Area and Perimeter for Ssc Exams

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Chapter-30

Mensuration I

(Area and Perimeter)

Mensuration

Mensuration is the science of measurement of the

length of lines, areas of surfaces and volumes of

solids. Its knowledge is of immense use to the

surveyor, architect and engineer. In this chapter,

we will deal with the perimeter and area of the

plane figures.

Perimeter

The perimeter of a figure is the sum of length of

all its sides. This is measured in the units of

length. For example, cm, m, etc.

Area

The area of any figure is the amount of surface

enclosed within its bounding lines. This is

measured by the number of square cm or square

metres (or other units of square measures) it

contains.

Measures of Area

Area, being the product of two linear measurements

(see in the following pages), is expressed in square

units of linear measure. The following will give

you the measures of area in the Metric System:

(i) Square Centimetre: A square centimetre is

the amount of surface enclosed within a square of

which each side is one centimetre in length.

The term square metre, square hectometre,

square kilometre etc are used in the same sense.

(ii) Square Metre: The area of a region formed

by a square of side 1 metre is called a square metre

and is written as 1 m2. The basic unit of length in SI system is a metre

(1 m) and the basic unit of area in SI system is a

square metre (1 m2).

Since,

1 m = 100 cm

1 m2 = 1 m × 1 m = 100 cm × 100 cm

= 10000 cm2

Thus, 1 m2 = 10000 cm2.

(iii) Square Decimetre: The area of a region

formed by a square of side 1 decimetre (1 dm) is

called a square decimetre and is written as 1 dm2.

Since,

1 m = 10 dm

1 m2

= 1 m × 1 m = 10 dm × 10 dm = 100 dm2

Also, 1 dm = 10 cm

1 dm2 = 1 dm × 1 dm = 10 cm × 10 cm = 100 cm2

(iv) Square Millimetre: The area of a region

formed by a square of side 1 millimetre (1 mm) is

called a square millimetre and is written as 1 mm2.

Since 1 cm = 10 mm

1 cm2 = 1 cm×1 cm = 10 mm × 10 mm

= 100 mm2

Also, 1 m2 = 10000 cm2

1 m2 = 10000 mm × 100 mm = 1000000 mm2

(v) Square Decametre or an Are: The area of a

region formed by a square of side one decametre (1

dam) is called a square decametre or an arc is

denoted by 1 dam2 or 1a .

Since 1 dam = 10 m

1 dam2

= 1 dam × 1 dam = 10 m × 10 m = 100 m2

Thus, 1 dam2 = 100 m2 or 1 are = 100 m2

(vi) Square Hectometre or Hectare: The area of

a region formed by a square of side 1 hectometre (1

hm) is called a square hectometre or a hectare and

is denoted by 1 hm2 or 1 ha.

Since, 1 hm = 100 m

1 ha = 1 hm × 1 hm = 100 m × 100 m

= 10000 m2

Also, 1 hm = 10 dam

1 ha = 10 dam × 10 dam

= 100 dam2 = 100 a.

Thus, 1 ha = 10000 m2 or 1 ha = 100 a.

(vii) Square Kilometre: The area of a region

formed by a square of side 1 kilometre (1 km) is

called a square kilometre and is written as 1 km2

.Since, 1 km = 1000 m

1 km2 = 1000 m × 1000 m = 1000000 m2

Since 1 ha = 10000 m2. Therefore,

1 km2 =10000

1000000 ha = 100 ha

Also, 1 km2 = 10000 are



502 Concep t o f A r i t hm e t i c

K KUNDAN

Ex. 1: F i n d t h e a r e a , i n h e ct a r e, o f a f i el d

w h o s e l e n g t h i s 2 4 0 m a n d b r e a d t h

1 1 0 m .Soln: We have, Length of the field = 240 m

Breadth of the field = 110 m.

Area of the field = (240 × 110) m2

= 26400 m2 =10000

26400 hectare

[ 10000 m2 = 1 hectare]

= 2.64 hectare

Ex. 2: W h a t w i l l h a p p e n t o t h e a r ea o f a

r e c t a n g l e i f

(i ) i t s l en g t h i s d ou b l ed a n d br ea d t h i s t r e b l ed .

(i i ) i t s l en g t h a n d b r ea d t h a r e

d o u b l e d ?

S o l n : Let l cm and b cm be the length and

breadth of the rectangle. Further, let A be

the area of the rectangle. Then,

A = l × b ...... (i)

Solved Examples

We have introduced various standard units of

area. Each can be converted into others as given

below.

Conversion of Units

Units of Length Units of Area

1 cm = 10 mm 1 cm2 = (10 × 10) mm2

= 100 mm2

1 m = 10 dm 1 m2 = (10 × 10) dm2

= 100 dm2

1 dm = 10 cm 1 dm2 = (10 × 10) cm2

= 100 cm2

1 m = 100 cm 1 m2 = (100 × 100) cm2

= 10000 cm2

1 dam = 10 m 1 dam2 = (10 × 10) m2

= 100 m2 = 1 are

1 hm = 100 m 1 hm2 = (100 × 100) m2

= 10000 m2

= 1 hectare

1 km = 10 hm 1 mm2 = (10 × 10) hm2

= 100 hm2

1 km = 1000 m 1 km2 = (1000 × 1000) m2

= 1000000 m2

Rectangle and Square

(i) Rectangle

A rectangle is a four-sided figure having all its

angles right angles. The page of this book,the faces

of a brick, the floor of a room are rectangles, for

their opposite sides are equal and their angles are

right angles.

The sides of a rectangle are usually called its

length and breadth. In the rectangle, the length

and breadth are uneqaual.

( i ) Perimeter of a rectangle = 2 (Length +Breadth)

( ii) Area of a rectangle = (Length × Breadth)

From this formula we get,

(a) Length =Breadth

Area

(b) Breadth = Length

Area

( i i i ) Diagonal of a rectangle

= 22 (Breadth)(Length)

ABC is a right-angled triangle.

AC = 22 BCAB

(ii) Square

A square is a four-sided figure having all its angles

right angles and all its sides are equal. In a square,

the length is equal to the breadth.

( i ) Perimeter of a square = 4 × length

= 4 × one of its sides

( ii) Area of a square = side × side = (side)2

From this formula we get,Side = Area

( i i i ) Diagonal of a square = )Area(2

If ABCD be a square, then

AC = 22 BCAB = 22 ABAB = AB2

[ AB = BC]

AC = 2AB

From the above, we have

AB =2

AC

Area = (AB)2 =2

1(AC)2

Hence, the area of a square is half the square

of its diagonal.



503Mensu r a t i o n I (A rea an d Per i me t e r )

K KUNDAN

or, 28922 y x ..... (i)

And, Perimeter = 46 cm

or, 2 (x + y ) = 46 cm

or, x + y = 23 .... (ii)Now, x + y = 23

or, (x + y )2 = 232

or, x 2 + y 2 + 2xy = 529

or, 289 + 2xy = 529

or, 2xy = 529 - 289

or, 2xy = 240

or, xy = 120

Hence, Area = xy = 120 cm2.

Ex. 6: T h e l e n g t h o f a r e ct a n g l e ex c e ed s i t s

w i d t h b y 8 c m a n d t h e a r e a o f t h e

r e c t a n g l e i s 2 4 0 s q c m . F i n d t h e

d i m en s i o n s o f t h e r e ct a n g l e .

S o l n : Let the breadth of the given rectangle be x

cm. Then, length = (x + 8) cm.

Now, Area = 240 cm2

or, length × breadth = 240or, (x + 8) x = 240

or, x 2 + 8x – 240 = 0

or, x 2 + 20x – 12x – 240 = 0

or, x (x + 20) – 12 (x + 20) = 0

or, (x + 20) (x – 12) = 0

or, x = 12 or x = –20

But x cannot be negative. So, x = 12.

Hence, breadth = 12 cm and length

= (12 + 8 =) 20 cm.

Ex. 7: T h e c os t o f p l o u g h i n g a r ec t a n g u l a r

f i e l d a t 8 5 p a i s e p er s q u a r e c en t i m et r e

i s R s 6 2 4 .7 5 . F i n d t h e p er i m et e r o f t h e

f i e l d i f i t s si d e s a r e i n t h e r a t i o 5 : 3 .

S o l n : Let the length and the breadth of the

rectangular field be 5x cm and 3x cm in

length. Then,Area = 5x × 3x = 15x 2 cm2 .... (i)

It is given that the cost of ploughing the

rectangular field at the rate of Rs100

85 is

Rs 624.75.

Area of the rectangular field

= msq perRate

cost Total

=85

10075.624

100

85

75.624

= 735 cm2 .... (ii)

From (i) and (ii), we get15x 2 = 735 or x 2 = 49 or x = 7

Hence, the sides of the rectangular field

are 5x = (5 × 7) = 35 cm and 3x = (3 × 7 =)

21 cm respectively.

Perimeter = {2 (35 + 21) =} 112 cm

(i) We have,

New breadth = 2l ; New length = 3b

A1 = New Area

= 2l × 3b = 6 (l × b ) = 6A

[Using (i)]Hence, the area of the new rectangle is 6

times the area of the old rectangle.

