusted with the responsibility as a national apex institution for open and distance education to implement distance education activities across the country to meet the educational and training needs of the states at elementary level. The DEP-SSA at national level aims at the capacity building of functionaries such as Master Trainers, Coordinators of BRCs/CRCs, faculty of DIETS and SCERTs and evolve a sustainable training system for elementary school teachers through Open and Distance Learning (ODL) mode. The National Council of Educational Research and Training (NCERT) has came out with National Curriculum Frame Work (NCF)-2005 which emphasizes on constructivist pedagogy for transaction of learning experiences. Subsequently, NCERT has revised the text books adopting the principles of constructivist pedagogy. Further this new pedagogy demands that teacher in the class room should display the role of facilitator instead of playing the role of knowledge distributor which is also very typical and complex. In this context, the teacher must know how to adopt the new paradigm for effective transaction of learning experiences in the classrooms? Secondly, learning outcome is an important indicator of teaching-learning process and quality education at all level. Hence, learning achievement of students are not up to the level of expectation particularly in Language (English), Mathematics, Science at national level. Therefore, Government of India and various state governments are working on the proposition of Learning Enhancement Programme (LEP) focusing on Language (English), Mathematics, and Science through SSA. It is quite essential to improve the teaching competencies of the teachers in Mathematics at elementary level with appropriate interventional strategies. The present module has made a small effort in this direction. The specific objectives of this module are: • To improve teaching skills of teachers in Mathematics at elementary level.

• To help the teachers to follow the principles of constructivist pedagogy for effective transaction of mathematical concepts, facts and principles in the class room by

• To enhance the professional competencies of teachers in developing interest and curiosity of children towards learning Mathematics at elementary level.

I hope that at the end of training programmes, this module would enable the teachers to absorb necessary skills and competencies for better transaction of teaching learning experiences in Mathematics at upper primary level.

I would like to express my gratitude to Department of School Education Literacy, MHRD, Govt. of India for sponsoring DEP-SSA to improve the professional competency of functionaries associated with SSA. I take this opportunity to express my thankfulness and gratitude to Prof. V.N. Rajasekharan Pillai, Vice Chancellor, IGNOU & Chairman, Advisory Committee, DEP-SSA for his constant support, encouragement and able guidance throughout the year to carry out project activities for accomplishment of its goals and objectives. I also take this opportunity to express my gratitude to the unit writers and the experts involved in preparation of this module. My sincere gratitude goes to Dr. Mohit Mohan Mohanty Ex-Reader, SCERT, Orissa and Sh. Madan Mohan Mohanty for their hard labour in designing, developing and editing the module. I express my heartful thanks Prof. C.B. Sharma and Dr. Eisha Kannadi, School of Education, IGNOU for language editing of the present document. I am also especially thankful to my colleagues of DEP-SSA for their coordination in the development of this training module and my thanks are also due to all the support staff working in this project towards the completion of this assignment. I look forward to receive constructive suggestion for the improvement of this training module. December 2009 Project Director DEP-SSA, IGNOU


Volume I

Block 1 Number System

Unit 1 Numbers and Numerals

Unit 2 Number Line and Operations on Numbers

Unit 3 Data and Its Graphical Representation

Block 2 Mathematics In Daily Life

Unit 4 Percentage and Its Applications

Unit 5 Simple and Compound Interest

Unit 6 Ratio and Proportion

Unit 7 Time and Distance

Volume II

Block 3 Introduction to Algebra

Unit 8 Algebraic Expression and Operations

Unit 9 Factorization

Unit 10 Algebraic Equations

Block 4 Geometrical Shapes and Figures

Unit 11 Introduction to Geometrical Figures and Shapes

Unit 12 Construction of Geometrical Figures

Unit 13 Perimeter, Area and Volume

Unit 14 Symmetry

ABOUT THE VOLUME

The main focus of this module is to equip the teachers with required skills and competencies to apply principles of constructivist pedagogy while transacting learning experience in Mathematics to students. The first and foremost step towards the achievement of this objective is to acquaint the teachers and resource persons with transactional strategies of mathematical concepts and principles in conformity to principles of constructivist pedagogy. Secondly, this module is intended to empower the teachers to be free from the tyranny of traditional approach of teaching Mathematics in which abstract concepts are usually presented to the students in an authoritarian way and to adopt activity approach, play way method and creating social situation which stresses the presentation of concrete experiences. This volume comprises two blocks 3 and 4. The briefs of each unit have been presented below:

Unit-8 explains algebraic expressions and related concepts briefly and discuss about fundamental difference between variable and constant; like and unlike terms; and perform different operations on algebraic expressions i.e addition, subtraction, multiplication and division and their properties.

Unit-9 defines the concept of factor and factorization and explain factors and factorization; and discuss about factorization of binomials, trinomials and factorize the polynomials of different forms.

Unit-10 In this unit the most basic form of algebraic equations, i.e. the linear equation has been extensively discussed as this constitute a part of the Mathematics curriculum of the upper primary classes. This unit defines write linear algebraic equations with the help of patterns; solve linear algebraic equations in one variable; solve real life problems using algebraic equations; and solve linear equations in two variables.

Unit-11 explains different types of geometrical figures and their basic components i.e. point, plane, line and line segment and curve. This unit further distinguish between open and closed geometrical figures and draw closed geometrical figures of specific shapes like, triangle, quadrilateral, rectangle, square and circle and explain the relationship between them.

Unit-12 In this unit we have discussed the algorithm i.e. the sequential steps of construction of different geometrical figures and construct different plane geometrical figures like circles, line segments, angles of specific measures, triangle and quadrilaterals using ruler and compasses which can be used to help children construct geometrical figures correctly and accurately.

Unit-13 In this unit we have discussed the concepts of perimeter, area and volume and their application in our day-to-day life. This unit also describes how to measure the area of closed plane figures using appropriate units of measuring area and distinguish a three

dimensional body from a plane geometrical figure and determine the surface area and volume of regular geometrical bodies.

Unit-14 Understanding the symmetrical figures and differentiating between symmetrical and asymmetrical figures are the key concepts discussed in this unit. Further, the properties of two major types of symmetry which are commonly experienced have also been discussed.


Volume II

Block 3 Page No.

INTRODUCTION TO ALGEBRA

UNIT 8

Algebraic Expression and Operations 1

UNIT 9

Factorization 29

UNIT 10

Algebraic Equations 51

UNIT 8 ALGEBRAIC EXPRESSION AND OPERATIONS

Structure

8.1 Introduction

8.2 Objectives

8.3 Algebraic Expression and Related Concepts

8.3.1 Variable and Constant

8.3.2 Terms of an Algebraic Expression

8.3.3 Product, Factor and Coefficient

8.3.4 Like and Unlike Terms

8.3.5 Classification of Algebraic Expressions

8.4 Operations on Algebraic Expressions

8.4.1 Addition

8.4.2 Subtraction

8.4.3 Multiplication

8.4.4 Division

8.5 Unit Summary

8.6 Glossary

8.7 Answers to Check Your Progress

8.8 Assignments

8.9 References

8.1 INTRODUCTION

Consider the following examples:

i) Meena has 4 books, she purchased 3 more books. So she has now 4+3=7 books.

ii) There were 4 friends and each of them contributed Rs.25.00 for the Relief Fund. So the total contribution of the four friends were Rs.25×4= Rs.100.00.

Let us suppose that Meena had p no. of books and she purchased q no. of books. Can we find the total books she had after the purchase? We cannot tell a single number as we could when we added the two numbers in the above example. But all we can say is that Meena had (p+q) no. of books after the purchase. Here, p and q can stand for any two definite numbers.

Similarly, in the second example if we assume that each contributed m rupees then the total contribution of the four friends were 4 × m or 4m rupees. Moving a step ahead, if we suppose there were n no. of men contributing m rupees each, then the total contribution would be m × n or mn rupees.

The above two examples where numbers were involved in expressing the quantities and in the process of calculation, come under the domain of Arithmetic, a branch of Mathematics that deal with concrete numbers. But, when we used letters like ‘p’ and ‘q’ in the first and ‘m’ and ‘n’ in the second example we are going beyond the domain of Arithmetic and are in a more generalized domain called Algebra.

In other words Algebra is a branch of Mathematics in which principles of Arithmetic are generalized by using letters as symbols representing numbers. Letters are used to represent any kind of number. Thus, with the help of letters as representation of numbers, number relation can be generalized. Letters are also used to represent an unknown number/quantity. Thereby, various word problems can be expressed as symbolic statements. Thus, the main feature of Algebra is to use letters to represent numbers or quantities in a general situation rather than only in a particular case (as in Arithmetic).

8.2 OBJECTIVES

After studying this unit, you will be

able to:

• explain algebraic expressions and categorize them;

• differentiate between variable and constant;

• distinguish between like and unlike terms; and

• perform different operations on algebraic expressions.

8.3 ALGEBRAIC EXPRESSION AND RELATED CONCEPTS

Ayushi purchased some books for Rs. x from the shop. She also purchased note books for 5 rupees less than the cost of books and pencils for Rs. 8. How much money Ayushi had to give to the shopkeeper?

The Origin of Algebra

The word “Algebra” is taken from an Arabic word ‘al-jabr ’ (meaning ‘reunion’) used in a mathematical treatise entitled “Al-Kitāb al-mukhtaṣar fī hīsāb al-ğabr wa’l-muqābala” (Arabic for “The Compendious Book on Calculation by Completion and Balancing” ) written by the Persian mathematician Muhammed ibn Musa al Khwarizmi of Baghdad in 820 A.D. Diophantus, the famous Greek mathematician living in Alexandria in the 3rd. Century A.D is regarded as the ‘Father of Algebra’ for his seminal work entitled “Arithmetica”.

Can you solve Ayushi’s problem?

You mostly come across such type problems in daily life. Let us try to solve this problem.

Cost of books = Rs. x

Cost of note books = Rs. (x−5)

Cost of pencils = Rs. 8

Then, total cost = Rs. x + Rs. (x−5) + Rs. 8

= Rs. {x + (x−5) + 8}

Here, x is a letter representing a number which we do not know and 5 and 8 are distinct numbers.

Thus, x + (x – 5) +8 is an example of an algebraic expression.

On simplifying it, you can find

{ x + (x-5) + 8} = x + (x – 5) + 8

= x + x − 5 + 8

= 2 x + 3

(2x + 3) is also known as an algebraic expression. The exact value of this expression can be determined if we know the number represented by x.

For example: If x = 2,

Then, the total cost = Rs. (2x+3) = Rs. (2×2+3) = Rs. (4 + 3) = Rs. 7.00.

When, x = 10,

Then, the total cost = Rs. (2x+3) = Rs. (2×10+3) = Rs. 23.00.

When, x = 31

Then, the total cost = Rs. (2x+3) = Rs. (2×31+3) = Rs. 65.00.

Here we can see that for different values taken for ‘x’, the value of the given expression changes i.e. it varies with the change in the value of x.

We have been using letters to represent numbers from earlier times. Here, we will move systematically from arithmetic to algebra and learn how the operations of addition, subtraction, multiplication and division are worked on letters representing numbers. When letters represent numbers they are called letter numbers. Since they represent numbers they follow all the rules and properties of addition, subtraction, multiplication and division of number.

Any number of group of numbers made by the use of fundamental operation is called algebraic expression. This can also be termed as –“A number or group of numbers (including letter numbers) in which fundamental operations are used to put them together, the result obtaining is called an algebraic expression”For example: x+3, a+5, 15m, 3y, a+b+c etc. are algebraic expressions. Some examples of algebraic expressions

Statement Algebraic Expression

5 added to ‘x’ x+5

9 subtracted from ‘m’ M−9

‘n’ multiplied by 4 4n

‘p’ divided by q q

p

Here, we see according to the statement, operations of addition, subtraction, multiplication or division is applied on variables resulting in an expression. Thus you can say that,“ A combination or group of number (s) and literal number (s) connected by one or more of the signs +, −−−−, × and ÷ is known as algebraic expression.’’

The position of an expression on a number line

In the expression x + 7, ‘x’ is a literal number and 7 is a number. Suppose ‘x’ has a position shown by the point ‘x’ on the number line.

x may be anywhere on the number line but it is definite that the x + 7 represents the point M which is 7 units to the right of x. Similarly, x −5 represents a point 5 units to the left of X.

What about the position of 3x and 3x + 4?

The position of 3x will be point Q, the distance of Q from the origin being four times the distance of X from the origin. The position of R for 3x + 4 will be 4 units to the right of Q.

−3 −2 −1 0 +1 +2

0

X M

7

x 0

X

8.3.1 Variable and Constant

In algebraic expressions letters are used to represent some numbers. These letters are actually numbers in disguise and we can take any number for such literal number (s) used in the expression. As we have seen that in calculating the total cost of the books, the note books and the pencils bought by Ayushi, the total cost was Rs.7.00 when x was taken as 2 and it became Rs.23.00 when x was taken as 10 and so on.

Let us consider another example.

We know that the perimeter of a rectangle = 2 (length + breadth) = 2 (lab).

Check Your Progress

Notes : a) Space is given below for your answer.

b) Compare your answer with the one given at the end of this unit.

E1. Write three algebraic expressions using the literal numbers ‘x’ and ‘y’. You may use any real number if you like. [Notes: b) is not applicable.]

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E2. Write the algebraic expression in each of the following cases :

(i) x multiplied by – 7 …………………………………………………

(ii) 15 added to 3p ……………………………………………………..

(iii)7 subtracted from 2m ………………………………………………

(iv) –p divided by 5 …………………………………………………….

E3. Match the following:

(i) 20 more than a number a x − 19

(ii) 25 less than a number r 100 − z

(iii)Sum of 55 and a number q r − 25

(iv) z less than 100 55 + q

(v) x more than 19 a + 20

E4. Take the literal number ‘x’ and the number 5 and form expressions using not more than one of the four signs used for number operations.

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Here both ‘l’ and ‘b’ are literal numbers. They take values independent of each other. That is, the value that ‘l’ takes does not depend on the value that ‘b’ takes.

The literal numbers ‘l’ and ‘b’ in the example above take on different values depending upon the size of the rectangle. But, in each case, the perimeter can be expressed as 2 (l + b).

Similarly, if the length of one side of a square is represented by the literal number ‘a’, then its perimeter ‘p’ is equal to 4a.

Thus, we see that ‘l’ and ‘b’ in respect of a rectangle and ‘a’ in respect of a square can represent any positive number.

For that ‘l’ and ‘b’ in respect of a rectangle and ‘a’ in respect of a square are known as variables.

In the algebraic expression, 4x + 5, x is a variable and its value is not fixed.

When x = 1, then 4x + 5 = 4 × 1 + 5 = 9

When x = 2, then 4x + 5 = 4 × 2 + 5 = 13

When x = 3, then 4x + 5 = 4 × 3 + 5 = 15

Well, can the number of angles of a triangle be anything other than 3. Definitely not. Hence, the number of angles of a triangle is a fixed number and thus, it is a constant. In other words, all real numbers are constants. So we cam say:

9, −3, 5

3, 3 , ∆ etc. are some examples of constants.

In algebraic expression 3x+9, x is a variable, 3 and 9 are the constants.

A symbol which does not have any fixed value for it, but may be assigned any value according to the requirement is known as a variable.

The letters x, y, z, p, q, r are usually taken to represent variables.

A symbol having a fixed numerical value is known as a constant.

Algebraic expressions are formed from variables and constants.

8.3.2 Terms of an Algebraic Expression

A term is a number (constant), a variable, a combination (product or quotient) of numbers and/or variables, contained as a part is an expression separated from other by ‘+’ or ‘ −’ sign. The sign before a term belongs to the term itself.

In the algebraic expression 2x + 3xy – 7

there are 3 terms, namely + 2x, + 3xy and −7 Though there exists no sign before 2x (as it is the first term in the expression), it is to be considered to have a ‘+’ sign before it.

8.3.3 Product, Factor and Coefficient

When two or more numbers (may be constant or variables) are multiplied together, the result is known as the product.

We are well acquainted with the product of two constants (real numbers).

For example: 2x 2 = 6 (which is a number different from 2 or 3).

Here 6 is the product and we know that 3 and 2 each is a factor of 6.

When two variables are multiplied, what is the product?

Suppose, we multiply ‘p’ and ‘q’, what is the product?

We write the product of ‘p’ and ‘q’ as p × q or simply pq.

Similarly, 2 × p = 2p

In fact the term 3xy can be exactly divided by each of 1, 3, x, y, 3x, 3y, xy and 3xy and so the factors of 3xy are 1, 3, x, y, 3x, 3y, xy and 3xy.

Similarly, the factors of 7xy are 1, 7, x, y, 7x, 7y, xy and 7xy.

The numerical factor in a term is known as the coefficient of the product of the remaining factors. It is also said as the coefficient of the term itself.

For example:

i) In 5xy, 5 is the co-efficient of the term. 5 is also said as the coefficient of xy.

ii) In 15 − p + 3q2, −1 is the coefficient of p and 3 is the coefficient of q2.

8.3.4 Like and Unlike Terms

When terms have the same algebraic factors they are like terms and when terms do not have their algebraic factors exactly the same they are unlike terms.

For example, in the expression 4pq−7p+3pq−9, look at the terms +4pq and +3pq. The factors of +4pq are 4, p and q. The factors of + 3pq are 3, p and q. Here the algebraic factors in both terms are ‘p’ and ‘q’ are the same and hence they are like terms. On the other hand the terms +4pq and −7p do not have the same algebraic factors. So they are unlike terms. Similarly, the terms +3pq and −9 are unlike terms. Also the terms −7p and −9 are unlike terms.

The resolution of any algebraic expression into its constituent terms and again resolving each term into its factors can conveniently demonstrated in a diagram as shown below called a tree diagram.

Expression: 3xy − 5

Terms: 3xy −5

Factors: 3 x y

Check Your Progress



E5. Identify variables and constant in the expression x + y + 7.

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E6. Write four expressions each having 3 terms.

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E7. What are the coefficients of x in the following expressions?

(i) 7x−5y (ii) 9−2x+7y (iii) y2x−y (iv) 3z−7xz

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E8. Distinguish like and unlike terms in the following expressions:

(i) 5x+7y (ii) −6ab+21ba (iii) 2pq−2p (iv) mn2 −3mn2

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E9. Identify the terms and factors in each of the following expressions.

(i) −6ab−3b (ii) mn+n (iii) 7x+9y2 (iv) −2p+3

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‘Mono’ stands for one; ‘Bi’ is for two;

‘Tri’ is for three, and ‘Poly’ is for many.

8.3.5 Classification of Algebraic Expressions

We have already seen that the expression have one or more than one terms.

Depending upon the number of terms contained in an expression, it is classified into different categories.

The classification is done in the following manner.

i) Monomial: An expression with only one term is called a monomial.

For example: –3x, 5xy, 7x2yz, −4m, 2z2, 5 etc.

ii) Binomial: An expression which contains two unlike terms is called a binomial.

For example: 2a+3b, x2−3y2, 2xy−z, m−7 etc.

iii) Trinomial: An expression which contains three terms is called a trinomial.

For example: a+b+c, 2×2+y+9, p+q+3, 5x2−2x+1 etc.

Irrespective of the number of terms, the expression is known as a polynomial.

Some Special Feature of a Polynomial

• Any expression with one or more terms is called a polynomial. Hence a monomial, a binomial and a trinomial are all polynomials. 2x, 3x2−4y2, z2−3z+8, 3p2q2−3r2s2 etc. are examples of polynomials.

• Here, the index of the variable in each term need to be whole number like ax2, 5z3, x2y2, 4 etc.

• Terms like x1/2, 5x −2, p1/3, a−3/4 etc. cannot find place in a polynomial. Hence, a algebraic expression like ax−1/2 + by+c is not a polynomial.

Degree of Polynomial

The degree of a polynomial is the degree of its highest degree term.

For example:

i) In expression x4+x2+xy2−5 the highest degree term is x4 and hence the degree of term x4 is 4. Hence the degree of polynomial is 4.

ii) In expression 3x4+5x2y3+7xy2; the degree of term 3x4 is 4, the degree of term 5x2y3 is 2+3 =5 and the degree of term 7xy2 is 1+2 =3. Since, the highest degree term is 5x2y3 and its degree is 5, therefore the degree of the given polynomial is 5.

iii) 5 too is an expression and it can be written as 5x0 as x0 = 1. Hence the degree of the given polynomial (5 or 5xo) is zero (0).

Check Your Progress



E10. Identify monomials, binomials and trinomials from the following algebraic expression:

(i) xyz (ii) 2

1a2b2c2 (iii) x2−2y2+z2

(iv) 2x+5 (v) 7−5t (vi) ab−2a−b

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E11. Which of the following expressions are polynomials?

(i) 1−x (ii) x+ x (iii) z−2

1 (iv) t2+ 3 t+5

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E12. Write the degree of each of the following polynomials:

(i) 100 (ii) 3−p2+p3 (iii) 3xy2+7x3y5+7 (iv) 2−x

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E13. Write the degree of each of the following polynomials:

(i) 7x (ii) −p2q (iii) 7m2n2 (iv) −3x2y3z7 (v) 3

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8.4 OPERATIONS ON ALGEBRAIC EXPRESSIONS

We are well acquainted with the fundamental operations on numbers. Those operations are Addition , Subtraction, Multiplication and Division. Algebra deals with literal numbers which are numbers in disguise. Hence, the four fundamental operations can also be applied in Algebra.

While discussing the operations in algebra, it should be done in two phases, first, the operations on literal numbers and then the operation under an expression.

8.4.1 Addition

a) Addition of Literal Numbers

Suppose, we have 200g of ground nuts in a packet. Then how much ground nuts will be there in 3 such packets ?

The total quantity of ground nuts in the 3 packets

= (200+200+200)g = 600g.

The total quantity of ground nuts could also be determined as: 3 × 200g = 600g (Because repeated addition of a number is the same as multiplication).

If the 3 packets would contain 200g, 100g and 50g, then the total quantity (200+100+50)g = 350g.

Suppose there are 3 packets of oil which contain ‘x’ litres, ‘y’ litres and ‘z’ litres. What is the total quantity of oil contained in those 3 packets?

Total quantity of oil contained in the 3 packets = (x+y+z) litres.

Here the sum cannot be written as a single literal number.

But what will be the total quantity if we take a number of pouches each containing ‘x’ litres?

Total quantity of 3 pouches = x+x+x = 3 × x litres.

Total quantity in 4 pouches = x+x+x+x = 4 × x litres and so on.

A brief way of writing the above result is as follows: 3 × x = 3x

4 × x = 4x

So we can say:

Total quantity in 7 pouches = 7x litres.

How do we add 4x and 3x? 4x + 3x = x+x+x+x+x+x+x = 7x.

All the rules of addition on numbers also apply here.

i) Addition is commutative

That is, x + y = y + x

ii) Addition is associative

That is, x + (y+z) = (x+y)+z

iii) Existence of additive identity

Additive identity is 0.

That is, x + 0 = 0 + x = x.

iv) Existence of additive inverse

x and –x are additive inverse of each other

i.e. x + (−x) = 0

b) Addition of Algebraic Expressions

If we have 3 baskets each containing some apples, some bananas and some pineapples and we put them together, how do we speak about the fruits put together in a new basket?

If the 1st basket contains 5 apples and 8 bananas,

the 2nd basket contains 8 apples and 3 pineapples

and the 3rd basket contains 7 bananas and 2 pineapples, then

When those are put together in one basket, we count the 3 varieties of fruits separately. Thus, there are (5+8) apples, (8+7) bananas and (3+2) pineapples.

That is, when we put the fruits together, the result comes to 13 apples, 15 bananas and 5 pineapples. Exactly the same process is followed in addition of algebraic expressions; we put like terms together and combine them.

For example: Let us add 3x + 7, 2y-3, and 4x+y.

Solution:

The sum = (3x+7) + (2y−3) + (4x+y)

= 3x+7 + 2y−3 + 4x+y

= (3x+4x) + (2y+y) + (7−3) (Placing the like terms together)

= 7x+3y+4 (Adding the like terms)

For convenience we may write the 3 expressions one below the other such that like terms remain in one column. Such as:

3x + 7

(+) 2y − 3

(+) 4x + y

7x + 3y + 4 c) Properties of Addition

All properties of addition of numbers also apply to the addition of expressions.

i) Commutative property of addition

Addition of expressions is commutative

e.g., (3x+5) + (2y−1) = (2y−1) + (3x+5)

ii) Associative property of addition

Addition of expression is associative.

e.g.,. {(x+3) + (2y−1)} + (2x−y)

= (x+3) + {(2y−1) + (2x−y)}

iii) Existence of additive identity

Additive identity is zero (0).

e.g.,. (3x+2)+0 = 0+(3x+2) = 3x +2

iv) Existence of additive inverse

Any algebraic expression has its own additive inverse. The terms of the additive inverse have just the opposite signs to their counterparts in the algebraic expression.

e.g., (x+2) and – (x+2) are additive inverse of each other.

Check Your Progress



E14. Add together: 2x+7, 3x+xy-1, 3xy−5y+3 and 2y−xy+x

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E15. The present age of Amit’s mother is 4 times that of Amit. The age of Amit’s grand father is 15 years more than the sum of the ages of Amit and his mother.

Taking Amit’s age is ‘x’ years, determine the sum of the ages of Amit, his mother and his grand father.

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8.4.2 Subtraction

In the set of integers we know that: 5−3 = 2

and 3−5 = −2

The same principle also works in Algebra i.e., 5x−3x as well as 3x−5x also give us valid results.

How do we subtract?

The basic principle of subtraction is, a number subtracted from the same number gives zero, such as:

5−5 = 0 and 8−8 = 0

Similarly, a term or a monomial subtracted from itself yields zero.

e.g., x−x = 0 and y−y = 0

Now let us workout the following subtraction.

5x−3x = (x+x+x+x+x) – (x+x+x)

= x+x+x+x+x+−x−x−x (brackets are removed)

= (x−x) + (x−x) + (x−x) +x +x

= 0+0+0+2x = 2x

3x−5x = (x+x+x) – (x+x+x+x+x)

= x+x+x−x−x− −x−x−x

= (x−x) + (x−x) + (x−x) −x –x−x

= 0+0+0−x−x−x

= 0 – 3x

In brief we can do the work in the following way:

5x−3x = (5−3) number of x’s = 2x;

3x−5x = (3−5) number of x’s = −2x

Here, we observed that, like in addition, subtraction can only take place between two like terms.

What is 2x−3y equal to?

As 2x and 3y are not like terms, the subtraction of 3y from 2x, does not give any other result than 2x − 3y.

Subtraction of Expressions

The process of subtraction is very much similar to that of addition. Some examples are discussed below.

For example: Subtract 3x + y − 1 from 5x + 4. Solution:

(5x+4) – (3x+y−1) = 5x+4−3x−y+1

= (5x−3x) – y +4+1 (Placing the like terms together).

= 2x – y + 5

Alternative method:

5x +4

(−) 3x+y −1

− − +

2x −y +5

Note:

i) In the first method, while opening the second pair of brackets, the signs of all the terms of the expression to be subtracted were changed, similarly in the 2nd method the signs of all the terms of the expression to be subtracted are changed and written below the respective terms.

ii) The result of subtraction of two like terms is a term like the terms taken for subtraction.

iii) Subtraction of two unlike terms does not yield a separate result.

Check Your Progress



E16. Take the problem given in E15. Find by how many years Amit's mother is younger than Amit’s grand father?

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8.4.3 Multiplication We discuss multiplication also in two phases. The first phase is multiplication of literal number and second phase is multiplication of expressions.

a) Multiplication of Literal Numbers

Here we find a new experience which we did not have with numbers.

If we multiply several number of x’s, what result do we get? Let us see

x × x = x2 (product of 2x’s)

Note that the number of x’s multiplied is written above x.

The number 2 in x2 is known as the index or exponent of x.

We read x2 as square of x or (x-square in short).

In x2, ‘x’ is said as the base and 2 is known as the index. Similarly,

Product of how many x’s The result written as The result read as

Product of 3x’s x3 x to the power 3 or x

cube

Product of 4x’s x4 x to the power 4

Product of 10x’s x10 x to the power 10

b) Product of Two Different Literal Numbers

What about the product of x and y?

x × y does not give a different result. We only write the product briefly as x × y or xy

Similarly x × x × y x y = x3 × y2 = x3y2

c) Properties of Multiplication

i) Commutativity of multiplication

Multiplication of literal numbers are commutative.

e.g. p × q = q × p i.e. pq = qp 2 × q = q × 2

But we would not say that 2q = q2

q2 is not considered as a symbolic way of writing q × 2.

ii) Associativity of multiplication

Multiplication is associative.

e.g., (x+y) + z = x+(y+z)

iii) Existence of multiplicative identity

Multiplicative identity in case of literal numbers is same as in case of numbers, i.e. 1.

e.g. x × 1 = 1 × x = x

iv) Distributive property

Multiplication distributes over addition and subtraction.

e.g. (3+4)x = 3x + 4x = 7x This provides a shorter method for addition and subtraction of like terms.

d) Multiplication of Two Monomials

We shall first learn to multiply two terms. Commutativity and associativity of multiplication help us to rearrange factors and place them any where else we like for convenience. Thus, x × y × 4 × x can be rearranged as 4 × x × x × y, so that all x’s remain together and the numerical factor remains at the beginning.

We adopt the following procedure for multiplying two terms.

• Bring all the numerical factors to the beginning and multiply them to get a single number.

• Put the same factors together.

Now observe,

i) x × 8y = x × 8 × y = 8 × x × y = 8xy

ii) 8x × 3y = 8 × x ×3 × y = 8 × 3 × x × y =24xy

iii) 5x × (−3y) = 5 × x × (−3) × y = 5 × (−3) × x × y = −15xy

iv) x2 × x3 = (x × x) (x ×x×x) = x×x ×x×x×x = x5

v) 5x × 4x2 = (5 × 4) × (x × x2) = 20 × (x×x×x) = 20×x3 = 20x3

vi) 9x × (-5xyz) = (9 × −5) × (x × xyz)

= − 45 × (x × x × yz)

= − 45x2yz

Here at first the numerical coefficients are multiplied together and then the literal factors are multiplied in different groups of the same literals.

