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Integrated Syllabus

FOUNDATION & OLYMPIAD CLASS - VIII

FOUNDATION & OLYMPIAD MATHEMATICS CLASS - VIII

www.bmatalent.com

Published by:

Brain Mapping Academy #16–11–16/1/B, First Floor, Farhat Hospital Road, Saleem Nagar, Malakpet, Hyderabad– 500 036 Andhra Pradesh, India. ✆ 040–65165169, 66135169 E–mail: [email protected] Website: www.bmatalent.com

Brain Mapping Academy ALL RIGHTS RESERVED C

No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher.

Publication Team Authors: Y.S. Srinivasu Design & Typing: P. S. Chakravarthi & Syed Ashraf Ali

ISBN: 978-81-907285-3-9

Disclaimer Every care has been taken by the compilers and publishers to give correct, complete and updated information. In case there is any omission, printing mistake or any other error which might have crept in inadvertently, neither the compiler / publisher nor any of the distributors take any legal responsibility. In case of any dispute, all matters are subject to the exclusive jurisdiction of the courts in Hyderabad only.

Prefac e Speed and accuracy play an important role in climbing the competitive ladder. Students have to integrate the habit of being able to calculate and function quickly as well as efficiently in order to excel in the learning culture. They need to think on their feet, understand basic requirements, identify appropriate information sources and use that to their best advantage. The preparation required for the tough competitive examinations is fundamentally different from that of qualifying ones like the board examinations. A student can emerge successful in a qualifying examination by merely scoring the minimum percentage of marks, whereas in a competitive examination, he has to score high and perform better than the others taking the examination. This book provides all types of questions that a student would be required to tackle at the foundation level. It will also help the student in identifying the pattern of questions set for various competitive examinations. Constant practice and familiarity with these questions will not only make him/her conceptually sound, but will also give the student the confidence to face any entrance examination with ease. Students are advised to go through every question carefully and try to solve it on their own. They should also attempt different methods and alternate processes in reaching the desired solution and seek their teacher’s help if required. Valuable suggestions as well as criticism from the teacher and student community are most welcome and will be incorporated in the ensuing edition.

Publisher

CONTENTS

1.

Number System ............................. 07

2. Exponents and Radicals ............... 46 3. Sets ................................................ 65 4. Mensuration – I ............................. 93 5. Polynomials – I .............................. 126 6. Linear Equations .......................... 148 7. Inequalities – I .............................. 177 8. Arithmetic .................................... 207 9. Plane Geometry – I ...................... 231 10. Coordinate Geometry – I ............ 275 Answers ....................................... 328

IIT Foundation & Olympiad Explorer VIII

Mathematics / Class -

er Chapter

MENSURRTION-I

4 1 SYNOPSIS

I. Areas and Dimensions of Plane Figures 1. Triangle (i) Equilateral Triangle A

a

a

B Area A

=

=

but h =

A =

2

1 2

a

h a 2

D

C

base height

1

b h

2

3 a 2 1 2

3 3 2 a a= a 2 4

(ii) Right - Angled Triangle Area A

=

= but h = A = 4. Mensuration - I

1 base height 2 1 2

h

bh d

2

A

d

2

b

1 b( d 2 2

B

b2 ) 93

b

C

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(iii) Acute - Angled Triangle s s

Area (A) =

Where S =

a

b 2

a s

b s

A

c

h

c

B

S is the semi - perimeter of the triangle

A =

1 b bh = 2 2

a2

a2

b2 2b

D

C

2

c2

(iv) Obtuse Angled Triangle A b c B Area (A) =

Where S =

A=

s s a

a s

b 2

a

C

b s c

c

c2

1 b 2 a bh = 2 2

a2 b2 2b

2

2. Quadrilaterals (i)

Square

a

a

d

(i)

Area =

(side)2

(ii)

Side (a) =

(iii)

Perimeter(s) = 4 side = 4a

a

a = a2

area

(iv) Diagonal (D) = 4. Mensuration - I

2 side

2

2 94

Area

1.414

Area

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(ii) Rectangle

d

b

l (i)

Area (A) = length

(ii) Area (A) = l d 2

breadth = l b l2 = b .