(ii) We have,

New length = 2l , New breadth = 2b

A1 = New Area

= 2l × 2b = 4 (l × b ) = 4A

[Using (i)]

Hence, the area of the new rectangle is 4

times the area of the old rectangle.

Ex. 3: W h a t w i l l h a p p en t o t h e a r e a o f a

s q u a r e i f i t s s i d e i s ( i ) d o u b l ed ( i i )

h a l v e d ?

So l n : (i) Let the side of a square be x cm. Then,

Area = A = x 2 cm2 ....(i)

When the side is doubled, then,

side of the new square = 2x cm. A

1 = Area of the new square

= (2x 2) cm2 = 4x 2 cm2 = 4A

[Using (i)]

Thus, if the side is doubled, then area

becomes 4 times.

(ii) When the side is halved, then,

Side of the new square =2

x cm.

A2 = Area of the new square

= A4

1

4

1cm

2

222

x

x

[Using (i)]

Thus, if the side is halved, then area

becomes one-fourth.Ex. 4: T h e l e n g t h o f a r e c t a n g l e i s t w i c e i t s

b r ea d t h . F i n d t h e d i m e n s i on s o f t h e

r e ct a n g l e i f i t s a r ea i s 2 8 8 s q cm .

Soln: Let the breadth of the given rectangle be x

cm.

Then, length = 2x cm [Given].

Area = 2x × x = 2x 2

or, 2x 2 = 288

[ Area = 288 sq cm (given)]

or, x 2 = 144

or, x = 12 cm

Hence, length = 24 cm, and breadth 12

cm.

Ex. 5: I f t h e d i a g o n a l o f a r ec t a n g l e i s 1 7 cm

l o n g a n d t h e p er i m et e r o f t h e r e ct a n g l e

i s 46 cm . F i nd t he a r ea o f t h e r ec t ang l e.

Soln: Let the length and breadth of the given

rectangle be x cm and y cm respectively.

Th en ,

Diagonal = 17 cm

or, 1722 y x




K KUNDAN

Ex. 8: T h e l en g t h a n d b r e a d t h o f a

r e ct a n g u l a r f i e l d a r e i n t h e r a t i o 3 : 2 .

I f t h e a r ea o f t h e f i e l d i s 3 4 5 6 c m 2 ,

f i n d t h e c o st o f f en c i n g t h e f i el d a t R s

3 . 5 0 p e r cm .Soln: Let the length and breadth of the

rectangular field be 3x and 2x cm

respectively. Then,

Area of the rectangular field

= (3x × 2x ) cm2 = 6x 2 cm2

Also, area of the rectangular field

= 3456 cm2 (given)

6x 2 = 3456

or, x 2 =6

3456

or, x 2 = 576

or, x = 576 = 24

Length = (3 × 24) cm = 72 cm

Breadth = (2 × 24) cm = 48 cm Perimeter of the field

= 2 × (length + breadth)

= [2 × (72 + 48)] cm = 240 cm

Rate of fencing = Rs 3.50 per cm

Cost of fencing = Rs (240 × 3.50)

= Rs 840

Ex. 9: F i n d t h e a r ea o f a r e ct a n g u l a r p l o t o n e

s i d e o f wh i c h i s 4 8 cm and i t s d i a g o n a l

5 0 c m .

Soln: Let the other side be x cm. Since ABC is

a right triangle, therefore,

AC2 = AB2 + BC2

or, 502 = 482 + x 2

or, x 2 = (50)2 – (48)2

or, x 2 = (50 + 48) (50 – 48)

or, x 2 = 98 × 2or, x 2 = 142

or, x = 14

Thus, the other side of the rectangle is

14 cm

Area of the rectangle = (48 × 14) cm2

= 672 cm2

Ex. 10: The a r ea o f squa r e ABCD i s 16 cm 2 . F i nd

t h e a r e a o f t h e s q u a r e jo i n i n g t h e m i d -

p o i n t s o f t h e s i d e s.

Soln: We have,

Area of square ABCD = 16 cm2

Each side of square ABCD

= 16 cm = 4 cm

In APS, we have

AP = AB2

1 = 2 cm and

AS = AD2

1

= 2 cm

Also, PS2 = AP2 + AS2

[Using Pythagoras theorem]

PS = 243244 22

Thus, eac h side of square PQRS is of

length 24 cm.

Area of the square PQRS

= 224 cm2 = 32 cm2

Ex. 11: T h e l en g t h o f a r e ct a n g u l a r f i el d i s

i n c r e a se d by 5 0% a n d i t s br e a d t h i s

d e c r e a s ed b y 5 0 % t o f o r m a n e w

r e ct a n g u l a r f i e l d . F i n d t h e p er c en t a g e

c h a n g e i n t h e a r ea o f t h e f i e l d .Soln: Let x and y denote the length and breadth

of the given field. Then, its area is given

by A = xy

Increase in length

= 50% of the original length of x

=2100

50 x x

New length = x x

x 2

3

2

Decrease in breadth

= 50% of the original breadth

= 50% of y =2100

50 y y

New length =22

y y y

A1 = Area of the new field

= xy y

x 4

3

22

3

Change in Area

= xy xy xy 4

1

4

3AA1

Percentage change in the area

= 1004

1

100A

AA1

xy

xy = 25%

Hence, there is 25% decrease in the area.

Ex. 12: A r e ct a n g u l a r c ou r t y a r d i s 3 m 7 8 cm

l o n g a n d 5 m 2 5 cm b r o a d . I t i s d e s i r ed

t o p a v e i t w i t h s q u a r e t i l e s of t h e s ame

s i z e. W h a t i s t h e l a r g es t s i z e o f t h e

t i l e t h a t c a n b e u s ed ? A l s o , f i n d t h e

n um ber o f s u c h t i l e s .




K KUNDAN

Soln: Clearly, the size of the tile should be a

factor of both the length and breadth of

the courtyard. Therefore, the size of the

largest size tile should be the HCF of 378

cm and 525 cm.

7

213

633

1893

3782

7

213

1053

5255

Now,

378 = 2 × 33 × 7 and 525 = 5 × 32 × 7

Hence, HCF of 378 and 525 is 3 × 7 = 21

Thus, the size of the largest size square

tile = 21 cm

Area of a tile

= (side)2 cm2 = (21)2 cm2 = 441 cm2

Area of courtyard = (378 × 525) cm2

= 198450 cm

2

Now,

Number of tiles =tileaof Area

courtyardof Area

=441

198450 = 450

Ex. 13: The f l o or o f a r ec t a n g u l a r h a l l i s t o b e

c ov er e d w i t h a c a r p et 1 5 0 cm w i d e. I f

t h e l en g t h a n d b r ea d t h o f t h e h a l l a r e

2 0 m a n d 1 8 m r e sp ec t i v el y , f i n d t h e

c os t o f t h e ca r p et a t t h e r a t e of R s 1 0

p e r me t r e .

Soln: We have,

Area of the hall = (20 × 18) m2 = 360 m2

Width of the carpet = 150 cm = 1.5 m

Length of the carpet

= 5.1

360

carpettheof Width

halltheof Area = 240 m

Rate of the carpet = Rs 10 per metre.

Cost of the carpet = Rs (240 × 10)

= Rs 2400.

Ex. 14: F i n d t h e h e i g h t o f t h e wa l l w h o se l en g t h

i s 4 m a n d w h i c h c a n b e co v er e d b y

2 4 0 0 t i l es o f si z e 2 5 cm b y 2 0 cm .

Soln: Area of a tile = (25 × 20) cm2 = 500 cm2.

Area of 2400 tiles = (2400 × 500) cm2

= 1200000 cm2

=10000

1200000 m2

]m1cm10000[ 22

= 120 m2.

Let the height of the wall be h metres.

Th en ,

area of the wall = 4h m2.

Since 2400 tiles completely cover the wall.

Therefore,

Area of the wall = Area of 2400 tiles

or, 4h = 120

or,4

120

4

4

h

[Dividing both sides by 4]

or, h = 30

Hence, the height of the wall is 30 metres.

Ex. 15: F i n d t h e p er i m et e r o f a r e ct a n g u l a r f i el d

w h o s e l e n gt h i s f o u r t i m e s i t s w i d t h

a n d wh i c h h a s a n a r e a eq u a l t o 3 0 9 7 6

cm 2 .

Soln: Let the width of the field be b cm. Then,

Length of the field = 4b cm.

Area of the field = (b × 4b ) cm2 = 4b 2 cm2

But, area of the field is given as 30976

cm2.

4b 2 = 30976

or, 77444

309762 b

or, b 2 = (88)2or, b = 88

Length of the field = 4b cm

= (4 × 88) cm = 352 cm

Width of the field = b cm = 88 cm

Hence,

Perimeter of the field

= 2 (length + breadth)

= 2 (352 + 88) cm = 880 cm

Ex. 16: A 5 m w i d e l a n e wa s p a v ed w i t h b r i c k s

o f s i z e 2 0 cm b y 1 5 cm . I f t h e r a t e o f

b r i c k s wa s R s 7 5 0 p e r t h o u s a n d a n d i f

b r i c k s w o r t h R s 4 9 5 0 0 w e r e u s ed f o r

p a v emen t s , f i n d t h e l e n gt h o f t h e l a n e.