See to the multiplication in (4) above.

x2 × x3 gives the result x5.

It can be seen that index 5 in the result is equal to the sum of the indices in the 2 factors multiplied. Thus we see that x2 × x3 = x2=3 i.e. (xm × xn ) = xm+n

e) Multiplication of Three or More Monomials

For examples:

i) 5x × 3y × 9z = (5x × 3y) × 7z = 15xy × 7z = 105 xyz

ii) 4xy × 5x2y2 × 6x3y3

= (4xy × 5x2y2) × 6x3y3

= (4 × 5 × x × x2 × y × y2) × 6x3y3

= 20x3y3 × 6x3y3

= 20 × 6 × x3 × x3 × y3 × y3

= 120x6 × y6

= 120x6y6

At first we multiply two monomials and multiply the result with the third monomial. We use this method to multiply multiple monomials.

f) Multiplication of a Monomial by a Binomial

For examples:

i) Multiply monomial 5x and the binomial (10x + 2)

5x × (10x + 2)

Using distributive law, we get

5x × (10x + 2) = (5x × 10x) + (5x × 2)

= (5 × 10 × x × x) + (5 × 2 × x)

= 50x2 + 10x

ii) Multiply: 5xy and y3 + 3

5xy × (y3 + 3) = (5xy × y3) + (5xy × 3) (by distributive property)

= (5 × x × y × y3) + (5 × 3 × x × y)

= 5xy4 + 15xy

g) Multiplication of a Monomial with a Trinomial

For example: Multiply 9x and 6x2 + 5x + 8

9x (6x2+5x+8) = (9x×6x2) + (9x×5x) + (9x+8) (by distributive property)

= 54x3 + 45x2 + 72x

Multiply each term of trinomial by a monomial and add the result.

h) Multiplication of a Polynomial by a Polynomial

Case I: Multiplication of a binomial by another binomial:

For example: Multiply x + y and 2x + 3y

(x+y) (2x+3y) = x (2x + 3y) + y (2x + 3y) (by distributive property)

= (x × 2x) + (x × 3y) + (y × 2x) + (y × 3y)

= (2 × x × x) + (3 × x × y) + (2 × x × y) + (3 × y × y)

= 2x2 + 3xy + 2xy +3y2

= 2x2 + 5xy + 3y2

Case II: Multiplication of a binomial by a trinomial:

For example: Multiply x+7 and x2+2x+5

(x + 7) × (x2 + 2x + 5) = x× (x2 + 2x + 5) + 7 × (x2 + 2x + 5) (by distributive property)

= x3 + 2x2 + 5x + 7x2 + 14x + 35

= x3+ (2x2 + 7x2) + (5x + 14x) + 35

= x3 + (2 + 7)x2 + (5 + 14) x + 35

= x3 + 9x2 + 19x + 35

8.4.4 Division

a) Dividing a Monomial by a Monomial

Working Rule:

Step I: Write the dividend as numerator and divisor as denominator.

Step II: Simplify the fraction by cancelling common factors from the numerator and the denominator both.

For example:

i) Division of 21xy by 7x = x

xy

7

21= 3y

ii) 9x2y ÷ −3xy = xy

yx

3

9 2

− = −3x

iii) 5

858

x

xxx =÷ =

xxxxx

xxxxxxxx

×××××××××××

= x × x × x (Out of 8 x’s of the numerator, 5 get cancelled. Hence the product of 3x’s are left)

= x3

The index of x in the result = 3 which is equal to 8−5. Thus we can say that x8 ÷ x5 = x 8−5

In general (xm÷ xn = xm−n).

b) Dividing a Polynomial by a Monomial

Working rule: Divide each term of the polynomial (i.e. the dividend) by the monomial and simplify each fraction.

For example:

i) Division of 10a3 − 15a2b by −5a2 .

10a3 − 15a2b ÷ −5a2 = 2

2

2

3

5

15

5

10

a

ba

a

a

−−

−

= −2a + 3b

ii) Division of 12x3y − 8x2y2 + 4x2y3 by 4xy.

12x3y − 8x2y2 + 4x2y3 ÷ 4xy = xy

yx

xy

yx

xy

yx

4

4

4

8

4

12 32223

+−

= 3x2 − 2xy + xy2

c) Dividing a Polynomial by a Polynomial

We divide the dividend by the divisor in the long division process as is done with numbers.

Example 1: Divide 3x2 + 7x − 6 by 3x − 2.

Step-I: Arrange the terms in the dividend and the divisor in descending order of the degrees of their terms. (Here it is already in descending order).

Step-II: Divide first term of dividend by the first term of division to get the first term of quotient.

Here x

x

3

3 2

= x which is the first term of

Step-III: Multiply quotient x with each term of divisor (3x−2) and write the result below the dividend and subtract.

x

xxx 67323 2 −+−

xx 23 2 −

Step-IV: Write the remainder 9x of step 3 and take the 3rd term of the dividend down. Thus, you get 9x−6.

x

xxx 67323 2 −+−

xx 23 2 −

− +

9x − 6

Step-V: Repeat the process from Step 2 to 4 taking 9x − 6 as new dividend 3 (3x−2) = 9x − 6

32 67323

+

−+−x

xxx

xx 23 2 −

− +

9x − 6

9x − 6

− +

0

Hence, quotient = x + 3

remainder = 0

i.e. (3×2 + 7x − 6) ÷ (3x − 2) = x + 3

Example 2: Divide 3 x 3 + 16x2 + 21x + 20 by x + 4.

Solution:

Hence, quotient = 3x2 + 4x + 5 remainder = 0

Example 3: Divide 6x4 − 5x3 + 2x2 − 7 by 2x2 − x + 1

Solution:

Hence, quotient = 3x2 − x − 1

remainder = − 6

Check Your Progress



E17. Subtract 5x − 6y − 12 from the sum of 3x − 2y + 9 and –y +12.

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E18. What must be added to x4 − x3 + x2 + x + 3 to obtain x4 + x2 − 1.

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E19. Multiply (2x + 3) and (x + 4).

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E20. Multiply 3x2 + 2x and 2x3 − 7x + 8.

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E21. Divide 6m2 − 17m −3 by m − 3.

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E22. Divide 6x2 + 7xy − 3y2 by 2x + 3y.

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Care to be taken during teaching:

8.5 UNIT SUMMARY

1. Algebra is generalized arithmetic where letters are used to represent numbers and are known as literal numbers or literals.

Some common mistakes which the children make are the following:

• While they add two terms 2x and 3 they write the result as 2x+3 =5.

• A term ‘x’ is considered to have no coefficient i.e. the coefficient of x is zero. As such they write 4x + x + 3x = 7x.

• Some children write 5x + 4x = 9x2, the child has added the coefficients of the like terms and added the exponents also.

• Some children write 2x + 7y = 9xy. In this case the child has added the coefficient and multiplied the variables.

• Some children write 5 (x − 2) = 5x – 2

• If x = 2 and y = 3 given, then some write xy = 23. x and y in xy have been considered as two digits.

The teacher has to make an effort to remove the misunderstanding of the children which leads to commit such errors.

Check Your Progress



E23. Try to make a list of 5 other common errors which the students make in solving exercises in algebra.

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Some of the literal numbers are variables, some are constants.

Every number is a constant i.e. 3 cannot be anything else but 3, but ‘x’ can be assigned different values in different situations.

2. A group of numbers and literals connected by one or more of the signs +, −, × and ÷ is known as an algebraic expression.

3. A term is a part of an expression which is separated from other part by a ‘+’ or ‘ −’ sign. It may be a constant (i.e. a number), a variable or a combination of constants and variables (in the way of multiplication or division).

4. An expression containing one term is a monomial.

An expression containing two terms is a binomial.

An expression containing three terms is a trinomial.

An expression with any number of terms is a polynomial.

A polynomial could be a monomial, a binomial or a trinomial.

Note: The index of every variable contained in an expression is always a whole number.

5. If a term contains one variable, the index of that variable becomes the degree of the term.

E.g. the degree of 3x5 is 5.

If there are more than one variable in the term, then the degree of the term is the sum of the indices of the variables.

E.g. the degree of the term 5x2y3z is 2 + 3 + 1 = 6.

6. Degree of a polynomial is the highest of the degrees of the terms contained in it.

7. The constants (numbers) and the variables that are multiplied to form a term are known as the factors of it.

8. The numerical factor is usually known as the coefficient of the term. As such 6 is the coefficient of the term 6x2y3.

9. Like terms are the terms in which literal factors are the same and they occur same number of times.

If the terms do not have their literal factor exactly the same, those are known as unlike factors.

10. Operations of addition, subtraction, multiplication and division are applied on polynomials (expressions).

Like terms when added or subtracted give a result which is a term like the terms added or subtracted.

When unlike terms are added or subtracted, the result becomes an expression while each of the unlike terms added or subtracted continues to remain as a term. While adding polynomials commutative and associative property are utilized to get the like terms together and add them.

While subtracting polynomial ‘p’ from a polynomial ‘q’ we add ‘q’ and ‘– p’.

8.6 GLOSSARY

Literal Numbers : Letters used to represent numbers.

Variable : A literal which represents several different numbers.

Constant : A fixed number.

Expression : May be seen in the unit summary.

Like Terms : -do-

Unlike Terms : -do-

Monomial : -do-

Binomial : -do-

Trinomial : -do-

Polynomial : -do-

8.7 ANSWERS TO CHECK YOUR PROGRESS

E2. (i) −7x (ii) 3p + 15 (iii) 2m – 7 (iv) 5

p−

E3. (i) a+20 (ii) r−25 (iii) 55+q (iv) 100−z (v) x+19

E4. x +5 or x – 5 or 5+x or 5-x or 5x or 5

x

or

x

5

E5. Variable – x, y. Constant − 7

E6.

E7. (i) 7 (ii) −2 (iii) y2 (iv) − 7z

E8. (i) unlike term (ii) like term (iii) unlike term (iv) like term

E9. (i) −6ab and 3b; factors of 7x are 7 and z factors of +9y3 are 3,6,2. Write for other bits.

E10. (i) monomial (ii) monomial (iii) trinomial (iv) binomial (v) binomial (vi) trinomial

E11. (i) (iv)

E12. (i) 0 (ii) 3 (iii) 8 (iv) 1

E13. (i) 1 (ii) 3 (iii) 7 (iv) 12 (v) 0

E14. Sum = 3x − 2y + 9 + (−y + 12)

E15. 10x + 15 years

E16. x + 15 years

E17. 3y − 2x + 33

E18. x3 − x − 4

E19. 2x2 + 11x + 12

E20. 6x5 + 4x4 − 21x3 + 10x2 + 16x

E21. 3x − y; R = 0

E22.

8.8 ASSIGNMENTS

Q1. “The present age of Shivang’s father is three times that of Shivang. Taking Shivang’s present age as ‘x’ years, write an expression the sum of their ages after five years.

Q2. Write the expression in each of the following cases:

(i) 9 added to 2p (ii) 15 subtracted from 2z

(iii) y multiplied by −9 (iv) r divided by s

Q3. Separate the constant term from the expression which contains it.

(i) 5 (ii) 2n (iii) −xy2 (iv) 4x+5

(v) 12az2 (vi) 2p−20

Q4. Find the coefficient:

(i) of pq in −3aqp (ii) of x2 in p2x2y2

(iii) of mn in – mn (iv) of 12 in − 12t2

Q5. Write numerical coefficient:

(i) of x in 3x + 5 (ii) of p

(iii) of − 3p2q3 (iv) of −8xy3 Q6. (a) Give one example of:

(i) a monomial (ii) a binomial (iii) a trinomial

(b) Which of them is a polynomial?

Q7. Write the degree of each of the following polynomials:

(i) 8x3+3x+7 (ii) 9−t3 (iii) 8 (iv) 7t−√8

Q8. Write the like terms together in separate groups:

(i) 9x2, xy, −3x2, x2 and 2xy

(ii) ab, −a2b, −3ab, 5a2b and −8a2b

(iii) 7m, 8m, −5mn, −m and 3n

Q9. Add 2x2 + 3x + 5 and 3x2 − 4x − 7.

Q10. Add 2t2 + t−1, 3t−5−3t2 and 1−3t−3t2.

Q11. Subtract −3x+8y from −6x−8y.

Q12. Subtract 2x2 + 3y2−3 from x2−7y2 + 1.

Q13. Subtract 5x− 6y − 12 from the sum of 3x− 2y + 9 and –y + 12.

Q14. What must be added to a2 + ab + b2 to obtain 2a2 + 3ab?

Q15. What should be subtracted from 2x+8y+10 to get −3x+7y+16?

Q16. Multiply x + y by x3 − xy2 − y3.

Q17. Multiply 6xy2 − 7x2 y2 + 10x3 by −3x2 y3.

Q18. Divided 8x2 + 4x− 60 by 2x − 5.

Q19. Divided 6x2 y2 − 7xyz − 3y2 by 3xy + z.

Q20. The area of a rectangle is 6m2 – 4m – 10n2 square units and its length is 2m + 2n units. Find its breadth.

8.9 REFERENCES

• Johnson, R.E. (et.al.) (1961): Modern Algebra, First Course, Addison-Wesley Publishing Company Inc., USA.

• Russel Donald, S. (1961): Elementary Algebra, Allyn and Bacon Inc., Bostan.

• Miglani, R.K. and Singh, D.P. (2004): Teaching of Mathematics at Elementary level Part I and II……

• Bansal, R.K. (2007): Middle School Mathematics, Selina Publisher, New Delhi.

• ---- (-): OMT 101, Preparatory Course in General Mathematics, IGNOU, New Delhi.

• --- (2008): Mathematics: A Text Book for Class VI, VII, VIII, NCERT, New Delhi.

UNIT 9 FACTORIZATION

Structure

9.1 Introduction

9.2 Objectives

9.3 Concept of Factor and Factorization

9.3.1 Identifying the Common Factor(s)

9.4 Factorization of Binomials

9.4.1 Difference of Perfect Square Terms

9.4.2 Sum or Difference of Two Perfect Cube Terms

9.4.3 Factorization of Some Special Expressions

9.4.4 Factorization of Expressions Containing Four Terms

9.5 Factorization of Trinomials of the Form ax2+bx+c

9.6 Factorization of Perfect Square Polynomials of Six Terms

9.7 Unit Summary

9.8 Glossary


9.10 Assignments

9.11 References

9.1 INTRODUCTION

In the set of natural numbers, we found some of the numbers to be prime and the rest (except 1) to be composite (1 is neither a prime number nor a composite number). Each composite number can be written as the product of 2 or more prime numbers. When a number is written as the product of other prime numbers, such as 6 = 2 × 3, 2 and 3 each is said to be a prime factor of 6. In general, if a number is written as the product of other numbers, each of the other numbers is known as a factor of the first number. Here the factors may not be prime.

When an algebraic expression is resolved into the product of two or more expressions, these expressions are called factors of the former expression.

For example: x2 + (a+b) x + ab = (x+a) (x+b)

Thus, (x+a) and (x+b) are factors of the expression x2 + (a+b) x + ab.

The process of finding the factors of an expression is called factorization.

9.2 OBJECTIVES

After studying this unit, you will be able to:

• explain factors and factorization; and

• factorize the polynomials of different forms.

9.3 CONCEPT OF FACTOR AND FACTORIZATION

Amit, Meetu and Sohana had 6 chocolates each. They distributed those among their friends. Amit distributes equally among 3 friends. Meetu among 6 of her friends and Sohana among 2 of her friends. How did they distribute?

AMIT MEETU SOHANA

F1 F2 F3 F1 F2 F3 F4 F5 F6 F1 F2

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0

0 0

2 × 3 = 6 1 × 6 = 6 3 × 2 = 6

From this we can see that 6 chocolates can be distributed in three ways. In other words 6 can be the result of multiplying a pair of number in three ways. That is the numbers involved in getting 6 as the product are 1, 2, 3 and 6. These numbers are called the factors of 6. Therefore, factors of 6 = {1, 2, 3, 6}.

Again, when 2 and 3 are multiplied we get 6. In this case we call 6 as the multiple of the factors 2 and 3. Similarly, 6 is also a multiple of 1 and 6. In other words, if ‘a’ is a factor of ‘b’, then we say ‘b’ is a multiple of ‘a’.

It may be observed that from among the four factors of 6, only 2 and 3 are prime numbers. Hence those are called the prime factor of 6.

From the above discussion, we can say that the process of expressing a number as the product of two or more prime numbers is called the process of factorization.

Here one activity is given to clarify the concept of factor. The teacher should also develop such type of other activities for same concept.

• Factors of monomial

Let us examine the following products:

14 = 2 × 7, Here 2 and 7 are two factors of 14.

7x = 7 × x, So 7 is a factor of 7x, x is also a factor of 7x

14x = 2 × 7 × x, So 2, 7 and x are factors of 14x

Similarly, 12y = 2 × 2 × 3 × y. Here, 2, 3 and y are factors of 12y.

Following the same process, we can say x2y = x × x × y, where x and y are two factors of x2y.

Check Your Progress



E1. Determine the set of factors of 30.

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E2. Determine the prime factors of 30.

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E3. What is the total number of factors of 12?

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E4. How many multiples does 2 have?

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9.3.1 Identifying the Common Factor(s)

We know the distributive law of multiplication in algebraic expressions.

It is a (b + c) = ab + ac.

It is the same to say that ab + ac = a (b+c) (9.1)

Thus ‘a’ and ‘b+c’ each is a factor of ‘ab + ac’.

We can see that the first term on the left of (9.1) i.e. ‘ab’ has the factors ‘a’ and ‘b’. Similarly, the factors of the second term ‘ac’ are ‘a’ and ‘c’.

The factor which is common in both the terms ‘ab’ and ‘ac’ is ‘a’ as can be seen below:

ab + ac = a × b + a × c

= a (b + c)

Here, ‘a’ is the common factor and in the process of factorization we say that ‘a’ has been taken common.

Thus ‘taking common’ is found to be a process of factorization for which we have to identify the common factor(s) of the terms of the expression which is being factorized. Its implication is that whenever the terms of an expression have a common factor, taking common process is to be first applied. This can be realized at a later stage.

Example 1: Factorize the following:

i) 3x2 + xy

ii) 2x2y + 6xy2 – 4xy

Solution:

i) 3x2 + xy = 3× x × x + x × y

Check Your Progress



E5. Determine the factors of the following:

(i) 25 (ii) 54m (iii) 12xy2

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= x (3 × x + y)

= x (3x + y)

Note: The first step shows what are the factors of each term. Thereby, the common factor existing in them could be clear. In the second step, common factor has been taken and the remaining factors are kept enclosed in a pair of brackets. In the third step, the first inside the pair of brackets i.e. 3 × x has been simplified and written as 3x.

Thus, we have 3x2 + xy = x (3x + y).

How do we get the terms to be written inside the pair of brackets after the common factor ‘x’ is taken out?

We divide the 1st term 3x2 by ‘x’ and 3x which is written as the 1st term inside the pair of brackets.

Then again we divide the 2nd term xy by ‘x’ and get ‘y’ which is written as the 2nd term inside the pair of brackets.

Thus, we can say:

ii) 2x2y + 6xy2 = 4xy

It can be seen that the common factor is ‘2xy’ (as 2 is the common factor of the coefficients of the 3 terms).

∴ 2x2y + 6xy2 – 4xy = 2xy (x + 3y – 2).

9.4 FACTORIZATION OF BINOMIALS

We have already seen how to factorize binomials by identifying common factors of the two terms of the binomial expression like

ax ± bx = x (a ± b),

or, like a2 ± ab = a (a ± b).

In the first expression ‘a’ is the common factor while in the second ‘b’ is the common factor between the two terms of the expression.

Taking common is the practice of division of each term of the expression by the common factor.

Let us consider factorizing some standard binomials like the one expressed as the difference of two terms, each of which is a perfect square that is in the form of a2 − b2.

9.4.1 Difference of Perfect Square Terms

When the binomial expression is in the form of difference of two perfect square terms, like a2 − b2 it can easily be seen that there is no common factor between the two terms.

In their first reaction the students will think that this expression cannot be factorized.

Let us just have an intelligent guess. Can a − b be a common factor? Suppose this is a factor of a2 − b2, then we may write the expression using terms as

a (a − b)

and b (a − b).

From these two terms we get a2 − ab and ab − b2. When we add these terms we get a (a − b) + b (a − b) = a2 − ab + ab − b2 = a2 – b2. From this guessing we get the vital clue for factorizing the expression a2 – b2.

Let us, then, factorize the expression as follows:

a2 – b2 = a2 − ab + ab − b2 (since adding and subtracting ‘ab’ does not disturb the value of the expression)

= (a2 − ab) + (ab − b2) (pairing the terms)

= a (a − b) + b (a − b) (taking common factors)

= (a − b) (a + b). (taking ‘a − b’ as the common factor)

Thus, we get a2 – b2 = (a − b) (a + b) (9.2)

This is used as a standard formula for factorizing a binomial expressed in the form of difference of two perfect squares. While solving problems we need not always factorize following the steps as shown above. We can just apply this formula in solving the problems of factorization as can be seen in the following examples:

Can we factorize a2 + b2? Try it yourself.

Example 2: Factorize 4x2 − 9y2

Solution: 4x2 − 9y2 = (2x)2 − (3y)2 (expressed in the form a2 – b2)

= (2x + 3y) (2x – 3y) (applying the formula)

Example 3: Factorize: 9x3 – 16xy2

Solution:

Can the binomial be expressed in the form of a2 – b2?

Definitely not. Now recall what you have studied earlier.

Sameer said:

Taking common is the first process of factorization. Here ‘x’ is the common factor in both the terms of the binomial. We should take ‘x’ common.

9x3 − 16xy2 = x (9x2 – 16y2)

Is the process of factorization over?

By writing 30 = 3×10 is the process of factorization over?

‘No’ says Leena.

‘Why Not’? asks Mandira

‘Because, 10 can be further factorized, says Leena’.

So Leena writes: 30 = 3 × 2 × 5

Can any of the factors be further factorized? Definitely not.

So the factorization is over.

Note: After factorizing a polynomial into 2 factors by applying any process, we should check if any of the factors can be further factorized. If yes, continue.

Now let us take the above questions.

9x3 – 16xy2 = x (9x2 – 16y2)

= x {(3x) 2 – (4y) 2}

= x (3x + 4y) (3x − 4y)

It can now be realized that without application of the process of taking common, the binomial could not be expressed in the form a2−b2 and as such the formula a2−b2 = (a+b) (a-b) could not be applied.

So we apply the process of taking common first if there exists anything common. Example 4: Factorize 16 – x4.

Solution: Is there any common factor? No.

Can the terms be expressed as perfect square terms? Yes.

So let us proceed:

16 – x4 = (4) 2 – (x2) 2

= (4 + x2) (4 − x2)

Can any of the factors be further factorized?

Yes, 4-x2 can be further factorized into (2 + x) (2 − x).

Thus, 16 – x4 = (4) 2 – (x2) 2

= (4 + x2) (4 − x2)

= (4 + x2) (2 + x) (2 − x).

9.4.2 Sum or Difference of Two Perfect Cube Terms

Let us try to factorize a3 + b3 first.

We can express a3 + b3 in terms of (a + b) 3.

We know that (a + b)3 = a3 + b3 + 3a2b +3ab2.

= a3 + b3 + 3ab (a + b) (Taking 3ab common from the last two terms)

From this we can calculate

a3 + b3 = (a + b) 3 − 3ab (a + b)

Check Your Progress



E6. Factorize the following:

(i) 3x + 6 (ii) 7x2 − 5x (iii) 6x2y + 4xy2

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(i) x2 − 9y2 (ii) 4x2 – 11 – 9x4 (iii) 9x3 – 16xy4 (iv) 16x4−1

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= (a + b) {(a + b)2 − 3ab} (Taking ‘a − b’ common from both the terms)

= (a + b) {a2 + b2 + 2ab − 3ab} (Expanding (a + b)2)

= (a + b) (a2 + b2 − ab)

Thus, a3 + b3 = (a + b) (a2 − ab + b2) (9.3)

By expanding (a − b) 3 and following exactly similar steps as we did for determining the factors of a3 + b3 we can factorize a3 − b3. But a simpler way to find the factors of a3 − b3 is to replace ‘b’ (+b to be exact) by ‘−b’ in the formula for a3 + b3. In either way we get,

a3 − b3 = (a − b) (a2 + ab + b2) (9.4)

If the binomial can be expressed as the sum of two perfect cube terms, the formula (9.3) is applicable for determining the factors and when the binomial can be expressed as the difference of two perfect cube terms the formula (9.4) is used for factorizing the binomial.

Example 5: Factorize

i) 8x3 + 1

ii) 27 − x6

Solution:

i) 8x3 + 1 = (2x)3 − 13 (Here, 2x in place of ‘a’ and 1 in place of ‘b’)

= (2x + 1) {(2x)2 − (2x) (1) + 12}

= (2x + 1) (4x2 − 2x + 1)

ii) 27 − x6 = (3)3 − (x2)3 (Here, 3 in place of ‘a’ and x2 in place of ‘b’)

= (3 − x2) {(3)2 + (3) (x2) + (x2)2}

= (3 − x2) (9 + 3 x2 + x4)

9.4.3 Factorization of Some Special Expressions

Example 6: Factorize x4 + 4y4

Can the binomial be expressed in the form a2 – b2? No.

Can it be expressed in the form a3 + b3? No.

Let us see what happens when it is expressed in the form of a2 + b2

We know that a2 + b2 = (a+b)2 – 2ab. Applying the above formula we get, x4 + 4y4 = (x2)2 + (y2)2

= (x2 + 2y2)2 – 2 (x2) (2y2)

= (x2 + 2y2)2 – 4x2y2

= (x2 + 2y2)2 – (2xy)2 (Now it is in the form of a2−b2)

= (x2 + 2y2 + 2xy) (x2 + 2y2 − 2xy)

Example 7: Factorize a4 + a2b2 + b4.

Note: It is not a binomial. Yet the formula applied for the above problem can be used in factorizing this expression.

Solution: a4 + a2b2 + b4 = a4 + b4 + a2b2

= (a2)2 + (b2)2 + a2b2

= (a2 + b2)2 – 2(a2) (b2) + a2b2

= (a2 + b2)2 – 2a2b2 + a2b2

= (a2 + b2)2 – a2b2

= (a2 + b2)2 – (ab)2 (Occurs in the form of a2 − b2)

= (a2 + b2 + ab) (a2 + b2 − ab)

Example 8: Factorize a6 – b6.

Note: The binomial can be expressed in the form a2 – b2, and can also be expressed in the form a3 – b3 which way should we express it?

Solution:

Taken in the form a2 −−−− b2 Taken in the form a3 −−−− b3

a6-b6 = (a3)2 – (b3)2 a6-b6 = (a2)3 – (b2)3

= (a3+b3) (a3−b3)

= (a+b) (a2−ab+b2) × (a−b) (a2+ab+b2)

= (a+b) (a−b) (a2+ab+b2) (a2−ab+b2)

= (a2−b2) {(a2)2 + (a2) (b2) + (b2)}

= (a+b) (a−b) {( a2 + b2)2 − (ab)2}

= (a+b) (a−b) (a2 + b2+ ab) (a2 + b2+ ab)

9.4.4 Factorization of Expressions Containing Four Terms

When there are four terms in an expression, then factorization is facilitated by grouping the terms. The four terms in the expression may be divided into two groups taking 2 terms in each group or taking 3 terms in one and 1 term in the other group as per the requirement.

Let us now discuss some examples of 2+2 grouping.

Example 9: Factorize x3 + 3x2 + x + 3.

In how many ways can we form 2 + 2 groups out of the four terms in the expression?

There are possibly three ways of such groupings:

i) 1st and 2nd terms in one group and the rest in the other group.

ii) 1st and 3rd terms in one group and the rest in the other group.

iii) 1st and 4th terms in one group and the rest in the other group.

Solution:

i) x3 + 3x2 + x + 3 = (x3+ 3x2 ) + (x + 3) (Grouping as per (i) above)

= x2 (x + 3) + 1 (x + 3)

= (x + 3) (x2 + 1) (x+3 is taken common from

Check Your Progress




(i) x3 + 8 (ii) 27x3y3 – 1 (iii) 4 + x4

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both the groups).

ii) x3 + 3x2 + x + 3 = (x3 + x ) + (3x2 + 3 ) (Grouping as per (ii) above)

= x (x2 + 1) + 3 (x2 + 1)

= (x2 + 1) (x + 3)

iii) x3 + 3x2 + x + 3 = (x3 + 3) + (3x2 + x) Can we further proceed for factorizing the expression in this form of grouping?

Give reasons why it is not possible to factorize with this way of grouping the terms of this expression.

Example 10: Factorize ax2 + bx2 + a2 − b2.

Solution: ax2 + bx2 + a2 − b2 = x2 (a + b) + (a + b) (a − b)

= (a + b) {x2 + (a − b)}

= (a + b) (x2 + a − b)

Note: The 2nd group has been factorized by application of the formula

a2 − b2 = (a + b) (a − b)

• Using (3 + 1) Grouping

Example 11: Factorize 4x2 − 9y2 + 4x + 1.

Note: You can try with all the 3 ways of (2 + 2) groupings. You will not be able to get common factors by employing any of these three ways of grouping.

Then, let us examine whether it is possible to factorize this expression by applying (3 + 1) groupings.

Solution: 4x2 − 9y2 + 4x + 1 = (4x2 + 4x + 1) − 9y2 (following ‘3+1’ groupings)

= (2x)2 + 2 × 2x × 1 + (1)2} – (3y)2

= (2x + 1)2 – (3y)2 (occurs in the form of a2 − b2)

= (2x + 1 + 3y) (2x + 1 − 3y)

This form of grouping is not applicable to all types of expressions with four terms. When the term of such expressions fulfil certain conditions, then only (3 + 1) grouping facilitates factorization. You can check those conditions in the above example.

1st Check: Are there 3 perfect square terms in the expression? Ignore the signs of the terms for this check.

Ans. Yes. Those are 4x2, 9y2 and 1.

2nd Check: Are two of them of the same sign and 3rd one of opposite sign?