d2

b2

(iii) Perimeter (S) = 2 (length + breadth) = 2(l + b) (iv) Diagonal

(D) =

l2

b2

(iii) Parallelogram

h b (i)

Area (A) = base height = bh

(ii) b =

A h

(iii) h =

A b

(iv) Rhombus d2 dl

Area (A)

4. Mensuration - I

=

l 2

product of its diagonals

=

l 2

dl d2 95

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(v)

Mathematics / Class VIII

Trapezium b

h

h

a

1 (a + b)h 2

---

---------

---------

(vi) Trapezoid ---

h

b

H

Area (A) =

h a

a bh

height

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

=

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

Area (A)

1 = (sum of parallel sides) 2

H

c

CH

2 A trapezoid can be divided into two triangles as indicated by dotted lines. The area of each of these triangles is calculated, and the result added to find the area of trapezoid. 3. Circle 2

(i)

Area of circle (A) =

r2 =

d 4

Where r = radius d = diameter of a circle

r

(ii)

Circumference = 2 r =

(iii)

Radius =

area

(iv) Diameter = 4. Mensuration - I

d = 3.l4d

circumference

96

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SOLVED EXAMPLES Example 4.1 The difference of the area of the circumscribed and the inscribed squares of a circle is 35 sq.cm. Find the area of the circle.

X

Solution: Let the radius of the circle be x cm diameter = 2x cm Now, diagonal of the inscribed square = 2x cm Area of the inscribed square =

diagonal 2 2

=

(2x)2 = 2x2 sq. cm 2

Again side of the circumscribed square = 2x cm Area of the circumscribed square = (2x)2 sq.cm = 4x2 sq.cm So, by the problem, 4x2 - 2x2 = 35 or

2x2 = 35 x2 =

Area of circle =

4. Mensuration - I

35 2 22 35 x 2 = 7 2 = 55 sq.cm.

98

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Example 4.2 The sides of a triangle are a = l3 cm, b = l 4 cm, c = l5 cm, the sides a and b are the tangents to a circle , whose centre lies on the third side. Find the circumference of the circle. Note:

l. Tangent is a line touching the circle at one point. 2. Radius is perpendicular to the tangent. c

A

B

0

Solution: The centre O of the circle lies an AB and let r cm be the be the radius of the cirlce. Since radius is perpendicular to the tangent at the point of contact. The area of

l x l3 x r sq.cm and 2

BOC

Hence the total area of the

ABC

BOC

=

l l .l3. r .l4. r 2 2

=

l x r x 27 sq.cm 2

Again if S = semi-perimeter =

AOC

l x l4 x r sq.cm 2

AOC

sq.cm

l3 l4 l5 cm 2

= 2l cm The area of

ABC =

s(s a)(s

b)(s

c) =

2l(8) (7) (6) = 84 sq.cm

Comparing these two,

l x r x 27 = 84 2 or

r=

56 9

cm

Circumference of the circle = 2 r 4. Mensuration - I

99

2

22 7

56 l = 39 cm 9 9

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Example 4.3 The two adjacent sides of a parallelogram are 5 cm and 4 cm respectively and if the respective diagonal is 7 cm, then find the area of the parallelogram? Solution: Required area = 2 where

s =

a

s(s a)(s

b D 2

= 2 8(8

5)(8

= 2 8 3 4

b)(s

D)

5 4 7 =8 2 4)(8

7)

8 6 = l9.6 sq.cm

Example 4.4 A 5l00 sq.cm trapezium has the perpendicular distance between the two parallel sides 60 cm. If one of the parallel sides be 40 cm then find the length of the other parallel sides. Solution: Let the length of the unknown parallel side be 'x' cm. A =

or

5l00 =

or

l (a + b) h 2 l (40 + x) x 60 2

l70 = 40 + x Required other parallel side = l70 - 40 = l30 cm Example 4.5

How many metres of a carpet 75 cm wide will be required to cover the floor of a room which is 20 metres long and l2 metres broad? Solution: Length required =

length of room breadth of room width of carpet

Length required =

20 l2 = 320 m 0.75

Example 4.6 How many paving stones each measuring 2.5 m x 2 m are required to pave a rectangular courtyard 30 m long and l6.5 m wide?