Soln: We have,

Rate of bricks = Rs 750 per thousand

Total cost of bricks = Rs 49500

Number of bricks =750

49500 thousand

= 66 thousand

= 66000

Area of one brick = (20 × 15) cm2

= 300 cm2

Area covered by 66000 bricks

= 66000 × 300 cm2

= 19800000 cm2

=10000

19800000 m2 = 1980 m2

22 m1cm 10000

Hence, area of the lane = 1980 m2

.It is given that the width of the lane is 5 m.

Length of the lane =

width

Aream

=5

1980 m = 396 m




K KUNDAN

Ex. 17: The l en g t h a n d b r e a d h o f a p l a y g r o u n d

a r e 7 5 m 2 0 c m a n d 3 4 m 8 0 c m ,

r e s p ec t i v e l y . F i n d t h e c os t o f l e v el l i n g

i t a t R s 1 . 5 0 p e r s q u a r e m e t r e . H o w

l o n g w i l l a b oy t a k e t o g o t h r e e t i m es r o u n d t h e f i el d , i f h e w a l k s a t t h e r a t e

o f 1 . 5 m / s ec .

Soln: We have,

Length of the playground

= 75 m 20 cm = 75.20 m

Breadth of the playground

= 34 m 80 m = 34.80

Area of the playground

= 75.20 × 34.80 m2 = 2616.96 m2

Cost of levelling = Rs 2616.96 × 1.50

= Rs 3925.44

Perimeter of the playground

= 2(length + breadth)

= 2(75.20 + 34.80) m

= (2 × 110) m = 220 m.

Total distance to be covered by the boy = 3 (Perimeter of the playground)

= 3 × 220 m = 660 m

Speed of the boy = 1.5 m/sec.

Time taken by the boy =5.1

660 sec

Speed

Distance Time

= 440 sec =60

440 minutes

=3

22 minutes

= 7 minutes 20 secondsEx. 18: The c a r p e t f o r a r o om 6 . 6 m b y 5 . 6 m

cos t s Rs 3 9 6 0 a n d i t w a s made f r om a

r o l l 7 0 c m w i d e . Fi n d t h e c os t o f t h e

c a r p e t p e r me t r e .

Soln: We have,

Area of the carpet = 6.6 × 5.6 m2

= 36.96 m2.

Width of the roll = 70 cm = 0.7 m

Length of the roll =Width

Area =

7.0

96.36

= 52.8 m

Cost of the carpet = Rs 3960.

Cost of the carpet per metre

= Rs 8.52

3960 = Rs 75.

Hence, the carpet costs Rs 75 per metre.

Ex. 19: T h e a r ea o f a r ec t a n g u l a r f i e l d i s

c a l c u l a t e d t o b e 2 0 0 m 2 wh en i t s s i d es

a r e m ea s u r ed w i t h a f a u l t y m et r e r o d .

I f t h a t me t r e r o d i s a ct u a l l y 0 . 9 0 me t r e

l o n g , f i n d t h e co r r e c t a r e a o f t h e f i el d .

Soln: Let the actual length and breadth of the

rectangular field be l and b metres

respectively.

The faulty metre-rod measures 0.90 metre

as 1 metre.

It will measure 1 metre as90.0

1 metre

It will measure l metres as l 90.0

1

metres.

Thus, according to the faulty metre-rod

the length of the field is90.0

l metres.

Similarly, breadth of the field measured

by the faulty rod =90.0

b metres.

Area of the field =81.090.090.0

lb b l

But, the faulty metre-rod measures the area

of the field as 200 m2.

20081.0

lb

or, lb = 200 × 0.81 m2

or, lb = 162 m2

Hence, the correct area of the field is 162

m2.

Areas of Paths

Ex. 20: A r e ct a n g u l a r g r a s sy l a w n m ea s u r i n g

3 0 m b y 2 8 m i s t o be su r r o u n d ed

ex t er n a l l y b y a p a t h w h i c h i s 2 m w i d e .

F i n d t h e co s t o f l ev el l i n g t h e p a t h a t

t h e r a t e o f R s 5 p er s q u a r e me t r e .

Soln: Let ABCD be the grassy lawn, and let

PQRS be the external boundaries of the

path.

We have,

length AB = 30 m, breadth BC = 28 m Area of lawn ABCD = 30 × 28 m2

= 840 m2

Length PQ = (30 m + 2 m + 2 m)

= 34 m

Breadth QR = 28 m + 2 m + 2 m

= 32 m

Area PQRS = 34 × 32 m2 = 1088 m2




K KUNDAN

Now, Area of the path

= Area PQRS - Area of the lawn

= (1088 - 840 =) 248 m2

Cost of levelling the path

= Rs (248 × 5) = Rs 1240Ex. 21: A g r assy p l o t i s 80 m × 60 m . Two c r oss

p a t h s ea c h 4 m w i d e a r e c o n st r u c t ed

a t r i g h t a n g l e s t h r o u g h t h e c en t r e of

t h e f i e l d , s u ch t h a t e a c h p a t h i s

p a r a l l e l t o o n e o f t h e s i d e s o f t h e

r e ct a n g l e . Fi n d t h e t o t a l a r e a u s ed a s

p a t h . A l so , f i n d t h e co st o f g r a v el l i n g

t h em a t R s 5 p er s q u a r e me t r e .

Soln: Let ABCD and EFGH be the cross paths.

We have, AB = 80 m and BC = 4 m

Area of path ABCD = (80 × 4) m2

= 320 m2

Again, EF = 60 m and FG = 4 m

Area of path EFGH = (60 × 4) m2

= 240 m2

Clearly, area PQRS is common to both the

paths.

We have,

Area PQRS = (4 × 4 =) 16 m2

Total area used as path = Area of path

ABCD + Area of path EFGH - Area PQRS

= (320 + 240 - 16) = 544 m2

We have, rate of gravelling the path

= Rs 5 per square metre

Total cost of gravelling the path

= Rs (5 × 544) = Rs 2720

Ex. 22: Ca l cu l a t e t he a r ea o f t he shaded r eg i on

s h own i n t h e f i g u r e g i v en b e l ow :

S o l n : We have,

Area of the rectangle ABCD

= (60 × 48) m2 = 2880 m2

Area of a square at one of the corners

= (8 × 8) m2 = 64 m2

Area of the four squares

= 4 × 64 m2 = 256 m2

Hence, required area = Area of therectangle ABCD – Area of the four squares

= (2880 - 64 =) 2816 m2

Ex. 23: Ca l c u l a t e t h e a r e a o f t h e f i g u r e gi v e n

b e l o w .

S o l n : Complete the rectangles ABCQ and DEFR

by drawing dotted lines NQ and PR.

Now,

Area of rectangle ABCQ = (12 × 3) m2

= 36 m2

Area of rectangle DEFR = (12 × 3) m2

= 36 m2

Area of rectangle PNQR = (9 × 4) m2

= 36 m2

[ PN = PH + HM + MN = (3 + 3 + 3) m

and, QN = AQ – AN = (12 – 8) m]

Area of rectangle LMHK = (6 × 3) m2

= 18 m

2

Hence, required area

= 36 m2 + 36 m2 + 36 m2 + 18 m2

= 126 m2

Ex. 24: A t a b l e cover , 4 m × 2 m , i s sp r ead on

a m e et i n g t a b l e . I f 2 5 cm o f t h e t a b l e

c ov er i s h a n g i n g a l l a r o u n d t h e t a b l e ,

f i n d t h e c os t o f p o l i s h i n g t h e t a b l e t o p

a t R s 2 . 2 5 p e r s q u a r e me t r e .

Soln: To find the cost of polishing the table top

we have to find its area for which we

require its length and breadth.

We have,

Length of the cloth = 4 m

Breadth of the cloth = 2 m




K KUNDAN

Since 25 cm width of cloth is outside the

table on each side. Therefore,

Length of the table = (4 – 2 × 0.25) m

= 3.5 m

Breadth of the table = (2 – 2 × 0.25) m = 1.5 m

Area of the top of the table

= (3.5 × 1.5) m2

It is given that the cost of polishing the

table top is at the rate of Rs 2.25 per square

metre. Therefore, cost of polishing the top

= Area × Rate per sq metre

= Rs (3.5 × 1.5 × 2.25)

=

4

9

2

3

2

7

= Rs 11.81

Ex. 25: T h e r e i s a s q u a r e f i e l d w h o s e s i d e i s

4 4 m . A s q u a r e f l ow e r b e d i s p r e p a r ed

i n i t s c en t r e l e a v i n g a g r a v el p a t h a l l

r o u n d t h e f l ow er b ed . T h e t o t a l c os t o f l a y i n g t h e f l ow er b ed a n d g r a v e l l i n g t h e

p a t h a t R s 2 . 7 5 a n d R s 1 . 5 0 p er s qu a r e

me t r e r e spec t i v e l y , i s Rs 4904 . F i nd t h e

w i d t h o f t h e gr a v e l p a t h .

Soln: Let the width of the gravel path be x

metres. Then, each side of the square

flower bed is (44 - 2x ) metres.