Ans. Yes. 4x2 and 1 are both +ve and 9y2 is –ve.

The two perfect square terms with the same sign and the non-square term are grouped together. Within this group check if (T2)

2 = 4 × T1 × T3 test holds good for this group.

9x2-4y2+6x+1

Check Your Progress




(i) 3a + 4b + 15ac + 20bc

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(ii) x3 + 2x − 5x2 − 10

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(iii) 10m2n = 9 + 6m + 15mm

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(iv) 9x2 − 4y2 + 6x + 1

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(v) 4z2 − 16x2 − y2 + 8xy

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9.5 FACTORIZATION OF TRINOMIALS OF THE FORM ax2+bx+c

Care should be taken to express the trinomial in descending order of powers.

ax2 + bx + c is a second degree trinomial. It may be of two forms.

i) Perfect square form like a2 ± 2ab + b2

ii) Not a perfect square form

Test for Perfect Square Form:

i) Perfect Square Trinomials

In a2 ± 2ab + b2

The 1st term (T1) = a2

The 2nd term (T2) = ± 2ab

The 3rd Term (T3) = b2

(T2)2 = (± 2ab)2 = 4a2b2 = 4 × T1 × T3

Thus, we see that: (T2)2 = T1 × T3

If this relation exists among the terms then it is of perfect square form like a2 ± 2ab + b2 which gives (a ± b)2.

Thus, we get a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b)

And a2 − 2ab + b2 = (a − 1)2 = (a − b) (a − b).

Example 12: Factorize 4x2 − 12xy + 9y2.

Solution: In 4x2 − 12xy + 9y2.

T1 = 4x2, T2 = − 12xy, T3 = 9y2

4 × T1 × T3 = 4 × 4x2 × 9y2 = 144x2y2 = (− 12)2 = (T2)2

Thus, (T2)2 = 4 × T1 × T3

∴ the polynomial can be expressed in the form a2 − 2ab + b2

4x2 − 12xy + 9y2 = (2x)2 – 2 × 2x × 3y + (3y)2

= (2x − 3y)2 = (2x − 3y) (2x − 3y)

Example 13: Factorize 25x3y – 20x2y + 4xy.

Note: Recall that 1st process of factorization is taking common. We find that ‘xy’ is a common factor in all the 3 terms. So first we take common.

Solution: 25x3y – 20x2y + 4xy = xy (25x2 – 20x + 4)

= xy {(5x)2 – 2 × 5x × 2 + (2)2}

= xy (5x – 2)2

= xy (5x − 2) (5x − 2)

Note: After ‘xy’ was taken common we could check that the terms in the bracketed factor stands the test (T2)

2 = 4 × T1 × T3 in order to cast it in the form of a2 − 2ab+b2.

ii) Factorization of Polynomials in the Form of ax2 + bx + c (which is not a perfect square) i.e. in which (T2)

2 ≠ 4 × T1 × T2

As we have already seen factorization is the reverse process of multiplication.

Let us consider the multiplication of (x + a) with (x + b).

(x + a) (x + b) = x (x + b) + a (x + b)

= x2 + bx + ax + ab

= x2 + ax + bx + ab

= x2+ (a + b) x + ab

∴ We can say that: x2 + (a + b)x + ab = (x + a) (x + b)

In this factorization can you observe any relationship between the coefficient of the second term and the third term?

If we want to resolve a polynomial of the form into two x2 + bx + c into two factors, say x + m and x + n, then necessarily, m + n = b and m × n = c as we have seen from the above deduction. Using this relationship we can factorize this type of polynomials/expressions.

Example 14: Factorize x2 + 5x + 6.

Here the coefficient of the second term (T2) is +5, and the third term (T3) is +6.

Then we have to find out a pair of numbers which when added gives +5 and when multiplied gives +6.

The pair of numbers are 2 and 3 or 3and 2. Taking the pair of numbers in any one order let us try to factorize the polynomial.

Solution: = x2 + (2 +3) x + (2 × 3) (Resolving Co-efficient of T2 and the third term (T3) into sum and product of 2 and 3)

= x2 + 2x + 3x + (2 × 3)

= (x2 + 2x) + {3x + (2 × 3)}

= x (x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

Hence, x2 + 5x + 6 = (x + 2) (x + 3).

Note: The important point in the factorization of this type of polynomial is to determine a pair of numbers which when added gives the coefficient of the second term and when multiplied gives the third term.

In the above example both the numbers had +ve sign. Let us factorize another polynomial of the same form.

Example 15: Factorize x2 − 7x + 10.

We have to find out a pair of numbers whose sum is −7 and whose product is +10.

It seems that the two numbers are 2 and 5 in any order. Will the two numbers have same sign or differing signs i.e. one positive and one negative? Clearly, here both the numbers have to be −ve because while the sum is negative, the product is positive. The product of two numbers with unequal signs can never be positive.

Solution: x2−7x+10 = x2 + {(-2) + (−5)} x + (− 2 × −5)}

= (x2 − 2 x) + {−5 x + (− 2) × (− 5)}

= x (x − 2) + (− 5) (x − 2)

= (x − 2) (x − 5)

Example 16: Factorize x2 − 4x − 12.

Note: We want two numbers whose sum is − 4 i.e. the co-efficient of the 2nd term, and whose product is −12 i.e. the 3rd term.

Solution:

To find the 2 numbers –

Ignoring the sign of the 3rd term we get 12.

We write all pairs of numbers whose product is 12.

Possible pairs are (1, 12), (2, 6) and (3,4)

Since the sign of the product (i.e. 3d term) is –ve then the two numbers cannot have the same sign. Again, since the sum of the two numbers is −4, then it clear that the two numbers are −6 and +2.

You can examine the suitability of the other two pairs of numbers fulfilling the two conditions.

Thus, x2 − 4x − 12 = x2 − 6x + 2x − 12

= (x2 − 6x) + (2x − 12)

= x (x − 6) + 2(x − 6)

= (x − 6) (x + 2)

Instead of breaking −4x as ‘− 6x + 2x’, if you write it as ‘+2x − 6x’.

Will you have any trouble in factorizing the polynomial?

Check Your Progress




(i) x2 + 12x + 32

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(ii) x2 − 8x – 48

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(iii) 35 − 12x + x2

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(iv) x2 − 10x + 24

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(v) (x − 5) (x + 2) − 30 (Hint: open the brackets)

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So far, we have been taking trinomials of the form ax2+bx+c where a = 1.

Now we will be taking trinomials of the form ax2 + bx + c where ‘a’ is some integer different from 1 or 0. Let us take an example.

Factorize 3x2 − 2x − 8.

Here for breaking the middle term, we need two numbers whose sum = coefficient of the middle term = −2. But, here the product of the two numbers = 3rd term × coefficient of the 1st term

= − 8 × 3 = − 24.

Procedure of factorization being same as followed in the previous example, we have to find two numbers differing in signs (since the product is −ve) whose sum is − 2.

Without considering the signs the possible pairs whose product is 24 are (1, 24), (2, 12), (3, 8), and (4, 6).

From this it becomes evident that the pair of numbers whose sum is −2 and product is −24 is (+4, −6).

Thus, 3x2 − 2x − 8 = 3x2 − 6x + 4x − 8

= (3x2 − 6x) + (4x − 8)

= x (x − 2) + 4 (x − 2)

= (x − 2) (3x + 4)

Check Your Progress



E11. Find a pair of numbers whose:

(i) Sum = − 5 and product = 4

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(ii) Sum = − 5 and product = − 24

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(iii) Sum = + 2 and product = − 48

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9.6 FACTORIZATION OF PERFECT SQUARE POLYNOMIALS OF SIX TERMS

We know that:

1. a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2

2. a2 + b2 + c2 + 2ab − 2bc − 2ca = (a + b − c)2

3. a2 + b2 + c2 − 2ab − 2bc + 2ca = (a − b + c)2

4. a2 + b2 + c2 − 2ab + 2bc − 2ca = (a − b −c)2

Hence, polynomials occurring in the above forms can also be factorized.

Example 17: Factorize: 9x2 + 4y2 + 1 + 12xy + 4y + 6x.

E12. Frame at least one polynomial for each pair of numbers as given in E11.

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(i) 2x2 + 7x + 6

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(ii) 6x2 + 2x – 8

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(iii) 4x2 − 11x + 6

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Solution: 9x2 + 4y2 + 1 + 12xy + 4y + 6x

= (3x)2 + (2y)2 + (1)2 + 2 × 3xy × 2y + 2 × 2y × 1 + 2 × 3x × 1 Occurs in the form 1

(3x + 2y + 1)2 = (3x + 2y + 1) (3x + 2y + 1)

9.7 UNIT SUMMARY

1. Factorizing a number means expressing it as the product of prime numbers.

2. Factorizing a polynomial means expressing it as the product of other polynomials.

3. After factorizing a polynomial into two factors, we must check if any of the factors can be further factorized.

4. Different processes of factorizations :

• For binomials, some standard forms of factorization are:

i) a2 − b2 = (a + b) (a − b)

ii) a3 − b3 = (a + b) (a2 ab + b2)

Check Your Progress




(i) 4x2 + 9y2 + 25z2 + 12xy − 30yz − 20zx

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(ii) 16x2 + 9y2 + z2 − 24xy − 6yz + 8zx

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iii) a3 − b3 = (a − b) (a2 + ab + b2)

• For trinomials in the form ax2+bx+c

iv) when the trinomial can be expressed in the form of a complete square of a binomial.

v) when it cannot be expressed as a complete square.

For polynomials having four terms : Grouping : (2+2) groups and (3+1) groups.

• For polynomials having 6 terms: Application of formulae.

vi) a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2

vii) a2 + b2 + c2 + 2ab − 2bc − 2ca = (a − b − c)2

viii) a2 + b2 + c2 − 2ab − 2bc + 2ca = (a − b + c)2

ix) a2 + b2 + c2 − 2ab + 2bc − 2ca = (a − b − c)2

9.8 GLOSSARY

Factor : When a number is expressed as the product of other numbers, each of these numbers is known as a factor.

If the factor is a prime number, it is known as a prime factor.

Multiple : If a number ‘a’ is a factor of a number ‘b’ then ‘b’ is known as a multiple of number ‘a’.


E1. {1, 2, 3, 5, 6, 10, 15, 30}

E2. 2, 3, 5

E3. Four

E4. Identify many.

E5. (i) 25 =5 × 5 (ii) 54 = 2 × 3 × 3 × 3

(iii) 12xy2 = 2 × 2 × 3 × x × y × y

E6. (i) 3 (x + 2) (ii) x (7x − 5) (iii) 2xy (3x + 2y)

E7. (i) (x + 3y) (x − 3y) (ii) (2x + 1) (2x − 1)

(iii) (1 + 3x2) (1 − 3x2) (iv) x (3x + 4y2) (3x − 4y2)

(v) (4x2 + 1) (2x + 1) (2x − 1)

E8. (i) (x +2) (x2 − 2x + 4) (ii) (3xy − 1) (9x2y2 − 3xy + 1)

(iii) (x2 + 2 + 2x) (x2 + 2 − 2x)

E9. (i) (3a + 4b) (1 + 5c) (ii) (x2 + 2) (x − 5)

(iii) (2m + 3) (5m + 3) (iv) (3x + 1 + 2y) (3x + 1 − 2y)

(v) (2z + 4x − y) (2z − 4x + y) E10. (i) (x + 4) (x + 8) (ii) (x − 12) (x + 4)

(iii) (5 − x) (7 − x) (iv) (x − 4) (x − 6)

(v) (x − 8) (x + 5)

E11. (i) − 1 and − 4 (ii) − 8 and + 3 (iii) + 8 and − 6

E13. (i) (2x + 3) (x + 2) (ii) 2 (x − 1) (3x + 4) (iii) (x − 2) (4x − 3)

E14. (i) (zx + 3y − 5z) (2x + 3y − 5z) (ii) (4x − 3y + z) (4x − 3y + z)

9.10 ASSIGNMENTS

Factorize the following:

Q1. 144 2 ++ xx

Q2. 4x4 + 1

Q3. 4224 5105 yyxx +−

Q4. 48116 x−

Q5. 54 3243 yyx −

Q6. 22 9)2(4 zyx −+

Q7. 22 )23()45( bayx −−−

Q8. 44 814 yx +

Q9. 12072 −− xx

Q10. 49429 2 +− xyx

Q11. 13 24 +− xx

Q12. 6xy − 15x + 8y − 20

Q13. 45281249 22 −+−− yxyx

Q14. xyxyyx 484211294964 22 +−−++

Q15. 222224 222 abxyabyxbayx −+−++

9.11 REFERENCES

• Textbook of Class V, VI, VII, VIII published by Gujarat State School Textbook Board, Gandhinagar.

• Textbook of Class VI, VII, VIII published by CBSE - New Delhi.

• http://www.inmymind.org/math/math-factorization.html

• http://www.en.wikiversity.org/wiki/primary-mathematics:methodforfactoring

• Textbooks of Class VI, VII, VIII published by NCERT.

• http://sitesgoogle.com/site/mrotmsted/home/factors

UNIT 10 ALGEBRAIC EQUATIONS

Structure

10.1 Introduction

10.2 Objectives

10.3 Concept of an Algebraic Equation

10.4 Solving Linear Equations

10.4.1 Properties of Equations

10.4.2 Solving Word Problems

10.4.3 Degree of an Equation

10.4.4 Linear Equations in Two Variables

10.4.5 Solution of Linear Equations in Two Variables

10.5 Unit Summary

10.6 Glossary


10.8 Assignments

10.9 References

10.1 INTRODUCTION

Searching for the unknown is one of man’s primordial instincts. All the branches of knowledge show diverse paths in man’s eternal quest to reveal the unknown. Mathematics, as one of the most systematic branch of knowledge, has developed tools to accurately reveal the unknown under varying conditions. One such tool is the algebraic equation.

Equations have infinitely wide range of utility. Mathematical equations are designed in appropriate ways for solving very simple daily life problems to accurately calculate complex phenomena of atoms and space travels.

We have already discussed the nature of algebraic expressions using letters instead of numerals for representing various statements. These expressions are in the form of polynomials with varying number of terms and degrees. The symbolic statement equating two polynomials under certain conditions becomes an equation. One equation may have one or (or, variable) several unknown entities (or, variable). Solving the equation means to determine the value(s) of the unknown(s).

In this unit the most basic form of algebraic equations, i.e. the linear equation has been extensively discussed as this constitute a part of the Mathematics curriculum of the upper primary classes.

10.2 OBJECTIVES


• write linear algebraic equations with the help of patterns; • solve linear algebraic equations in one variable; • solve real life problems using algebraic equations; • write linear equations in two variables; and • solve linear equations in two variables.

10.3 CONCEPT OF AN ALGEBRAIC EQUATION

When the measure of a certain quantity is unknown, but some relation connecting that measure is known, then by using a literal number to represent the unknown quantity and the relations available, a statement of equality can be prepared. This statement of equality is known as an equation.

To consolidate this idea, we can perform the following activities of developing patterns. Patterns are the best way to know how to use letters for numbers while trying to write algebraic equations.

Example 1: To find the pattern for even numbers

1 × 2 = 2, 2 is an even number



Μ


Μ

From observing these examples, we can say that when an integer is multiplied by 2, it yields an even number. In general, if ‘n’ is any integer, then 2n represents an even number.

In this context we may face a question like this: “Which integer multiplied by 2 gives the even number 132?” Most common method of arriving at the result is to divide the number by 2. But let us look into the other method.

If, suppose, the required number is x, then the above trend helps us to write a statement relating x with its corresponding even number 132. That is: “Twice of x is equal to 132.”

or symbolically, 2x = 132 (10.1)

This is one of the simple forms of algebraic equations. Similarly, patterns can be developed using match sticks. Match stick patterns can be used to help discovering ways to represent certain type of numbers in the form of shapes using match sticks. Example 2: Making simple patterns of match sticks.

Take two matchsticks and form ‘L’ from them as shown in the figure below,

(a) (b) (c)

We see that two matchsticks are needed to form one ‘L’. Another ‘L’ can be made by adding two more matchsticks. Proceeding in this manner we can complete the following table:

Table 10.1: No. of patterns with no. of match sticks required

Number of L’s formed

1 2 3 4 5 6 7 8 − x n

Number of matchsticks required

2 4 6 8 10 12 14 16 − 58 2n

It can be seen from the table above that the number match sticks required to make a pattern containing ‘n’ number of L’s = 2n. There are 58 match sticks and we want to know how many L's can be made using all the 58 match sticks.

If, suppose, x number of L's can be made out of these match sticks. Then from the pattern observed from the table we know that 2 x numbers of match sticks are required. Now we have the relation i.e.

2 x = 58 (10.2)

This is an equation with ‘x’ representing the number of L’s.

Example 3: Another matchstick pattern is:

1 2 3 4

What we see in the pattern above can be put into the following table.

Table 10.2: No. of closed figures with no. of match sticks required

Sl. No.

No. of closed figures in the shape

No. of matchsticks and the pattern

1. 1 6 = 1 × 5 + 1

2. 2 11 = 2 × 5 + 1

3. 3 16 = 3 × 5 + 1

4. 4 21 = 4 × 5 + 1

- - -

x x × 5 + 1 = 5 x + 1

n n n × 5 + 1 = 5n × 1

Here also ‘n’ is the variable and can take any value from 1,2,3,4, ……… so on.

In a certain shape of the above pattern 61 match sticks are used and we want to know the number of closed figures as shown in the pattern.

If there are x number of closed figures in the particular shape using 61 match sticks, then we get the relation between x and 61 to be

5x + 1 = 61 (10.3)

It is an equation with ‘x’ representing the unknown.

We have got three equations from the 3 examples discussed before. Those are:

i) 2x = 132 (where ‘x’ represents an unknown integer).

ii) 2x = 58 (where ‘x’ represents the number of L’s contained in the shape).

iii) 5x + 1 = 61 (where ‘x’ represents the number of closed figures in a shape).

Each of them is known as a linear or first degree or simple equation.

10.4 SOLVING LINEAR EQUATIONS

In the Equation (3.1) given above, 2x = 132, what is the value of x? It is given that ‘x’ is an integer. So we go on trying with each of the integers 0, 1, 2…. for ‘n’ till we get the value of 2x = 132. It can be seen by trial that when x = 61 only at that point 2x = 132. For no other value of x the equation holds good.

Let A be 5 years younger than B. If B’s age is x years, then from the given condition A’s age is (x – 5) years. If A is 10 years old, then the whole relation can be expressed in the following manner.

i.e., x – 5 = 10 (10.4)

This is an equation in the variable x. This equation (10.4) is satisfied for what value / values of x? This can be found out by taking different values of x. Taking x = 0, 1, 2,

Check Your Progress



E1. For the given series 1, 3, 5, 7, ……………… find

(i) General rule (ii) nth term (iii) 50th term

................................................................................................................

................................................................................................................

................................................................................................................

E2. Complete the table using the given formula n = 7m – 3 where ‘m’ would be given values 1, 2, 3…… to 10.

................................................................................................................

................................................................................................................

................................................................................................................

………. we will see that the above equation (10.4) is satisfied for only one value of x, i.e., for x = 15.

That is if B’s age is 15, than A’s age = 15 – 5 = 10, which is true?

For no other value of x, the above equation is satisfied.

In the following equations the variables are identified while the rest are numbers and constant:

a) x + 20 = 50 (x is the variable)

b) n – 5 = 20 (n is the variable)

c) 25 m = 100 (m is the variable)

d) 8y

= 5 (y is the variable)

• An equation is a condition of equality imposed on a variable.

While 5y + 4 = 14 is an equation marked by an equality condition that 4 added to 5 times y is equal to 14, neither 5y + 4 > 14 nor 5y + 4 < 14 is an equation as the conditions are not of equality.

It can be easily observed that in the middle of each equation there is the equality sign (=) and there are algebraic expressions (polynomials) to its left and right. In the equation 4 x + 5 = 10 – x.

The left hand side (or commonly written as L.H.S.) = 4 x + 5, and

The right hand side (or commonly written as R.H.S.) = 10 – x.

• The equation (linear) is satisfied only for a definite value of the variable.

For example, the Equation (10.1): 2x = 132 is satisfied only by the value 61 of the variable x. Similarly, the equation x – 5 = 10 is satisfied only by the value 15 of the variable x.

Can all the statements showing equality relations be called equations?

Let us examine the following statement of equality

2 (x − 3) = 2 x – 6

an algebraic equation involves variable (unknown entity) and constants, either numerals or literal numbers.

Try taking 1, 2, 3, 4 …….. for x which value of x satisfies the statement of equality? You can see that every value of x, satisfies the equation. In such a case, the statement showing equality is not an equation. It is an identity. Some examples of familiar algebraic identities are:

i) (a + b)2 = a2 + 2ab + b2

ii) (a − b)2 = a2 − 2ab + b2

iii) a2 − b2 = (a + b) (a − b)

iv) (a + b)3 = a3 + 3a2b +3ab2 + b3

v) (a + b)3 = a3 + 3a2b +3ab2 + b3

vi) (a + b)3 = a3 + 3a2b +3ab2 + b3

vii) (a − b)3 = a3 − 3a2b +3ab2 − b3

Check Your Progress



E3. 5 times a number added to 10 gives 40 as result. Write the equation.

................................................................................................................

E4. After 15 years, Sheela’s age will become four times that of her present age.

................................................................................................................

E5. Pick out the solution / solutions of equations given below from the values given in the bracket at the right of each equation.

(a) p – 5 = 5 (0, 10, 5, –5)

(b) 5m = 60n (10, 5, 12, 15)

(c) 72

=q (7, 2, 10, 14)

................................................................................................................

E6. Complete the table and by inspection of the table find the solution to the equation m + 10 = 16

m 1 2 3 4 5 6 7 8 9 10

m+10

10.4.1 Properties of Equations

The process that we have followed above for solving the equation is known as process of trial . The process is quite tedious and time consuming. As you have seen that after 60 unsuccessful trials (trying the numbers 1 to 60) we got the successful trial i.e. we could get the number 61 which satisfies the equation. In order to make the process of solving the equation we need to understand the basic rules/properties of equations.

The physical representation of the concept of an equation is a beam balance having two pans on its either sides − some material on one pan and the weight on the other such that the beam remains horizontal indicating that the weights on both the pans are equal.

Supposing the weight on the left pan is ‘a’ and the weight on the right pan is ‘b’. In the balanced condition we have a = b.

i) If we put a weight ‘c’ on both the pans in what state will be beam remain? Horizontal or tilted to one side?

Surely horizontal.

Thus we get: a + c = b + c

ii) In the original state of balance, that is when we had a = b.

Had we taken away weight ‘c’ from both sides, in what state would be beam remain?

Definitely horizontal.

Thus we get: a – c = b – c

iii) Again as we were in the original state of balance, had we made the material on the left pan 5 times and the weight on the right pan also 5 times, the beam would still remain horizontal.

Thus we get: 5a = 5b

Similarly a × c = b × c

iv) In original state of balance, had we taken one-third of the materials on the left and one third of the weight on the right, what would happen?

Surely, the beam would remain balanced.

Thus we get: 3

b

3

a =

Similarly c

b

c

a = when c ≠0

Corresponding to these properties of weighing in a common balance, there are four rules of equality.

When a, b, c…. are all real numbers, and

i) if a = b, then adding the same number c to both the sides of the equation does not disturb the equality.

That is, a + c = b + c ……….. (A)

ii) Similarly, subtracting the same number c from both the sides of the equation a = b keeps the equality in tact.

That is, a – c = b – c ……….. (B)

iii) Multiplying a number c to both the sides of the equation a = b

we get a × c = b × c ……….. (C)

iv) Dividing both the sides of the equation a = b by a non-zero real number c, we get

c

b

c

a = (c ≠ 0) ………. (D)

These four rules help in solving linear equations in precise manner without resorting to any trial and error method which was quite lengthy and tedious.

Let us see how these rules are applied in solving the linear equations.

Example 4: Solve x – 3 = 8.

Note: Solving an equation means finding the value of ‘x’ and to do this we should have no number left on the left side of the equation.

Solution: x – 3 = 8

⇒ x – 3 + 3 = 8 + 3 ……… (Rule A)

(We choose 3 to be added because addition of 3 can cancel out – 3 from the left.)

⇒ x = 11 (Ans.)

Example 5: Solve x – 7 = 3.

Solution: x – 7 = 3

⇒ x − 7 + 7 = 3 + 7 ……… (Rule A)

⇒ x = 10 (Ans.)

Example 6: Solve 3 x + 2 = 14.

Note: There are 2 terms on the L.H.S. 3x and 2. Which is to be removed first?

From the LHS first the numerical term that has been added (i.e. 2 in this case) to the term containing the variable (i.e. 3x) is to be eliminated first. After the addition of 2 is neutralized, the multiplication of 3 is then neutralized.

Solution: 3x + 2 = 14

⇒ 3x + 2 − 2 = 14 – 2 ……… (Rule B)

⇒ 3 x = 12

⇒ 3

12

3

x3 = ……… (Rule D)

⇒ x = 4

In practice to be brief we do not show the step where the application of the rule is shown. Solution of Example 1 is shown below.

With application of Rule In practice

x – 3 = 8 x – 3 = 8

⇒ x – 3 + 3 = 8+3 ⇒ x = 8 + 3

The four rules of equality can otherwise be stated as the rule of transposition of terms in an equation.

Rules of Transposition

In an equation, in which a term is transposed from one side to the other side (from LHS to RHS or vice versa), the operation is reversed. That is on transposing a term from one side to another:

i) Addition changes to subtraction (y + 3 = 4 ⇒ y = 4 - 3 )

ii) Subtraction changes to addition ( p − 5 = 1 ⇒ p = 1 + 5)

iii) Multiplication changes to division (3x = 21 ⇒ x = 3

21 )

iv) Division changes to multiplication (32x

= 4 ⇒ 2x = 4×3)

Example 7: Solve 82

5x3 =−

Note: During finding out the value of 2

5x3 − with x being known, the order of working

out the operations are multiplication, subtraction and division. Hence, in finding

out the value of x with 2

5x3 − being known the order of reversing the operation

will be division by 2, subtraction of 5 and multiplication by 3.

Thus, we first remove 2, then 5 and then 3.

Solution: 2

53 −x = 8

⇒ 3x – 5 = 8 × 2 =16

⇒ 3 x = 16 + 5 = 21

⇒ x = 3

21= 7 (Ans.)

How to get rid of fractions while solving an equation?

It is convenient to solve an equation when it is free of fractions. If the equation involves fraction then how to remove the fraction without disturbing the equality.

If b

a=

d

cis an equality then how to make if fraction free? We apply Rule C in this case.

b

a =

d

c

⇒ b

a × bd =

d

c × bd (As we want the denominators ‘b’ and ‘d’ to be

cancelled we multiply both the sides by ‘bd’)

⇒ a × d = c × b

∴ Thus we find: b

a =

d

c results in a × d = c × b

i.e. denominator of the left gets multiplied by the numerator of the right and denominator of the right gets multiplied by the numerator of the left.

Thus, we get b

a =

d

c⇒ a ×d = c × b

This is known as the rule of cross multiplication.

Example 8: 2

1

5

32 +=− xx

Solution:

We cross multiply to make the equation fraction free.

Thus we get, (2x − 3) × 2 = (x + 1) × 5

⇒ 4 x − 6 = 5 x + 5

⇒ 4 x − 5 x = 5 + 6 (Terms containing unknown are brought to the left and the constant form taken to the right)

⇒ − x = 11

⇒ x = − 11 (Ans.)

Making use of the rule of cross multiplication, the equation of the type

,kdcd

bax =++

where a, b, c, d and k are numbers, and cx + d 0, can be reduced to the

form of a linear equation.

Example 9: 8

7

45=

− x

x

Solution:

The first phase of our work is to make it fraction free by using the rule of cross multiplication.

Thus: 8

7

45=

− x

x

⇒ x × 8 = 7 × (45 − x)

⇒ 8 x = 315 – 7 x

⇒ 8 x + 7 x = 315

⇒ 15 x = 315

⇒ x = 15

315= 21

Hence, the solution is correct.

Example 10: 7

12

32

25 =++

x

x

Note: The first phase of the work of solution is to make the equation fraction free. Then the systematic method of applied.

Solution: 7

12

32

25 =++

x

x

⇒ (5 x + 2) × 7 = 12 × (2 x +3) (by cross-multiplication)

⇒ 35x + 14 = 24 x + 36

⇒ 35x – 24 x = 36 – 14 (Terms containing unknown are brought to left and the constant form taken to the right)

⇒ 11x = 22

⇒ x = 11

22= 2

Verification of the Correctness of the answer:

If original equation is satisfied by the value of x obtained from the solution, the solution is correct. How to check?

LHS = 7

12

34

210

322

225

32

25 =++=

+×+×=

++

x

x and it is equal to the RHS.

∴ The answer is correct.

Check Your Progress



E7. (i) 9u = 81.

................................................................................................................

................................................................................................................

(ii) p − 20 = 30.

................................................................................................................

................................................................................................................

(iii)3y = 33.

................................................................................................................

................................................................................................................

(iv) 208

=k.

................................................................................................................

................................................................................................................

E8. (i) 2x − 3 = 7.

................................................................................................................

................................................................................................................

(ii) 2y + 9 = 4.

................................................................................................................

................................................................................................................

E9. (i) 2

3

2

5

3

−=+x.

................................................................................................................

................................................................................................................

(ii) 7

17

7

3 =+ x .

................................................................................................................

................................................................................................................

(iii) 183

2 =x.

..........................................................................................................

...............................................................................................................

(iv) 4035

=+ xx.

................................................................................................................

................................................................................................................

................................................................................................................

E10. (i) .3

4

63

8 −=+x

x

................................................................................................................

................................................................................................................

................................................................................................................

(ii) .8

5

73

32 =++

x

x

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

10.4.2 Solving Word Problems

The first step of solving word problems relating to various real life situation is to transfer the situation to a symbolic statement. The statement so obtained takes the shape of an equation.