4. Mensuration - I

100

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Solution: Number of teles requied =

length breadth of courtyard length breadth of each tile

=

30 l6.5 = 99 2.5 2

Example 4.7 A hall-room 39 m l0 cm long and 35 m 70 cm broad is to be paved with equal square tiles. Find the largest tile so that the tiles exactly fit and also find the number of tiles required. Solution: Side of largest possible tile = H.C.F of length and breadth of the room = H.C.F of 39.l0 and 35.70 m = l.70 m Also, number of tiles required length breadth of room = (H.C.F of length and breadth of the room)2 =

39.l0 35.70 l.70 l.70

= 483

Example 4.8 A rectangular grassy plot is ll2 m by 78 m. It has a gravel path 2.5 m wide all round it on the inside. Find the area of the path and the cost of constructing it at Rs. 2 per square metre? Solution: Area of path = (l2 78) - (l07 73) = 925 sq.m Cost of construction = rate x area = 2 x 925 = Rs. l850 Example 4.9 A square field of 2 sq. kilometers is to be divided into two equal parts by a fence which coincides with a diagonal. Find the length of the fence. Solution: Area of square = 2 km2 Diagnol =

2 2 km = 2 kilometeres.

Hence length of the fence = 2 km 4. Mensuration - I

101

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Other diagonal = 2 x

Now, area =

(36.5)2

55 2

2

= 48 cm

l l (product of diagonals) = x 48 x 55 = l320 sq.cm 2 2

Example 4.13 A hall whose length is l6 m and breadth twice its height takes l68 m of paper 2 m wide for its four walls. Find the area of the floor. Solution: Let the breadth = 2h m, then height = h m Area of walls = 2(l6 + 2h) h sq.m Area of paper = l68 x 2 sq.m 2(l6 + 2h)h = l68 x 2

(8 + h) h = 84

On solving h = 6, -l4; -l4 is not acceptable h = 6 and breadth = l2 Area of floor = l6 x l2 sq.m = l92 sq.m Example 4.14 The length of a rectangle is increased by 60%. By what per cent should the width be decreased to maintain the same area? Solution: Let the length and breadth be x and y then its area = xy New length = x

l60 l00

8x 5

As the area remains the same, the new breadth of the rectangle =

Decrease in breadth = y % decrease in breadth =

3y 5y = 8 8

xy = 5y 8x 8 5

3y l00 75 l = = 37 % 8 y 2 2

Example 4.15 If the length of a rectangle increases by l0% and the breadth of the rectangle decreases by l2% then find the % change in area. Soluton: Let length = l00 units and breadth = l00 units 4. Mensuration - I

103

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Then Area = l00

Mathematics / Class VIII

l00 = l0000 square units

New length = ll0 units New breadth = 88 units New Area = ll0 88 = 9680 sq. units % change in area =

320 l00 = 3.2% l0000

Example 4.16 (a) (b) (c)

What is the relation between a circle and an equilateral triangle which is inscribed in the circle? What is the relation between an equilateral triangle and a circle inscribed in a circle? An equilateral triangle is circumscribed by a circle and another circle is inscribed in that triangle. Find the ratio of the areas of two circles?

Solution: (a)

The area of a circle circumscribing an equilateral triangle of side x is

(b)

The area of a circle inscribed in an equilateral triangle of side x is

(c)

From the above, we can say that the required ratio =

3 =

x2 :

l2

x2 3

x2 . l2 x2

l l : =4:l 3 l2

Example 4.17 The front wheels of a wagon are 2 m in circumference and the back wheels are 3 m feet in circumference. When the front wheels have made l0 more revolutions than the back wheels, how many metres has the wagon travelled? Solution: Suppose the back wheel has made x revolutions. Front wheel has made (l0 + x) revolutions. 3 x=2 x=2

(l0 + x) x l0

x = 20

The wagon has travelled 3 x = 60 . Example 4.18 Amar drew a square. He then erased it and drew a second square whose sides were 3 times the sides of the first square. By what percent was the area of the square increased? 4. Mensuration - I