Now, area of the square field

= (44 × 44 =) 1936 m2

Area of the flower bed = (44 – 2x )2 m2

Area of the gravel path = Area of the

field - Area of the flower bed

= 1936 – (44 – 2x )2

= 1936 – (1936 – 176x + 4x 2)

= (176x – 4x 2) m2

Cost of laying the flower bed = (Area of

the flower bed) (Rate per sq m)

=100

275)244( 2 x

=2

)244(4

11

x = 2)22(11 x

Cost of gravelling the path

= (Area of the path) × (Rate per sq m)

= )44(6100

150)4176( 22 x x x x

It is given that the total cost of laying the

flower bed and gravelling the path is Rs

4904

11 (22 – x )2 + 6 (44x – x 2 ) = 4904

or, 11 (484 – 44x + x 2) + (264x – 6x 2)

= 4904

or, 5x 2 – 220x + 5324 = 4908

or, 5x 2 – 220x + 420 = 0

or, x 2 – 44x + 84 = 0

or, x 2 – 42x – 2x + 84 = 0

or, x (x – 42) – 2 (x – 42) = 0

or, (x – 2) (x – 42) = 0

or, x = 2 or x = 42

But x 42, as the side of the square is 44

m. Therefore, x = 2.

Hence, the width of the gravel path is 2

metres.

Ex. 26: The l en g t h a n d b r e a d t h o f a r ec t a n g u l a r

f i e l d a r e i n t h e r a t i o o f 7 : 4 . A p a t h 4

m w i d e r u n n i n g a l l a r o u n d o u t s i d e i t

h a s a n a r ea o f 4 1 6 m 2 . Fi nd t he l eng t h

a n d b r e a d t h o f t h e f i el d .

Soln: Let the length and breadth of the field be

7x and 4x metres respectively. Then,

Area of the field = (7x × 4x ) m2

= 28x 2 m2

Length of the field (including path)

= (7x + 8) m

Breadth of the field (including path)= (4x + 8) m

So, Area of the field and path together

= (7x + 8) (4x + 8) m2

Area of the path

= [(7x + 8) (4x + 8) – 28x 2] m2

= (88x + 64) m2

It is given that the area of the path is 416

m2

88x + 64 = 416

or, 88x = 416 - 64

or, 88x = 352

or, x = 4m

Hence, length of the field = 7x = 28 m

breadth of the field = 4x = 16 m

Ex. 27: A c h e ss b o a r d c o n t a i n s 6 4 e qu a l

s q u a r e s a n d t h e a r e a o f ea c h s q u a r e i s 6 . 2 5 cm 2 . A b o r d e r r o u n d t h e b oa r d

i s 2 c m w i d e. Fi n d t h e l en g t h o f t h e

s i d e o f t h e c h e s s b o a r d .

Soln: Let the length of the side of the chess

board be x cm. Then,




K KUNDAN

area of 64 squares = (x – 4)2

(x – 4)2 = 64 × 6.25

or, x 2 – 8x + 16 = 400

or, x 2 – 8x – 384 = 0

or, x 2 – 24x + 16x – 384 = 0

or, (x – 24) (x + 16) = 0

or, x = 24 cm

Triangle

A figure bounded by the three sides is called a

triangle.

(i ) Area of a triangle =2

1 × Base × Height

ABC is the given triangle. Let BDEC be the

rectangle on the same base BC and on the

same height AF. Since AF is perpendicular

to BC, each of the figures ADBF, AECF is a

rectangle.

Now triangle ABF =2

1 rectangle ADBF

Now triangle ACF =2

1 rectangle AECF

By adding, we get

ABF + ACF

=2

1rectangle ADBF +

2

1rectangle AECF

ie ABC =2

1 rectangle BCED

=2

1 × BC × CE

=2

1 × BC × AF [As CE = AF]

=2

1 × Base × Height

Hence the area of a triangle is equal to half

the product of the base and the height.

From the above, we have

Base = Height

Area2

and Height = Base

Area2

( ii) Area of a Triangle when its sides are given

(Hero’s Formula) = ))()(( c s b s a s s

Where, is the area of the triangle; and a ,

b , c are its sides and s =2

1(a + b + c ) =

semi-perimeter of the triangle.

Thus, from half of the sum of the three

sides subtract each side separately. Multiply

the half sum and the three remainders

together. The square root of the product will

be the area of triangle.

Equilateral TriangleA triangle whose all the three sides are equal is

called equilateral triangle.

In an equilateral triangle ABC, a = b= c

s =2

a a a = a

2

3

Hence,

( i ) Area =

a

a a

a a

a a

2

3

2

3

2

3

2

3

=2222

3 a a a a =

2

4

3a

=43 (Side)2

( ii) Height =a

a 2

4

32

Base

Area2

= (Side)2

3

2

3a

Right-angled Triangle

A triangle having one of its angles equal to 90° is

called right-angled triangle.

The figure ABC is right-angled tri angle, angle

B being a right angle ie of 90°. Here, BC is the

base of the triangle, AB is the height of the triangle.




K KUNDAN

AC, the side opposite to the right angle, is called

the hypotenuse .

In case of a triangle ABC right-angled at B, AC2

= AB2 + BC2. This is known as the Pythagoras

Theorem. It may be stated in words thus:In a right-angled triangle the square described

on the hypotenuse is equal to the sum of the

squares on the other two sides.

Let b be the base, p be the perpendicular and h

be the hypotenuse of a right-angled triangle. Then,

( i ) Perimeter = (b + p + h )

( ii) Area =2

1× Base × Height =

p b

2

1

( i i i ) Hypotenuse = 22 p b

Isosceles Triangle

An isosceles triangle is one which has two of its

sides equal. Its third side is usually called the

base.

Let ABC be an isosceles triangle such that AB =

AC = b units and BC = a units.

Area of ABC

=22 (Base)

4

1side)(EqualBase

2

1

Isosceles Right-angled Triangle

For an isosceles right-angled triangle, each of

whose equal side is a , we have

( i ) Hypotenuse = a a a 222

( ii) Perimeter = a a a 2222

( i i i ) Area =2

1× Base × Height

=2

2

1

2

1a a a

Solved Examples

Ex. 28: T h e a r e a o f a r i g h t -a n g l e d t r i a n g l e i s

5 0 m 2

. I f o n e o f t h e l e g s i s 2 0 m , f i n d t h e l e n g t h o f t h e o t h e r l eg .

Soln: In a right-angled triangle, if one side is

the base, then the other side is its altitude

or height.

Let the given leg be the base. Then, the

other leg is the altitude.

Here, Area of the triangle = 50 m2

One leg of the triangle = 20 m

The other leg of the triangle

= Height of the triangle

=

20

502

Base

Area2m = 5 m

Ex. 29: F i n d t h e a r e a o f a n i s o sc el e s r i g h t -

ang l ed t r i a n g l e , i f one o f t he equa l s i des

i s 2 0 cm l o n g .Soln: We know that in an isosceles right angled

triangle, any one of the two sides which

are at right angle can be taken as the base

and the other perpendicular side is the

altitude.

There fore ,

base = 20 cm and altitude = 20 cm

So, area of the given triangle

=22 cm200cm2020

2

1

Ex. 30: The a r ea o f a t r i a n g l e i s eq u a l t o t h a t

o f a s q u a r e who s e ea c h s i d e mea s u r e s

6 0 m et r es . F i n d t h e si d e o f t h e t r i a n g l e

w h o se co r r e s p on d i n g a l t i t u d e i s 9 0

m e t r e s .Soln: We have,

Area of the square = (60 × 60) m2

= 3600 m2

Area of the square = 3600 m2

Altitude of the triangle = 90 m

Side of the triangle

= AltitudeingCorrespond

Area2

=

90

36002m = 80 m

Ex. 31: T h e b a se o f a t r i a n g u l a r f i e l d i s t h r ee

t i m e s i t s a l t i t u d e . I f t h e c os t o f

c u l t i v a t i n g t h e f i e l d a t R s 2 4 . 6 0 p e r

hec t a r e i s Rs 332 . 10 , f i n d i t s base and h e i g h t .

Soln: Let the altitude of the triangular field be x

metres.

Then, base = 3x metres (given).

Area =2

1(Base × Height)




K KUNDAN

=2

1(3x × x )

= msq 2

3 2

x ...... (i)

It is given that the cost of cultivating the

field at the rate of Rs 24.60 per hectare is

Rs 332.10.

Area =60.24

10.332

Rate

cost Total

= 13.5 hectares

= (13.5 × 10000) sq m.

[ 1 hectare = 10000 sq m]

= 135000 sq m ....... (ii)

From (i) and (ii) , we have

1350002

3 2 x

or,321350002 x

or, x 2 = 90000

or, x = 300

Hence, height = 300 m and

base = 3x = 900 m.

Ex. 32: F i n d t h e a r ea o f a r i g h t -a n g l e d t r i a n g l e

w i t h h y p o t e n u se 2 5 cm and b a s e 7 cm .

Soln: Let ABC be the right-angled triangle with

base

BC = 7 cm and

hypotenuse AC = 25 cm.

Using Pythagoras Theorem, we have

AC2 = AB2 + BC2

or, (25)2 = AB2 + 72

or, AB2 = 252 – 72 = 625 – 49 = 576

or, AB = 24.

Hence, area of ABC =2

1 (Base × height)

=

2

1 (7 × 24) = 84 cm2.