The algorithm involved in solving a verbal/word problem is:

1. Understand the problem.

2. Check a variable for the unknown to be determined.

3. Identify the relation contained in the problem.

4. The relation can be used to write an equation.

5. Solve the mathematical equations obtained in step (4) using the rules for solving the equation.

6. Verify the correctness of the solution and interpret it in the context of step (1).

How does a relation give an equation? See the examples below.

i) Twice of a number is 12. The number being taken as x, the relation gives 2x = 12.

ii) 3 times a number (taken as x) is greater than half of it by 5. Thus we get,

3x −2

x = 5.

⇒ 2x = 64

We begin with a simple example.

Example 11: The sum of two numbers is 74. If one of the numbers is 10 more than the other then what are the numbers?

Solution: We have a puzzle here. We do not know either of the two numbers, we have to find them. We are given two conditions.

i) One of the numbers is 10 more than the other. ii) Their sum is 74.

If the smaller number is taken to be x, the larger number is 10 more than x, i.e. x + 10. The second condition gives us a relation that the sum of these two numbers x and x + 10 is 74.

Writing this relation in the form of an equation,

We get, x + (x + 10) = 74

⇒ 2 x + 10=74

⇒ 2 x = 74 – 10 (transposing 10 to RHS)

⇒ 2 x = 64.

Dividing both sides by 2, we get x = 32.

∴ The smaller of the two numbers is 32.

The other number is x + 10= 32 + 10 = 42.

The desired numbers are 32 and 42 for checking the correctness; we add 32 and 42 and see whether sum comes to be 74which has been given. Hence the answer obtained is correct.

Example 12: A is 24 years older than B. 10 years back; the age of A was 5 times the age of B. Find the present ages of A and B.

Solution: Let the present age of B be x years.

Then, age of A = (x + 24) years.

10 years back, age of A was x + 24 – 10 = x + 14

Age of B was x – 10

According to given condition

A’s age was 5 times B’s age

⇒ x + 14 = 5 (x – 10)

⇒ x + 14 = 5 x – 50

⇒ x – 5 x = – 50 – 14

⇒ – 4 x = – 64

⇒ x = 164

64 =−−

.

Therefore, present age of B = 16 years.

And, present age of A = x +24 = 16 + 24 = 40 years.

10.4.3 Degree of an Equation

Till now in this unit we have come across the equations of the type

ax + b = c …………… (i)

In addition we may also have the equations of the type

ax2 + bx + c=0 …………… (ii)

and ax3 + bx2 + cx + d=0 …………… (iii)

in all the equations above, a, b, c, d are all constants

Check Your Progress



E11. After 15 years, Sheela’s age will become four times that of her present age. What is her present age?

................................................................................................................

................................................................................................................

E12. An amount of money is divided between Rama and Reema such that Rama gets Rs.10 more than Reema. If the shares of Rama and Reema are in the ratio 5: 3, then find the total amount and the amounts each of them got.

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

E13. The length of a rectangular plot of land exceeds its breadth by 4 m. If the perimeter of the plot of land is 84 m, then find its length and breadth.

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

We see that in all the above equations the term containing highest exponent of x are different.

In equation (i), the highest exponent of the variable x is 1.

In equation (ii), the highest exponent of the variable x is 2

In equation (iii), the highest exponent of the variable x is 3.

For simplicity we can say that the highest exponent of the variable occurring in an equation is the degree of the algebraic equation.

Therefore, degree of equation (i) is 1. The equation whose degree is 1 are called linear equations or first degree equations.

Degree of equations (ii) is 2. The equations whose degree is two are called quadratic equations or second degree equations.

Degree of equations (iii) is 3. The equations whose degree is three are called cubic equations or third degree equation.

Types of Equations

Depending on their degree, the equations can be of different types. Some of the types and the general forms of equations in one variable are as follows: Type Degree General Form

1. Linear one ax + b = 0

2. Quadratic two ax2 + bx + c = 0

3. Cubic three ax3 + bx2 + cx + d = 0

4. Biquadratic four ax4 + bx2 + c = 0

5. Quartic four ax4 + bx3 + cx2 + dx + e = 0

(General Form) All the equations above are equations in one variable.

10.4.4 Linear Equations in Two Variables

A linear equation in one variable of the type 5x + 7 = 17 has x as the only variable in it. Thus the value of x is its solution. That is only one number can satisfy the equation when taken for ‘x’. An equation in two variables, such as y = 7x – 3 has many solutions. For instance, x = 2 and y = 11 is a solution of y = 7x – 3. We denote the solution as (2, 11) where the first number 2 in the parenthesis represents the value of x and the second number 11 in the parenthesis represents the value of y.

Thus, (1, 4) is a second solution of the equation. You may check whether x = 1 and y = 4 satisfy the given equation or not. We can get infinitely many solutions for an equation of this kind.

The solution of an equation in two variables is ordered pairs (x, y) which satisfy the given equation.

The equation given above is a linear equation in two variables as each of the two variables ‘x’ and ‘y’ has the exponent 1.

Even the equation 3x = 6 can be written in the form 3x + 0y = 6.

Thus, a linear equation in one variable can also be written as a linear equation in two variables.

What are then the solutions for it?

Treating the equation as 3x + 0.y = 6, the solutions are (2, −3), (2, −2), (2 − 1), (2, 0)…….. i.e. we take anything for y. You may check if the solutions noted above satisfy the given equation or not. Also you can check whether x can take any value other than 2 in the equation 3x + 0.y = 6.

10.4.5 Solution of Linear Equations in Two Variables

As we have already seen, one linear equation in two variables gives infinitely many solutions. Conventionally, ‘x’ is treated as the independent variable and ‘y’ is treated as the dependent variable in the equation. We are free to take any value for ‘x’ and correspondingly we get a value for ‘y’.

Thus, we can have all real numbers as the value of ‘x’ and the value of ‘y’ will also be all the real numbers.

If we have a second equation of this kind then we also get infinitely many solutions for it.

The two sets of solutions we get may have different possibilities.

a) There may be a unique common solution for both the equations.

b) There may not be any common solution in them.

c) All the solutions of one may be the same as the solutions of the other.

The following examples will make the above facts clear.

Example 13: Let us consider a pair of equations:

2x – 3y = 1 … (i)

and 3x + y = 7 … (ii)

Some of the solution of the Eq (i) are shown in the table below.

i) 2x – 3y = 1 ⇒ − 3y = 1 – 2x ⇒ 3y = 2x – 1 ⇒ y = 3

1x2 −.

Thus, ‘y’ is made the subject of the equation.

A.

x −3 −2 −1 0 1 2 3 4 5

y 16 13 10 7 4 1 −2 −5 −8

Some of the solutions of the Eq. (ii) are shown in the table below.

B.

x −3 −2 −1 0 1 2 3 4 5

y 16 13 10 7 4 1 −2 −5 −8

It can be seen that (2, 1) is a common solution for both the equation.

We may find out many other solution for each of the two equations. No common solution can be further available.

Thus the two equations (i) and (ii) have a unique common solution.

We say (2, 1) is the solution of the pair of simultaneous equations i.e. 2x – 3y = 1 and 3x + y = 7.

Example 14: Solve:

2x – 3y = 1 … (i)

4x – 6y = 2 … (ii)

Solution: Now to find the solutions of the 2 equations:

2x – 3y = 1 ⇒ − 3y = 1 – 2x

⇒ 3y = − 1 + 2x

⇒ y = 3

1x2 −

Some of the solutions are:

C.

x −3 −2 −1 0 1 2 3 4

Y 3

7− 3

5− −1 3

1− 3

1+ 1 3

5

3

7

4x – 6y = 2 ⇒ − 6y = 2 – 4x

⇒ 6y = − 2 + 4x

⇒ y = 6

14 −x

D.

x −3 −2 −1 0 1 2 3 4

Y 3

7− 3

5− −1 3

1− 3

1+ 1 3

5

3

7

It can be seen that all the solutions are common to both the equations.

Thus, those equations have no unique common solution.

Example 15: 2x – 3y = 1

4x – 6y = 2

Solution: For each (i), some of the solutions you have already got in table (A) and table (C) as well.

You try to find as many solutions for equation (ii). No common solution can be available. Let us try to find the conditions under which the 3 pair of equations behave differently.

1. Taking the 1st pair of equations:

i) 2x – 3y = 1⇒ 2x – 3y – 1 = 0

ii) 3x + y = 7 ⇒ 3x + y – 7 = 0

Assuming equation (i) to be in the form

a1x + b1y + c1 = 0

a1 = 2, b1 = +1, c1 = −7

It can be seen that:

3

2

a

a

2

1 = , 1

3

b

b

2

1

+−= = −3

Thus, 2

1

2

1

b

b

a

a ≠

So we conclude:

The pair of equation a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have unique

solution if 2

1

2

1b

b

a

a≠ .

Such a pair of equation are known as consistent and independent.

2. Taking the second pair of equations:

i) 2x – 3y = 1 ⇒ 2x – 3y – 1 = 0

ii) 4x – 6y = 2 ⇒ 4x – 6y – 2 = 0

Assuming equation (i) to be in the form a1x + b1y + c1 = 0.

We get a1 = 2, b1 = −3, c1 = −1

Assuming equation (iii) to be in the form a2x + b2y + c2 = 0

We get a2 = 2, b2 = −6, c2 = −2

2

1

4

2

a

a

2

1 == , 2

1

6

3

b

b

2

1 =−−= ,

2

1

2

1

c

c

2

1 =−−=

Thus we see that: 2

1

2

1

2

1

c

c

b

b

a

a ==

So we conclude that:

The pair of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 have infinitely

many solutions if 2

1

2

1

2

1

c

c

b

b

a

a == .

Taking the 3rd pair of equations:

i) 2x – 3y = 1 ⇒ 2x – 3y – 1 = 0

ii) 4x – 6y = 5 ⇒ 4x – 6y – 5 = 0

Assuming equation (i) to be in the form a1x + b1y + c1 = 0.

We get a1 = 2, b1= −3, c1 = −1.

Assuming equation (ii) to be in the form a2x + b2y + c2 = 0.

We get a2 = 4, b2= −6, c2= −5.

Thus 2

1

4

2

a

a

2

1 == , 2

1

6

3

b

b

2

1 =−−= ,

5

1

5

1

c

c

2

1 =−−=

It can now be seen that: 2

1

2

1

2

1

c

c

b

b

a

a ≠=

So we conclude that:

The pair of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 will have no

solutions if 2

1

2

1

2

1

c

c

b

b

a

a ≠= . Such a pair of equation are known as inconsistent.

Check Your Progress



E14. Find which of the following equations have x = 3 and y = 1 as a solution.

(i) 3x + y =10

................................................................................................................

................................................................................................................

................................................................................................................

(ii) 2x − 3y = 3

................................................................................................................

................................................................................................................

(iii) x + y − 4 = 0

................................................................................................................

(iv) 2x + 3y − 7 = 0

................................................................................................................

................................................................................................................

................................................................................................................

E15. Four linear equations are given below:

(i) x + y = 0 (ii) 2 (x−1) + 3y = 4 (iii) y = 0 (iv) x = 0

(a) Write at least four solutions for the linear equations (i) and (ii).

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

(b) Write equations (iii) and (iv) in two variables and then write 4 solutions for each of them.

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

E16. Identify which of the following pairs of equations give (a) unique solution (b) no solution (c) infinitely many solutions.

(i) 3x + y =7 and x − y = 2

................................................................................................................

................................................................................................................

................................................................................................................

10.5 UNIT SUMMARY

1. A statement showing equality of two algebraic expressions that contains some unknown (i.e. variable) is known as an equation.

a) If the exponent of the variable 1 at the greatest and is one variable only, it is known as linear equation in one variable.

b) I the exponent of each of the 2 variables contained in the equation is 1 at the greatest, thus it is known as the linear equations in 2 variables.

2. A linear equation is one variable can be solved by trial or by a systematic method.

3. A linear equation in 2 variables give infinitely many solutions.

4. A pair of linear equations in 2 variables (a) give unique solution if those are consistent and independent; (b) give infinitely making solutions if those are consistent and dependent (c) give no solution if those are inconsistent.

5. Word problems can be reduced to equations and solved.

10.6 GLOSSARY

Solution of an equation : The value of the unknown contained in the equation. A pair of equations in 2 variables are consistent and

independent if 2

1

2

1

b

b

a

a ≠ . A pair of equation in two

variables are consistent and dependent if 2

1

2

1

2

1

c

c

b

b

a

a == .

(ii) 4x − 2y = 5 and 2x − y =1

................................................................................................................

................................................................................................................

................................................................................................................

(iii) 2x + 3y = 1 and 6x + 9y = 3

................................................................................................................

................................................................................................................

................................................................................................................

Inconsistent : A pair of equations in 2 variables are inconsistent if

2

1

2

1

2

1

c

c

b

b

a

a ≠= .


E1. (i) 1 = 2 × 0 + 1

3 = 2 × 1 + 1

5 = 2 × 2 + 1

7 = 2 × 3 + 1

----------------

----------------

General rule is 2 m − 1 where m = 1, 2, -----

(ii) nth term is 2n − 1

(iii) 50th term is 99.

E2.

M 1 2 3 4 5 6 7 ---

n = 7m-3 4 11 18 25 32 39 46 ----

E3. 5x + 10 = 40.

E4. x + 15 = 4x

E5. (a) 10

(b) 12

(c) 14

E6. 11, 12, 13, 14, 15, 16, 17, 18, 19, -----

The solution is m = 6.

E7. (a) u = 9 (b) P = 50

(c) y = 11 (d) K = 160

E8. (a) x = 5 (b) y = 2

12− or – 2.5

E9. (a) x = −12 (b) x = 2

(c) x = 27 (b) x = 75

E10. (a) x = −2/3 (b) x = 11.

E11. Sheela’s present age is 5 years.

E12. Reema’s share Rs.15 and Rama’s share Rs.25.

E13. Length = 23m Breadth = 19m

E14. Only in equation (i)

E15. (a) (i) (1, −1), (2, −2) (3, −3), (4, −4)

(ii) (1, 3

4), (2,

3

2), (3, 0), (4, −

3

2)

(b) (iii) 0. x + y = 0 …. (−1, 0), (0, 0), (1, 0), (2, 0)….

(iv) x + 0.y = 0 …. (0, −1), (0, 0), (0, 1), (0, 2

1)…..

E16. (i) Unique solution

(ii) No Solution

(iii) Infinitely many solutions

10.8 ASSIGNMENTS

Q1. Which of the following is an equation?

(i) 2 (x + 3) = 3 (x – 1) + 9 – x

(ii) 3 (x – 1) = x + 2

Q2. Try with the values / ordered pairs given in the pair brackets and find the solutions of the equation.

(i) 2x – 6 = 0 (−1, 0, 2, 3, 4)

(ii) 3x – y = 8 [(−1, 5), (0, −8), (1, −3), (1, −5), (2, −2), (3, 1)]

Q3. Solve using the rules of equalities.

7

12

4

53 +=− xx

Q4. Solve: 8

5

5

12 =+−

x

x

Q5. By increasing the speed of cycling from 6 km per hour to 8 km per hour. Ramesh could cover a certain distance in 30 minutes less time. Taking the distance covered as x km, write an equation in x and solve it.

Q6. (a) construct 3 equations starting with x = 4 and (b) write an equation in the form ax + b = c whose solution is – 2.

Q7. Sheem bought some plane paper copies and some ruled paper copies. If the ruled paper copies are 6 more than twice the plane paper copies and the total number of copies bought are 30, find the number of ruled paper copies she bought.

10.9 REFERENCES

• Textbooks for Classes VI, VII and VIII published by NCERT in the year, 2006.

Block 4 GEOMETRICAL SHAPES AND FIGURES

UNIT 11

Introduction to Geometrical Figures and Shapes 81

UNIT 12

Construction of Geometrical Figures 111

UNIT 13

Perimeter, Area and Volume 151

UNIT 14

Symmetry 181

UNIT 11 INTRODUCTION TO GEOMETRICAL FIGURES AND SHAPES

Structure

11.1 Introduction

11.2 Objectives

11.3 Basic Geometrical Figures

11.3.1 Geometrical Figures in Our Surroundings

11.3.2 Fundamental Components of Geometrical Figures

11.3.3 Closed and Open Geometrical Figures

11.4 Geometrical Shapes of Three Dimensional Objects Found in Our Environment

11.5 Unit Summary

11.6 Glossary


11.8 Assignments

11.9 References

11.1 INTRODUCTION

Let us look at the familiar objects, animals and people around us. We can easily recognize the hills and plains, trees and plants, animals and birds, men, women and children that are around us. We can also differentiate between mango from banana, cats from dogs, cows from buffalos, one person from another. How can we do this? Colour, shape or form, size and weight, are some of the major distinguishing features by which we try to recognize them. Of all the characteristics, shape is the most crucial identifying feature. Every object or class of objects have a typical shape. For example, the shape of banana leaves are completely different from palm leaves or leaves of any other plant. Similarly, cows have unique forms that differentiate it from other animals.

Studying the fundamental shapes has engaged human intellect from the time immemorial. The branch of Mathematics that studies the characteristics of fundamental shape is called ‘Geometry’.

The word ‘Geometry’ is derived from the Greek word Geometron, where ‘Geo’ means earth and ‘metron’ means measurement. This indicates that Geometry probably developed due to the need to measure land, when the boundaries of agricultural land

were to be demarcated and re-demarked. To handle the problem of determining the shape of a field, suitable shapes were to be designed.

Study of Geometry started in India in connection with construction of altars vedi/ havan kunds. Some names of ancient Indian Mathematician who made important contribution to Geometry are Baudhayana (850 B.C.), Aryabhatta (47A.D.), Brahmgupta (598A.D.) and Bhaskaracharya (114-1185). Baudhayana had calculated and said that square of the diagonal of a rectangle is equal to sum of square of two sides of it. Later the same theorem was stated for a right angle triangle by Pythagoras (550 B.C.). We know a rectangle has two right-angled triangles in it. Hence, the theorem stated for a right-angled triangle is just a modified version of the theorem stated for a rectangle.

Successful effort for putting geometrical facts together in a logical and systematic manner was made by Euclid, a Greek Mathematician around 300 B.C. His work were published in thirteen volumes entitled ‘Elements’ which is considered not only as the most comprehensive and fundamental treatise in geometry, but also as one of the most influential books across the centuries.

11.2 OBJECTIVES


• explain what are geometrical figures and what are the basic components of them;

• describe the fundamental components of geometry - point, plane, line and line segment and curve;

• distinguish between open and closed geometrical figures;

• recognize and draw closed geometrical figures of specific shapes like, triangle, quadrilateral, rectangle, square and circle;

• identify the different shapes and explain the relationship between them; and

• differentiate between one dimensional, two dimensional and three dimensional shapes.

11.3 BASIC GEOMETRICAL FIGURES

The understanding of geometrical figures are essential to the study the shapes of various objects we see around. The idea in introducing geometrical figures is to make you aware of their utility in real life situations and their practical usefulness.

11.3.1 Geometrical Figures in Our Surroundings

We always observe and use different objects, in our daily life. Looking around, we find various shapes in nature, for example, a tree, a building, walls and floor, roof of a

building, coins, playground, a log of wood etc. Some of them form part of a flat surface and others form a part of the space. On closely observing them, some are found to be similar, others are found to be dissimilar. The boundary of some of the shapes are made of straight edges and others of curved. If we look at a building in the Figure 11.1, we will find shapes which have three and four straight edges. There can also be seen a shape whose boundary is curved.

Figure 11.1

In studying such objects, we come across, different geometrical shapes.

Activity 1: Given are four geometrical figures: , , , and .

Using these figures draw the pictures of at least four familiar objects. The sizes of the figures may be enlarged or minimized as per the requirement of the picture or the objects. For example see the picture of a house.

Figure 11.2

Check Your Progress



E1. List some of the common flat shapes which you are familiar with.

................................................................................................................

................................................................................................................

E2. Look at the cartoon in Figure 11.3. Which part of the cartoon has four straight edges? How many corners are there in that part?

Figure 11.3

................................................................................................................

................................................................................................................

................................................................................................................

E3. Which part of the cartoon has a curved boundary? How many corners does it have?

................................................................................................................

................................................................................................................

................................................................................................................

E4. Which part of the cartoon has 3 corners? How many straight edges are the in its boundary?

................................................................................................................

................................................................................................................

................................................................................................................

A

B

C

11.3.2 Fundamental Components of Geometrical Figures

The terms like point, line and line segment and plane are not new to you. They are building blocks of geometry. In this unit, we deal with these terms and their properties which will help us in understanding the concepts related to various geometrical facts.

Point: The term ‘point’ is undefined. But for practical purposes we put a very small visible dot to represent a point. A point is usually named by a capital letter like A, B, C etc. and read as point A, point B etc. so that one point can be easily distinguished from another (Figure 11.4).

Figure 11.4

• The points which lie on the same plane are called coplanar.

• Three or more points lying in a line are known as collinear. Points A, B, C shown above are non-collinear.

• A unique plane is determined by three non-collinear points.

• The intersection of two lines is a point.

Plane: For a person working in a field has different meanings for the term ‘plane’. A cultivator shows some of his agricultural land as plane. He only means the field is levelled and there is no ups and downs, humps and ditches in the field.

But in mathematics, the term ‘plane’ is undefined. The surfaces of a blackboard, a wall, a floor, a page of a book are examples of planes. From such examples of a plane, we can derive some of its characteristics.

The characteristics of a plane:

• A plane is collection of infinite number of points.

• When a line is drawn on it, every point on the line lies in the plane.

• If two points are taken on a plane, then the line joining them lies entirely in the plane.

• A plane has infinite extensions, thus plane can be extended in all directions infinitely.

Line: The term ‘line’ is also an undefined term. Some characteristics of a line which help us in working with it are the following:

• Two points determine a unique line.

• A line is a set of infinite number of points.

• Between any two points in a line, howsoever close they may be, there exists another point. This character expresses the fact that a line is continuous and straight in nature.

• A line has infinite extension in either direction. As such it does not have any finite length nor any end points.

• A line has one dimension only.

• If two lines lie in one plane, they are called coplanar.

• Two lines, not lying in a plane, are skew lines.

• Two coplanar lines may intersect each other or may not intersect each other.

• Two parallel lines are two lines in the same plane which never meet.

• Two non-parallel lines meet at one point only.

• We cannot draw a complete line on a sheet of paper.

• So, we draw a portion of a line and put arrow heads on its two ends which indicate the infiniteness of its extension. Figure 11.5 shows a line that contains two points A and B.

A B

Figure 11.5

This line is represented by the symbol AB.

• A line segment is a part of a line which has two fixed end point and a finite length.

• The line segment with A and B as end points is represented by the symbol AB .

Line segment: The line segment AB is the set of points A − B and all points between them. A and B are the end points of AB . The length of a line segment is the distance between its end points.

Activity 2: Mark two points A and B on a sheet of paper. Draw several lines through point A and several lines through point B (as many as you can).

Draw a line passing through point A and B.

Figure 11.6 Can you draw another line through A and B?

I am sure after all your efforts, you will say ‘no’. Thus, we observe that one and only one line passes through two different points. Draw four pairs of straight lines with the following characteristics:

i) Two lines which intersect each other.

ii) Two lines which when further extended to the right, intersect each other.

iii) Two lines which when extended to the left, intersect each other.

iv) Two lines which when extended on either side, do not meet.

Which pair of lines that you have drawn are parallel?

• When two lines l and m in a plane intersect each other at a point, the point is known as the point of intersection.

• If two lines do not intersect each other at any point, such lines are called parallel lines.

• Distance between two parallel lines at every part of them is the same.

• The pair of lines in (iv) you have drawn is a pair of parallel lines.

Concurrent lines: Three or more lines in a plane are said to be concurrent if all of them pass through the same point. The point is called the point of concurrence. P is the point of concurrence in Figure 11.7(i).

A

B

(i) (ii) Concurrent line Non-concurrent line Figure 11.7

Point of concurrence of three or more lines is also the point of intersection.

s

P

r n m

s

n

m

r

Check Your Progress



E5. From the Figure 11.8 find:

(i) Name the points of concurrence.

(ii) Name the points of intersection.

A B

C

D

E

F

t

k s

n

m

r

Figure 11.8

............................................................................................................

Example 1: Look at the Figure 11.9 and answer the questions given below:

(i) (ii) (iii)

Figure 11.9

i) How many lines are drawn through point P in Figure 11.9(i)? Can any more lines be drawn through?

ii) Name the line that contains both P and Q in Figure 11.9(ii).

iii) Which are collinear points in Figure 11.9(iii)? Solution: From figure we get the following:

i) Four lines are drawn passing through point P in Figure 11.9(i)? Yes more lines can be drawn.

ii) The line is PQ.

iii) P, R and T are collinear.

P Q

P P

R

S

Check Your Progress



E6. Draw as many lines as possible passing through at least 3 of the points given below. Indicate the points which are collinear.

A B C

F

G

E

D

................................................................................................................

................................................................................................................

Curve: ‘Curve’ is line which is not straight. You recognize figure of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10... and letter A, B, C, D, E, F, G...Z.

In the figure of numbers and letters some parts of certain numbers of the figure are straight and some parts are not straight. The part that is not straight is a curve. In the letter B, how many parts are straight and how many parts are curve?

Activity 3: Draw some lines on a paper without lifting the pencil from the paper and without using a ruler. This may result in a figure of the kind shown below:

(i) (ii) (iii) (iv) (v)

Figure 11.10

Some curves cross each other and some curves do not. If a curve does not cross itself then it is called simple curve.

Could you identify which are the simple curves in the figures above?

11.3.3 Closed and Open Geometrical Figures

A child of class-V knows all letters of the English alphabet.

From among the letters A, B, C, O, U, V it can be seen that C, U, V, W ... are open where as B, D, O are closed figures.

Some open and closed geometrical figures are shown below. Identify which are open and which are closed figures.

Figure 11.11(i)

Figure 11.11(ii)

What is that makes a figure open or closed?

In a figure in which the lines/curves of any diagram do not end at the starting point such a figure is called an open figure. You can easily verify that all the diagrams in Figure 11.11(i) are open figures.

Look at the diagrams in Figure 11.11(ii). If you start moving on the boundary lines/curve, can you reach at the starting point again? Try this with any of the five diagrams in Figure 11.11(ii). In each case, you will find that you reach the starting point without any break on the boundary line/curve. Such figures are called closed figures.

• Look at the figures shown below:

(i) (ii) (iii) (iv)

Figure 11.12 We find that all these figures are closed and made of line segments.

Now look at following figures:

(i) (ii) (iii) (iv) (v)

Figure 11.13

We find out that all these figures are made of line segments and are examples of open figures. Let us discuss the nature of some common yet basic open and closed geometric figures.

An angle is the most common and the simplest of the open geometric figures.

To understand the concept of an angle, first we have to understand what a ray is.

(i) (ii) (iii)

Figure 11.14

To get an idea of a ray of light, consider a beam of light emerging from a torch or a candle.

It is easy to observe that path traced by a light is a part of line which extends infinitely in one direction from a starting point.

To deal with geometrical shapes we should have an idea of how a ray of light and a line are related to each other. Look at Figure 11.14 (iii), the line AB has a point P in it. At P the line AB is divided into 2 parts and each part represents a ray. The two rays are PA andPB. P is the common end point.

Angle: Angle is one of the open geometrical figures. There are some common features we observe in all angles. Two different rays with a common end point form an angle.

You might have seen a box. At every corner of a box, we find three-edges meeting. Two of the three edges constitute an angle.

At the corner of each page of a book two edges meet constituting an angle.

An angle seen at the corner A of the box shown in the figure below.

Figure 11.15

How many angles are seen at the corner B in the Figure 11.15?

A P B

A

D C

B

E F

G

A

C

B B

C

B

D

E

A

A

(i) (ii) (iii)

Figure 11.16

(i) (ii)

Figure 11.17

Each of the angles shown above is made up of 2 rays having same initial point. Do the angles shown in the above figure look alike? Definitely not. The angles appear to be different. What is it due to? The reasons would be clear when we measure the angles.

• Terms related to an angle

Vertex: The common initial point of the rays forming an angle is its vertex. O (plural ‘vertices’) is the vertex of an angle made by rays OA and OB as shown in Figure 11.18.

Figure 11.18 Arms: The two rays that form the angle are known as the arms of the angle. OA and OB are the two arms of ∠AOB.

An angle ABC is, therefore, defined as the union of two rays BA and BC when A, B, C are non collinear points and symbolically written as ∠ABC.

Depending on the measure of the angle, those are classified into five types as shown in the following figure:

A

B O

(i) (ii) (iii)

(iv) (v)

(vi) (vii)

Figure 11.19

• If a ray takes half turn then that ray moves through an angle of 3600/2=1800. See Figure 11.19(i). Such an angle is formed by two opposite rays. An angle of measure 1800 is called a straight angle.

• If a ray takes a quarter turn (1/4 turn), that means the ray has moved through an angle of 3600/4=900. An angle of measure 900

is called a right angle [Figure 11.19(ii)].

• If a ray makes one full rotation then it comes back to its original position. If a complete turn into 360 equal parts then each part is called one degree and we adopt this as one basic unit of measurement of an angle and it is known as a degree. The unit ‘degree’ is denoted by a small circle. Thus the angle formed by one turn (complete rotation) = 3600.

• If a ray does not move at all then that ray has moved through an angle 00. An angle of measure 00 is called zero angle [See Figure 11.19(vii)].

• An angle which is greater than a right angle but less than 1800 is called an obtuse angle [See Figure 11.19(iii)].

A

A

A

B B B 0

0

A

B

B

0

0 A

A A

0 0 360o

A ’

• An acute angle is one whose measure is greater than 00 but less than 900 [See Figure 11.19(iv)].

• An angle whose measure is greater than a straight angle but less than 3600, is called a reflex angle [See Figure 11.19(v)].

Example 2: How many angles are shown in the Figure 11.20? Name them (exclude the reflex angle).

Figure 11.20

Solution

Six angles are formed namely ∠AOB, ∠BOC, ∠COD, ∠AOC, ∠BED, ∠AOD. On Measuring it can be seen that ∠AOB, ∠BOC, ∠COD, ∠BOD are acute angles, ∠AOC is a right angle, ∠AOD is obtuse angle.

0

D C B

A

Check Your Progress



E7. List the situation where an angle is found in your surrounding.