104

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Example 4.21 ABCD is a rectangle with sides AB = x and AD = y. E is the mid point of DC. Then find the area of the shaded portion? A B

Soluton: Area of

D l ABE = x AB x 2

E

C

from E on AB =

l xy. 2

Example 4.22 The area of the larger square is a2 and that of the smaller square is b2. Then

Area of the shaded portion find Area of the l arg er square . Solution: a 2 b2 a2

=l-

b2 a2

= l

b a

l

b . a

Example 4.23 What will be the perimeter of a rectangle if its length is 3 times its width and the length of the diagonal is 8 l0 cm? Solution: 9B2 + B2 = (8 l0 )2 = 640

B2 = 64

8 10

B=8

B

3B Perimeter = 8B = 64 cm. 4. Mensuration - I

106

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Example 4.24 What will be the ratio of the circumference to the diameter of a circle if its original radius is tripled? Solution: 2 r = Ratio of the circumference to the diameter of the circle = 2r radius.

= Independent of the

Example 4.25 There are two 2-metre wide cross roads in a lawn l50 m by l20 m dimensions. One of the roads is parallel to the length and the other is parallel to the breadth. If it costs Rs. 2 per sq. metre for levelling the road, what would be the cost involved ? Solution: Area of the road = l50 x 2 + l20 x 2 - 2 x 2 = 300 + 240 - 4 = 536 sq m. The cost for levelling the road = 536 x 2 = Rs. l072 Example 4.26 What is the area of the triangle in which two of its medians 9 cm and l2 cm long intersect at right angles? Solution: Area of the triangle = 2 x

l x l2 x 6 = 72 sq.cm 2

Example 4.27 If one leg of an isosceles right-angled triangle is increased by 6 cm and that of the other leg decreased by 4 cm, then the area of the triangle decreases by 24 sq cm. Find the length of the leg of the original triangle. Solution: Let x be the length of the leg of the right-angled isosceles triangle, originally.

X

X Its area =

4. Mensuration - I

l 2 x 2

107

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The area of the new triangle =

l (x - 4) (x + 6) 2

X 4

X+6 l 2 l x = (x - 4) (x + 6) + 24 2 2 x2 = x2 + 2x - 24 + 48

x = -l2.

Example 4.28 What is the ratio of the heights of two isosceles triangles which have equal vertical angles, and of which the areas are in the ratio of 9 : l6? Solution: Let the heights be hl and h2. Let the bases be Bl and B2 respectively. l Bl hl 9 2 = l l6 B2 h 2 2 Let

Bl B = 2 = k, say hl h2 h l2 h 22

9 l6

hl h2

3 4

Example 4.29 Four horses are tethered at four corners of a square plot of side l4 metres (m) so that the adjacent horses can just reach one another. There is a small circular pond of area 20 m2 at the centre. Find the ungrazed area. Solution: Total area = l4 x l4 = l96 m2 Grazed area =

r2 4

x 4=

r2 = 22 x 7 (r = 7) = l54 m 2

Ungrazed area = (l96 - l54) = 42 m2 4. Mensuration - I

108

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CONCEPTMAP======================== Plane Figures: A flat surface, which any two points are joined by a straight Iiiii! lying entirezy on the surface

.-----------------11 in

Aeute angled triangle

Circle:

4 I

r-;Jt

-----------.

P=a+b+c

Circumferem:e = 211T or !!d

A= )s<s-a)(s-b)(s-c)

V

Area =

0

ltr

V ad

aSquare

P =a+a + a+a

= !rd'

= 411

A = (side)' = a

A N

a

Radius=

6 L

d.=.J2(a't

•s

Diametre cuc ueru:e

=o:../2

..tl'loJu

=.J?i.area)

Jlpr•

..

having

,

Q

t ll n 1

U'

Semi-cdrde

'--r-

} r

P=2r+11'T = r(x + 2) r

r

A=.!.lll"' 2

P=2l +2b ='llJ.+b)

.. I

£

A

r

A=lb

d=.Jf +b'

A D

.. li

= t.Jti.' -l'