Ex. 33: The l e n g t h o f t h e s i d e s f o rm i n g r i g h t -

a n g l e o f a r i g h t -a n g l e d t r i a n g l e a r e 5 x

cm an d (3x – 1 ) cm . I f t h e a r ea o f t h e

t r i a n g l e i s 6 0 cm 2 , f i n d i t s h y p o t e n u se.

Soln: Let ABC be a right-angled triangle with

right-angle at B. Let AB = 5x and BC

= 3x – 1.

Th en ,

Area of ABC =2

1(Base × Height)

or, 60 =2

1(AB × BC)

or, 60 =2

1× 5x (3x – 1)

or, 120 = 5x (3x – 1)

or, 24 = x (3x – 1)

or, 3x 2 – x – 24 = 0

or, 3x 2 – 9x + 8x – 24 = 0

or, 3x (x – 3) + 8 (x – 3) = 0

or, (x – 3) (3x + 8) = 0

or, x – 3 = 0 or, 3x + 8 = 0

or, x = 3

or, x =3

8

or, x = 3 [ x 3

8 ]

AB = 5x = (5 × 3 =) 15 cm and

BC = (3x – 1) = (3 × 3 – 1 =) 8 cm.

Now, AC2 = AB2 + BC2

[By Pythagoras Theorem]

or, AC2 = (15)2 + (8)2

or, AC2 = 289

or, AC = 17 cm.

Hence, hypotenuse = 17 cm.

Ex. 34: The p er i m et e r o f a r i g h t -a n g l e d t r i a n g l e

i s 60 cm . I t s hy po t enues i s 25 cm . F i nd

t h e a r e a o f t h e t r i a n g l e.

Soln: Let ABC be the given right-angled triangle

such that base = BC = x cm and

hypotenuse AC = 25 cm.




K KUNDAN

Now, perimeter = 60 cm

or, AB + BC + AC = 60

or, AB + x + 25 = 60 cm

or, AB = 35 – x .

By Pythagoras Theorem, we haveAB2 + BC2 = AC2

or, 222 25)35( x x

or, 0600702 2 x x

or, 0300352 x x

or, 030015202 x x x

or, 0)15)(20( x x

or, x = 20 or x = 15

If x = 20, then AB = 35 – x = 15 and

BC = x = 20

Area = AB)(BC2

1

=2150cm15)(20

2

1

If x = 15, then AB = 35 – x = 20 and

BC = x = 15

Area = AB)(BC2

1

= .cm150)2015(2

1 2

Hence, area = 150 cm2

Ex. 35: T h e a r e a o f a r i g h t - a n g l ed t r i a n g l e i s

6 0 0 s q cm . I f t h e b a s e o f t h e t r i a n g l e

ex c eed s t h e a l t i t u d e by 1 0 cm , f i n d t h e

d i m e ns i o n s of t h e t r i a n g l e .

Soln: Let the altitude of the given triangle be x cm long.

Then, base = (x + 10) cm.

Now, Area = 600 cm2

or,2

1(Base × Height) = 600

or,2

1(x + 10)x = 600

or, x 2 + 10x = 1200

or, x 2 + 10x – 1200 = 0

or, x 2 + 40x – 30x – 1200 = 0

or, x (x + 40) – 30 (x + 40) = 0

or, (x + 40) (x – 30) = 0

or, x = 30 or x = – 40But x cannot be negative. So, x = 30.

So, base = x + 10 = (30 + 10 =) 40 cm.

Since the triangle is right-angled.

Therefore,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

or, (Hypotenuse)2 = 402 + 302

or, (Hypotenuse)2 = 2500

or, Hypotenuse = 50 cm

Hence, the dimensions of the given

triangle are Base = 40 cm, Altitude = 30

cm and Hypotenuse = 50 cm.

Ex. 36: F i n d t h e p e r i m et e r o f a n eq u i l a t e r a l

t r i a n g l e w h o s e a r ea i s 3 4 cm 2 .

S o l n : Let each side of the triangle be a cm. Then,

its area is .4

3 2a

344

3 2 a

or, 163

4342 a

or, a = 4.

Hence, perimeter of the given triangle

= 3a cm = (3 × 4) cm = 12 cm.

Ex. 37: I f ea c h s i d e o f a n e qu i l a t e r a l t r i a n g l e i s i n c r e a se d b y 2 cm , t h e n i t s a r e a

in c reases by 3 3 cm 2 . F i nd t h e l eng t h

o f e a ch s i d e a n d i t s a r e a .

Soln: Let ABC be an equilateral triangle of side

a cm.

Th en ,

A1 = Area of ABC =

2

4

3a cm2 ......(i)

Let DEF be the new equilateral triangle of

side (a + b ) cm. Then,

A1 = Area of DEF =

2)(4

3b a cm2 .... (ii)

It is given that

33AA 12

or, 334

3)2(

4

3 22 a a

[Using (i) and (ii)]

or, )12(4

3)2(

4

3 22 a a

or, 12)2( 22 a a

or, 1244 22 a a a

or, 4a = 8

or, a = 2

So, length of each side of ABC = 2 cm.

And, area of ABC =2)2(

4

3 = 3 cm2.

Ex. 38: F i n d t h e a r e a o f a n i s os c el e s t r i a n g l e

h a v i n g t h e b a s e 6 cm a n d t h e l e n gt h

o f e a ch e q u a l s i d e 5 cm .




K KUNDAN

Soln: We know that,

Area of an isosceles triangle

=2

1 × Base ×

22 (Base)

4

1side)(Equal

Here, base = 6 cm, equal side = 5 cm.

Area of the given triangle

=222 cm(6)

4

1(5)6

2

1

= 2cm9253

= 2cm163 = 12 cm2

Ex. 39: The base o f an i soscel e s t r i a n g l e i s 12

cm and i t s p er i m et e r i s 3 2 cm . F i n d i t s

a r e a .

Soln: We have, base = 12 cm and

perimeter =32 cm.

Let the length of each of the two equal

sides be b cm. Then,Perimeter = 32 cm

or, 2b + 12 = 32

or, 2b = 32 – 12

or, 2b = 20

or, b = 10

Thus, we have

Base = 12 cm and equal side = 10 cm.

Area of the given triangle

=2

1 × Base ×

22 (Base)4

1side)(Equal

=222 cm(12)

4

1(10)12

2

1

=

2

cm361006 = 2cm646

= 6 × 8 cm2 = 48 cm2

Alternative Method:

Let the length of the two equal sides be x

cm. Then,

Perimeter = 32 cm

or, 2x + 12 = 32

or, 2x = 32 – 12

or, 2x = 20

or, x = 10

Thus, the sides of the given triangle are a

= 10 cm, b = 10 cm and c = 12 cm, and 2s

= 32 cm.

s = 16 cm,

s – a = (16 – 10) cm = 6 cm,s – b = (16 – 10) = 6 cm and

s – c = (16 – 12) cm = 4 cm.

Area of the triangle

= ))()(( c s b s a s s

= 2cm46616

= 2cm6664

= 222 cm68

= (6 × 8) cm2

= 48 cm2

Ex. 40: F i nd t he per i me t e r o f an i soscel e s r i gh t -

a n g l ed t r i a n g l e h a v i n g a n a r ea o f 2 0 0

cm 2 .

Soln: Let ABC be an isosceles right-angled

triangle with right angle at B such that

AB = BC = a cm. Then,

Area of ABC =2

1(Base × Height)

=2

1(a × a ) =

2

2a

or, 200 =2

2a [ Area = 200 cm2 (given)]

or, a 2 = 400

or, a = 20

Now, AC2 = AB2 + BC2

[By Pythagoras Theorem]

or, AC2 = a 2 + a 2

or, AC2 = 2a 2

or, cm.a2AC Hence,

perimeter = AB + BC + AC

= 2a + a 2

= 40 + 20 2 [ a = 20]

= 40 + 20 × 1.41

= 68.2 cm.

Ex. 41: The a r ea of an i soscel e s t r i a n g l e i s 60

cm 2 a n d t h e l en g t h o f ea c h o n e o f i t s

eq u a l s i d e s i s 1 3 cm . F i n d i t s b a s e .

Soln: Let ABC be the given isosceles triangle in

which AB = AC = 13 cm. Draw AD

perpendicular from A on BC. Let BC = 2x

cm.




K KUNDAN

Then, BD = DC = x cm.

In ABC, we have

222 BDADAB [ By Pythagoras Theorem]

or, 222 AD13 x

or, 222 16913AD x x .

Now, area = 60 cm2

or, 60AD)(BC2

1

or, 6016922

1 2

x x

or, 60169 2 x x

or, 3600)169( 22 x x

or, 03600169 24 x x

or, 0)25)(144( 22 x x

or, x 2 = 144 or x 2 =25

x = 12 or x =5

Hence, Base = 2x = 24 cm or 10 cm

Ex. 42: T h e p er i m et e r o f a t r i a n g u l a r f i el d i s

1 4 4 m and t h e r a t i o o f t h e s i d es i s 3 :

4 : 5 . F i n d t h e a r ea o f t h e f i e l d .

Soln: Let a , b , c be the lengths of the sides of

the triangular field. Then,

a : b : c = 3 : 4 : 5

or a = 3x , b = 4x and c = 5x .