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

Triangle

Triangle is a closed figure formed by three line segments.

If A, B, C are three non-collinear points then the union of the three line segmentsAB , BC and CA is the triangle ABC.

In Figure 11.22(i), A, B, C are three points which do not lie in a line. Let us draw the line segments.

AB , BC andCA . We get the ∆ABC as in Figure 11.22(ii).

(i) (ii)

Figure 11.22

The line segment AB , BC and CA are known as the sides of the triangle ABC and the points A, B and C are known as its vertices (vertices is plural of vertex). A symbol ∆ is used to represent a triangle. Every triangle contains three angles known as the angles of the triangle. The angles of ∆ABC are ∠BAC, ∠ACB and ∠ABC, in short, ∠A, ∠C and ∠B. Triangle is one of the most common figures in geometry. It has wide application in

A

B C

A

B C

E8. Name the vertex and arms of the following angles:

A B

P

Q

N

M

R

S

0

0

0

0

(i) (ii) (iii) (iv)

Figure 11.21

................................................................................................................

................................................................................................................

art, architecture and engineering. Triangles are put into different categories basing on the relative lengths of their sides.

(i) (ii) (iii)

Figure 11.23

• A triangle in which no two sides are of equal length is known as a scalene triangle such as ∆ABC [Figure 11.23(i)].

• Triangle that has two sides of equal length called isosceles triangle, such as ∆PQR in Figure 11.23(ii).

• There are some triangles which have all three sides of equal lengths. Such triangles are known as equilateral triangles such as ∆MNP in Figure 11.23(iii).

Triangles are also classified according to the measurements of their angles.

• A triangle in which one angle is a right angle i.e. 900 is a right angled triangle.

• A triangle in which one angle is an obtuse angle is an obtuse angled triangle.

• A triangle in which all the three angles are acute angles is an acute angled triangle.

• The sum of all the three angles in a triangle is 1800.

(i) (ii) (iii) (iv)

Figure 11.24

Being a closed figure triangle has an interior and an exterior. The triangle divides all the points of the plane into three parts.

A

B C

P

Q R N

M

P

A A A A

B B B B C C C C

Interior Triangle Triangle Region

P P S R

T

R Q

M

• The part of plane which consists of all points such as P inside the triangle is called the interior of the triangle [Figure 11.24(i)].

• The part of plane which consists of all points such as R, S and T form the triangle itself [Figure 11.24(ii)].

• The interior of the triangle ABC together which the triangle itself is known as the triangular region ABC [Figure 11.24(iii)].

• The part of the plane which consist of all points such as M, N and Q which neither lie on the triangle nor in its interior is called the exterior of the triangle [Figure 11.24(iv)].

Quadrilateral

A quadrilateral is a closed geometrical figure often found around us. Sitting in a classroom of we observe, surface of table, the floor, the roof and the walls and doors of the room which are all quadrilaterals.

Some figures are given below. Identify which of them are closed figures of four line segment.

(i) (ii) (iii)

(iv) (v) (vi)

Figure 11.25

Figure 11.25(i), (ii) and (vi) are simple quadrilaterals formed by four line segments. Figure 11.25(iii) and (v) are not closed figures as there are two open ends of the curve

forming the figure. Figure 11.25 (iv) is a closed figure but not a quadrilateral because line segmentAB , CD intersect each other at points other than their end points and form two triangles.

If A, B, C and D are four co-planer points, such that no three of them lie in a line and no two of the line segments AB , BC , CD , DA intersect each other at any point other than their end points, then the closed figure formed by AB , BC , CD and DA is a quadrilateral named ABCD.

As in the case of triangles, the four points, A, B, C and D are known as the vertices and the line segments,AB , BC , CD and DA are known as four sides of a quadrilateral ABCD. The pair of sides which intersect each other are known as consecutive sides of the quadrilateral ABCD. The pair of sides that do not intersect entered are opposite sides.AB , CD are one pair of consecutive sides andAD , BC are one pair of opposite sides. Try to find out other pairs of consecutive sides and opposite sides.

A and D are successive vertices and A and C are opposite vertices. A quadrilateral has four angles in it such as ∠ABC, ∠BCD, ∠CDA, ∠DHE are the four angles of the quadrilateral ABCD.

The sum of the measures of the angles of every quadrilateral is 3600, no matter what is the shape of quadrilateral.

Example 3: Examples of the quadrilaterals with following characteristics are given below:

• Opposite sides are parallel [Figure 11.26(i)].

• All four angles are right angles [Figure 11.26(ii)].

• All the sides are of equal length and all angles are right angles [Figure 11.26(iii)].

• All four sides are of equal length and opposite sides are parallel [Figure 11.26(iv)].

• One pair of opposite sides are parallel [Figure 11.26(v)].

(i) (ii) (iii)

(iv) (v)

Figure 11.26

• Various kinds of quadrilaterals

Trapezium

A quadrilateral in which one pair of opposite sides are parallel is a trapezium [Figure 11.26(v) and 11.27(i)).

Parallelogram

Check Your Progress



E9. Measure of the angle indicated by ‘X’ in each of the following.

62o

110o

Xo

125o

74o

81o

55o

X

30o

(i) (ii)

Figure 1.27

................................................................................................................

................................................................................................................

A D

B C

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. If a quadrilateral ABCD is a parallelogram, then AB || DC and AD || BC should be parallel. The other properties of a parallelogram are:

i) Opposite sides are of equal length, and ii) Opposite angles are of equal measure.

Figure 11.28

AD and BC are of equal length.

AB and CD are of equal length.

∠A and ∠C are of equal measure.

∠B and ∠D are of equal measure.

Look around and you will find out objects which have some of the surface in the shape of a parallelograms.

Rhombus

A rhombus is a parallelogram with all its sides of equal length. But every parallelogram is not a rhombus [Figure 11.26(iii)].

The other properties of it are:

• Opposite sides are parallel, and • Opposite angles are of equal measure.

Thus, a rhombus is also a parallelogram with specific property that all its sides are of equal length.

Rectangle

A quadrilateral in which all angles are right angles is a rectangle. Can we also tell that a parallelogram with one of its angles is 900 is also a rectangle?

.

A

C D

B

Figure 1.29

Other properties of a rectangle are:

i) Opposite sides are parallel. Thus, AB || CD and AD || BC ii) Opposite sides are of equal length. Thus AB = CD, AD = BC.

Therefore, each rectangle is a parallelogram with the specification that all angles of it are right angles.

But is every parallelogram a rectangle?

Square

Figure 11.30

If all the four sides of a quadrilateral are of equal length and all angles are right angles, then it is a square. We can also define it as a rectangle in which all sides are of equal length. Thus all squares are rectangles. But every rectangle is not a square. The categories of quadrilaterals and relation between them can be visualized from the diagram below (Figure 11.31).

90o

A B

C D

Figure 11.31

• If one pair of consecutive sides with one common end points are of equal length and the other pair of consecutive sides are of equal length, then it is known as a Kite .

Quadrilateral

Trapezium

Parallelogram

Rectangle Rhombus

Square

Angles are right angled

All sides are equal

All sides are equal All angles are

right angles

Second paid of opposites sides are parallel

One pair of opposite sides are parallel

The quadrilateral seen in Figure 11.31(ii) has AD = DC and AB = BC. Thus, ABCD is a kite.

(i) (ii)

Figure 11.32

Polygons

The word polygon is derived from two Greek words ‘poly’ and ‘gon’. Poly means 'many' and gon means 'angle'. Polygons are given special names according to the number of sides they have.

Some figures are shown below, look at these figures. These include a triangle, a quadrilaterals, a pentagon (closed figure with five sides) etc.

(i) (ii) (iii) (iv)

Figure 11.33

• A closed plane figure made of a number of line segments is called a polygon.

• In general a polygon having ‘n’ number of sides is known as n-gon. The simplest polygon is the one which has three sides and it is a triangle. The names of some of the polygons according to the number of sides they have are listed below:

Polygons No. of Sides No. of Angles No. of Vertices

Triangle 3 3 3

A

A

B B

C C D

D

Quadrilateral 4 4 4

Pentagon 5 5 5

Hexagon 6 6 6

Heptagon 7 7 7

Property of the angles of a polygon

The sum of measures of all interior angles = (n−2) × 1800 where ‘n’ is the number of sides.

In a triangle, n = 3.

So the sum of all interior angles = (3−2) ×1800 = 1800

In a quadrilateral, n = 4.

So the sum of all interior angles = (4−2) × 1800 = 360.0

We use the round shape in many ways. It is easier to roll a heavy steel tube than to drag a much lighter cube or cuboids. A circle is a simple closed curve that is not a polygon.

Regular polygons

(i) (ii) (iii) (iv)

Figure 11.34

The Figure 11.34 shows 4 polygons each having the length of their sides equal and measures of their angles equal. Such polygons are known as regular polygons.

Angles of a regular polygon

Sum of all interior angles = (n−2) × 1800 (property of a polygon).

=> n × each interior angle of a regular polygon = (n−2) × 1800 (as all angles are equal in measure).

=> each interior angle = n

180)2n( 0×−

There is an interesting property of a polygon. As the number of sides of a regular polygon increases, the polygon looks more and more like a circle.

Circle

Circle is one of the most appealing geometrical figure we come across in our everyday life situation. It is used frequently in the designs of public buildings, gardens, machinery, vehicle, toys etc.

In our environment, you will find many things which are round like a wheel, a bangle, and finger ring etc.

A circle is a simple closed curve that is not a polygon. A circle is defined as a closed curve on a plane (flat surface) such that every point on the closed curve is at a fixed distance from a fixed point contained in the same plane. This fixed point is known as the centre of the circle.

(i) (ii) (iii) (iv)

Figure 11.35

• Chord: A chord of a circle is a line segment that joins two points of the circle. In Figure 11.35(iii), A and B are two points on the circle. Thus AB is a chord of the circle.

• Diameter: Diameter of a circle is twice the radius. It we represent a circle’s diameter by d and its radius by r then d = 2r.

A chord of a circle that passes through the centre is a diameter of the circle. In Figure 11.35(iv) AB is a diameter.

A

B

O A

B

C

r rO

P

rr

Q O P

Q

S

Check Your Progress



E10. In the figure below, one vertex of ∆ABC is the centre O of the circle ABC. Answer the following:

B

C O

Figure 1.36

(i) How many vertices of the triangle are on the circle?

(ii) What are BO and CO called?

(iii) What is BC called?

(iv) What kind of triangle is OBC? Why?

11. Indicate whether each of the following statements is true or false:

(i) Every diameter of a circle is also a chord.

(ii) Every chord of a circle is also a diameter.

(iii) Two chords of a circle will always intersect each other.

(iv) The diameter of a circle is twice its radius.

................................................................................................................

................................................................................................................

11.4 GEOMETRICAL SHAPES OF THREE DIMENSIONAL OBJECTS FOUND IN OUR ENVIRONMENT

In our study so far, we have been dealing with figures which are contained in a plane. These can easily be drawn on our notebook or on a black board.

We have learnt about some simple plane figures like triangle, rectangle, square etc. Such figures we consider to be of two dimensions as they extend in two directions, length-wise or breadth-wise.

Now we will consider some figures which do not lie in a plane.

Whenever we look around usually most of the objects that we see are bodies that occupy some part of the space, such as a building, a book, a table etc. We call those objects as solid objects or three dimensional (3D) objects. It is because of the fact that they have extensions in left to right direction, top to bottom direction and found to back direction.

Let us try to understand what is meant by dimension?

• From our understanding of dimension, we can say that an object which has extension in one way but has no breadth nor thickness, then it is one dimensional body, such as a line, a line segment and a ray.

• If an object has extensions in two ways only, then it is two dimensional body. All plane figures are two dimensional. Each surface of a box is of two dimensions.

• If an object has extensions in three directions, it is a body of three dimensions. A box, a ball, a piece of stone are of three dimensions.

Three dimensional shapes

Some solid shapes which we generally see in day-to-day life are shown in Figure 11.36.

There are three dimensions because of the fact that they have extensions in left to right direction, top to bottom direction and front to back direction. In other words these objects have length, breadth and height or thickness.

This ball is a sphere This ice cream is a cone This can is a cylinder

Figure 11.37

11.5 UNIT SUMMARY

In this unit we have focused on the following points:

1. The terms point, line and plane are undefined yet are regarded as the fundamental concepts of geometry along with the terms like line segment, curve etc.

2. The geometrical figures can either be open or closed. While angle is an example of an open figure, circle, triangle, quadrilaterals, polygons are examples of closed figures.

3. Angle defined as the union of two rays with one common initial point viz., vertex can be categorized according to its measurement. It may be a right angle, or an acute angle or an obtuse angle depending on whether its measure is equal to one right angle, less than one right angle or more than it respectively.

4. Triangles may be of various types depending upon the length of their sides (equilateral, isosceles, or scalene) or relative measures of angles (right angled, acute angled or obtuse angled).

This box is a cuboid This playing dice is a cube

This is the shape of a pyramid

Check Your Progress



E12. List names of five things which resemble a sphere, a cone, a cuboid, a cylinder or cube.

................................................................................................................

................................................................................................................

................................................................................................................

5. Quadrilaterals under different conditions may be of different types like trapezium, parallelogram, rectangle, rhombus or a square.

6. Closed figures constitute of line segments as their sides are called polygons. Instances of polygons are triangle, quadrilaterals, pentagons, hexagons etc.

7. When the sides of a polygon are of equal length and the measures of its angles are also equal, then the polygon is called a regular polygon.

8. Circle is the most complete closed plane figure. The circle and its interior together constitute the circular region.

9. Most of the objects that are around us are three dimensional i.e. having length, breadth, thickness or height.

11.6 GLOSSARY

Undefined terms in Geometry : ‘point’, ‘line’, ‘plane’ and ‘space’ are undefined terms in

Geometry.

Coplanar : Lying in a plane.

Collinear : Lying in a line.


E2. arms, no corner

Ex3. stomach eyes, no corner

E4. face

E5. (i) B and D (ii) A, C, E, and F.

E8. (i) O - OA, OB (ii) O, OP, O, OQ

(iii) O, OM, ON (iv) O, OR, OS

E9. (i) 630 (ii) 300

E10. (i) 2, (ii) each is a radius

(iii) chord (iv) isosceles because has two equal sides (OB=OC)

E11. (i) True (ii) False (iii) False (iv) True

11.8 ASSIGNMENTS

Q1. Name the initial point of the following rays.

(i) PQ (ii) CP (iii) YZ

Q2. (i) How many rays are seen in Figure 11.38? (ii) How many pairs of opposite rays are there in Figure 11.38?

Figure 11.38

Q3. Name the rays in figure shown below whose initial point is (i) O (ii) P.

Figure 11.39

Q4. How many angles are seen in Figure 11.40?

Figure 11.40

0

H

G F

D

C B

E

A

T P O Q

0

D C

B

A

Q5. Which of the following degree measures represent (i) acute angle (ii) right angle (iii) obtuse angle?

(i) 500 (ii) 1100 (iii) 750 (iv) 900 (v) 1350

Q6. Measures of three angles are given below:

State in which case the angle can be possible be those of a triangle.

(i) 300, 400, 1100 (ii) 590, 120, 1090

(iii) 450, 600, 730 (iv) 530, 330, 830

Q7. If in a triangle

(i) Two angles are 300 and 600. Find the third angle?

(ii) Two angles are 350 and 750. Find the third angle?

Q8. Find the third angle of a triangle of which two angles are 1040 and 300.

Q9. The radius of circle is 5cm. What is its diameter?

Q10. Write true or false for the following statements.

(i) Diameter is the longest chord of a circle.

(ii) All radii of a circle are equal.

(iii) All chords of a circle are equal.

(iv) All diameter of a circle are equal.

Q11. Classify the triangle as scalene, isosceles or equilateral if their sides are of lengths:

(i) 5cm, 5cm, 5cm.

(ii) 7cm, 12cm, 13cm

(iii) 4cm, 4cm, 5cm.

(iv) 5cm, 7cm, 9cm.

Q12. Classify the triangle as acute, obtuse or right angled, if the measures of their angles are:

(i) 500 400 900 (ii) 1200, 300, 300 (iii) 700, 600, 600

Q13. State whether the following statements are true or false.

(i) Every rectangle is square.

(ii) Every square is parallelogram

(iii) Every parallelogram is square.

Q14. Name the types of the following triangles:

(i) Triangle with length of sides 7cm, 8cm and 9cm.

(ii) PQR such that PQ = QR = PR = 6cm.

(iii) ABC with AB = 8cm. AC = 7cm. BC = 7cm.

Q15. Match the following:

Particulars in a Triangle Type of Triangle

(i) 3 sides of equal length (a) Scalene triangle

(ii) 2 sides of equal length (b) isosceles triangle

(iii) No 2 sides are of equal length (c) Acute angled triangle

(iv) All three angles are acute (d) Right angled triangle

(v) One angle is a right angle (e) Equilateral

11.9 REFERENCES

• NCERT books for class 6, 7, 8.

UNIT 12 CONSTRUCTION OF GEOMETRICAL FIGURES

Structure

12.1 Introduction

12.2 Objectives

12.3 Construction of Lines and Angles

12.3.1 Drawing a Line Segment of a Given Length

12.3.2 Drawing the Perpendicular Bi-sector of a Given Line Segment

12.3.3 Bisection of a Given Angle

12.3.4 To Copy a Given Angle

12.3.5 Construction of an Angle of Measure 600, its Multiples, Sub- multiples and their Combinations

12.3.6 Construction of a Line Parallel to a Given Line

12.3.7 Drawing Perpendicular to a Given Line

12.3.8 Division of a Given Line Segment into Different Number of Equal Parts

12.4 Construction of Triangles

12.4.1 Construction of Triangles When Lengths of 3 Sides are Given (S-S-S Type)

12.4.2 Construction of Triangles When Lengths of Two Sides and the Measure of the included Angle are Given (S-S-A Type)

12.4.3 Construction of Triangles When Two Sides and an Angle Opposite to One of them are Given (S-S-A Type)

12.4.4 Construction of Triangles When the Measures of Two Angles and Length of the Included Side are Given (A-S-A Type)

12.4.5 Construction of Triangles When Length of One Side and Measures of Two Angles (One adjacent and the other opposite it) are Given (S-A-A Type)

12.5 Construction of Quadrilaterals

12.5.1 Construction of Quadrilaterals When the Lengths of Four Sides and on Diagonal are Given

12.5.2 Construction of Quadrilaterals When the Lengths of Three Sides and Two Diagonals are Given

12.5.3 Construction of Quadrilaterals When Lengths of Four Sides and Measure of One Angle are Given

12.5.4 Construction of Quadrilaterals When the Lengths of Three Sides and the Measures of Two Included Angles are Given

12.5.5 Construction of Quadrilaterals When the Measures of Three Angles and the Lengths of Two Included Sides are Given

12.5.6 Construction of Special Quadrilaterals

12.6 Unit Summary

12.7 Glossary


12.9 Assignments

12.10 References

12.1 INTRODUCTION

You may have observed various geometrical figures during the course of your study of geometry. Observing such figures, sometimes you may be interested in constructing such figures and for that purpose you have to know properties of those figures and use of geometrical instruments. In this unit we have discussed the algorithm i.e. the sequential steps of construction that can be used to help learners construct geometrical figures correctly and accurately. Once you acquire the skill of constructing simple geometrical figures, it will help you in drawing further complex figures.

12.2 OBJECTIVES

After going through this unit, you will be able to:

• know the use of various geometrical instruments; • explain the algorithm i.e. sequence of construction of different geometrical

figures; and

• construct different plane geometrical figures like circles, line segments, angles of specific measures, triangles and quadrilaterals using ruler and compasses.

12.3 CONSTRUCTION OF LINES AND ANGLES

For geometrical construction the ruler and the compass are only two geometrical instruments recommended by Euclid. So the construction which is done with the help of ruler and compass is called Euclidean construction. Protractor and setsquares can also be used for construction. But for accuracy and geometrical logic ruler and compass are recommended by Euclid.

Before making use of the drawing instruments, each of the drawing instruments such as a scale, a pair of compass, setsquares, divider and protector should be described properly. What kind of drawing can be made using them and how to handle them should be discussed in detail and demonstrated on the blackboard by the teacher.

The teacher should try to help the children in developing the correct concepts regarding various parts of different geometrical figure. He/she should discuss the algorithm of construction of each geometrical figure. Then s/he has to demonstrate the method of construction to the learners on the blackboard with the help of geometrical instruments. After the demonstration the students will be asked to do the construction in their notebook following the steps of the algorithm. Of course the teacher is required to supervise their activities and give provide appropriate support.

• Use of Compasses

A pair of compass is used to mark the length of a given measure and to draw arcs and circles. The pencil which is inserted in the compass should not be used to draw lines or curves and mark points. Check the screw of the compass before using it (Figure 12.1).

Figure 12.1

i) Construction of an arc

The arc of the circle is the collection of all the points lying at a given distance (r = radius) from a given point C (the centre). We can construct an arc of a given radius at a given point on a plane following the steps given below:

Step I: Set the compass in such a manner that the distance between the metallic point and the pencil point is equal to the desired radius (Figure 12.2).

Figure 12.2

Step II: Mark the point on your notebook which is required to be taken as the centre of the arc to be drawn. Name the point as ‘O’.

Step III: Place the metallic point of the compass at ‘O’ (Figure 12.3).

Figure 12.3

Step IV: Keep the pencil point on the paper. Rotate the arm of the compass carrying the pencil slowly to draw an arc (Figure 12.4).

O

Figure 12.4

We can draw our desired arc of a given radius.

ii) Construction of a circle

Use of an indigenous compass is felt when we need to draw a circle on the playground. We follow a process as shown in the Figure 12.5.

Figure 12.5

Process: AB is a rope. End A is kept fixed under the foot of the teacher/ child. A small stick is to be tied up at the other end B. A pupil/another child will hold the stick and move forward keeping the stick tightly close to the ground. After completion of one round of movement we will get a circle on the ground.

We can construct a circle of a given radius with the help of our compass. The steps to be followed are:

Step I: Open the legs of the compass for the required radius (Figure 12.6).

Check Your Progress



E1. How many arcs can you draw with the centre ‘O’?

................................................................................................................

................................................................................................................

................................................................................................................

AB

Figure

12.6

Step II: Mark a point to be taken as centre of your notebook with a sharp pencil and name it as ‘O’.

Step III: Place the metallic point of the compass on ‘O’ (Figure 12.7).

Step IV: Slowly turn the arm carrying around ‘O’ and complete the movement in one instant (Figure 12.8).

O

Figure 12.7 Figure 12.8

12.3.1 Drawing a Line Segment of a Given Length

A line segment is a part of a line having two end points. So it has a definite length.

We can draw a line segment of given length by using (i) scale (ii) ruler and compass. Let us go for drawing a line segment.

Work for you: To draw a line segment of length 5.2 cm.

i) Using scale

Follow the steps as given below:

Step I: Take the scale and place it on the page of your note book (Figure 12.9).

Figure 12.9

Step II: Mark two points on the paper that are 5.2 cm apart and are very close to the graduated edge of the scale. Name them as P and Q (Figure 12.10).

Check Your Progress



E2. Think and Say: How many circles you can draw with ‘O’ as centre and of

(i) a given radius?

(ii) Different radii (plural of radius)?

................................................................................................................

................................................................................................................

................................................................................................................

0 1 82 11 10 976543

0 1 82 11 10 976543

P Q

Figure 12.10

Step III: Join the points P and Q. Now PQ is the required line segment (Figure 12.11).

Figure 12.11

Note: Take care that:

ii) Using ruler and compasses

Step I: Draw a line L, mark a point on it and name it as ‘P’ (Figure 12.12).

Step II: Set your compass to have the radius equal to 5.2cm (Figure 12.13).

• While marking the points you should look vertically above the scale. • You do not look obliquely down the scale. (Show it through a drawing)

Q P

L

P

0 1 82 11 10 976543

Figure 12.13

Step III: draw an arc of radius 5.2 cm with ‘P’ as centre which cuts the line L at a point the eight of. Name the point as ‘Q (Figure 12.14).

Figure 12.14

Step IV: Draw another arc with Q as centre and having the same radius as before which cut the line L towards the left of ‘Q; you will see that the point of intersection coincides with P (Figure 12.15).

Figure 12.15 12.3.2 Drawing the Perpendicular Bisector of a Given Line Segment

Construction: To draw a perpendicular bisector of a given line segment AB.

Teacher’s Activity (For all following constructions)

• Demonstrates the construction on the blackboard using scale and compasses. • Moves step by step asking the students to observe and follow him/her step by step. • Divides the students into different groups of 5-6 students.

L P

Q

L

P Q

• Provides each group an instruction sheet in which the algorithm of construction is written.

• Asks the students to read the algorithm and discuss among themselves. • Asks them to do the construction work on their notebook following the steps of the

algorithm. • Supervises learners’ work and provides academic support. Checks the students’

work and corrects if necessary.

Read, Discuss and Construct

1. Construct the line segment AB of the given length with the help of a scale, pencil and compass (Figure 12.16).

Figure 12.16 2. With ‘A’ as centre and radius equal to more than half of AB draw two arcs using compasses on each side of AB (Figure 12.17).

Figure 12.17

A B

A B

Verification: The students will be asked to measure AR, RB and ∠PRB and will be asked to find out the relation between them and AB.

3. With ‘B’ as centre draw two more arcs with the same radius cutting the two former arcs (Figure 12.18).

Figure 12.18

4. Name the points of intersection of the arcs as P and Q.

5. Draw a straight line through points ‘P’ and ‘Q’ which cuts AB at R (Figure 12.19).

Figure 12.19

6. PQ is the required perpendicular bisector of AB .

B A

B A R

P

Q

12.3.3 Bisection of a Given Angle

Construction: To bisect a given angle ∠AOB.

Check Your Progress



E3. What would happen if the radius taken for constructing arcs in step 2 is less than half of AB?

................................................................................................................

................................................................................................................


1. Draw a ray OB .

2. Construct any angle ∠ AOB with the help of a scale and pencil (Figure 12.20).

A

B O

Figure 12.20

3. Taking ‘O’ as centre and with any suitable radius draw an arc which cuts arms OA and OB and name the points of intersections as ‘P’ and ‘Q’ (Figure 12.21).

A

B O P

Q

Figure 12.21

4. With P as centre and radius more than half of PQ, draw an arc in the interior of ∠AOB (Figure 12.22).

A

B O P

Q

Figure 12.22

Assignment: Asks the students to construct two or three angles and bisect each of them following the same steps.

12.3.4 To Copy a Given Angle

5. With Q as centre and with same radius as in step 3, draw another arc, which cuts the previous arc. The point of intersection be names as R (Figure 12.23).

A

B O

P

Q

Figure 12.23

6. Join points O and R and produce to get the ray OR . Now, OR is the bisector of ∠ AOB (How can you verify the result?) (Figure 12.24).

A

B O P

Q R

Figure 12.24

To copy a given angle on a given line at a given point on it we have to follow some steps as it would be clear from the following example.

Construction: Let ∠POQ be a given angle whose measure we do not know. ∠POQ is to be copied at A on AB . (A is the given point and AB is the given line)

12.3.5 Construction of an Angle of Measure 600, its Multiples, Sub- multiples and their Combinations


1. Draw any angle POQ∠ with a scale and pencil (Figure 12.25).

P

Q O

Figure 12.25

2. Draw a ray AB (Figure 12.26).

B A

Figure 12.26 3. With O as centre and with any suitable radius (say ‘r’) draw and arc to cut the arms OP and OQ. Name the points of intersection as X and Y (Figure 12.27).

P

Q O

Y

X Figure 12.27

4. Now taking A of AB as centre, draw an arc with radius equal to r to cut to the right of A at a point, say C (Figure 12.28).

B A C Figure 12.28

i) To construct angles of measures 600, 300 and 150

After teacher’s demonstration, the students may be asked to construct individually or in groups following the steps given below.


1. Draw a ray OB (Figure 12.31).

B O

Figure 12.31

5. With C as centre and a radius equal to XY, draw an arc to cut the arc drawn in step 4. Name the point of intersection as D (Figure 12.29).

B A C

D

Figure 12.29

6. Draw ray AD (Figure 12.30).

B A C

D

Figure 12.30

7. Now, let the students measure the angles ∠BAD and ∠POQ and find out the relation between the two measures. Each student may be asked to draw three different angles and construct three other angles of equal measure.

2. With O as centre and with any suitable radius draw an arc which cuts →

OB. The point of intersections be named as P (Figure 12.32).

P B O

Figure 12.32

3. With P as centre and with the same radius as in step 2, draw another arc which cuts the previous arc. Name the point of intersection as Q (Figure 12.33).

B O P

Q

Figure 12.33

4. Draw OA through Q. AOB∠ , so formed, is of measure ο60 (Figure 12.34).

B O

A

ο60

Figure 12.34

5. To construct an angle of measure ο30 bisect AOB∠ following the required

algorithm. Let OR be the bisecting ray of AOB∠ . Thus BOR∠ , so formed, as the

angle of measure ο30 . (Which other angle in the same figure has the measure ο30 (Figure 12.35)?

B O

A

ο30

R

Figure 12.35

6. To construct an angle of measure ο15 , bisect BOR∠ by OS . BOS∠ , so formed is

the angle of measure ο15 (Figure 12.36).

B O

A

ο15

R

S

Figure 12.36

From the above figure we have learned how to construct angles which are half, one-fourth or one-eighth of an angle measuring 600. We can also combine these small angles measuring parts of 600 so that several angles as discussed below can be constructed.

• Measure of ∠AOB = 600

• Measure of ∠BOR = ½ measure of ∠AOB = 600/2 = 300

• Measure of ∠BOS = ½ measure of ∠BOR = 300/2 = 150

• Measure of ∠AOS = 300 + 150= 450

• Angles of measure 7½0 and 22½0 can also be constructed, bisecting the angles of measure 150 and 450 respectively.

ii) Construction of angles of measure 1200, 900, 1050, 750 and 450

Teachers activity initiating a construction is to be followed by following activities by students in groups or individually.


1. Draw a ray AB (Figure 12.37).

Figure 12.37 2. Taking A as centre with a suitable radius, draw an arc which cuts AB and name the point of intersection as P (Figure 12.38).