A

= Seetor

£

s

Paralle"Joaram l=271r_! 86d' 1178 or 180

y

P=2l+2h =2(l+b)

A=bh

b

P= xr9 +2r 180 llrl'l" 1 A=-- =-lr 360 2

Obtwle angled triangle

P= a +b +c a

Trapellli-

Cit)Ellipse: b

A= .Js<s- a) (ft-b) (s-a)

a -+ T7Uijor CJXill b -+ minor I1Zis

P=a+b+c+d

b

h h A= (a+h)lo a

P= 2(a'+b') A=xab

or

h

a

c

Trapeoroid

Sum of the &idea

A= (H+h)a+bh+CH 2

Composite

.6-. @)j))

Area

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

Outer .Area. -Inner area

A.+A..,. +A,.. Area ofoquare -Area of four BoctOl'B

IIT Foundation & Olympiad Explorer

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BASIC PRACTICE l.

What is the area of the shaded region? 15 cm

8 cm

2.

12 cm In the figure drawn below, its dimensions are given in metres. (a) Find the perimeter of the figure in terms of C. (b) If C = 3, find the total cost to fence the garden if the cost of fencing is o 7 per metre. 9 c 12 6

3.

c

(5 + c) What is the area of triangle ABC? A

D 4 cm C

4.

B The figure below is made up of an isoceles triangle and a semi-circle. Find the perimeter of the figure.

8 cm

4. Mensuration - I

111

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5.

6.

Mathematics / Class VIII

The figure is made up of two identical small circles and a large circle with diameter 28 22 , express the shaded area as a fraction of the total area. cm. Taking 7

28 cm The ratio of the area of A to the area of D is 4 : 5 and the ratio of the area of B to the area of C is l : 3. Find the ratio of the area of A to the area of C. A B

D C

7.

The figure is made up of 2 big identical quarter circles of radius 8 cm and a small quarter circle. Find the perimeter of the figure.

3 cm

8.

The figure is made up of 2 rectangles 3 cm 3 ctn

7r cm 4r cm

3 ctn

3 cm

9.

(a) What is the perimeter of the big rectangle? (b) What is the area of the small rectangle? The figure shows a trapezium PQRS. Given that RQ = PQ = 8 cm and the area of triangle OPQ is l2.8 cm2, what is the area of triangle OPS?

4. Mensuration - I

112

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s

R

0

l0.

Mathematics / Class VIII

8 cm

8 cm p Q By what percentage will the area of a square change if its side increases by l0%?

ll.

The wheel of a cycle covers 660 metres by making 500 revolutions. What is the diameter of the wheel (in cm)?

l2.

In the given figure, what is the minimum distance, in metres, that a person would have to walk to go from the point A to a point on the side BC? A

B

C

Note : The diagram is not drawn to scale. l3. l4.

Find the ratio of the area of a square to that of the square drawn on its diagonal? A wire bent in the form of a square enclosed an area of l2l sq cm. If the same wire is bent so as to form a circle, then find the area enclosed?

l5.

All the three quadrilaterals ADEC, ABIH and BCGF are squares and ABC = 90o. If the area of ADEC = x2 and area of AHIB = y2 (x2 > y2), then find the area of BCGF? D

E H

A

I

B

F 4. Mensuration - I

113

C

G

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24.

Find the area of quadrilateral ABCD given below.

25.

Find the area of the shaded portion. Given that each circle is having a radius of 2 cm.

26.

PQRS is the diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Find the area of the shaded portion.

R2 P

27.

Q

R Rl

S

In the given figure B 90o , AC is diameter of a semicircle and with BC as radius, a quarter circle is drawn, find the area of the shaded portion.

4. Mensuration - I

115

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FURTHER PRACTICE l.

Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centres at P and R respectively, and S is the centre of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circle and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semicircles with diameter AB that cannot be grazed by the horses is nearest to:

S

A

A) 20 2.

B) 28

R

B

C) 36

D) 40

A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the length OP and AB respectively. Suppose APB = 60o then the relationship between h and b can be expressed as: A) 2b2 = h2

3.