Now, perimeter = 144 m

or (3x + 4x + 5x ) = 144 m

or 12x = 144

x = 12

144 = 12

Thus, the sides of the triangle are

a = (3 × 12) m = 36 m,

b = (4 × 12) m = 48 m and

c = (5 × 12) m = 60 m

Now, s = )(2

1c b a

or

s =2

1 × (36 + 48 + 60) = 72

s – a = (72 – 36) = 36,

s – b = (72 – 48) = 24 and

s – c = (72 – 60) = 12

Area of field = ))()(( c s b s a s s

= 2m12243672

= 2222 m2626626

= 246 m26

= 63 × 22 = 864 m2

Ex. 43: F i nd t he per cen t age i n c r ease i n t he a r ea

o f a t r i a n g l e i f i t s e a ch s i d e i s d o u b l ed .

Soln: Let a , b , c be the sides of the old triangle

and s be its semi-perimeter. Then,

s =2

1 (a + b + c ).

The sides of the new triangle are 2a , 2b

and 2c . Let s be its semi-perimeter. Then,

s =2

1 (2a + 2b + 2c ) = a + b + c = 2s .

Let and be the areas of the old and

new triangles respectively. Then,

= ))()(( c s b s a s s and

= )2)(2)(2( c s b s a s s

= )22)(22)(22(2 c s b s a s s

[ s = 2s ]

= 4))()(( c s b s a s s

Increase in the area of the triangle

= – = 4 – = 3.

Hence, percentage increase in area

=

100

3 = 300%.

Ex. 44: T h e l e n gt h s of t h e si d e s of a t r i a n g l e

a r e 5 c m , 1 2 c m a n d 1 3 c m . Fi n d t h e

l e n g t h o f p e r p en d i c u l a r f r o m t h e

oppos i t e ve r t e x t o t he s i de wh ose l eng t h

i s 1 3 cm .

Soln: Here a = 5, b = 12 and c = 13.

s =2

1 (a + b + c )

=2

1 (5 + 12 + 13) = 15

Let A be the area of the given triangle.

Th en ,

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

or, A = )1315)(1215)(515(15

or, A = 2cm231015 ....(i)

Let p be the length of the perpendicular

from vertex A on the side BC. Then,




K KUNDAN

Area =2

1 × Base × Height

or, A = 2

1

× (13) × p ..... (ii)

From (i) and (ii), we get

2

1 × 13 × p = 30

13

60 p cm.

Hence, the length of the perpendicular

from the opposite vertex to the side whose

length is 13 cm is

13

60 cm.

Ex. 45: A f i e l d i n t h e f o r m o f a p a r a l l el o g r am

ha s o n e o f i t s d i a g o n a l s 4 2 m l o n g a n d

t h e p e r p en d i c u l a r d i s t a n c e o f t h i s

d i a g o n a l f r o m e i t h e r o f t h e ou t l y i n g

v e r t i c e s i s 1 0 m 8 dm (s ee F i g ) . F i n d

t h e a r ea o f t h e f i el d .

Soln: We have, AC = 42 m and DL = BM = 10 m

8 dm = 10 m 80 cm = 10.8 m.

Area of the field = 2 × Area of ACD

=2m10.842

2

12

= 453.6 m2

Ex. 46: ABCD i s a squa r e. F i s t h e m i d -po i n t o f

A B a n d B E i s o n e t h i r d o f B C . I f t h e

area of the F B E i s 1 0 8 s q cm , f i n d

t h e l e n g t h o f AC.

Soln: Let the side of the square be x cm. Since

F is the mid-point of AB.

So, BF = AF =2

x cm.

Also, BE is one third of BC.

BC =3

x cm

Now, Area of FBE = 108 cm2 (given)

or, 108BE)(FB2

1

or, 108322

1

x x

or, x 2 = 1296

or x 2 = 12 × 12 × 9

or, x = 12 × 3 = 36 cm.

In ABC, we have AC2 = AB2 + BC2

222 3636AC

or, 2362AC

or, 236AC

Hence, 236AC cm.

Parallelogram

A parallelogram is a four-sided figure whose

oppostie sides are parallel. Thus ABCD is a

parallelogram in which AB||DC and AD||BC.

Area of parallelogram ABCD = Base × Height

Let CDEF be a rectangle on the same base DC

and of the same height FC. Then since

parallelograms on the same base and of the same

height are equal in area.

Parallelogram ABCD = Rectangle CDEF

= CD × FC

= Base × Height

Hence area of a parallelogram is equal to the

product of its base and height.

Area = Base × Height

From the above we have,

( i ) Base of a parallelogram = Height

Area

or, Side of a parallelogram

= altitudeingCorrespond

Area

( ii) Height of a parallelogram =Base

Area

or, Altitude of parallelogram

= sideingCorrespond

Area




K KUNDAN

Solved Examples

Ex. 47: T h e ba s e o f a p a r a l l e l o g r am i s t h r i c e

i t s h ei g h t . If t h e a r e a i s 8 7 6 cm 2

, f i n d t h e b a s e a n d h ei g h t o f t h e

p a r a l l e l o g r a m .

Soln: Let the height of the parallelogram be x

cm. Then, base = 3x cm.

Area of the parallelogram

= (x × 3x ) cm2 = 3x 2 cm2

But, area of the parallelogram is given as

867 cm2.

3x 2 = 867

or, x 2 = 289

17289 x

Thus, height = 17 cm and base

= (3 × 17) cm = 51 cm.

Ex. 48: I n t h e f i g u r e gi v e n b el o w , A BCD i s a

p a r a l l e l o g r am , CM AB and BL AD .(i ) I f AB = 1 6 cm , A D = 1 2 cm a n d

CM = 10 cm , f i n d BL .

(i i ) I f AD = 1 0 cm , CM = 8 cm a n d

BL = 12 cm , f i n d AB.

Soln:(i) We have, base AB = 16 cm and

altitude CM = 10 cm.

Area of parallelogram ABCD

= Base × Altitude

= (16 × 10) cm2 = 160 cm2 .....(i)

Now, taking AD as the base, we have


= Base × Altitude

= (12 × BL) cm2 .....(ii)

From (i) and (ii), we have

12 × BL = 160

cm13.3312

160BL

(ii) We have, AD = 10 cm, BL = 12 cm


= Base × Height

= 10 cm × 12 cm = 120 cm2 ....(iii)

Now, taking AB as the base, we have

Area of parallelogram ABCD = AB × CM

= (AB × 8) cm2 ....(iv)

From (iii) and (iv), we get

AB × 8 = 120

or, AB =8

120 cm

or, AB = 15 cm.

Ex. 49: A f i e l d i n t h e f o r m o f a p a r a l l el o g r am

h a s b a se 1 5 d m a n d a l t i t u d e 8 d m .

F i n d t h e co st o f w a t e r i n g t h e f i e l d a t

t h e r a t e o f 5 0 p a i s e p e r s q u a r e me t r e .

Soln: We have,Base = 15 dm = (15 × 10) m

[1 dm = 10 m]

= 150 m

Altitude = 8 dm = (8 × 10) m = 80 m

Area of the field = (150 × 80) m 2

= 12000 m2

Rate of watering the field

= 50 paise per square metre

= Re2

1 per square metre

Cost of watering the field

= Rs

2

112000 = Rs 6000.

Ex. 50: I n t h e f i g u r e gi v e n b el o w , A BCD i s a parallelogram. DL AB , AB = 8 cm a nd

A D = 5 c m . I f t h e a r e a o f t h e

p a r a l l e l o gr am i s 2 4 cm 2 , f i n d A L .

Soln: We have, base = AB = 8 cm

Area = 24 cm2

Area = Base × Altitude

or, 24 = 8 × DL

or, DL =8

24cm = 3 cm.

Now, in ALD, we have

AD = 5 cm, DL = 3 cm.

By Pythagoras theorem, we have

AD2 = AL 2 + DL 2

or, 52 = AL 2 + 32

or, AL 2 = 52 – 32 = 25 – 9 = 16

or, AL 2 = 42

or, AL = 4

Rhombus

A rhombus is a parallelogram all of whose sides

are equal. In a rhombus the diagonals bisect each

other at right angles. Thus in the rhombus ABCD,

AB = BC = CD = DA and




K KUNDAN

AO = OC, BO = OD

BOC = 90°

Area of the rhombus ABCD

= 2BCD = 4BOC

= 4 ×2

1 × BO × OC

=2

AC

2

BD2

=2

1 × BD × AC

=2

1 (Product of diagonals)

Hence, the area of rhombus is equal to the half

of the product of its diagonals.

Solved Examples

Ex. 51: F i n d t h e a r ea o f a r h o m b u s w h o s e

d i a g o n a l s a r e o f l en g t h s 2 0 cm and 1 8

cm .

Soln: Area of the given rhombus

=2

1 × Product of diagonals

=

1820

2

1cm2 = 180 cm2.

Ex. 52: The a r e a o f a r h om bu s i s 7 2 cm 2 . If i t s

p er i m et e r i s 3 2 cm , f i n d i t s a l t i t u d e.