Figure 12.38

3. With P as centre, draw an arc, with the same radius (as in step 2), which cuts the former arc and name of the point of intersection as Q (Figure 12.39).

Figure 12.39

4. With Q as centre, draw another arc of the same radius cutting the first arc and name the point of intersection as R.

5. Draw the ray AR . ∠BAR so formed is the angle of measure 1200 (Figure 12.40)

B A

B A

P B A

Q

Figure 12.40

6. Draw the ray AQ .

7. Bisect ∠QAR as follows:

Let AS be the bisector of ∠QAR.

∠BAS so formed is the angle of measure 900 (600+300) (Figure 12.41).

Figure 12.41 8. Bisect ∠RAS to get an angle of measure 1050 (900 + 150) (Figure 12.42).

Figure 12.42

P B A

Q

ο120

P B A

Q

ο90

R

S

P B A

Q

ο105

R

S T

12.3.6 Construction of a Line Parallel to a Given Line

i) Through an external point

Construction: XY be a given line and P be an external point. Construct a line

through P and parallel to XY .


1. Draw a straight line XY and mark a point P outside XY . Now XY is the given line and P is the given point (Figure 12.43).

Figure 12.43

2. Take any point Q on XY and draw QP (Figure 12.44).

Y X

P

Check Your Progress



E4. Construct the angles with following measures after writing the steps of construction:

(i) 72

1 0 (ii) 222

1 0 (iii) 1500 (iv) 972

1 0 (v) 1350

................................................................................................................

................................................................................................................

................................................................................................................

E5. State the measures of at least 3 angles (not discussed so far) which you can construct using ruler and compass. Give the steps of construction and construct those angles.

................................................................................................................

................................................................................................................

................................................................................................................

Figure 12.44

3. At P construct ∠QPB =∠XQP in such a way that B will be placed on the same

side of QP as of Y.

Now, PB is the required line through P parallel to XY (Figure 12.45).

Figure 12.45

Verification: Students may be asked to verify the parallelism of PB and XY using a pair of set squares.

Teacher’s Discussion: The teacher will discuss the following points with the students.

• Since ∠BPQ and ∠XQP are alternate angles and their measures are equal, hencePB

is parallel toXY . • Instead of drawing an angle alternate to ∠XQP, we could have drawn an angle

corresponding to ∠PQY at P.

ii) At a given distance

Construction: XY be a given straight line. A line is to be drawn parallel to

XY and at a given distance (say 4 cm) from it.


1. Draw a straight lineXY and P be any point on it. XY is the given line (Figure 12.46).

Figure 12.46

2. Draw PQ perpendicular to XY by constructing an angle of measure 900 at P (Figure 12.47).

Y X

P

Q

Y X P

B P

Y X

Q

Figure 12.47

3. From PQ cut PR of length 4 cm (Figure 12.48).

Figure 12.48

4. Through R draw ST perpendicular toPQ (Figure 12.49).

Figure 12.49

Thus ST is the required line parallel to XY at a distance of 4 cm from it.

Verification: Students may be asked to verify the parallelism of ST with XY using a pair of setsquares.

Y X P

Q

Y X P

Q R

Y X P

Q

R S T

Teacher’s Discussion: The following facts and points are to be discussed with the children.

• Measure of ∠YPR = measure of ∠QRT = 900. • Since ∠YPR and ∠QRT are corresponding angles and they are of equal measure,

hence ST is parallel toXY .

12.3.7 Drawing Perpendicular to a Given Line

i) At a point lying on it

Construction: AB is a given straight line and P be a point on it. Draw

perpendicular toAB at P.


1. Draw a straight lineAB and mark a point P on it (Figure 12.50).

Figure 12.50

2. With P as centre and with a suitable radius, draw two arcs which cutAB at Q and R such that P lies between Q and R (Figure 12.51).

Figure 12.51

3. With centre at Q and with radius more than half of QR draw an arc above AB .

Check Your Progress



E6. Construct a parallel line to a given line XY through an external point P at a distance of 6 cm from it.

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

B A P

B A P Q R

4. With centre at R and with same radius as in step –3, draw another arc which cuts the previous arc. Name the intersection point as S (Figure 12.52).

Figure 12.52

5. DrawPS .

6. Now PS is perpendicular to AB (Figure 12.53).

Figure 12.53

Verification: Students may be asked to measure ∠BPS.

ii) From a point not lying on it

Construction: AB is a given straight line and P be any point outside AB Draw

a perpendicular from P toAB .


1. Draw a straight lineAB and mark a point P outsideAB (Figure 12.54).

Figure 12.54

2. With P as centre and with a suitable radius draw an arc cuttingAB at two points and name them as Q and R (Figure 12.55).

B A P Q R

S

B A P Q R

S

B A

Figure 12.55

3. Taking Q and R as centre and with radius more than half of QR draw two arcs

which cut each other on the side of AB opposite to P. 4. Name the intersection point as S (Figure 12.56).

Figure 12.56

5. DrawPS which cutsAB and name the point as T (Figure 12.57).

Figure 12.57

Now PS is the required perpendicular toAB .

Verification: Ask the students to measure ∠PTB.

Discuss the following points with students:

• Measure of ∠PTB = 900, hence PS is perpendicular toAB .

B A Q R

B A Q R

S

B A

P

Q R

S

T

• The distance of point P from .PTAB = . • Point T is called as the foot of the perpendicularPT.

12.3.8 Division of a Given Line Segment into Different Number of Equal Parts

Construction: Divide a line segment PQ into three equal parts.


1. Draw a line segment PQ (Figure 12.58).

Figure 12.58

2. With P as centre draw an arc of suitable radius which cuts the line segment PQ at E (Figure 12.59).

Figure 12.59

3. With E as centre draw an arc which cuts the arc drawn in step 2, at point D. Draw ray PR to form angle QPD (Figure 12.60).

Figure 12.60

Check Your Progress



E7. Students may be asked to construct perpendiculars as (i) and (ii).

................................................................................................................

................................................................................................................

................................................................................................................

Q P

Q P E

Q P E

D

R

4. Taking Q as centre, draw an arc with the same radius (as in step 3) which cuts PQ

at F and extended towards the lower side ofPQ (Figure 12.61).

Figure 12.61

5. With centre F and radius equal to ED draw an arc which cuts the arc (drawn in step 4) and the point of intersection be named as G (Figure 12.62).

Figure 12.62 6. Draw the ray QG.

7. Taking P as centre and with any suitable radius draw 2 (3-1) small arcs successively which cut PR and name the points obtained as K and L (Figure 12.63).

Figure 12.63

8. With Q as centre and with the same radius (as in step 7) draw small arcs

successively which cuts QG and the points are named as M and N of intersection (Figure 12.64).

Q P E

D

R

F

G

K

L

P E

D

R

F

G Q

P E

D

R

F Q

Figure 12.64

9. Join the point K (1st point of PR) with N (last point of QG) and L (2nd point of

PR) with M (1st point of QG).

10. Name the intersection point of KN and LM with PQ as X and Y respectively (Figure 12.65).

Figure 12.65

Now, PX, XY and YQ are equal in length. Verification: The children will be asked to measurePX , XY andYQ . Discuss the following with the children. • PX = XY = YQ = 1/3 PQ.

Hence PQ is divided into 3 equal parts. • Following the same algorithm, we can divide a line segment into any number of

equal parts.

Q P E

D

R

F

G

K

L R

M

N

Q P E

D

R

F

G

K

L R

M

N

Y X

Check Your Progress



E8. Draw circles of radii 2.5 cm, 5 cm and 3.7 cm with centre at ‘O’.

................................................................................................................

................................................................................................................

................................................................................................................

E9. Draw a line segmentAB of length 4 cm. Taking A and B as centers draw two circles E1 with radius 2 cm. Whether the two circles intersect? If so, at how many points?

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

E10. Draw a lineXY . Construct AB of length 5 cm on it and BC of length 3 cm towards Y.

................................................................................................................

................................................................................................................

................................................................................................................

E11. Verify the limits of AC as the sum of length of BA and BC.

................................................................................................................

................................................................................................................

12.4 CONSTRUCTION OF TRIANGLES

A triangle has 3 sides and 3 angles. We need at least three independent measures to construct a triangle uniquely. The possible combinations for unique construction of a triangle are

• Lengths of three sides (S-S-S)

• Lengths of two sides and the measure of included angle (S-A-S)

• Lengths of two sides and measure of an opposite angle (S-S-A)

• Measures of two angles and the length of the included side (A-S-A)

• Length of one side and measures of any two angles (S-A-A)

You can observe from the above combinations that all possible combinations of lengths of three sides and measures of three angles are included except one, that the measures of the three angles (A-A-A). You can easily verify that when the measures of three angles are given, infinite number of triangles can be constructed proportionately varying the lengths of the sides. In other words, only the measures of three angles do not define a triangle uniquely.

Let us see how we can draw a triangle when the minimum requirements as given above are met.

E12. Construct the following angles using rule and compass as (a) 22.5°(b) 37.5°.

................................................................................................................

E13. Draw ∠QPR of measure 75° in which QR = 5 cm. Through R draw

RS parallel toPQ .

................................................................................................................

................................................................................................................

................................................................................................................

E14. Write the steps of the construction of an angle measuring 67.5°.

................................................................................................................

................................................................................................................

................................................................................................................

Before actually asking students to draw the triangle or any geometrical figure, it is always advisable to encourage students to analyze the problem individually or in groups and come out with the steps of construction. You may, then, summarize the steps with proper sequence; then the students can proceed to draw the figures individually. For practice, the students may be asked to frame problems, analyze them for solution and then actually draw the figures.

12.4.1 Construction of Triangle When Lengths of 3 Sides are Given (S-S-S Type)

Construction: Construct ∆ ABC where AB = 4 cm, BC = 6 cm and AC= 5cm.


1. Draw BC of length 6 cm (Figure 12.66).

Figure 12.66

2. Taking B as center and with radius 4 cm draw an arc on the upper half plane of BC (Figure 12.67).

Figure 12.67

3. Taking C as centre and with radius 5 cm, draw another arc which cuts the former arc and the point of intersection is named as A (Figure 12.68).

C B 6 cm

C B 6 cm

B 6 cm

A

Figure 12.68

4. Draw AB and AC (Figure 12.69).

Figure 12.69

Now ABC is the required triangle.

Note: The work can be started taking any one of the 3 sides first. 12.4.2 Construction of Triangles When Lengths of Two Sides and the Measure of

the included Angle are Given (S-S-A Type)

Construction: To construct ∆ PQR when PQ = 5 cm, QR = 7 cm, and measure of ∠PQR = 300.


1. Draw QR = 7 cm (Figure 12.70).

Figure 12.70

2. At Q on QR draw QX such that the measure of ∠XQR = 300 (Figure 12.71).

Figure 12.71

3. With Q as centre and radius 5 cm, draw an arc which cuts QX and the point intersection be marked as P (Figure 12.72).

C B 6 cm

5 cm 4 cm

A

R Q 7 cm

R Q 7 cm

X

ο30

Figure 12.72 4. Join PR. PQR is the required triangle (Figure 12.73).

Figure 12.73

Note: Instead of drawing QR first we can draw PQ first and follow the same algorithm for the completion of the construction.

12.4.3 Construction of Triangles When Two Sides and an Angle Opposite to One of Them are Given (S-S-A Type)

Construction: To construct ∆ XYZ whose XY = 6 cm, XZ = 4 cm and ∠XZY= 450.


1. Draw XZ = 4 cm (Figure 12.74).

Figure 12.74

2. At Z drawZP such that measure of ∠XZP = 450 (Figure 12.75).

R Q 7 cm

X

ο30

P

R Q 7 cm

X

ο30

P

5 cm

Z X 4 cm

Figure 12.75

3. Taking X as centre with radius 6 cm draw an arc which cuts ZP and the point of intersection is named as Y (Figure 12.76).

Figure 12.76

4. Join XY (Figure 12.77).

Z X 4 cm

P

ο45

ο45

Figure 12.77

Z X 4 cm

ο45

Z X 4 cm

P

ο45

Now XYZ is the required triangle.

Note:

• If we draw ZX first, it should be drawn in the same manner as it appears in the analysis figure.

• We could have drawnZY to start with. Try by yourself.

This construction may also be done in other methods. Ask the students to probe the alternative methods.

Note: While drawing constructions of triangle with such data (SSA), you will find in some cases, the arc drawn in step (iii) cuts ZPat 2 points giving two possible situation, for Y. If we name them as Y1 and Y2 then we will get two triangles XY1Z and XY2Z.

12.4.4 Construction of Triangles When the Measures of Two Angles and Length

of the Included Side are Given (A-S-A Type) Construction: Construct ∆ABC whose ∠ABC = 600, ∠ACB = 450 and BC = 6cm


1. Draw BC of length 6 cm (Figure 12.78).

Figure 12.78

2. On BC at B constructBX taking the measure ∠XBC = 600 (Figure 12.79).

Figure 12.79

3. At C draw CY such that ∠YCB measures 450 and CY intersects (Figure 12.80).

C B 6 cm

C B 6 cm

X

ο60

C B 6 cm

X

ο60

Y

4. Name the intersection point of BX and CY as A (Figure 12.81).

Figure 12.81

Now ∆ ABC is the required triangle.

The triangle can also be constructed by drawing either AC first or BA first. Ask the students to develop the steps of construction and draw the triangle.

12.4.5 Construction of Triangles When Length of One Side and Measures of Two Angles (One adjacent and the other opposite it) are Given (S-A-A Type)

Construction: Construct ∆PQR whose QR = 7 cm, measure of ∠PRQ = 750 and ∠QPR = 600


1. Calculate measure of ∠PQR = {1800 − (measure of ∠P + measure of ∠R)}.

2. Now we know the length of QR, and measures of ∠PQR and ∠PRQ. This becomes the construction of the form ASA (section 12.4b above) and the triangle can be constructed using the same algorithm of ASA construction (Figure 12.82).

C B 6 cm

X

ο60

Y

ο45

A

Figure 12.80

Figure 12.82

Note: Measure of ∠PQR can be determined using ruler and compasses too as follows:

• On XY , construct ∠YXO = ∠R, and ∠XYO = ∠P. • ∠XOY automatically become equal to ∠PQR.

Let us recapitulate

1. The fundamental triangles are of the following kinds:

(i) S-S-S (ii) S-A-S (iii) S-S-A (iv) A-S-A (v) S-A-A.

Construction of first four of them have independent processes whereas the 5th one is changed into ASA type after determining the unknown angle. The students should be made through with the algorithms of the first 4 types of triangles. These algorithms will also help to construct all types of quadrilaterals. No new algorithm would be needed.

2. For the construction of special triangles 3 measurements of less than 3 parts would be necessary which must include at least the length of one of its sides.

i) For isosceles triangle 2 measurements are necessary (not 2 angles).

ii) For equilateral triangle only one measurement is required (not one angle).

iii) For right angle triangle 2 measurements are required (but not 2 angles).

R Q 7 cm

P

ο45 ο75

ο60

12.5 CONSTRUCTION OF QUADRILATERALS

A quadrilateral has ten elements such as 4 sides, 4 angles and 2 diagonals. A unique convex quadrilateral (none of whose angles are of measure greater than 1800) can be drawn when any five independent measurements are given.

A quadrilateral can be divided into 2 triangles by drawing a diagonal. The measurements given make (i) both the triangles fundamental (SSS/SAS/SSA/ ASA/AAS) or (ii) one of the triangles fundamental.

Check Your Progress



E15. Construct an isosceles triangle ABC in which AB = AC, BC = 5 cm, and measure of angle ABC = 750.

................................................................................................................

................................................................................................................

................................................................................................................

E16. Construct a right angle triangle XYZ, right angled a Y, when XY = 4.8 cm, and YZ = 6.5 cm.

................................................................................................................

................................................................................................................

................................................................................................................

E17. Write the algorithm of construction of an isosceles right triangle PQR whose hypotenuse PR = 7 cm.

................................................................................................................

................................................................................................................

................................................................................................................

In case (i), we construct both triangles following the algorithm already discussed in the last section.

In case (ii), after completing the triangle which is fundamental, the other triangle becomes fundamental by itself and then it can be constructed.

12.5.1 Construction of Quadrilaterals When the Lengths of Four Sides and on Diagonal are Given

Construction: Construct a quadrilateral ABCD in which AB = 6 cm, BC = 6cm, CD = 4.5 cm, AD = 5.5 cm and AC = 8 cm (Figure 12.83).

Figure 12.83

The quadrilateral ABCD is divided into two triangles by the diagonal AC i.e. ∆ ABC and ∆ ACD. As per the given data both the triangles are of S-S-S type. Hence, the quadrilateral can be constructed by constructing the two triangles following the steps as elaborated in the previous section.

Note: We can start with either of the 5 line segments AB, BC, CD, DA, AC. Preferably we start with AC.

12.5.2 Construction of Quadrilaterals When the Lengths of Three Sides and Two Diagonals are Given

Construction: Construct a quadrilateral PQRS whose PQ = 4 cm, QR = 6cm, RS = 6.5 cm, PR = 7.5 cm and QS = 7 cm (Figure 12.84).

Figure 12.84

It can be seen in the figure that both the triangles PQR and SQR are of S-S-S type.

A

B

6cm

5.5 cm

8 cm

D

4.5 cm

C 6cm

P

Q

4cm

5.5 cm S

6.5 cm

R

7.5cm

7cm

i) Any one of the triangles can be constructed first following the appropriate algorithm.

ii) The other triangle can be constructed by following the appropriate algorithm.

12.5.3 Construction of Quadrilaterals When Lengths of Four Sides and Measure of One Angle are Given

Construction: Construct a quadrilateral KLMN whose

KL = 4 cm, LM = 6.5 cm, MN = 7.2 cm, KN = 5 cm and measure of ∠ LMN = 900 (Figure 12.85).

Figure 12.85

By drawing the diagonal LN, the quadrilateral is divided into 2 triangles i.e. ∆ LMN and ∆ KLN.

∆ LMN is of S-A-S type and can be constructed following the appropriate algorithm.

After constructing ∆ LMN, we see that ∆ KLN is now of S-S-S type. Completing the construction of ∆ KLN, we get the quadrilateral KLMN.

12.5.4 Construction of Quadrilaterals When the Lengths of Three Sides and the Measures of Two Included Angles are Given

Construction: Construct a quadrilateral ABCD in which

AB = 4.5 cm, BC = 6 cm, CD = 6.2 cm, measure of angle ABC = 750 and

measure of angle BCD = 600.

By drawing the diagonalBD , it can be seen that triangle BCD is of S-A-S type fundamental triangle. But triangle ABD does not have adequate data.

Thus, ∆ BCD has to be constructed first (Figure 12.86).

K

L

4cm

N

7.2 cm

M 900

6.5cm

5cm

Figure 12.86

After triangle BCD is completed, we can draw the angle ABC. Now for triangle ABD, BD is available and ∠ABD = ∠ABC − ∠CBD is also available. Hence, triangle ABD also becomes S-A-S type fundamental triangle. Now triangle ABD can be constructed following SAS algorithm.

Note:

i) We do not measure ∠DBC and subtract it from the measurement of ∠ABC. But after completing of triangle DBC, ∠CBA = 750 is drawn. Thus we got ∠ABD.

ii) While drawing triangle BCD, after getting points B and D, you may not drawBD . Drawing BD in the analysis diagram helped us to decide the procedure to be followed in constructing the quadrilateral.

12.5.5 Construction of Quadrilaterals When the Measures of Three Angles and the Lengths of Two Included Sides are Given

Construction: Construct a quadrilateral MNOP whose MN = 4.2 cm, NO = 5.5 cm measure of ∠PMN = 1050, ∠MNO = 750 and measure of ∠NOP = 600 (Figure 12.87)

Figure 12.87

Analysis: Draw the diagonalMO .

Now triangle MNO is found to be S-A-S type fundamental triangle. So, it can be constructed. After completing triangle MNO, MO, ∠NMO and ∠NOM are available.

Now we can get the measure of ∠PMO = ∠PMN − ∠NMO and measure of ∠POM = ∠NOP − ∠NOM.

Thus, triangle MOP becomes ASA type fundamental triangle and hence can be constructed.

A

B C

D

4.5cm

6cm

6.2cm

750 600

O 5.5m

750

P

600

1050 M

N

4.2cm

Note:

i) After triangle MNO is completed by drawing ∠NMP = 1050, AND ∠NOP = 600, we got the ∠OMP and ∠MOP.

ii) While constructing triangle MNO, after getting points M and O, MO may NOT be drawn. By constructing ∠NMP and ∠NOP, we get the point P.

12.5.6 Construction of Special Quadrilaterals

For the construction of special type of quadrilaterals less than 5 numbers of measurements may suffice.

For the construction of

i) a trapezium only 4 measurements are required. ii) a parallelogram only 3 measurements are required.

iii) a rhombus only 2 measurements are required.

iv) a rectangle only 2 measurements are required.

v) a square only 1 measurement is required.

Students may be asked to draw each of these special type of quadrilaterals selecting appropriate measures in each case.

12.6 UNIT SUMMARY

This unit has presented some algorithms for construction of different geometrical figures which a teacher should know and follow them while teaching for effective learning. The teacher should demonstrate the construction following different algorithms so that the pupils can observe and understand the steps of the construction. For the effective teaching-learning of the construction of geometrical figures the teacher has to follow the following steps:

• Write the problem (question) on the board. • Draw the analysis are given and show all given measurements by its side.

• Analyze the figure and discuss with the learners so that they can know and understand the algorithm of the construction.

Check Your Progress



E18. Construct a parallelogram KLMN whose KL = 5.2 cm, LM = 6.4 cm and LN = 7.3 cm.

................................................................................................................

................................................................................................................

................................................................................................................

E19. Construct a rectangle PQRS where PQ = 4.7 cm and QR = 7.1 cm.

................................................................................................................

................................................................................................................

................................................................................................................

E20. Construct a quadrilateral ABCD whose AB = 4.5 cm, BC = 3.5 cm, CD = 5 cm, ∠B = 450, and ∠C = 1500.

................................................................................................................

................................................................................................................

................................................................................................................

• Demonstrate the construction on the board and let all students observe it.

• Divide the students into small groups (5/6 pupils).

• Provide an Instruction Sheet showing due appropriate algorithm to each group.

• Give instruction to the learners to start the construction in their own notebook using ruler and compasses only.

• Supervise the students’ activities and give academic support to the students who need.

• Correction work should be done first by the group members and then by the teacher.

• Suggest different processes of the construction.

• Discuss some important facts and points on the construction with the students.

• Give some assignments to the learners for practice and reinforcement.

• Correction of the class and home assignments.

12.7 GLOSSARY

Fundamental Triangle : A triangle in which only 3 independent measure needed out of the given lengths of 3 sides and measures of 3 angles.


E2. (i) one (ii) innumerable.

12.9 ASSIGNMENTS

1. Measures of 3 sides are given below. Point out in which of the cases the construction of triangle is not possible.

(a) 7cm, 5cm, 6.2 cm (b) 6cm, 5cm, 11cm

(c) 5.2cm, 8.5cm, 5.2cm (d) 4cm, 10cm, 5cm

2. Construct an angle of 67.50 using ruler and compasses only.

3. Draw a line PQ and R is a point on it. Draw RS = 4.5 cm and perpendicular to PQ . Through S, draw a line parallel to PQ.

4. Construct an equilateral triangle ABC which side is 6.7 cm long. At a draw a perpendicular onBC .

5. Construct a rhombus PQRS each side of which is 5.2 cm and diagonal is 6.5 cm.

6. A quadrilateral KLMN has KL = 5.3 cm, LM = 6.7 cm, ∠K = 1050, ∠L = 850 and ∠K =500. Write the algorithm of the construction.

12.10 REFERENCES

1. Mathematics for class VII of CBSE and ICSE course.

2. Mathematics for class VIII of CBSE and ICSE course.

3. Mathematics for class IX of CBSE and ICSE course.

UNIT 13 PERIMETER, AREA AND VOLUME

Structure

13.1 Introduction

13.2 Objectives

13.3 Perimeter and Area of Closed Figures

13.3.1 Concept of Perimeter

13.3.2 Perimeter of Closed Figures

13.4 Area of Two Dimensional Figures

13.4.1 Concept of Area and its Use in Daily Life Situations

13.4.2 Determination of Area of Closed Figures

13.5 Surface Area and Volumes of Three-dimensional Bodies

13.5.1 Recognition of Three-dimensional Shapes

13.5.2 Calculating the Surface Area and Volumes of Specific Kinds of Three-dimensional Bodies

13.6 Unit Summary

13.7 Glossary


13.9 Assignments

13.10 References

13.1 INTRODUCTION

We all have houses, big or small and each house is situated in a certain premises. We always demarcate our premises so that it gets separated from the premises of others. To demarcate a certain premises, either we raise a fence or build a compound wall. To determine the materials necessary for raising the fence or building the compound wall, we need the length of the line of demarcation drawn all around the premises. Likewise, to build a house or planning a garden we need to know exactly the space we have and the amount of space we need for the purpose. For all these, we have to calculate the length of the boundary of the plot or premises and the space it occupies. In other words we have to determine the perimeter and the area of the plot. In addition to these two, we may also need to calculate the quantity of bricks that we need to construct the boundary wall or house. For this purpose we have to know how to calculate volume of a brick or the walls.

In this unit we are going to discuss the concepts of perimeter, area and volume and their application in our day-to-day life.

13.2 OBJECTIVES


• explain what is perimeter of a closed figure and knows the way to measure it; • explain the concept of area related to a closed geometrical figure;

• measure the area of closed plane figures using appropriate units of measuring area;

• distinguish a three dimensional body from a plane geometrical figure; and

• determine the surface area and volume of regular geometrical bodies.

13.3 PERIMETER AND AREA OF CLOSED FIGURES

13.3.1 Concept of Perimeter A piece of metallic wire is taken and is bent to form a certain shape as shown in the Figure 13.1 care being taken that both the ends are brought together. Put it on a piece of paper and draw the outline.

Then the wire is stretched and made straight and another shape is formed with both the end remaining together. Again, draw the outline of it on a piece of paper. Each figure has a boundary line. In Figure 13.1 the boundary line has straight parts. The lengths of the straight parts could be measured and the sum of their lengths gives us the length of the boundary line. But in case of Figure 13.2, the length of the boundary line cannot be measured so easily.

The boundary line in each case represents the wire. Hence, we could stretch the wire straight and measure it length. In this way we could determine the lengths of the boundary lines of the two figures.

Thus, we come to know that the length of the boundary line of a geometrical plane figure drawn on a plane is known as its perimeter.

We build a compound wall surround in our premises or we raise a fence all around to keep well demarcated from other’s premises. The length of the compound wall or the length of the fence is the same as the perimeter of the premises. Practically, everybody uses the concept of perimeter in his day to day line.

We can define perimeter of a closed figure as the measure of its boundary.

13.3.2 Perimeter of Closed Figures

Figure 13.1

Figure 13.2

A

B C

P

Q R

Let us take specific geometrical plane figures and determine their perimeter. a) Triangles

Let us consider how to determine the perimeter of a triangle which may be a scalene or an equilateral triangle.

i) The boundary line is ∆ ABC in Figure 13.3 has 3 parts. The parts are AB, BC and CA.

∴ The perimeter of ∆ ABC = AB + BC + CA (i.e. the sum of the lengths of 3 sides).

Thus, the perimeter of any triangle = sum of the lengths of its three sides. ii) Triangle PQR in Figure 13.4 is an equilateral triangle. Hence AB = BC = CA.

∴ Perimeter = AB + BC + CA

= AB + AB + AB = 3AB

Thus, perimeter of an equilateral triangle = 3×length of each side.

b) Perimeter of a rectangle

Let us take some matchsticks and arrange them in shape of rectangles as shown in Figure 13.5 below:

(i) (ii) (iii)

Figure 13.5

Look at the figures shown and complete the table below as done in respect of Figure 13.5(i).

Fig. No. of Sticks

along the longer sides

No. of sticks

along the shorter sides

Total No. of sticks

used

Length of back

stick

Length of the boundary line

(i) 2×2 = 4 2×1 = 2 4 + 2 = 6 4 cm 6×4 cm=24 cm

Figure 13.3

Figure 13.4

A

B C

D a

a a

a

(ii)

(iii)

The length of the boundary line in each figure gives the perimeter of it.

In a rectangle, the longer side is known as its length and let us denote it as ‘l’. Similarly, let us denote its breadth as ‘b’ as shown in Figure 13.6. Perimeter of the rectangle ABCD

= AB + BC + CD + DA = AB + DC + BC + AD = b+b+l+l = 2b + 2l = 2 (l+b)

Let us go back to the results obtained in the activity performed earlier.

In case of Figure 13.5(i)

Length (made up of 2 sticks) = 2 × 4cm = 8 cm Breadth (made up of 1 stick) = 1 ×4 cm = 4 cm

∴ Perimeter = 2 (l+b) = 2 (8cm + 4 cm) = 2 ×12 cm

= 24 cm

Thus, this result tallies with the result obtained earlier. c) Perimeter of a square

In a square, as you know, all sides are of equal length.

Let the length of each side be denoted as ‘a’, then The perimeter = sum of lengths of 4 sides = 4a

d) Perimeter of other rectilinear figures Fill in the table given below by taking the measures of sides in each Figure 13.8.

(i) (ii) (iii)

5cm

3cm 3cm

3cm 3cm

5cm

J

K

L M

N

P

5cm

3cm 3cm

5cm A

B C

D 5cm

2cm

3cm

2cm

5cm

E

F

G H

I

A

B C

D l

b

Figure 13.6

Figure 13.7

R U

V S

T

(iv)

Figure 13.8 In respect to Figure 13.8(iv), measure the sides using a scale and calculate its perimeter.

