of Textbook Production and Curriculum Research, Pune. …targetpublications.org/media/catalog/product/pdf/7th... · · 2017-12-215.3 cm ii. 6.7 cm iii. 3.8 cm Ans: i. 5.3 cm Line

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

STD. VII Mathematics

Printed at: Repro India Ltd., Mumbai

P.O. No. 58247

Written as per the syllabus prescribed by the Maharashtra State Bureau

of Textbook Production and Curriculum Research, Pune.

10740_11610_JUP

PREFACE Preparing this Mathematics book was a rollercoaster ride. We had a plethora of ideas, suggestions and decisions to ponder over. However our basic premise was to keep this book in line with the new, improved syllabus and provide students with an absolutely fresh material. To begin with, let us look at this book as a ‘powerful concept building tool’. We want this book to act as a facilitator for students to deeply understand mathematical concepts presented in the class VII book by the Maharashtra State Education Board. The understanding of these concepts would eventually help students, link textual problems with their daily life and comprehend its application for future use. Every chapter covers a multitude of solved examples related to the topic. These examples are textual as well external practice problems, so as to reinforce the topic’s understanding within the reader. The part of Formative Assessment covers Activity Based Questions from the text book. We’ve partially solved these questions and added additional ones for practice sake. Every chapter ends with an Assessment Test. This test stands as a testimony to the fact that the child has understood the chapter thoroughly. The Multiple Choice Questions included in this test facilitate students to prepare for competitive examinations. All the diagrams are neat and have proper labelling. The book has a unique feature that all the constructions are as per the scale. With absolute trust in our work, we hope, our holistic efforts towards making this book are paid off if students understand mathematics conceptually rather than just focusing on the problem solving part. This text would definitely act as a reference point for the same. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! From, Publisher

Contents

No. Topic Name Page No.

Part One

1. Geometrical Constructions 1

2. Multiplication and Division of Integers

23

3. HCF and LCM 28

4. Angles and Pairs of Angles 47

5. Operations on Rational Numbers 64

6. Indices 78

7. Joint Bar Graph 88

8. Algebraic Expressions and Operations on them

99

Miscellaneous Problems : Set 1 108

No. Topic Name Page No.

Part Two

9. Direct Proportion and Inverse Proportion

117

10. Banks and Simple Interest 127

11. Circle 137

12. Perimeter and Area 144

13. Pythagoras’ Theorem 157

14. Algebraic Formulae – Expansion of Squares

164

15. Statistics 172

Miscellaneous Problems : Set 2 178

1

Chapter 1: Geometrical Constructions

1. Angle bisector : A ray which divides an angle in two equal parts is called as angle bisector of that angle. Example: In the adjacent figure, is ray BM the bisector of ABC? (Textbook pg. no. 1) Ans: mABM = m CBM = 20 Ray BM is the bisector of ABC. 2. Perpendicular Bisector of a Line Segment: A line which divides a line segment in two equal parts and is perpendicular to it is called the

perpendicular bisector of that line segment. Example: 1. Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1) i. How will your verify that CD is the perpendicular bisector? m CMS = __________ ii. Is l(PM) = l(SM)? Ans: Here, mCMS = 90 Also, l(PM) = l(SM) = 2cm line CD is the perpendicular bisector of seg PS. 1. The three angle bisectors of a triangle are concurrent

i.e. they pass through a same point. In the adjacent figure, the angle bisectors of A, B

and C pass through the common point I. 2. The lengths of the perpendiculars from the point of

concurrence to the sides are equal. In the adjacent figure IP, IQ and IR are perpendicular

to AB, BC and AC respectively. l(IP) = l(IQ) = l(IR) = 1.2 cm.

Summative Assessment

The property of the Angle Bisectors of a Triangle

P S4 cm

M

C

D

C

A

B

M

A

R

C

P

I

Q B

Let’s Recall

Let’s Study

1. Geometrical Constructions

2

2

Std. VII: Mathematics 1. The perpendicular bisectors of the sides of a triangle are concurrent. In the figures below, the perpendicular bisectors pass through the common point C 2. Point C is equidistant from the three vertices of the triangle i.e. l(PC) = l(QC) = l(RC). Example: Use a ruler to draw an acute angled triangle and an obtuse angled triangle. Draw perpendicular

bisectors of each side of the two triangles. Note your observation. (Textbook pg. no. 2) Ans: The perpendicular bisectors of the sides of both triangles are concurrent. 3. Position of the point of concurrence of the perpendicular bisectors of the sides of triangle: i. Acute angled triangle: In the interior of the triangle ii. Right angled triangle: On the hypotenuse iii. Obtuse angled triangle: In the exterior of the triangle.

The property of Perpendicular Bisectors of the sides of a Triangle

P

S

Q T

R

U

C

C

US

P

QTR

3


1. The point of concurrence of angle bisectors of a triangle is called the incentre and is shown by the letter ‘I’ 2. The point of concurrence of perpendicular bisectors of a triangle is called circumcentre and is shown

by the letter ‘C’. 1. Draw line segments of the lengths given below and draw their perpendicular bisectors: i. 5.3 cm ii. 6.7 cm iii. 3.8 cm Ans: i. 5.3 cm Line AB is the perpendicular bisector of seg PQ.

ii. 6.7 cm Line UV is the perpendicular bisector of seg ST.

iii. 3.8 cm Line ST is the perpendicular bisector of seg LM.

