integrated syllabus (free sample) · detailed solutions for all problems of iit foundation &...

UNIQUE ATTRACTIONS●

● Cross word Puzzles

● Graded Exercise

Basic Practice■

Further Practice■

Brain Works■

● Multiple Answer Questions

● Paragraph Questions

� Simple, clear and systematic presentation

� Concept maps provided for every chapter

� Set of objective and subjective questions at the

end of each chapter

� Previous contest questions at the end of each

chapter

� Designed to fulfill the preparation needs for

international/national talent exams, olympiads

and all competitive exams

CLASS - VIII

IIT F

oundatio

n &

Olym

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rer - M

ath

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ss - VIII

FOUNDATION OLYMPIAD&

IntegratedSyllabus

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Rs. 85Detailed solutionsfor all problems

of IIT Foundation &Olympiad Explorer

are available in this book

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MATHEMATICSCLASS - VIII

FOUNDATION & OLYMPIAD

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Published by:

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

C Brain Mapping AcademyALL RIGHTS RESERVEDNo 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 andpublishers to give correct, complete and updated information. In case there is any omission, printing mistake or anyother error which might have crept in inadvertently,neither the compiler / publisher nor any of thedistributors take any legal responsibility.

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PrefaceSpeed 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.

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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 .......................................328www.bm

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IIT Foundation & Olympiad Explorer Mathematics / Class - VIII

4. Mensuration - I © Brain Mapping Academy93

SYNOPSIS

I. Areas and Dimensions of Plane Figures

1. Triangle

(i) Equilateral Triangle

Area A = 12 × base× height

= 12 × b× h

but h = 32

a

∴ A = 12 ×

32

a× a = 34

a2

(ii) Right - Angled Triangle

Area A = 12

× base× height

= 12

bh

but h = 2 2d b−

∴ A = 12

b( 2 2d b− )

A

B CD

a ah

a2

a2

A

hd

Bb C

1

Chapter

MENSURATION-I

Chapter

4

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(iii) Acute � Angled Triangle

Area (A) = ( )( )( )s s a s b s c− − −

Where S = a b c

2+ +

S is the semi - perimeter of the triangle

A = 12

bh =

22 2 22b a b c

a2 2b

+ −⋅ −

(iv) Obtuse Angled Triangle

Area (A) = ( )( )( )s s a s b s c− − −

Where S = a b c

2+ +

A = 12

bh =

22 2 22b c a b

a2 2b

− −−

2. Quadrilaterals

(i) Square

(i) Area = (side)2 = a2

(ii) Side (a) = area

(iii) Perimeter(s) = 4× side = 4a

(iv) Diagonal (D) = ( )22 side 2 Area 1.414 Area= × =

b

A

c

B a C)

A

h

B CD

a

a a

a

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(ii) Rectangle

(i) Area (A) = length × breadth = l× b

(ii) Area (A) = l 2 2d − l = b . 2 2d b−(iii) Perimeter (S) = 2 (length + breadth)

= 2(l + b)

(iv) Diagonal (D) = 2 2b+l

(iii) Parallelogram

(i) Area (A) = base× height

= bh

(ii) b = A

h

(iii) h = A

b

(iv) Rhombus

Area (A) = 1

2× product of its diagonals

= 1

2× d1 × d2

d b

l

b

h

d2

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(v) Trapezium

Area (A) = 12

(sum of parallel sides) × height

= 12

(a + b)h

(vi) Trapezoid

Area (A) = ( )H h a bh CH

2

+ + +

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

3. Circle

(i) Area of circle (A) = π r2 = 2d

4π

Where r = radius

d = diameter of a circle

(ii) Circumference = 2 π r = π d = 3.14d

(iii) Radius = area

π

(iv) Diameter = circumference

π

- - -- - - - - -

- - -- - - - - -

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

- - -

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

- -

h

b a c

H

• r

h

b

a

h

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Example 4.1

The difference of the area of the circumscribed and the inscribed squares of a circle is 35sq.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 = 2diagonal

2

= 2(2x)

2 = 2x2 sq. cm

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 = 35

2

Area of circle = 2xπ = 22 35

7 2× = 55 sq.cm.

SOLVED EXAMPLES

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Example 4.2

The sides of a triangle are a = 13 cm, b = 1 4 cm, c = 15 cm, the sides a and b are thetangents to a circle , whose centre lies on the third side. Find the circumference of thecircle.

Note: 1. Tangent is a line touching the circle at one point.