Figure Lengths of the sides Perimeter

(i) AB = CD = 3cm

BC = AD = 5cm

3cm + 3cm + 5cm + 5cm = 16cm

(ii) EF = EI = 5cm

FG = HI = 2cm,GH = 3cm

(iii) JK = KL = MN = NP = 3 cm

JP = LM = 5 cm

(iv)

e) Perimeter (otherwise known as the circumference) of a circle

Draw 3 different circles of radius 3 cm, 4 cm and 5 cm. Measure their boundary lines by putting a string along them. Your results may be as follows:

Circle Radius (r) Diameter (d) Boundary Length (l)

Ratio (d

l)

1st Circle 3 cm 6 cm 18.8 cm nearly 3.13

2nd Circle 4 cm 8 cm 25 cm nearly 3.12

3rd Circle 6 cm 12 cm 37.8 cm nearly 3.15

Observation: ‘ l’ here denotes the circumference of the circle.

Thus we see that d

l = 3.1 nearly in all cases. The result of

d

l is a constant and it

is denoted as

‘π’ and its approximate value is 7

22 or 3.14

Thus, circumference of a circle = πd or 2πr.

Check Your Progress



E1. Determine the perimeter of:

(i) ∆ ABC in which AB = 6 cm, BC = 3

2

AB and CA is 2 cm more than AB.

(ii) Equilateral ∆ ABC in which each side is of length 5cm.

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E2. Determine the perimeter of the rectangle ABCD in which AB = 12cm and BC is 2cm less than AB.

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E3. There is a rectangular field of length 28m and breadth 21m. Pillars are to be raised at intervals of 7m. If the first pillar is raised at one corner, then calculate:

(i) How many pillars are raised along the length?

(ii) How many pillars are raised along the length?

(iii) What is the total number of pillars to be raised all around?

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13.4 AREA OF TWO DIMENSIONAL FIGURES

13.4.1 Concept of Area and its Use in Daily Life Situations

There are two vacant rooms in a school numbered as 3, 4. Students of Class- VI are to take their mid-day meals. The monitor was asked to make arrangements for the students of his class in room number 3. He said “students of our class cannot be accommodated in Room No. 3, but they can be accommodated in Room No. 4”. What could be the difference between the two rooms so that one of those can accommodate the students of Class VI whereas the other one cannot accommodate it?

A general answer will be, “Room No. 4 is bigger than the other room”.

Now the question is, Room No. 4 is bigger in what sense? For each child some part of the floor is necessary so that s/he could sit and take his/her food. A part of the floor to be used by one child is shown in the Figure 13.9. It is square in shape with each side measuring 60 cm. it encloses some part of the floor (which is otherwise a plane). Let us see the two Figures marked as (i) and (ii). Every body says that Figure 13.10(ii) is bigger than Figure 13.10(i). What of the Figure 13.10(ii) is bigger than that of Figure 13.10(i)? It is the part (or region) of the page of the book enclosed by Figure 13.10 (ii) is greater than that enclosed by Figure 13.10(i).

A cultivator says, between two of his fields, one is larger and it yields more grain than the other. What of the field is larger? It is again the region of the ground enclosed within the boundary is larger. To be more specific, we say, the field that yields more grain has greater area.

Thus, the measure of the region (part of the plane) enclosed in a closed figure is expressed by the term area.

So, now we say that the area of Room No. 4 is larger than that of Room No. 3. The field that has more area yields more grain.

• Unit of measurement of area

To measure anything we first fix a unit to be used to measure it such as:

• To measure a length, we use a meter as a unit. • To measure the mass of a body, we use Kg as a unit.

Now similarly we are to fix a unit to measure the area.

60cm

60cm

(i) (ii)

Figure 13.9

Figure 13.10

Well, how do we fix a unit?

Let us analyze the units with which we are well acquainted.

To measure length, a metre is the unit we know.

Examine the scale used by shopkeepers for measuring cloth. It is nothing but a particular length which has been internationally accepted to be taken as a metre.

To know the weight of a body we use a weight measuring as one kg. Again, we see that the unit of weight measure is a particular weight measure which has been accepted as a Kg.

So the unit fixed for measuring an area of a closed region (part of a plan) should also be a specific closed region whose area has been accepted as a unit to measure the area of any closed region.

Look at the Figure 13.11 along side. It is a square with each side of length 1m. The measure of the region enclosed by this figure is named as a square metre and is written as 1 sqm or 1m2.

A square of which, the length of one side is 1 cm.

The measure of the region enclosed in this square is named as a square centimeter and is written as 1 sq cm or 1 cm2.

1 m2 is the unit used in measuring area.

1 cm2 is a sub-unit used for measuring smaller areas.

What is the relation between 1m2 and 1 cm2?

The Figure 13.12 shows a square of each side measuring 1m.

Horizontal lines are drawn to divide it into 100 equal parts horizontally and vertical lines are drawn to subdivide each horizontal division into 100 equal parts.

Thus, small squares each of side 1cm long are formed (which are marked as a, b, c, d). There will be 100 such small squares in a row and 100 such squares in a column after all the horizontal and vertical lines are drawn.

So the total number of small squares will be 100 × 100 = 10,000 in number

∴ 1m2 = 10,000 cm2

1m

1m

a

c

b

d

1cm

1m

1cm

Figure 13.11

Figure 13.12

Depending on the length of each side of the unit square we take, the unit of measuring area would be different. As we have just defined, the units may be 1m2 or 1 cm2. Similarly, if each side of the unit square measures 1 decimeter (or 1dm) or 1millimeter (or 1mm), then the unit of measurement of area will be 1dm2 and 1mm2 respectively.

You can easily verify the relationship between these metric units of measuring area.

1m2 = 100 dm2

= 10,000 cm2

= 10,00,000 mm2

These are the standard units of measurement of area used internationally. 13.4.2 Determination of Area of Closed Figures

a) Area of any irregular figure

i) Take a graph sheet. Check what the measure of each square in the graph sheet. If the length of the side of each square is 2mm, then each square of measure (2×2) sq mm = 4 sq mm.

ii) Draw a closed figure on the graph sheet as shown in Figure 13.13. Now it can be seen that some complete squares are included in the closed figure drawn. In addition to the complete squares, some parts of the squares also included.

We calculate the area of the figure according to following intervention.

a) All the complete square enclosed are counted and total of them is calculated by taking each as a unit square.

b) The figure may be found enclosing more than values of some squares. Count them all and calculate the total of them treating each as one unit square.

c) The figure may be found enclosing exactly halves of some squares count them all and calculate the total of them each as ½ unit square.

d) The figure may enclose less than halves of some squares. Ignore them.

Then the total number of unit squares enclosed is calculated and then the area of the figure is calculated.

iii) Calculation of the Area:

Figure 13.13

Prepare a Table as shown below:

Squares covered Number Number in unit squares

Area estimate in mm2

Complete Squares 31 31 31×4 = 124

More than half – squares11

11 11 11×4 = 44

Exactly half-squares

02 2 × ½ = 1 1×4 = 4

Less than half-square ignored

Total 172 sq mm = 1 sq cm, 72 sq mm

This method gives a fair estimate of the area of a closed figure.

Try the following activity:

a) Draw a circle of radius 4 cm on a graph paper and estimate its area following the process discussed above.

b) Check if your estimate comes to be 72 cm2 (or near about). It will be almost equal to 3 × r2 square units.

b) Area of a rectangle

a) Take a graph paper (or prepare one by yourself) with cm-squares on it.

b) Draw 3 rectangles on it in such a way that the sides of a rectangle coincide with the lines of the graph paper.

c) Complete the table as shown below:

Sl. no. of the

rectangle

No. of square units

in a row

No. of rows Total no. of square units

Area

(1) 3 2 3×2 = 6 6 cm2

(2)

(3)

From the table, we conclude that:

The area of a rectangle = (l ×b) sq. units (where ‘l’ denotes length and ‘b’ denotes breadth)

c) Area of a square

The rule to calculate the area of a square can also be derived in the same manner as above.

We can also say logically that:

A square is a rectangle having its length equal to its breadth.

Let there be a rectangle having its length = breadth

Then, the area of this rectangle = l×b = a×a = a2.

Thus, we get:

Area of a square = ‘a’ square units.

Where ‘a’ is the length of its side.

d) Area of a triangle

An activity for you:

a) Take a square paper (on prepare one) having sq. cm as the unit squares.

b) Draw the outlines of 3 triangles such that the vertices of them coincide with the corners of the squares in the graph sheet as shown in Figure 13.14.

c) Now prepare a table and complete as shown.

Sl. no. of the

triangle

No. of complete squares enclosed

No. of more than half squares enclosed

No. of exactly half

squares

Area measure

1 3 2 2 × ½ = 1 3+2+1 = 6 sq cm

Less than half squares have been ignored.

It can be seen from the figure that ‘base’ of the first triangle (BC) = 4 units

A

B D C

Figure 13.14

‘height’ of the first triangle (AD) = 3 units

And the area = 6 sq. units

Thus the area = 2

1 × base × height

[=2

1 × 4 × 3 = 6sq units)

And the result tallies with we have found out.

Area of a triangle = 2

1 base × height

e) Area of parallelogram

A parallelogram has both the pairs of opposite sides parallel.

One of the parallel sides is treated as base of it.

The distance between the base and the side parallel to it is considered as the height of the parallelogram as shown in Figure 13.15.

Parallelogram Base Base (b) height

ABCD BC PQ

ABCD CD MN

An Activity for you:

On a graph paper, draw a parallelogram taking BC and AD along two graph lines with a distance of 5 cm between them.

Take BC of length 8 cm and take AD of length 8 cm in such a way that AB and CD do not be along graph lines.

Determine the area by counting unit squares (in the manner as discussed above).

The area that you determined will be near about 40 sq cm.

Base × height = 8 ×5 = 40.

Thus we found that:

Area of a parallelogram = b×h,

M

B

A D

C

Height (h)

Q Figure 13.15

where ‘b’ is the length of the base and ‘h’ is the length of its height.

f) Area of a trapezium

Trapezium is a quadrilateral in which one pair of opposite sides are parallel (the other pair may or may not be parallel).

Trapezium with one pair of opposite sides parallel with a distance of 5 cm between them. Take one of the parallel sides as 11 cm and the other as 9cm.

Height of the Trapezium:

= Distance between the parallel sides

= 5 cm (in this case).

Now determine the area of the trapezium by counting the unit squares.

It will be near about 50 cm2.

The trapezium, in the Figure 13.16 has been divided into 2 triangles by drawing the diagonal BD.

For triangle ABD, base is AD and height is BQ.

∴ Area of the triangle ABD =2

1 AD × BQ

For triangle DBC, base is BC and height is DP

∴ Area of triangle DBC =2

1BC × DP

Hence area of the trapezium ABCD = Area of triangle ABD = Area of triangle DBC.

= 2

1 AD ×h +

2

1 BC × h

= 2

1 h (AD + BC)

= 2

1 × 5 (9+11)

= 2

1 × 5 × 20 = 50 cm2

A D

C P B

Q

Figure 13.16

The result obtained using the graph paper and through calculation tally.

Thus, the area of a trapezium = 2

1 h (sum of the

parallel sides).

g) Area of a general polygon

i) In Figure 13.17 the quadrilateral ABCD diagonal AC is drawn and perpendiculars from B and D to AC.

Ar (quad ABCD) = Ar (triangle ABC) + Ar (triangle ACD)

= 2

1 × base AC× height BQ+2

1 × base AC× height DP

=2

1 AC (BQ + DP)

ii) The Figure 13.18 ABCDE is a polygon of 5 sides. AD is one of its diagonals.

BP, CR and EQ are drawn perpendiculars to AD.

Perpendiculars BP, CR and EQ are known as offsets from the vertices to the diagonal AD.

Ar (ABCDE) = Ar (triangle ABP)

+ Ar (triangle CDR)

+ Ar (triangle DQE)

+ Ar (triangle AQE)

+ Ar (trap. BCRQ)

Thus, the area of ABCDE can be determined.

h) Area of a circle

To determine the area circle using graph paper, a circle is drawn on a graph sheet.

Then, by counting the unit squares, area can be determined.

The area of a circle is given by the following rule.

Area of a circle = πr2

Where π is approximately equal to 7

22 or 3.4.

A

B

D Q

C

A

B E

D

C R

Q P

Figure 13.17

Figure 13.18

Let us try to solve some word problems:

Example 1: A premises in the form of a rectangle of length 24 m and breadth 20 m is to be surrounded by 4 stretches of barbed wire leaving a gate of length 4 m. Determine the total length of the barbed wire necessary.

Solution:

The shape of the premises is a rectangle of which length = 20 m and breadth = 20 m.

The perimeter of the premises = 2 (l+b) = 2 (24 + 20) m = 88 m

Gate is of length 4 m.

Remaining length = 88 m – 4 m = 84 m.

Length of one stretch of wire = 84 m.

∴ Length of 4 stretches of wire = 84 m × 4 = 336 cm.

Example 2: How many rectangular tiles of length 15 cm and breadth 10 cm are required to cover:

a) A rectangular floor of length 6 m and breadth 5 m.

b) A floor in the shape of a square of side 4.5m.

Analysis:

i) What kind of figures the question (a) speaks of? It speaks of 2 rectangles, one rectangle is the floor and the other one is a tile.

ii) What measurements are given for the floor (rectangle), length = 6m and breadth = 5m.

iii) What is asked to be determined? The number of tiles required to cover the floor.

Hint for selection:

First we decide, which way should tiles be laid on the floor so these will exactly fit in (there will be no necessity for breaking). Then we calculate how many tiles can fit along the length of the floor and how many along the breadth. Then the number of tiles can be calculated.

Breadth of the floor i.e. 5m = 500 cm is divisible by 10 cm (breadth of the tile), but not divisible by 15 cm (length of the tile). Hence, the tile should be laid in the manner as shown at corner D, not as shown at corner A as shown in Figure 13.19.

A D

C B

5m 10cm

15cm

Not to be laid

to be laid

10m 15cm

Figure 13.19

Solution:

Number of tiles to be laid along AD = 15cm

600cm = 40

Number of tiles to be laid along AB = 10cm

500cm = 50

Thus, along the length of the floor:

The number of tiles to be placed in a row = 40

The number of rows of tiles to be placed = 50

∴ The total number of tiles required = 40 × 50 = 200

The same could be the method to solve bit (ii) of the question.

Alternative method:

Each tile covers some area out the total area of the floor.

If we can determine the area of the floor and the area of each tile, then we can determine the number of tiles required. Both the areas should be determined in the same unit.

Area of the floor = l× b = 600cm × 500cm = 3,00,000cm2

Area of each time = 15cm × 10cm = 150cm2

150 cm2 area is covered by one tile.

∴ The number of tiles required to cover 3,00,000 cm2 = 150

000,00,3

= 2000

Note: Sometimes, the data would be such that either, length-wise or breadth-wise the tiles do not exactly fit in. if the length of the floor was 8m instead of 6m, then neither if these measurement would be divisible exactly 15cm. In such cases the first method is not applicable.

Check Your Progress



E4. Determine the area of a rectangle having length = 18cm and breadth = 15cm. How many square sheets of paper each of length 4cm can you get by cutting the above mentioned sheet?

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E5. Determine the area of a square having each side equal to 9cm. If a triangular parts at corners obtained by joining the mid-points of consecutive sides are removed, find the area of the sheet left?

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E6. Determine the area of a circle of radius 5cm using a graph sheet. Then calculate the area using the rule taking π = 3.14. How much is the difference between the two results obtained?

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E7. In the Figure 13.20 shown below, all angles are right angles and lengths of different edges are noted in the figure.

Determine the area enclosed in Figure 13.20(i) and shaded area in Figure 13.20(ii).

2cm

2cm

2cm

2cm

2cm 2cm

2cm

3cm 5cm

3cm 5cm

(i) (ii)

2cm

8cm

4cm 5cm 5cm

(iii)

Figure 13.20

13.5 SURFACE AREA AND VOLUMES OF THREE- DIMENSIONAL BODIES

13.5.1 Recognition of Three-dimensional Shapes

Let us observe the Figure 13.21 that represents the shape of a piece of unbroken brick. The corners are named as A, B, C and D.

It can be seen from the figure that:

• It has some extension in the direction from A to B. We call this as the length.

• It has some extension in the direction from A to C. We call it as breadth.

• It has some extension in the direction from A to D. We call it as height.

Thus, a piece of brick has extensions (which can be measured) along 3 directions. We call them dimensions.

Hence, we say a piece of brick is a body having 3 dimensions.

In case of a piece of a brick, in each of the 3 dimensions (i.e. directions) it has definite measurement. There are some other shapes where definite measurement does not occur in each of the dimensions. Some examples are given below:

A piece of pie shown in Figure 13.22(i) has definite measurement along Left–Right extension. But it does

(i) (ii)

B

D

A

Figure 13.21

Figure 13.22

E8. Determine the area enclosed in Figure 13.20(iii)

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E9. A floor measures 8 m ×5.40 m (its measuring is: length is 8 m and breadth 5.40 m). There are carpets each of length 2 m and breadth 1.5 m. The carpets are to be laid in such a way, that the uncovered part of the floor should be the minimum. Determine the number of carpets required.

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not have a definite measurement along Front–Back extension not has definite measurement along Top–Bottom extension. It might have a maximum measurement along Front–Back and Top– Bottom directions.

Similarly, in case of a ball.

However, each of the bodies shown in Figure 13.22(i) and Figure 13.22(ii) have extensions in 3 dimensions. Thus bodies like a piece of brick, a piece of pipe or a ball are said to be bodies of 3 dimensions.

Now you can be tempted to say that each of the plane geometrical closed figures are of two dimensions.

Of course, a ‘line’ is of one dimension a point does not have any dimension.

• Geometrical names used for specific three-dimensional shapes

i) Cuboids: The shape of a piece of brick, a box, a book are common examples of cuboids. Cuboids have definite length (measurement in one dimension), breadth (measurement in another dimension), height or thickness (measurement in the third dimension).

ii) Cube: The dice used in the game of LUDO is in the shape of a cuboid. But its length, breadth and height are equal. This shape is known as cube. Thus, cube is a cuboid of which length = breadth = height.

iii) Cylinder: A pipe has two circular ends and there exists a certain length between its ends. The shape

of it is known as a cylinder shown along side in Figure 13.23.

iv) Cone: The upper part of a funnel (without the lower part of it), a joker’s cap have a geometrical shape as shown the Figure 13.24. Such a shape is known as a cone.

v) Sphere: A ball has the shape which is known as a sphere.

vi) Pyramid: The shapes of a pyramid and a cone are very much similar with the following differences.

a) The base of a cone is a circle whereas the base of a pyramid is a polygon as shown in Figure 13.25 (may be triangle, or a quadrilateral or a pentagon etc.)

Figure 13.23

Figure 13.24

Figure 13.25

A

D

C

E

F Q

P H

G

I

J

L

K

B

b) The lateral surface of a cone is curved having no edge on if gaining the vertex to any point on the base. But the lateral part of a pyramid contains plane surfaces in shapes of triangle the number of which is equal to the number of sides in the polygon forming the base of it.

Activity: Take a piece of thick sheet of paper and cut the shapes as shown in the Figure 13.26(i) (ii).

Figure 13.26

In Figure 13.26(i) AB = BC = CD = DE = HI = IJ = JK = KL = 3cm

AL = FG = 8cm

EF = HG = 6cm and angles are right angles.

Mark the mid-points of EF and HG as P and Q respectively.

Fold along BK, EH, PQ, BE and KH, till the edges: CB and BA coincide.

∴ LK and KJ coincide, and

AL and FG coincide.

Fix up the joined edges by adhesive tapes.

Similar work be done with the sheet shown in Figure 13.26(ii). Think and draw other designs of paper sheets to make a cube.

Now answer the following on the basis of your observation.

i) What is the shape formed from the sheet of paper in the shape of Figure 13.26(i)?

ii) What are the measurements along the three dimensions of it?

iii) What is the shape formed from the sheet of paper in the shape of Figure 13.26(ii)?

iv) What are the measurements along the three dimensions of it?

A B

C

E

D P F

G Q I H

J K L

(i) (ii)

Activity: Draw a circle of radius (say 6 cm) on a thick sheet of paper mark the centre as 0. Draw 2 radius namely OA and OB (with any angle between them).

Now cut out the circle along its boundary and also cut along the two radius.

Now you got two sheets each in the shape of a sector. Turn one of the two sheets as shown in the Figure 13.27 till the edges OA and OB come together. Join the edges by adhesive tapes.

Do similar exercise with the other sheet.

Each of them gives rise to the shape of a cone.

The cone obtained from the bigger sector has a larger circular base than the other. But look to the height of the vertex above the base. In which one of the two the vertex lies at a greater height?

Activity: From a thick sheet of paper cut about 6 isosceles triangles of the measurement shown in the Figure 13.28(i).

(i) (ii)

Figure 13.28

Now we join the triangles so that their equal edges will coincide:

i) If 2 of the triangular sheets are joined, will it give rise to a 3 dimensional shape? Definitely not.

ii) If 3 of the sheets are taken and joined what shape shall be formed? The shape formed is shown in the Figure 13.28(ii) and such a shape is known as a pyramid.

It is a triangular pyramid (as its base is a triangle).

Thus, pyramids with bases in the shapes of square, regular pentagons and regular hexagons can also be formed.

8cm 8cm

5cm

O

A

B Figure 13.27

3-dimensional bodies are also known as 3-dimentional bodies. Let us compare a closed plane figure with a 3-dimensional body.

i) Closed plane figure has corners (also known as vertices). Of course, a circle has no vertex as 3-D body also has some vertices. Of course a sphere or a cylinder has no vertex.

ii) Plane closed figures enclose a part of the plane in which it is drawn. A 3-D body encloses a part of the space in which it is kept.

iii) The boundary line (may contain line segments or curves) separates it from the remaining part of the plane. The surfaces which a 3-D body has, separate it from the remaining part of the space.

iv) The part of the plane enclosed within the boundary line is the area of the closed figure and measured in square units. The part of the space enclosed by the surfaces the 3-D body is the volume of the body.

Thus as we observe the following:

Name of the 3-D body

No. of vertices

No. of edges No. of surfaces

Basic measurement

Cuboid / cuboid

8 12

(all straight)

6 Length, breadth and height for a cube – only length

Cylinder Nil Two

(both circular)

03

1: curved

2: flat and circular

Radius, height (or length)

Cone 1 1

(circular)

2

1: Curved

1: Flat and circular

Radius, height

Sphere Nil Nil 1

Curved

Radius

Pyramid 4 at the least. May be 5 / 6….

6: at the least may be 8 / 10 / 12….

4 at the best, may be 5 / 6/

7.

Base of triangular surface and height of it. Height of the

pyramid.

13.5.2 Calculating the Surface Area and Volumes of Specific Kinds of Three-dimensional Bodies

a) Cuboid

Keep an unbroken piece of brick or a box before you. The Figure 13.29 represents such a shape.

Length, breadth and height of the cuboid are denoted as ‘l’, ‘b’ and ‘h’ respectively and shown along the edges which represent them.

You may observe:

i) The surfaces at the top and bottom are both rectangles of equal length (l) and equal breadth (b)

∴ The area of each = l×b = lb

ii) The surfaces to our left and to our right are both rectangles of equal length (b) and equal breadth (h)

∴ The area of each = b×h = bh iii) The surfaces of the front and the back are both rectangles of equal length (l) and

equal breadth (h).

∴ The area of each = l×h = lh

Lateral surface area = sum of the areas of the surfaces to the left, right, front and back front and back = bh + bh + lh + lh

= 2bh + 2 lh = 2h (b+l)

= 2h (l+b)

In case of a room whose shape is a cuboid, the lateral surfaces represent its 4 walls.

∴ Area of the walls of the room = 2 × height (length + breadth)

Total surface area = area of the lateral surfaces + areas of surfaces at the top and bottom

= 2h (l+b) + lb + lb

= 2hl + 2hb + 2lb

= 2lb + 2lh + 2bh

= 2(lb + lh + bh)

In case of a cube each surface is a square of length ‘a’ cm (assumed).

h l

b

Figure 13.29

A B 6

5 4 3 2

A B 6

5432

Hence, its area = a2 cm2.

∴ the total surface area = 6a2 cm2

• Volume of cuboids

Take a measuring cylinder which is marked indicating deciliters shown in Figure 13.30.

Pour water into it till it is filled to the level marked A shown in Figure 13.30(i).

Then drop a piece of stone into it. You will find that the level of water rises up to B shown in Figure 13.30(ii). Now the measuring cylinder shows 2 deciliters more

water (from 4 to

6) where frames these 2 deciliters of water.

The stone occupied some space inside the cylinder and the water displaced by it occupied the part of the cylinder from A to B.

Thus, it is seen that the piece of stone occupies some space.

The space occupied represents its volume.

• Unit to measure volume

You have seen that to measure the area of a plane figure we used the area of a plane figure of a specific shape (square) and specific size (side of length 1m or 1cm) as unit. Similarly to measure the volume of a 3-dimentional body, we use the volume of a 3-dimentional body of specific space (cube) and of specific size (each side of length 1m or cm or 1mm) as unit.

The volume of such a cube (each side of length 1m) is said to be 1 cubic meter and is written as cum or 1m3 see in Figure 13.31.

Let us think that the cube along planes parallel to its front surface, parallel to its right surface and parallel to its top surface at distances of 1cm. What will we get?

We will get 100 × 100 × 100 number of cubes, each side of which is of length 1cm.

The volume of each cube is 1 cu.cm = 1cm3

This is known as cubic unit.

1m

1m

(i) (ii) Figure 13.30

Figure13.31

There is another unit through which volume of liquid is measured. That is 1 litre (denoted as 1l ).

Equivalence between cubic unit and litre unit:

1l = 1000 cm3

1m3 = 1000 l

• Volume of a cuboid

Assume a cuboid of length = 5cm, breadth = 3 cm and height = 2 cm shown in Figure 13.32.

What is its volume in cm3?

It can be known if we can determine how many cubes of length 1 cm can be made out of it.

Let us cut along planes

i) parallel to the front surface; ii) parallel to the right surface; and iii) parallel to the surface at intervals of 1cm.

By the first cut, we will be getting 3 pieces of the following shape and size 3 such big blocks will be available as shown in Figure 13.33.

By the second cut (parallel to right surface) we will be getting 5 pieces (small) of the following shape and size from one big block as shown in Figure 13.34.

5×3 = 15 small

blocks from the 3 big blocks.

By the third cut (parallel to the top (surface), we will be getting 2 pieces (cibes of length 1cm) of the following shape and size from each of the small blocks as shown Figure 13.35.

15×2 = 30

cubes of length 1cm from 15 small blocks in total.

The volume of each small cube = 1cm3,

∴ The value of 30 small cubes = 30 cm3.

5cm

3cm

2cm

1cm

2cm

5cm

1cm 1cm

2cm

1cm 1cm

1cm

Figure 13.32

Figure 13.33

Figure 13.35

Figure 13.34

Thus, we see that:

The volume of a cuboid

= l ×b × h

(where ‘l’ represents its length, ‘b’ represents its breadth, ‘h’ represents its height.

• Volume of a cube

A cube each side of which is 3 cm long is shown in the Figure 13.36.

A cube is a cuboid in which l = b = h

Let us take each side to be length ‘a’ units.

Then the volume of it = a × a × a cubic units i.e. Volume of a cube = a3 units.

b) Cylinder

Total surface area of a cylinder:

The Figure 13.37 shows the shape of a cylinder.

It has two kinds of surfaces: A curved surface and 2 plane circular surfaces at its two ends.

The radius of it is ‘r’ (the radius of the circular surface) and its length (also called its height) is ‘l’.

Hence, the total surface area of a cylinder

= Area of the curved surface+ Area of the 2 circular surfaces.

Area of curved surface:

If the ends are separated and the body of it is cut open and stretched, it produces a rectangle.

For the rectangle obtained by stretching the curved surface: Length = circumference of the circular end = 2πr, Breadth = length (or height) of the cylinder,

= l (may be taken as h).

∴ Area of the curved surface = length × breadth = 2πrl (or 2πrh)

Area of the circular surfaces:

Area of 2 circles = 2πr2

3cm

3cm 3cm

r

l

Figure 13.36

Figure 13.37

∴ Total surface area = curved surface area + area of the 2 circular ends,

= 2πrl + 2πr2,

= 2πr (l + r).

More commonly the distance between two circular ends is considered as the height of the cylinder (assuming the cylinder stands vertical as trunk of a palm tree) and is denoted as ‘h’ instead of ‘l’. In that case

Area of curved surface c = 2πrh, and

Total surface area = 2πr (h+r)

• Volume of a cylinder

Let us analyze the rule described for calculating the volume of a cuboid.

Volume of a cuboid = l × b × h

= (l × b) × h

= area of the base × height

This rule applies to all bodies having a regular shape which has a uniform section from one end to the other. A cylinder is also a shape in which the section remains uniform (circular) from one end to the other.

Assuming the cylinder stands vertically; its base is a circle.

Hence, the volume of a cylinder = base area × height,

= πr2 × h = πr2h,

Thus, the volume of a cylinder = πr2h.

c) Sphere

A sphere has only one curved surface. The surface area and the volume of a sphere with radius ‘r’ (and diameter ‘d’) are given by:

Surface area = 4πr2h or πd2

Volume =3

4 πr3 or 6

πd3

d) Cone

Let the radius and height of a cone be denoted as ‘r’ and ‘h’ respectively.

O

A B

l h

r

Distance from the vertex 0 to any point in the circular edge B (or A) is known as the slant height and is denoted as ‘l’.

The Figure 13.38 shows that ‘h’, ‘r’ and ‘l’ form a right angle triangle with the angle between ‘h’ and ‘r’ as a right angle.

∴ l2 = r2 + h2

Curved Surface area of a cone = πrl

Total surface area of a cone = area of the curved surface

+ area of the circular base

= πrl + πr2 = πr2 (l+r)

Thus, Total surface area of a cone = πr (l+r)

Volume of cone

If we take a cylinder and 3 cones, all having the same radius and same height, then by water displacement method, it can be shown that:

The cylinder displaces as much water as 3 cones displace.

∴ Volume of a cylinder = 3 × volume of a cone (if they have equal radius and equal height).

Thus, the volume of a cone = 3

1 πr2h.

Figure 13.38

Check Your Progress



E10. ABCD is a rectangular sheet of paper in which P and Q are the mid-points of the sides AD and CD.