Something more

Let’s Practise : Practice Set 1

M

S

L O

T

3.8 cm

B

A

PM

Q 5.3 cm

V

U

SW

T

6.7 cm

4

4

Std. VII: Mathematics

2. Draw angles of the measures given below and draw their bisectors: i. 105 ii. 55 iii. 90 Ans:

i. 105° Ray BD is the angle bisector of ABC.

ii. 55°

Ray QS is the angle bisector of PQR.

iii. 90 Ray MT is the angle bisector of LMN.

3. Draw an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the

angle bisectors of each triangle. Where do the points of concurrence lie? Ans: The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

C

A

D

B

105Q

P

S

R

55

M N

L

T

K

L M

P

QN

I

A

D

E

I F

CB

5


4. Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?

Ans: The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on

the hypotenuse. 5. Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the

three houses. Which geometrical construction should be used to represent this? Explain your answer. Ans: The position of the toy shop which is equidistant from three houses can be found out by drawing the

perpendicular bisector of the sides of the triangle joining the three houses. The shop will be at the point of concurrence of the perpendicular bisectors.

Q

W

V

R

T

XC

S

P

U

DJ

F

HC

Maithili’sHouse

Ajay’s House

(Shop)

Shaila’sHouse

E

G

6

6


Try This

I. To construct a triangle given the lengths of its three sides: Example: Draw ABC such that l(AB) = 6 cm, l(BC) = 4 cm, l(AC) = 5cm Step 1 : As shown in the rough figure, draw seg AB of length 6 cm as the

base. Step 2 : Take a distance of 5 cm on the compass. Place the metal tip of

the compass on A, draw an arc on one side of AB. Step 3 : Take a distance 4 cm on the compass. Place the metal tip of the compass on B and draw an arc

such that it intersects the previous arc. Name the point C. Step 4 : Draw segments AC and BC to get the triangle. Note: Triangle can also be constructed by drawing arcs on other side of AB as shown below. 1. Draw ABC such that l(AB) = 4 cm, and l(BC) = 3 cm. (Textbook pg. no. 3) i. Can this triangle be drawn? ii. A number of triangles can be drawn to fulfil these conditions. Try it out. iii. Which further condition must be placed if we are to draw a unique triangle using the above

information?

Ans: i. ABC triangle cannot be drawn as length of third side is not given.

ii. For ABC to draw l(AC) l(AB) + l(BC) i.e., l(AC) > 4 + 3 i.e., l(AC) > 7cm

number of triangles can be drawn if l(AC) > 7 cm

iii. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

Construction of a triangle

5 cm 4 cm

6 cm B

C

A

Rough Figure

5 cm 4 cm

6 cm B

C

A

Rough Figure

5 cm 4 cm

6 cmA B

C

5 cm 4 cm

6 cm A B

C

Let’s Study

7


1. Draw triangles with the measures given below: i. In ABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm.

Ans: ii. In STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm.

Ans: iii. In PQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm.

Ans: 2. Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.

Ans:

5 cm

7 cm T

U

S

Rough Figure

4 cm

4.5 cm

6 cmQ

R

P

Rough Figure

3.8 cm

3.5 cm 3.5 cm

5 cm

P

Q R

3.5 cm

5 cm R

P

Q

Rough Figure

3.5 cm

3.5 cm

5.5 cm B

C

A

Rough Figure

4.2 cm

3.5 cm 4.2 cm

5.5 cmBA

C

5 cm 4 cm

7 cm TS

U

4.5 cm 3.8 cm

6 cmQP

R

Let’s Practise : Practice Set 2

8

8


3. Draw an equilateral triangle with side 6.5 cm. Ans: 4. Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene

triangle. Ans: i. Equilateral triangle LMN, l(LM) = l(MN) = l(LN) = 4 cm. ii. Isosceles triangle STU, l(ST) = l(TU) = 4 cm, l(SU) = 6 cm. iii. Scalene triangle XYZ, l(XY) = 4.5 cm, l(YZ) = 6.5 cm, l(XZ) = 5.5 cm. Note: Students may prepare and solve problems of their own.

6.5 cm

6.5 cm C

A

B

Rough Figure

6.5 cm

4 cm

4 cm N

L

M

Rough figure

4 cm

4.5 cm

6.5 cm Z

X

Y

Rough Figure

5.5 cm

4 cm

6 cm U

T

S

Rough Figure

4 cm

6.5 cm

4.5 cm 5.5 cm

X

Y Z

6.5 cm

6.5 cm

A

B C

6.5 cm

4 cm4 cm

4 cm M N

L

4 cm4 cm

6 cmS U

T

of Textbook Production and Curriculum Research, Pune. …targetpublications.org/media/catalog/product/pdf/7th... · · 2017-12-215.3 cm ii. 6.7 cm iii. 3.8 cm Ans: i. 5.3 cm Line

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