2. Radius is perpendicular to the tangent.

O

A

B

C

b = 14 cm

a = 13 cm

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 1

BOC2

∆ = × 13 × r sq.cm and 1

AOC2

∆ = × 14 × r sq.cm

Hence the total area of the ABC BOC AOC∆ = ∆ + ∆

= 1 1

.13.r .14.r2 2

+ sq.cm

= 1

2 × r × 27 sq.cm

Again if S = semi-perimeter= 13 14 15

2

+ + cm

= 21 cm

The area of ∆ ABC = s(s a)(s b)(s c)− − − = 21(8)(7)(6) = 84 sq.cm

∴ Comparing these two,

1

2 × r × 27 = 84

or r = 56

9 cm

∴ Circumference of the circle = 22 56

2 r 27 9

π = × × = 39 1

9 cm

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Example 4.3

The two adjacent sides of a parallelogram are 5 cm and 4 cm respectively and if therespective diagonal is 7 cm, then find the area of the parallelogram?

Solution:

Required area = 2 s(s a)(s b)(s D)− − −

where s = a b D 5 4 7

2 2

+ + + += = 8

= 2 8(8 5)(8 4)(8 7)− − −

= 2 8 3 4 8 6× × = = 19.6 sq.cm

Example 4.4

A 5100 sq.cm trapezium has the perpendicular distance between the two parallel sides60 cm. If one of the parallel sides be 40 cm then find the length of the other parallelsides.

Solution:

Let the length of the unknown parallel side be �x� cm.

A = 1

2 (a + b) h

or 5100 = 1

2 (40 + x) × 60

or 170 = 40 + x

∴ Required other parallel side = 170 � 40 = 130 cm

Example 4.5

How many metres of a carpet 75 cm wide will be required to cover the floor of a roomwhich is 20 metres long and 12 metres broad?

Solution:

Length required = lengthof room breadthof room

widthof carpet

×

∴ Length required = 20 12

0.75

× = 320 m

Example 4.6

How many paving stones each measuring 2.5 m × 2 m are required to pave a rectangularcourtyard 30 m long and 16.5 m wide?

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Solution:

Number of teles requied = length breadthof courtyard

length breadthof eachtile

××

= 30 16.5

2.5 2

×× = 99

Example 4.7

A hall-room 39 m 10 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 re-quired.

Solution:

Side of largest possible tile = H.C.F of length and breadth of the room

= H.C.F of 39.10 and 35.70 m

= 1.70 m

Also, number of tiles required

= 2

length breadthof room

(H.C.Fof lengthand breadthof the room)

×

= 39.10 35.70

1.70 1.70

×× = 483

Example 4.8

A rectangular grassy plot is 112 m by 78 m. It has a gravel path 2.5 m wide all round iton the inside. Find the area of the path and the cost of constructing it at Rs. 2 per squaremetre?

Solution:

Area of path = (12× 78) � (107× 73)

= 925 sq.m

Cost of construction = rate × area

= 2 × 925 = Rs. 1850

Example 4.9

A square field of 2 sq. kilometers is to be divided into two equal parts by a fence whichcoincides 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

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∴ Other diagonal = 2 ×

2

2 55(36.5)

2

− = 48 cm

Now, area = 1

2 (product of diagonals) =

1

2 × 48 × 55 = 1320 sq.cm

Example 4.13

A hall whose length is 16 m and breadth twice its height takes 168 m of paper 2 m widefor its four walls. Find the area of the floor.

Solution:

Let the breadth = 2h m, then height = h m

Area of walls = 2(16 + 2h) h sq.m

Area of paper = 168 × 2 sq.m

∴ 2(16 + 2h)h = 168 × 2 ∴ (8 + h) h = 84

On solving h = 6, �14; �14 is not acceptable

∴ h = 6 and breadth = 12

∴ Area of floor = 16 × 12 sq.m = 192 sq.m

Example 4.14

The length of a rectangle is increased by 60%. By what per cent should the width bedecreased to maintain the same area?

Solution:

Let the length and breadth be x and y then its area = xy

New length = x 160 8x

100 5

=

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

8x

5

= 5y

8

∴ Decrease in breadth = y � 5y

8 =

3y

8

∴ % decrease in breadth = 3y 100

8 y

×× =

75

2 = 37

1

2%

Example 4.15

If the length of a rectangle increases by 10% and the breadth of the rectangle decreasesby 12% then find the % change in area.

Soluton:

Let length = 100 units and breadth = 100 units

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Then Area = 100× 100 = 10000 square units

New length = 110 units

New breadth = 88 units

New Area = 110× 88 = 9680 sq. units

∴ % change in area = 320

10010000

× = 3.2%

Example 4.16

(a) What is the relation between a circle and an equilateral triangle which is inscribedin the circle?