A D

M Q

C B

P A D

Q

C B

P A D

Q

C B

P

(i) (ii) (iii)

Figure 13.39

In Figure 13.39(i) PMQD is a rectangle.

In Figure 13.39(ii) PQD is a triangle.

In Figure 13.39(iii) PQR is an equilateral triangle.

In Figure 13.39(i) rectangle PMQD is removed.

In Figure 13.39(ii) triangle PQD is removed.

In Figure 13.39(iii) the quadrilateral PRQD is removed.

Answer the following:

1. In which case the remaining part has the same perimeter as before?

2. In which case the remaining part has the smaller perimeter as before?

3. In which case the remaining part has quarter perimeter than before?

4. Give reason for each of your answer.

E11. A cube is made by suitably stacking 27 cubes of equal sizes.

(i) Determine the ratio between the total surface area of 27 cubes and the total surface area of the cubes formed.

(ii) If all the surfaces of excluding the base of the cube formed is painted while, then how many smaller cubes will have:

a) One surface white?

b) Two surfaces painted white?

c) Three surfaces painted white?

d) No surface painted white?

(iii) Total area of the surfaces painted white?

................................................................................................................

................................................................................................................

................................................................................................................

E12. On the top of a cuboid of length 10 cm, breadth 8 cm and height 5 cm, a cylindrical hole of radius 3.5 cm and height 3 cm is drilled. Determine (i) the total surface area of the remaining body (ii) volume of the remaining body?

................................................................................................................

................................................................................................................

................................................................................................................

13.6 UNIT SUMMARY

• Perimeter of a closed figure is the measure of its boundary

Perimeter of a closed figure is the distance covered along the boundary when we go round the figure once.

a) Perimeter of a polygon = sum of the lengths of its sides.

b) Perimeter of an equilateral triangle = 3 × length of each side.

c) Perimeter of a rectangle = 2 (l+b) where ‘l’ and ‘b’ represent the length and breadth of it.

d) Perimeter of a square = 4a where ‘a’ is the length of it.

e) Perimeter of a circle (known as its circumference) = 2πr.

Where ‘r’ represents its radius and π may be taken as 7

22 or 3.14.

• Area of a closed figure is the measure of region (i.e. the part of the plane on which it is drawn) enclosed in a closed figure. It is measured in square units like 1m2, 1cm2 etc.

1m2 = (100 × 100) cm2 = 10000 cm2.

a) Area of a rectangle = l × b.

b) Area of a square = a2.

c) Area of a parallelogram = b × h (where ‘b’ represents the base and ‘h’ represents the height of the parallelogram).

d) Area of a triangle = ½ bh.

e) Area of a Circle = πr2.

• Volume of a 3-dimensional body is the measure of the space enclosed in it.

Shapes Basic Measurements Volume

Cuboid Length, breadth and

height V = l × b × h

Cube Length (a) V = a3

Cylinder Radius (r), height (h) V = πr2h

Cone Radius (r), height (h) V = 3

1 πr2h

Sphere Radius (r) V = 3

4 πr3

13.7 GLOSSARY

Shape : As a figure drawn on a plane on the 3-dimensional body appears to us.

Size : The measure of the extensions of the figure or solid body in different directions. How big or how small a figure or a solid body is indicated by its size.


E1. 16 cm.

E2. 120 cm2 = 0.0120m2.

E3. (i) 5 (ii) 4 (iii) 14

E4. 270 cm2, 12

E5. 81cm2; 40.5 cm2.

E6. No answer is given.

E7. (a) 20 cm2 (b) 42 cm2

E8. 28 cm2

E9. 12

E10. (1) In Figure 13.39(i) perimeter remains the same.

(2) In Figure 13.39(ii) small perimeter.

(3) In Figure 13.39(iii) larger perimeter.

(4) In Figure 13.39(i), PD + DQ = PM + MQ. So no change in perimeter. In Figure 13.39(ii), PQ < PD + DQ (one side of a triangle is less than the sum of 2 sides). In Figure 13.39(iii), PQ is the hypotenuse of a right triangle. Hence PQ > PD, PQ > DQ. But PR = RQ = PQ as sides of an equilateral triangle.

∴ PR > PD and RQ > DQ.

Hence, PR + RQ > PD + DQ.

Thus, the perimeter is increased.

E11. (i) 3 : 1

(ii) (a) 9 (b) 12 (c) 4 (d) 2

E.12. (i) 406 cm2 (ii) 284.5 cm3

13.9 ASSIGNMENT

Answer the following questions.

Q1. From a rectangular sheet of paper of length 14 cm and bread 10 cm, squares of side 2 cm were cut from the corners and removed and a tray is formed by folding the parts of width 2 cm on each side. Determine the total surface area and the volume of the tray formed.

Q2. In the figure shown the angles at A, B, C and G are right angles. DEF is an equilateral triangle.

AG = BC = 10 cm. AB = 4 cm FG = CD = 2 cm.

Determine (i) perimeter (ii) area of the shape.

Q3. Two concentric circles of radii 7 cm and 3.5 cm are drawn. Determine the area enclosed between the circles.

Q4. Determine the area of a trapezium of which one of the parallel sides is 14 cm and each of the remaining sides is 6 cm long.

Q5. How many cubical blocks of which side measures 3 cm can be kept in a box with length, breadth and height of the inner space as 20 cm, 16 cm and 14 cm respectively? Determine the volume of the empty space left.

Q6. A rectangular sheet of card board was turned to form a cylinder with no area of it being waster. Determine the volumes of the 2 cylinders that can be formed in turning the sheet length-wise and breadth-wise.

Corners removed

Tray formed

A

B C D

E

F G

13.10 REFERENCES

• Textbooks published by National Council of Educational Research and Training in the year, 2008 for Classes VI, VII and VIII.

UNIT 14 SYMMETRY Structure

14.1 Introduction

14.2 Objectives

14.3 Concept of Symmetry

14.3.1 Symmetrical Figures and Asymmetrical Figures

14.3.2 Making Symmetrical Figures

14.4 Line of Symmetry

14.4.1 Figures with One Line of Symmetry

14.4.2 Figures with Two Lines of Symmetry

14.4.3 Figures with Multiple Lines of Symmetry

14.5 Reflection Symmetry

14.5.1 Application of Reflection Symmetry

14.6 Rotational Symmetry

14.6.1 Line Symmetry and Rotational Symmetry

14.7 Unit Summary

14.8 Glossary


14.10 Assignments

14.11 References

14.1 INTRODUCTION

Symmetry is quite a common term used in our day to day life. Symmetry is an important general concept, commonly exhibited in nature and is used almost in every field of activity. Artists, professionals, designers of clothing or jewellery, architects and many others make use of the idea of symmetry. The concept of symmetry helps the children to identify the patterns in various types of geometrical shapes and helps in relating the mathematical concepts in our immediate environment.

In our natural environment, we observe shapes with symmetry in various objects. So identifying the symmetrical figures from the environment helps us to relate the geometrical concepts with our life experiences.

Understanding the symmetrical figures and differentiating between symmetrical and asymmetrical figures are the key concepts discussed in this unit. Further, the properties of two major types of symmetry which are commonly experienced have also been discussed.

14.2 OBJECTIVES

After reading this unit, you will be able to:

• differentiate symmetrical and asymmetrical figures; • draw and prepare symmetrical figures;

• identify the lines of symmetry in any symmetrical figure; and

• differentiate between line symmetry and rotational symmetry.

14.3 CONCEPT OF SYMMETRY

Symmetry is a very common term which is used by all of us. When we see figures which have evenly balanced proportions, we say, they are symmetrical. Symmetry usually has two meanings: one conveys an imprecise sense of harmonious or aesthetically pleasing proportionality and balance which remains undisturbed through reflection or rotation to certain degrees. The second meaning is a precise and well defined concept of balance or ‘patterned self-similarity’ that can be demonstrated or proved according to the rules of a formal system like geometry or physics.

Now let us observe the Figure 14.1 below:

Figure 14.1

If we could fold each of these two pictures along the middle as shown through dotted lines in the two figures, we will find that the lines, curves and any other elements of one half will be exactly superimposed on the respective elements perfectly without any deviation. In other words, the pictures in the left and right halves match exactly. In such a case the picture is said to be symmetrical about the middle line.

Any square, rectangle, or circle is a symmetrical figure as each of the figures is a figure with evenly balance proportion.

14.3.1 Symmetrical Figures and Asymmetrical Figures

Let us observe the picture of Tajmahal below:

☺

This picture of architectural marvel is beautiful because of its symmetry. If we could fold such pictures in half such that the left and right halves matches exactly then the picture is said to have line symmetry.

Figure 14.2

Look at the Figure 14.2, the dotted line divides the shapes into two halves. If we fold it along the dotted line, then the left half will exactly cover the right half (Figure 14.2).

Figure 14.3(a) Figure 14.3(b)

But in the Figures 14.3(a) and 14.3(b), the dotted line divides the shapes into two parts. But if we fold it along the dotted line, the left half does not cover the right halves completely. So the two haves are not mirror halves. Such figures are asymmetrical figures.

In the symmetrical figures, there are two mirror halves.

The beehives, flowers, tree-leaves, religious symbols, rages and handkerchiefs everywhere we can find symmetrical designs. In every symmetrical design the line which can divide the figure in two halves such that the left and right halves match exactly is known as the line of symmetry.

In the symmetrical figures, we can see that the two haves of the figure are mirror image of each other. If we place a mirror on the fold (line of symmetry) then the image of one side of the figure will fall exactly on the other side of the paper.

14.3.2 Making Symmetrical Figures

To make a symmetrical figure, let us do one activity.

Activity 1: Making symmetrical figures from a drop of paint

• Let us take a sheet of paper.

Check Your Progress



E1. Draw two symmetrical figures and two asymmetrical figures from your own environment indicating the line of symmetry.

................................................................................................................

................................................................................................................

................................................................................................................

Check Your Progress



E2. Look at the Figure and say whether this figure is symmetrical? Give reasons for your answer.

Figure 14.4

................................................................................................................

................................................................................................................

................................................................................................................

................................................................................................................

• Fold it into half as shown in Figure 14.5.

• Open the fold and put a drop of paint on the middle line.

• Fold it twice and press it to spread the paint.

• Open it and you can see a beautiful pattern.

• Cut the pattern in such a way to get two similar mirror halves.

Figure 14.5

Symmetrical pictures can be prepared with the help of coloured strings. Such type of activities are very interesting in nature and easy to do. Let us do another activity. Activity 2

• Let us take a rectangular piece of paper.

• Fold the paper in half as shown in Figure 14.6.

• On one half, we can arrange short lengths of strings dipped in a variety of coloured inks or paints.

• Now press the two halves.

• We can get a figure on the rectangular paper.

• The figure will be a symmetrical figure.

Symmetrical figures can be made by the use of graph paper (square paper). We can prepare symmetrical figures on the square paper with the help of a scale and a pencil. Observe the following activity in.

Figure 14.6

Activity 3

• Take a square paper or prepare a square paper of your own.

• Take any line on the square paper (call it line of symmetry as shown in Figure 14.7)

• Now draw a figure on one side of the line of symmetry.

• Draw a similar figure towards the other side of the line of symmetry.

• Two diagrams on both sides of the line combined form a figure which is symmetrical about the middle line.

Symmetrical pictures can be made through various methods. We can use the paper cutting and paper folding activity for that. Particularly, during various occasions we used colour paper to decorate our house, school etc. We cut the colour paper with the help of scissors and get symmetrical shapes which looks very beautiful. Let us do another paper cutting and paper folding activity.

Activity 4

• Take a piece of coloured paper (Figure 14.8).

• Fold the paper into two half.

• With the help of a scissor cut the folded paper as you like.

• Unfold the paper and you will find a symmetrical shape.

From the activities described in this section, we may conclude that:

• In the symmetrical figures, there are two identical halves, i.e. one half of the figure can completely coincide with another half.

• Line of symmetry is very important to prepare symmetrical figures.

Check Your Progress



E3. Try to prepare some symmetrical figures with the help of oil paper. Write down the steps you followed to prepare the figure.

................................................................................................................

................................................................................................................

................................................................................................................

Figure 14.7

Figure 14.8

• In paper folding, ink blot activity and inked string activity the folded line of the paper (after pressing it) is taken as the line of symmetry.

• The line of symmetry is also known as the mirror line.

• The mirror line helps in drawing or preparing symmetrical pictures.

14.4 LINE OF SYMMETRY

Look at the following figures.

Figure 14.10 Figure 14.11 Figure 14.12

All the shapes are divided into two mirror halves by the dotted line. As described in section 14.3 of this unit, the mirror line (line of symmetry) helps in identifying the

Check Your Progress



E4. Here is a given Figure 14.9, where l is the line of symmetry. Complete the diagram to make it symmetrical.

l

Figure 14.9

................................................................................................................

................................................................................................................

................................................................................................................

symmetrical figures. The line of symmetry is also known as the axis of symmetry. If we place a mirror on the fold of the image (mirror line), then the image of one side of the picture will fall exactly on the other side of the picture.

14.4.1 Figures with One Line of Symmetry

Let us do an activity to understand the line of symmetry.

Activity 5

• In the instrument box, you will find two types of set squares and is 300 set square and the other is 450 set-square.

• Take the 300 set-square.

• Now take two such identical set-squares and place them side by side as shown in the Figure 14.13.

• Now we will draw the outline of the figure, made by the two set-squares.

• Now we shall get a figure as shown in the picture.

• Draw a triangle taking the smaller side of the triangle as the base.

• We will get the shape of a kite.

• Now tell how many lines of symmetry does the shape have.

You can observe that this figure is a symmetrical one; and there is only one line of symmetry in this figure. The line of symmetry is shown through dotted lines.

Check Your Progress



E5. Draw some figures which have only one line of symmetry. Draw the line of symmetry as well.

.................................................................................................................

.................................................................................................................

.................................................................................................................

Figure 14.13

14.4.2 Figures with Two Lines of Symmetry

In order to understand figures with two lines of symmetry now we have to do the following activity.

• Take a rectangular sheet of paper.

• Fold the paper sheet once length wise so that one half covers completely the other half.

• Press the paper along the folded line.

• Is this fold a line of symmetry? Give reasons for your answer?

• Now open it up and again fold along its width in the same way.

• You can get the second fold.

• Is the second is line of symmetry?

In this case, we get two lines of symmetry in one symmetrical figure. Through both the lines, the figure can be cut into two exactly equal halves, so that one half will completely cover the other half. Such figures are known as figures with two lines of symmetry.

Check Your Progress


b) Compare your answer with the one given at the end of thisunit.

E6. (i) Which English alphabets have (a) one line of symmetry, (b) two lines of symmetry?

........................................................................................................

........................................................................................................

........................................................................................................

Figure 14.14

E6. (ii) Write the English numerals (up to two digits) with two lines of symmetry.

........................................................................................................

........................................................................................................

........................................................................................................

Now we will do another, paper-cutting activity to get figures with two lines of symmetry.

Activity 6

• Take a rectangular piece of paper.

• Fold it once and then once again.

• Draw some design as shown in the picture.

• Cut the shape drawn and unfold the shape.

• How many lines of symmetry does the shape have which has been cut out?

• Create more such designs like this.

14.4.3 Figures with Multiple Lines of Symmetry

There are symmetrical figures which have more than two lines of symmetry. Consider an equilateral triangle whose sides are of equal length as shown in Figure 14.16.

We know that the line joining one vertex of the equilateral triangle to the mid point of the opposite side (median) is perpendicular upon that side and divides the triangle into two equal halves. There are three such medians.

You can now say how many lines of symmetry the figure has. Look at the lines of symmetry of an equilateral triangle shown through dotted lines in Figure 14.17.

We can get three lines of symmetry in this figure. The figure can be cut into two equal halves by any dotted line so that both the halves will completely cover the other half.

Now we may take another example:

Activity 7

• Take a square piece of paper.

• Fold it into half vertically as dotted in Figure 14.18.

• Fold it again into half horizontally.

• Now open out the folds and again fold the square into half along a diagonal.

• Again open it and fold it into two half along the other diagonal.

• Open out the fold.

Figure 14.15

Figure 14.16

Figure 14.17

Figure 14.18

How many lines of symmetry does the square shape have?

Can a rectangle have three lines of symmetry like that of a square? Give reasons for your answer.

Now we may consider a circle. How many lines of symmetry a circular shape has? Can you draw the lines of symmetry in a rectangular shape?

Any straight line passing through the centre and intersecting the circle into two opposite points is a line of symmetry which can divide the circular shape into two equal halves so that both the halves will exactly cover the other half. Try to draw the lines of symmetry in a circular shape as shown in Figure 14.19.

You will observe that the line of symmetry in a circular shape is its diameter. So how many lines of symmetry can be drawn? Definitely, there is no limit to the number of axes of symmetry for the circular shapes. In other words, we may say the circular figure has multiple lines of symmetry.

.

Figure 14.19

Check Your Progress



E7. Take a square piece of paper and sketch the following:

• A triangle with a horizontal line of symmetry but no vertical line of symmetry.

• A quadrilateral with both horizontal and vertical line of symmetry.

• A hexagon with exactly two lines of symmetry.

• A hexagon with six lines of symmetry.

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.................................................................................................................

.................................................................................................................

.................................................................................................................

Now we will discuss the line of symmetry in regular polygons. A polygon is said to be regular if all its sides are of equal length and its angle are of equal measure. Thus, an equilateral triangle is a regular polygon with three sides and a square is a regular polygon with four sides.

Similarly if the pentagon is regular as shown in Figure 14.20, naturally, its sides should have equal length and each of its angle should be of equal measure.

Similarly, a regular hexagon has all its sides equal and each of its angles measures 1200 as shown in Figure 14.21.

We may observe all the regular polygons, are symmetrical figures with multiple lines of symmetry. The interesting fact is that each regular polygon has as many lines of symmetry as it has sides.

Now observe the lines of symmetry from the following:

• The lines of symmetry are shown through dotted lines.

Figure 14.22: Equilateral triangle (three lines of symmetry)

1080

1080

1080 1080

1080

1200

1200

1200 1200

1200

1200

Figure 14.20

Figure 14.21

Figure 14.23: Square (four lines of symmetry)

We can observe the symmetry everywhere in our natural environment. The nature has plenty of things having symmetry in their shape. Try to explore them. We can observe also the symmetry in the designing of playing cards. Collect them and identify the lines of symmetry in any symmetrical figure.

14.5 REFLECTION SYMMETRY

In the previous section, we have already discussed the concept of line symmetry and line of symmetry. A figure has line symmetry if a line can be drawn dividing the figure into two identical parts. The line is known as line of symmetry. If we keep a mirror on the line of symmetry facing one half of the figure, then this half will be reflected in the mirror. The half and its image together shall look like the total figure. That means the image will exactly fit the other half. In such a case the figure is said to have reflection symmetry.

Figure 14.24: Regular pentagon (five lines of symmetry)

Figure 14.25: Regular hexagon (six lines of symmetry)

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E8. Can you draw a triangle which has:

(i) No line of symmetry?

(ii) Exactly one line of symmetry?

(iii) Exactly two lines of symmetry?

(iv) Exactly three lines of symmetry?

Draw a figure for each case.

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In this sense, an object or figure which is indistinguishable from its transformed image is called reflection symmetric, mirror symmetric. In other words, a figure has reflection symmetry if there is at least one line which splits the figure into two halfs so that one side is the mirror image of the other. Line symmetry, reflection symmetry, mirror symmetry, mirror-image symmetry, or bilateral symmetry is symmetry due to reflection along the line or axis of symmetry.

Now, let us consider a Figure 14.26. We can take the English letter A. Now using a mirror, we can see the image of A.

Figure 14.26

Here we can observe that the object and its image are symmetrical with reference to the mirror line. As shown in the Figure 14.4, if we fold the paper along the dotted line, the mirror line becomes the line of symmetry because the object and the image will coincide with each other. In this case, we can conclude that the image is the reflection of the object in the mirror line.

In reflection symmetry, the figure and its reflected image taken together is symmetrical in nature and the mirror line acts as the line of symmetry.

Let us consider another example.

• Take a spread sheet.

Object Image

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E9. Cite three examples of the reflection symmetry, where the object and its mirror image are symmetrical. Show the line of symmetry for each example.

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• Draw the figure XYZ on it.

• Find its mirror image X' Y' Z' with ‘l’ as the mirror line.

• Now compare the lengths of XY & X'Y'; YZ and Y'Z';

XZ and X'Z'.

• What are you observing?

• By using protractor, find out the measure of the

angles XYZ∠ and 'ZY'X'∠ . Are they same?

• Does reflection change the size of the angle?

• Now join XX', YY' and ZZ'.

• Use your protractor to measure the angles between the line l and XX', l and YY' and l and ZZ'.

You can observe that, the angle between the mirror line and the line segment joining a point and its reflected image are the same.

We see that there is no change in the length and angles of a figure in its reflection i.e. the length and angles of the figure and the corresponding lengths and angles of the image are the same.

The line of symmetry is closely related to mirror reflection. When dealing with mirror reflection, we have to take into account the left �� right changes in orientation.

14.5.1 Application of Reflection Symmetry

In our day to day life, we are using the concept of reflection symmetry without knowing the principles of it.

Z

Y

X X’

Y’

Z’ l

Figure 14.27

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E10. Take any square sheet of paper. Draw any figure on that, find out its mirror image and observe the reflection symmetry.

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Now we can discuss another example.

In Figure 14.28, there are two houses A and B. In front of the houses there is a road. Both the houses will be given electric connection, so that equal length of electric line will be required for both the house owners. Then what will be the position of electric pole?

So a point ‘X’ on the side of the road will be identified so that the length of AX and BX will be the same.

Here we can use the concept of reflection symmetry. Let A’ be the image of A in the mirror line (where the side of the row is treated as mirror line). If we can join A’ with B then it will intersect the linear line at the point X. So X is the point which is at equal distance from A and B.

Road

X

I A II

B

A

1

Figure 14.28

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E11. Give an example of the application of reflection symmetry and show it diagrammatically.

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E11. Give an example of the application of reflection symmetry and show it diagrammatically.

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14.6 ROTATIONAL SYMMETRY

The term “rotational symmetry” is associated with the word rotation.

We can see in Figure 14.29(i), that the blades of a fan rotate similarly the hands of the clock also rotate about a fixed point. In our surrounding, we can observe many objects which rotate some of them rotate clock-wise while some of them rotate anti-clock-wise.

Figure 14.29(i) Figure 14.29(ii)

Observe the picture given below.

Figure 14.30(i) Figure 14.30(ii)

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E12. Give two examples each (i) a clockwise rotation (ii) anti-clockwise rotation.

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In Figure 14.30(i) shows that a wheel is rotated. After the rotation, the position of the wheel is given in Figure 14.31(ii).

In both the figures the shape and the size of the wheel do not change and the wheel turns around a fixed point. In this case, the dot in both the figures implies the rotation of the wheel. This is the half turn of the object (i.e. wheel).

Which of the shapes below would look the same after a half-turn.

Activity 8

Now let us do an activity to understand the half turn.

• We can take any shape.

• Trace its outline on a sheet of paper.

• Now keep the shape on its outline and give it a half turn.

• See if the shape fits its outline.

Similarly, we can understand the ¼th turn of an object in Figure 14.31.

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E13. (i) There are some English alphabets which look the same after half a turn. Identify them.

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(i) Think of all 2, 3 digit numbers which look the same after a half urn.

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Figure 14.31

The wind mill is a concrete example of the rotational symmetry. The paper wind mill picture looks symmetrical, in that picture we do no find the line of symmetry. However, if we rotate it by 900 about a fixed point, the wind mill will look exactly the same. We say the wind mill has rotational symmetry.

Let us do an activity to prepare a wind mill.

Activity 9

1. Take a sheet of paper.

2. Fold it as shown in the Figure 14.32(i).

3. Cut out the lower rectangle point of the paper. The sheet of the paper now looks like a square.

4. Fold it along the red lines and then open the fold. Draw a circle on the sheet as shown in the picture.

5. Cut along the lines till you reach the circle. The paper will look like this.

6. Take a pin and make holes on the four corners as shown in the picture.

7. Now fold the corners in a manner that all the holes lie on top of each other.

8. Pass the pin through the holes and fix it in the stick. Now the wind mill is ready.

Follow up questions:

• Does the wind mill look the same on ¼ th turn?

• Does it look same on half turn?

Before turning

1st turn

2nd turn

3rd turn

4th turn

Before Turning After ¼th turn

Figure 14.32(i)

Figure 14.32(ii)

We observe that when an object rotates, its shape and size do not change. The rotation turns an object about a fixed point. The fixed point is the centre of rotation. The angle of turning during rotation is called the angle of rotation. A full turn means a rotation of 3600. Similarly, half a turn rotation means rotation by 1800; and quarter turn (¼ turn) is rotation by 900.

In order to understand the angle of rotation, we can do the following activity.

Activity 10

• Take a square with P as one of its corners.

• Find out the intersecting point of the diagonals and mark it as X.

Figure 14.33

The Figure 14.33(a) is the initial position. Rotation by 900 about the centre leads to Figure 14.33(b).

• Rotation again through 900 and you will get Figure 14.33(c).

• In this way, we can complete from quarter turns and the square reaches its original position.

P P P P P

(a) (b) (c) (d) (e)

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E14. Draw what the following shapes would look like on ¼ th turn and half a turn.

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From the above activity, we can say that the square has a rotational symmetry of order 4 about its centre (as after from 900 turns the square reaches its original position). Thus, a square with a rotational symmetry of order 4 about its centre. The centre of rotation is the centre of square, the angle of rotation is 900, the direction of rotation is clockwise and the order of rotational symmetry is 4.

From the above discussion we can conclude that a figure or an object has rotational symmetry when it can be rotated around a central point or a point of rotation less than 3600 and shall be identical to the original figure or object. In other words, an object with rotational symmetry is an object that looks the same after a certain amount of rotation less than 3600.

14.6.1 Line Symmetry and Rotational Symmetry

In the previous sub-sections of this unit, we have discussed that there are some shapes which have only line symmetry.

In this section, we have discussed the concept of rotational symmetry and figures with rotation symmetry. Similarly, some figures have both line symmetry and rotational symmetry.

Example 1: Think of a square. It is a figure with line symmetry. Can you say, how many lines of symmetry a square has? It has 4 lines of symmetry. Does the Figure 14.34 have any rotational symmetry. What is the order of rotational symmetry.

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E15. Take a equilateral triangle, find its centre of rotation, angle of rotation, direction of rotation and order of rotational symmetry.

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Figure 14.34

Example 2: Draw a equilateral triangle as Figure 14.35. It is a figure with line symmetry. It has 3 lines of symmetry. Does the equilateral triangle have any

rotational symmetry. Yes, it has rotational symmetry with order of rotation 3.

Example 3: We can take a circle. Circle is known as the most perfect symmetrical figure. It can be rotated around its centre through any angle and at the same time it has unlimited number of lines of symmetry. Every line through its centre form a line of (reflection) symmetry and it has rotational symmetry around the centre for every angle.

Figure 14.35

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E16. Name any two figures that have both line symmetry and rotational symmetry.

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E17. If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than one?

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14.7 UNIT SUMMARY

• Symmetry has plenty of application in our day to day life. We are applying symmetry in art, architecture, textile technology, decoration, geometrical reasoning, rangolis, designing, creation etc.

• A figure has line symmetry, if a line can be drawn dividing the figure into two identical parts. The line is called line of symmetry.

• A figure may have no line of symmetry, only one line of symmetry, two lines of symmetry or multiple lines of symmetry. Isolate triangle is an example of a figure with one line of symmetry. Similarly, rectangle and equilateral triangle have two and three lines of symmetry respectively.

• Regular polygons have equal sides and equal angles. They have multiple lines of symmetry.

• In reflection symmetry the figure and its reflected image both are symmetrical in nature and the mirror line acts as a line of symmetry.

• Mirror reflection leads to symmetry, under which the left-right orientation have to be taken care of.

• Rotation turns an object about a fixed point. The fixed point is the centre of rotation. The angle by which the object rotates is the angle of rotation.

• After a rotation, if an object looks exactly same then, we can say it has a rotational symmetry.

• In a complete turn, the number of lines an object looks exactly the same is called the order of rotational symmetry. The order of symmetry of a square is 4 and for an equilateral triangles 3.

• Some shapes have only one line of symmetry and some have only rotational symmetry and some have both the symmetry.

• Circle is the most perfect symmetrical figure because it can be rotated around its centre through any angle as well as it has unlimited lines of symmetry.

14.8 GLOSSARY

Symmetry : Axis of symmetry

Line symmetry

Reflection symmetry

Rotational symmetry


E6. (i) (a) A, C, D, E, I, M, T, U, V, W

(b) H, O, X

(ii) 88, 888

E8. (i) Yes, a scalene triangle

(ii) Yes, an isosceles triangle

(iii)No triangle is possible

(iv) Yes, equilateral triangle.

E13. (i) H, I, N, O, S, X, Z.

(ii) 88, 888

E16. Any two of the following:

Equilateral triangle, Square, Rectangle, Any regular polygon, Circle.

E17. Yes

14.10 ASSIGNMENTS

Q1. Make a list of objects in your surroundings which have line/reflection symmetry. Prepare sketches of five objects each having (i) one line of symmetry, (ii) two lines of symmetry, and (iii) more than three lines of symmetry.

Q2. Draw the sketches of leaves of plants and trees which are symmetrical. Identify the nature of symmetry (whether reflection or rotational) in each case. Indicate the axis/axes of symmetry in case of reflection symmetry. Do you find any leaf having rotational symmetry?

Q3. List the flowers around your locality and identify those having (i) rotational symmetry (ii) reflection symmetry, and (iii) both rotational symmetry. Indicate the axes of symmetry or the centre and angles of rotation as appropriate to the type of symmetry the object has.

Q4. Can an object have both reflection and rotational symmetries? Give at least 5 examples of such objects. Verify and state what should be the minimum number of axes an object having reflection symmetry to have rotational symmetry?

Q5. List the objects or figures having rotational symmetry of order 2 or more than 2.

14.11 REFERENCES

• Mathematics Textbooks for Classes VI and VII.

MathsModuleVol II

Documents

Transcript of MathsModuleVol II