0

P

B) 2h2 = b2

C) 3b2 = 2h 2

D) 3h2 = 2b2

In the figure below, the rectangle at the corner measures l0 cm x 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm? A

A) l0 cm 4.

C) 50 cm

D) 30 cm

Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side 'a' in the following steps: the original square of side 'a' is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side 'a' and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the later is: A) l : l

5.

B) 40 cm

B)

C)

2 :l

n(4 4n

)

D)

4n

n(4 ) In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with

4. Mensuration - I

116

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centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is AB at E such that

ODC =

: 3 . The line segment DE intersects

ADE. What is the ratio AE : AD? A

E

B 0

D A) l : 3 6.

B) l : 2

C C) l : 2 3

D) l : 2

Let Sl be a square of side 'a'. Another square S2 is formed by joining the mid-points of the sides of Sl. The same process is applied to S2 to form yet another square S3, and so on. If Al, A2, A3, ... be the areas and Pl, P2, P3, ... be the perimeters of Sl, S2, S3, ...., respectively, then the ratio: Pl P2 P3 ... equals: Al A 2 A 3 ... A)

7.

2(l

2) a

2(2

2) a

C)

2(2

2) a

D)

2(l 2 2) a

A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle? A) l6.665

8.

B)

B) l6.565

C) l5.465

The sides of a triangle are in the ratio of

D) l4.365

l l l : : . If the perimeter is 52 cm, then the 2 3 4

length of the smallest side is: 9.

A) 9 cm B) l0 cm C) ll cm D) l2 cm A, B and C start running at the same time and from the same point around a circular track of 70 metres radius. A and B run clockwise and C counter clockwise. If A meets C every 88 seconds and B meets C every ll0 seconds, then A meets B every seconds.

l0.

A) 22 B) l98 C) 440 D) 2l2 A pond l00 m in diameter is surrounded by a circular grass walk 2 m wide. How many square meters of grass is there on the walk?

ll.

A) 98 B) l00 C) 204 D) 202 The length of a rectangle is increased by 60%. By what percent would the width be decreased so as to maintain the same area? A) 37

l % 2

4. Mensuration - I

B) 60%

C) 75%

117

D) l20%

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24.

Mathematics / Class VIII

In the figure, ABCD is a square with side l0. BFD is an arc of a circle with centre C. BGD is an arc of a-circle with centre A. What is the area of the shaded region? 10

A

B

10

D

25.

26.

A) l00 - 50 B) l00 - 257 C) 50 - l00 D) 25 - l00 A steel wire bent in the form of a square of area l2l cm 2. If the same wire is bent in the form of a circle, then the area of the circle is: A) l30 cm 2 B) l36 cm2 C) l54 cm2 D) l68 cm2 The number of revolutions made by a wheel of diameter 56 cm in covering a distance of 22 : 7 A) 3l.25 B) 56.25 C) 625 D) 62.5 Semi-circular lawns are attached to the edges of a rectangular field measuring 42 m x 35 m. The area of the total field is: A) 38l8.5 m2 B) 83l8 m2 C) 58l3 m2 D) l358 m2 l.l km is

27.

28.

29.

30.

3l.

32.

33.

c

Use

A wire is in the form of a circle of radius 35 cm. If it is bent into the shape of a rhombus, what is the side of the rhombus? A) 32 cm B) 70 cm C) 55 cm D) l7 cm The cross-section of a canal is in the form of a trapezium. If the canal top is l0 m wide and the bottom is 6 m wide, and the area of cross-section is 72 m 2, then the depth of the canal is: A) l0 m B) 7 m C) 6 m D) 9 m A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square. A) 44 cm B) 45 cm C) 46 cm D) 48 cm A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running round it on the outside. Find the cost of gravelling the path at Rs. 4 per square metre. A) Rs. 2002 B) Rs. 2003 C) Rs. 2004 D) Rs. 2000 If the circumference and the area of a circle are numerically equal, then what is the numerical value of the diameter? A) l B) 2 C) 4 D) A wire is in the form of a circle of radius 35 cm. If it is bent into the shape of a rhombus, what is the side of the rhombus? A) 32 cm B) 70 cm C) 55 cm D) l7 cm

4. Mensuration - I

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44.