(b) What is the relation between an equilateral triangle and a circle inscribed in acircle?

(c) An equilateral triangle is circumscribed by a circle and another circle is inscribedin 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 2x

3

π

(b) The area of a circle inscribed in an equilateral triangle of side x is 2x

12

π.

(c) From the above, we can say that the required ratio = 2x

3

π :

2x12

π

= 1 1

:3 12

= 4 : 1

Example 4.17

The front wheels of a wagon are 2π m in circumference and the back wheels are 3π mfeet in circumference. When the front wheels have made 10 more revolutions than theback wheels, how many metres has the wagon travelled?

Solution:

Suppose the back wheel has made x revolutions.

∴ Front wheel has made (10 + x) revolutions.

⇒ 3 π x = 2 π (10 + x)

⇒ π x = 2 π × 10 ⇒ x = 20

∴ The wagon has travelled 3 π x = 60 π .

Example 4.18

Amar drew a square. He then erased it and drew a second square whose sides were 3times the sides of the first square. By what percent was the area of the square increased?

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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 thearea of the shaded portion?

Soluton: Area of ∆ ABE = 1

2 × AB × ⊥ from E on AB =

1

2xy.

Example 4.22

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

find Areaof theshaded portion

Areaof the larg ersquare .

Solution:

2 2

2

a b

a

− = 1 �

2

2

b

a =

b b1 1

a 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 10 cm?

Solution:

9B2 + B2 = 2(8 10) = 640 ⇒ B2 = 64 ⇒ B = 8

3B

B8 10

∴ Perimeter = 8B = 64 cm.

D

B

C

A

CE

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Example 4.24

What will be the ratio of the circumference to the diameter of a circle if its originalradius is tripled?

Solution:

2 r

2r

π = Ratio of the circumference to the diameter of the circle = π = Independent of the

radius.

Example 4.25

There are two 2-metre wide cross roads in a lawn 150 m by 120 m dimensions. One ofthe 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 = 150 × 2 + 120 × 2 � 2 × 2

= 300 + 240 � 4 = 536 sq m.

The cost for levelling the road = 536 × 2 = Rs. 1072

Example 4.26

What is the area of the triangle in which two of its medians 9 cm and 12 cm long inter-sect at right angles?

Solution:

Area of the triangle = 2 × 1

2× 12 × 6 = 72 sq.cm

Example 4.27

If one leg of an isosceles right-angled triangle is increased by 6 cm and that of the otherleg decreased by 4 cm, then the area of the triangle decreases by 24 sq cm. Find thelength 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 = 1

2x2

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The area of the new triangle = 1

2 (x � 4) (x + 6)

x 4−

x + 6

∴1

2x2 =

1

2(x � 4) (x + 6) + 24

⇒ x2 = x2 + 2x � 24 + 48 ⇒ x = �12.

Example 4.28

What is the ratio of the heights of two isosceles triangles which have equal verticalangles, and of which the areas are in the ratio of 9 : 16?

Solution:

Let the heights be h1 and h

2. Let the bases be B

1 and B

2 respectively.

∴1 1

2 2

1B h

21

B h2

× ×

× × =

9

16

Let 1

1

B

h = 2

2

B

h = k, say

∴2122

h 9

16h= ⇒

1

2

h 3

h 4=

Example 4.29

Four horses are tethered at four corners of a square plot of side 14 metres (m) so that the

adjacent horses can just reach one another. There is a small circular pond of area 20 m2 atthe centre. Find the ungrazed area.

Solution:

Total area = 14 × 14 = 196 m2

Grazed area = 2r

4

π ×

× 4 = π r2 = 22 × 7 (r = 7)

= 154 m2

Ungrazed area = (196 � 154) = 42 m2

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CONCEPT MAP

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BASIC PRACTICE

1. What is the area of the shaded region?

12 cm

8 cm

15 cm

2. 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 ` 7 per metre.

6

c

c

9

12

(5 + c)3. What is the area of triangle ABC?

A

B

C4 cm

3cm

9cm

D

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

8cm

7cm

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5. The figure is made up of two identical small circles and a large circle with diameter 28

cm. Taking 22

7π = , express the shaded area as a fraction of the total area.

28 cm6. 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 1 : 3. Find the ratio of the area of A to the area of C.

AB

CD

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

3 cm

8. The figure is made up of 2 rectangles

3 cm

3 cm

3 cm

3 cm

7r cm4r cm

(a) What is the perimeter of the big rectangle?

(b) What is the area of the small rectangle?