Soln: We have, perimeter of the rhombus

= 32 cm

4 (side) = 32 cm

[ Perimeter = 4 (side)]

side =4

32cm = 8 cm

Now, area of the rhombus = 72 cm2

or, (Side × Altitude) = 72

or, 8 × Altitude = 72

Altitude =8

72cm = 9 cm

Ex. 53: F i n d t h e a l t i t u d e of a r h o m b u s w h o s e

a r e a i s 3 6 m 2 a n d p e r i m et e r i s 3 6 m .

S o l n : We have, perimeter of the rhombus = 36

m and, area of the rhombus = 36 m2

Now, side of the rhombus

=436

4Perimeter m = 9 m

Altitude of the rhombus

=9

36

Side

Area m = 4 m.

Ex. 54: F i nd t he a r ea o f a r hom bus whose each

s i d e i s of l e n g t h 5 m a n d o n e of t h e

d i a g o n a l s i s o f l en g t h 8 m .

Soln: Let ABCD be a rhombus whose each side

is of length 5 m. Let the diagonals ACand BD intersect at O. Let AC be 8 m.

Since the diagonals of a rhombus bisect

each other at right angles. Therefore,

AOB is a right-triangle

or, AB2 = OA2 + OB2

[Using: Pythagoras theorem]

or, 52 = 42 + OB2

or, OB2 = 25 – 16

or, OB2 = 9

or, OB2 = 32

or, OB = 3 m

BD = 2 × OB = (2 × 3)m = 6m.

Hence, area of rhombus ABCD

=2

1 × Product of diagonals

=

68

2

1m2 = 24 m2

Ex. 55: I f t h e a r ea of a r h ombus be 48 cm 2 an d

o n e o f i t s d i a g o n a l i s 1 2 c m , f i n d i t s

a l t i t u d e .

Soln: Let ABCD be a rhombus of area 48 cm2

and diagonal BD = 12 cm.

Now,

Area = 48 cm2

or, 48BDAC2

1

or, 4812AC2

1

or, 6 × AC = 48

AC =6

48cm = 8 cm




K KUNDAN


each other at right angles.

cm4AC2

1OA

cm3BD2

1OB

Also, AB2 = OA2 + OB2

[Using Pythagoras Theorem]

or, AB2 = 42 + 32

or, AB2 = 16 + 9 = 25 = 52

or, AB = 5 cm

Since a rhombus is a parallelogram also.

Therefore,

Area of rhobmus = AltitudeAB2

1

or, Altitude52

148

or, Altitude = cm5

248 =

5

96cm = 19.2 cm

Ex. 56: I f t h e a r ea o f a r h ombu s b e 2 4 cm 2 an d

o n e o f i t s d i a g o n a l s be 4 cm , f i n d t h e

p e r i m e t e r o f t h e r h om b u s .

Soln: Let ABCD be a rhombus such that its one

diagonal AC = 4 cm. Suppose the diagonals

AC and BD intersect at O.

Now,

Area of rhombus ABCD = 24 cm2

or, 24BDAC2

1

or, 24BD42

1

or, 2 × BD = 24

or, BD = 12

Thus, we have, AC = 4 cm and BD = 12 cm

cm2AC2

1OA and

cm6BD2

1

OB


each other at right angle. Therefore, OAB

is a right triangle, right angled at O.

By Pythagoras theorem, we have

AB2 = OA2 + OB2

or, AB2 = 22 + 62 = 40

or, cm102cm40AB

Hence, perimeter of rhombus

cm108cm1024ABCD .

Ex. 57: I f t h e s i d e o f a s qu a r e i s 4 m and i t i s

c on v e r t e d i n t o a r h ombu s who s e ma j o r

d i a g o n a l i s 6 m , f i n d t h e ot h er d i a g o n a l

a n d t h e a r ea o f t h e r h om b u s .

Soln: Let AB = 4 m be the side of a square ABPQ

which is converted into a rhombus ABCD

such that diagonal AC = 6 m.


each other at right angle, therefore

3mAC2

1OA and AOB = 90°.

In OAB, we have

AB2 = OA2 + OB2

or, 42 = 32 + OB2

or, OB2 = 16 – 9

or, OB = m7

BC = 2OB = m72 .

Hence, area of rhombus ABCD

=BDAC

2

1

=22 m76m726

2

1

.

Trapezium

A trapezium is a four-sided figure having a pair of

opposite sides parallel. Thus ABCD is a trapezium

in which AB||DC.

Area of trapezium

=2

1 × Height × Sum of the parallel sides

=2

1 × (Distance between parallel sides) ×

(Sum of parallel sides)




K KUNDAN

Draw AE and BF perpendicular from A and B to

DC.

Trapezium ABCD

= ADE + Rectangle ABFE + BFC

=2

1 × DE × AE + AE × EF +

2

1 × BF × FC

=2

1 × AE (DE + 2EF + FC) [ BF = AE]

=2

1 × AE (DE + EF + FC + AB) [ EF = AB]

=2

1 × AE × (DC + AB)

=2

1 × Height × Sum of the parallel sides

Hence, the area of a trapezium is equal to the

product of half the sum of parallel sides and height.

Solved Examples

Ex. 58: F i n d t h e a r e a of a t r a p e zi u m w h o s e

p a r a l l e l s i d e s a r e o f l en g t h s 1 0 cm and

1 2 cm a n d t h e d i s t a n c e b et w e en t h em

i s 4 cm .

Soln: We have,

Area of the trapezium =2

1× (Sum of the

parallel sides) × (Distance between the

parallel sides)

=

41210

2

1

cm2

=

422

2

1cm2 = 44 cm2

Ex. 59: F i nd t he a r ea o f t he f i gu r e g i ven bel ow :

Soln: We have,

Area of the given figure

= Area of trapezium ABCD + Area

of trapezium CDEF

=

2cm101521

2

1

2

cm1224152

1

=22 cm1239

2

1cm1036

2

1

= 180 cm2 + 234 cm2 = 414 cm2

Ex. 60: T h e a r e a of a t r a p e zi u m i s 1 8 0 c m 2

a n d i t s h e i g h t i s 9 c m . I f o n e o f t h e

p a r a l l el s i d e s i s l o n g er t h a n t h e o t h e r

b y 6 cm , f i n d t h e t w o p a r a l l e l s i d es .

Soln: Let one of the parallel sides be of length x

cm. Then, the length of the other parallel

side is (x + 6) cm.

Area of the trapezium

=

96

2

1x x cm2

=

962

2

1x cm2

= (9x + 27) cm2

But the area of the trapezium is given as

180 cm2.

9x + 27 = 180

or, 9x = 180 – 27 = 153

or, x =9

153 = 17

Thus, the two parallel sides are of lengths

17 cm and (17 + 6) cm = 23 cm.

Ex. 61: I f t h e p e r i m e t e r o f a t r a p e zi um b e 5 2

cm , i t s n o n -p a r a l l el s i d e s a r e eq u a l t o

1 0 c m ea c h a n d i t s a l t i t u d e i s 8 c m ,f i n d t h e a r ea o f t h e t r a p e z i um .

Soln: We have,

Perimeter of the trapezium = 52 cm

or, Sum of the parallel sides + Sum of the

non-parallel sides = 52 cm

or, Sum of the parallel sides + 2 × 10 = 52

or, Sum of the parallel sides

= (52 – 20)cm = 32 cm

Altitude of the trapezium = 8 cm.

Area of the trapezium

=2

1× (Sum of the parallel sides) × Altitude

=

832

2

1cm2 = 128 cm2.

Ex. 62: T h e p a r a l l e l s i d es o f a t r a p ez i u m a r e

2 0 cm and 1 0 cm . I t s n o n -p a r a l l e l s i d e s

a r e bo t h equa l , each bei ng 13 cm . F i nd

t h e a r e a o f t h e t r a p ez i um .

Soln: Let ABCD be a trapezium such that AB =

20 cm, CD = 10 cm and AD = BC = 13 cm.




K KUNDAN

Draw CL || AD and CM AB.

Now, CL || AD and CD || AB.

ALCD is a parallelogram.

AL = CD = 10 cm and

CL = AD = 13 cm.

In CLB, we have

CL = CB = 13 cm

CLB is an isosceles triangle.

LM = MB = BL 2

1 = 10

2

1 cm = 5 cm

cm10cm10)(20

AL ABBL

Applying Pythagoras theorem in CLM,

we have

CL 2 = CM2 + LM2

or, 132 = CM2 + 52

or, CM2 = 169 – 25 = 144

or, CM = 144 = 12

Area of CLB = CMBL 2

1

=

1210

2

1cm2 = 60 cm2.

Area of parallelogram ALCD = AL × CM= (10 × 12)cm2 = 120 cm2.

Hence, area of trapezium ABCD = Area of

parallelogram ALCD + Area of CLB

= (120 + 60)cm2 = 180 cm2.

Ex. 63: T h e p a r a l l e l s i d es o f a t r a p ez i u m a r e

2 5 cm an d 1 1 cm , wh i l e i t s n o n -p a r a l l e l

s i d es a r e 1 5 cm a n d 1 3 c m . Fi n d t h e

a r e a o f t h e t r a p e zi um .

Soln: Let ABCD be a trapezium such that AB ||

DC, AB = 25 cm, DC = 11 cm, AD = 15 cm

and BC = 13 cm.

Draw CE || DA and CF AB.