Mathematics / Class VIII

In the figure, when the outer circles all have radii 'r', then the radius of the inner circle will be:

A)

2r

B) ( 2 l)r

C)

l 2r

D)

2 ( 2

l)r

BRAIN WORKS l.

Find the perimeter of the shaded portion.

2.

PQRS is the diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn with PQ and QS as diameters as shown in figure. Find the ratio of the area of the shaded region to that of the unshaded region.

p

3.

Q

R

s

In the given figure DC = CB = 4 cm and AE = 2 cm. Calculate the area of shaded portion, if DC AB.

4. Mensuration - I

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Mathematics / Class VIII

4.

Find the area of the shaded portion.

5.

The figure below shows two concentric circles with centre 'O'. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD? p

B

Q

A

0

C

s

D

R

6.

Four identical coins are placed in a square. For each coin, the ratio of area of circumference is same as the ratio of circumference to area. Then, find the area of the square that is not covered by the coins.

7.

AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is l5 cm, AC is l2 cm and BD is 9 cm, find the area of the quadrilateral ACBD. C A

8.

B

D The adjoining figures shows a set of concentric squares. If the diagonal of the innermost square is 2 units, and if the distance between the corresponding corners of any two successive squares is l unit, find the difference between the areas of the eight and the seventh squares, counting from the innermost square.

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l5.

Mathematics / Class VIII

Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three angles are not right angles. 25

24

25

32

What is the area of the plot (in m 2)? l6.

Euclid has a triangle in mind. Its longest side has length 20 and another of its sides has length l0. Its area is 80. What is the exact length of its third side?

l7.

A rectangular pool 20 metres wide and 60 metres long is surrounded by a walkway of uniform width. If the total area of the walkway is 5l6 square metres, how wide, in metres, is the walkway?

l8.

Consider a circle with unit radius. There are seven adjacent sectors, Sl, S2, S3 . . . S7, in the circle such that their total area is

l 8

of the area of the circle. Further, the area of

the jth sector is twice that of the (j - l)th sector, for j = 2, ... 7. Find the area of the sector Sl. l9.

A former has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?

20.

What is the number of distinct triangles with integral valued sides and perimeter as l4?

MULTIPLE ANSWER QUESTIONS l.

2.

3.

If each side an equilateral triangle is 4 3 cm then which of the following is/are true? A) If area is natural number B) Numerical value of area and perimeter are same/equal C) Perimeter s an irrational number D) All of these are correct If area of a triangle is l2 cm2, then which of the following pairs of base and height are possible? A) Base = l2 cm, Height = l cm B) Base = 6 cm, Height = 3 cm C) Base = 8 cm, Height = 3 cm D) Base = 4 cm, Height = 6 cm Area of any triangle will be doubled if A) base be doubled B) each side is doubled C) height is doubled D) Both base and height are doubled

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PARAGRAPH QUESTIONS

Passaee - I l.

Area of a rectangle is twice the area of a triangle with base a and height b. The figure below is made up of a square, a triangle and a rectangle. Find the area of the shaded region. 6 ctn 4 ctn 9 ctn

A) 32 m 2

B) 40.5 m 2

C) 52.5 cm 2

D) 72 m2

Passaee - II Area of a sector is

l. a.

r2

360o All the sides of a square are equal. The figure below consists of a square, a quarter circle and a semicircle. Find the area of the shaded portion. b. Find the perimeter of the shaded portion.

lO ctn

lO ctn

2.

The figure is made up of a rectangle and 2 similar semicircles. Find the perimeter of the shaded portion. 43 tn

2l tn

2l tn

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e

Simple, clear and systematic presentation

e

Concept maps provided for every chapter

e

Set of objective and subjective questions at the end of each chapter

e

Previous contest questions at the end of each chapter

e

Designed to fulfill the preparation needs for international/national talent exams, olympiads and all competitive exams

UNIQUE ATTRACTIONS ● ●

Cross word Puzzles



Graded Exercise ■

Basic Practice



Further Practice



Brain Works



Multiple Answer Questions



Paragraph Questions

Detailed solutions for all problems of IIT Foundation & Olympiad Explorer are available in this book

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