9. The figure shows a trapezium PQRS. Given that RQ = PQ = 8 cm and the area of triangleOPQ is 12.8 cm2, what is the area of triangle OPS?www.bm

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S R

P Q

8 cmO

8 cm

10. By what percentage will the area of a square change if its side increases by 10%?

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

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

B

A

C

Note : The diagram is not drawn to scale.

13. Find the ratio of the area of a square to that of the square drawn on its diagonal?

14. A wire bent in the form of a square enclosed an area of 121 sq cm. If the same wire isbent so as to form a circle, then find the area enclosed?

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

H

F G

IC

D

B

A

E

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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.

QSP

RR

1

R2

27. In the given figure oB 90∠ = , AC is diameter of a semicircle and with BC as radius, a

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FURTHER PRACTICE

1. Three horses are grazing within a semi-circular field. In the diagram given below, AB isthe diameter of the semi-circular field with centre at O. Horses are tied up at P, R and Ssuch 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 andOB. The horses tied at P and R can graze within the respective semi-circle and the horsetied at S can graze within the circle centred at S. The percentage of the area of the semi-circles with diameter AB that cannot be grazed by the horses is nearest to:

A P O R B

S

A) 20 B) 28 C) 36 D) 40

2. 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 hand b can be expressed as:

A) 2b2 = h2 B) 2h2 = b2 C) 3b2 = 2h2 D) 3h2 = 2b2

3. In the figure below, the rectangle at the corner measures 10 cm × 20 cm. The corner Aof the rectangle is also a point on the circumference of the circle. What is the radius ofthe circle in cm?

A

A) 10 cm B) 40 cm C) 50 cm D) 30 cm

4. Consider two different cloth-cutting processes. In the first one, n circular cloth piecesare cut from a square cloth piece of side �a� in the following steps: the original square ofside �a� is divided into n smaller squares, not necessarily of the same size; then a circle ofmaximum 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 theprocess ends there. The cloth pieces remaining after cutting the circles are scrapped inboth the processes. The ratio of the total area of scrap cloth generated in the former tothat in the later is:

A) 1 : 1 B) 2 : 1 C) n(4 )

4n

− π− π

D) 4n

n(4 )

− π− π

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

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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 : 3π . The line segment DE intersects

AB at E such that ∠ ODC = ∠ ADE. What is the ratio AE : AD?

O

A E B

CD

A) 1 : 3 B) 1 : 2 C) 1 : 2 3 D) 1 : 2

6. Let S1 be a square of side �a�. Another square S2 is formed by joining the mid-points of thesides of S1. The same process is applied to S2 to form yet another square S3, and so on. IfA1, A2, A3, ... be the areas and P1, P2, P3, ... be the perimeters of S1, S2, S3, ...., respectively,then the ratio:

1 2 3

1 2 3

P P P ...

A A A ...

+ + ++ + +

equals:

A) 2(1 2)

a

+B)

2(2 2)

a

−C)

2(2 2)

a

+D)

2(1 2 2)

a

+

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

A) 16.665 B) 16.565 C) 15.465 D) 14.365

8. The sides of a triangle are in the ratio of 1 1 1

: :2 3 4

. If the perimeter is 52 cm, then the

length of the smallest side is:

A) 9 cm B) 10 cm C) 11 cm D) 12 cm

9. A, B and C start running at the same time and from the same point around a circulartrack of 70 metres radius. A and B run clockwise and C counter clockwise. If A meets Cevery 88 seconds and B meets C every 110 seconds, then A meets B every _____ seconds.

A) 22 B) 198 C) 440 D) 212

10. A pond 100 m in diameter is surrounded by a circular grass walk 2 m wide. How manysquare meters of grass is there on the walk?

A) 98 π B) 100 π C) 204 π D) 202 π11. 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) 371

2% B) 60% C) 75% D) 120%

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

10

10

D C

BA

A) 100 � 50 π B) 100 � 257 π C) 50 π � 100 D) 25 π � 100

25. A steel wire bent in the form of a square of area 121 cm2. If the same wire is bent in theform of a circle, then the area of the circle is:

A) 130 cm2 B) 136 cm2 C) 154 cm2 D) 168 cm2

26. The number of revolutions made by a wheel of diameter 56 cm in covering a distance of

1.1 km is 22

Use7

π = :

A) 31.25 B) 56.25 C) 625 D) 62.5

27. Semi-circular lawns are attached to the edges of a rectangular field measuring 42 m × 35m. The area of the total field is:

A) 3818.5 m2 B) 8318 m2 C) 5813 m2 D) 1358 m2

28. 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) 17 cm

29. The cross-section of a canal is in the form of a trapezium. If the canal top is 10 m wideand the bottom is 6 m wide, and the area of cross-section is 72 m2, then the depth of thecanal is:

A) 10 m B) 7 m C) 6 m D) 9 m

30. 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

31. A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running round iton 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

32. If the circumference and the area of a circle are numerically equal, then what is thenumerical value of the diameter?

A) 1 B) 2 C) 4 D) π33. 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) 17 cm

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

A) 2 r B) ( 2 1)r− C) 1

2rD)

2

( 2 1)r+

BRAIN WORKS

1. 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 ratioof the area of the shaded region to that of the unshaded region.

P

R

S

Q

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

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4. Find the area of the shaded portion.

5. The figure below shows two concentric circles with centre �O�. PQRS is a square inscribedin the outer circle. It also circumscribes the inner circle, touching it at points B, C, Dand A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

C

QBP

RS D

OA

6. Four identical coins are placed in a square. For each coin, the ratio of area of circumferenceis same as the ratio of circumference to area. Then, find the area of the square that isnot covered by the coins.

7. AB is the diameter of the given circle, while points C and D lie on the circumference asshown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of the quadrilateralACBD.

C

B

D

A

8. The adjoining figures shows a set of concentric squares. If the diagonal of the innermostsquare is 2 units, and if the distance between the corresponding corners of any twosuccessive squares is 1 unit, find the difference between the areas of the eight and theseventh squares, counting from the innermost square.

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

32

25

25

24

What is the area of the plot (in m2)?

16. Euclid has a triangle in mind. Its longest side has length 20 and another of its sides haslength 10. Its area is 80. What is the exact length of its third side?

17. A rectangular pool 20 metres wide and 60 metres long is surrounded by a walkway ofuniform width. If the total area of the walkway is 516 square metres, how wide, inmetres, is the walkway?

18. Consider a circle with unit radius. There are seven adjacent sectors, S1, S2, S3 . . . S7, in

the circle such that their total area is 1

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

the jth sector is twice that of the (j � 1)th sector, for j = 2, ... 7. Find the area of the sectorS1.

19. A former has decided to build a wire fence along one straight side of his property. Forthis, he planned to place several fence-posts at 6 m intervals, with posts fixed at bothends of the side. After he bought the posts and wire, he found that the number of postshe had bought was 5 less than required. However, he discovered that the number ofposts he had bought would be just sufficient if he spaced them 8 m apart. What is thelength 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 14?

MULTIPLE ANSWER QUESTIONS

1. 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

2. If area of a triangle is 12 cm2, then which of the following pairs of base and height are possible?

A) Base = 12 cm, Height = 1 cm B) Base = 6 cm, Height = 3 cm

C) Base = 8 cm, Height = 3 cm D) Base = 4 cm, Height = 6 cm

3. 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

Passage - I

Area of a rectangle is twice the area of a triangle with base a and height b.

1. The figure below is made up of a square, a triangle and a rectangle. Find the area of theshaded region.

9 cm

6 cm

4 cm

A) 32 m2 B) 40.5 m2 C) 52.5 cm2 D) 72 m2

Passage - II

Area of a sector is o

2r

360

θΠ

All the sides of a square are equal.

1. The figure below consists of a square, a quarter circle and a semicircle.

a. Find the area of the shaded portion. b. Find the perimeter of the shaded portion.

10 cm

10 cm

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

43 m

21 m

21 m

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UNIQUE ATTRACTIONS●

● Cross word Puzzles

● Graded Exercise

Basic Practice■

Further Practice■

Brain Works■

● Multiple Answer Questions

● Paragraph Questions

� Simple, clear and systematic presentation

� Concept maps provided for every chapter

� Set of objective and subjective questions at the

end of each chapter

� Previous contest questions at the end of each

chapter

� Designed to fulfill the preparation needs for

international/national talent exams, olympiads

and all competitive exams

CLASS - VIII

IIT F

oundatio

n &

Olym

pia

d E

xplo

rer - M

ath

em

atic

s Cla

ss - VIII

FOUNDATION OLYMPIAD&

IntegratedSyllabus

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Detailed solutionsfor all problems

of IIT Foundation &Olympiad Explorer

are available in this book

CLASS - X

YOUR

COACH

India’s FIRST scientifically designed portalfor Olympiad preparation• Olympiad & Talent Exams preparation packages

Analysis Reports Previous question papers• •Free Demo Packages Free Android Mobile App• •

Get 15% discount on all packages by using the discount coupon code: KR157N

A unique opportunity to take about 50 tests per subject.

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