Clearly, AECD is a parallelogram.

Now, EB = AB – AE

= AB – DC [ AE = DC]

= (25 – 11) cm = 14 cm

Also, EC = AD = 15 cm. Thus, in ECB, we have

EB = 14 cm, EC = 15 cm and

BC = 13 cm

Let s be the semi-perimeter of the ECB.

Th en ,

1315142

1s cm = 21 cm.

Area of ECB

= ))()(( c s b s a s s

= )1321)(1521)(1421(21 cm2

= 86721 cm2

= 3232737 cm2

= 432 237 cm2

= 7 × 3 × 22 cm2 = 84 cm2 ....(i)

Also, Area of ECB =2

1(Base × Height)

= CF)(EB2

1

= )CF14(2

1 cm2

= (7 × CF) cm2 ....(ii)

From (i) and (ii), we get 7 × CF = 84

or, CF =7

84cm = 12 cm.

Area of parallelogram AECB

= Base × Height

= AE × EF

= (11 × 12) cm2 = 132 cm2

Now,

Area of trapezium ABCD = (Area of

parallelogram AECB) + (Area of ECB)

= (132 + 84) cm2 = 216 cm2

Note : We can find the area of trapezium

directly as follows:

We have, lengths of parallel sides

= 11 cm and 25 cm

Height of the trapezium = 12 cm

Area of trapezium

=21 × Height × (Sum of the parallel sides)

=

)1125(12

2

1 sq cm

=

3612

2

1 = [36 × 6] = 216 sq cm




K KUNDAN

Quadrilateral

A quadrilateral is a plane figure bounded by four

sides. Thus ABCD is a quadrilateral.

Area of quadrilateral ABCD

=2

1 (Length of a diagonal) × (Sum of the lengths

of perpendiculars from the remaining two

vertices on the diagonal).

Join DB. Draw AE and CF perpendiculars to DB

Quadrilateral ABCD = ABD + BDC

=2

1 × BD × AE +

2

1 × BD × CF

=2

1 × BD × (AE + CF)

Hence, the area of quadrilateral is equal to the

product of one diagonals and half the sum of

perpendiculars drawn on it from the other two

vertices.

Note: (i) The area of the quadrilateral can also be

found if the lengths of all its sides and

one diagonal is known, then the area of

each of the two triangles into which the

diagonal divides the quadrilateral can be

found.

( ii) This formula is al so appl icable to a

rectangle, a square, a parallelogram, a

rhombus or a trapezium for which wehave special formulae.

Solved Examples

Ex. 64: T h e d i a go n a l o f a q u a d r i l a t e r a l i s 2 0

m i n l e n gt h a n d t h e p e r p en d i c u l a r s t o

i t f r om t h e oppos i t e ver t i c e s a r e 8 . 5 m

a n d 1 1 m . F i n d t h e a r e a o f t h e

q u a d r i l a t e r a l .

Soln: In quadrilateral ABCD, we have AC = 20

m. Let BL AC and DM AC such that

BL = 8.5 m and DM = 11 m.


=2

1× AC × (BL + DM)

=

)115.8(20

2

1 m2

= (10 × 19.5) m2 = 195 m2.

Ex. 65: I n q u a d r i l a t er a l A BCD s h ow n i n f i g u r e g i v en b el o w , A B | | D C a n d A D A B .

A l so , AB = 8 m , DC = BC = 5 m . F i nd t he

a r e a o f t h e qu a d r i l a t e r a l .

Soln: Clearly, ABCD is a trapezium, and AD

= CE is its height.

We have,BE = AB – AE = AB – DC = (8 – 5) m

= 3 m.

In BCE, we have

BC2 = BE2 + CE2

or, 52 = 32 + CE2

or, CE2 = 25 – 9

or, CE2 = 16

or, CE = 16 m = 4 m.


=2

1 × (AB + DC) × CE

=

4)58(

2

1 m2 = 26 m2.

Ex. 66: F i n d t h e a r e a o f t h e q u a d r i l a t e r a l

ABCD, i n w h i ch AB = 7 cm , BC = 6 cm ,

CD = 12 cm , DA = 15 cm an d AC = 9 cm .

Soln: The diagonal AC divides the quadrilateral

ABCD into two triangles ABC and ACD.


= Area of ABC + Area of ACD

For ABC, we have

s =

2

97611 cm




K KUNDAN

Area of ABC = ))(()( c s b s a s s

= )911()711()611(11

= 24511

= 440 sq cm = 20.98 cm2

For ACD, we have

s =

2

1512918 cm.

Area of ACD

= )1518()1218()918(18

= 36918

= 18 × 3 cm2 = 54 cm2

Hence, area of quadrilateral ABCD

= (20.98 + 54) cm2 = 74.98 cm2.

Ex. 67: F i n d t h e a r ea o f a q u a d r i l a t e r a l A BCD

wh o s e s i d e s a r e 9 m , 4 0 m , 2 8 m an d 15 m r espec t i v e l y an d t he ang l e betw een

t h e f i r s t t w o s i d e s i s a r i g h t a n g l e .

Soln: Let ABCD be the given quadrilateral such

that ABC = 90° and AB = 9 m, BC = 40

m, CD = 28 m, AD = 15 m.

In ABC, we have AC2 = AB2 + BC2

[Using Pythagoras Theorem]

or, AC2 = 92 + 402 = 1681

or, AC = 41 m

Now, Area of ABC =2

1 (Base × Height)

=2

1 (AB × BC)

=2

1(9 × 40) m2 = 180 m2

In ACD, we have

AC = 41 m, CD = 28 m and DA = 15 m.Let a = AC = 41 m, b = CD = 28 m and c

= DA = 15 m. Then,

s =2

1 (a + b + c)

=2

1 (41 +28 + 15) = 42

Area of ACD = )()()( c s b s a s s

= )1542)(2842()4142(42

= 2714314 = 14 × 9 = 126 m2

Hence, Area of quadrilateral ABCD

= (Area of ABC) + (Area of ACD)

= (180 + 126) m2 = 306 m2

Regular Polygons

A polygon is figure bounded by more than four

straight lines. A polygon is said to be regular when

all its sides and angles are equal.

Polygon can be either convex or concave as

mentioned below.

A polygon in which none of its interior (internal)

angles is more than 180°, is known as a convex

polygon.

On the other hand, if at least one angle of apolygon is more than 180º then it is said to be

concave polygon.

We use the following terminology depending

upon the number of sides of a polygon.

Number of sides Polygon

5 Pentagon

6 Hexagon

7 Septagon

8 Octagon

9 Nonagon

10 Decagon

11 Undecagon

12 Dodecagon

(i) Area of a regular polygon of n sides having given

the length of a side and the radius of the inscribed

circle

=2

1 × number of sides (n ) × length of a side ×

radius of the inscribed circle

Let ABCDE be a regular polygon in which AB = a

and OG (the radius of the inscribed circle) = r

Join OA, OB, OC, OD, OE. Thus the polygon is

divided into as many triangles as its number of

sides.

Area of polygon

= Area of AOB × Number of sides of the

polygon

=2

1 × AB × OG × n = r a

n

2




K KUNDAN

(a) Hexagon

It can be easily seen that triangle AOB is

equilateral.

Area of equilateral triangle AOB =2

4

3a

Area of hexagon ABCDEF =2

4

36 a

= 2

33 2a

(b) Octagon

Here it will be seen that the radius of the inscribed

circle

OG = OH + HG = 2

a

+ HG

HG = KB =2

a

OG =

2

12

22a

a a

Area of octagon ABCDEFLM = r a n

2

= 122

2

12

2

8 2

a a

a

(ii) The area of a regular polygon of n sides having

given the length of one side and the radius of thecircumscribed circle

=2

1× number of sides × side

22

2

SideRadius

Let ABCDE be a regular polygon of n sides with

AB = a and OB (the radius of the circumscribed

circle) = R

Draw OG perpendicular to AB. Then OG is the

radius of the inscribed circle.

OG =

2222

2RGBOB

a

Area of a regular polygon

22

222

a R a

n ar

n

In the case of a hexagon, R = a and n = 6.

area of a regular hexagon

22

2

2

33

22

6a

a a a

(iii) Some Important Results

(A) Interior Angle of a Regular Polygon: Each

interior angle of a regular polygon of n sides is

equal to

º

1802

n

n or 180° – (Exterior Angle).

(B) Exterior Angle of a Regular Polygon: Each

exterior angle of a regular polygon of n sides is

equal to

º360

n .

(C) In a convex polygon of n sides, we have:

(i ) Sum of all interior angles

= (2n – 4) right angles

= (n – 2) × 180°

( ii) Sum of all exterior angles = 4 right angles

( i i i ) Number of diagonals of a polygon of n sides

=

n

n n

2

)1(

Some Particular cases:

Regular Polygon Internal Angle Tr iang le 60°

Quadrilateral 90°

Pentagon 108°

Hexagon 120°

Octagon 135°

Nonagon 140°

Decagon 144°




K KUNDAN

(D) Circum-circle of a Regular Polygon: A

regular polygon can be inscribed in a circle which

is known as the circum-circle or circumscribing

circle. The centre of this circle is also the centre

of the polygon and the radius is known as thecircum-radius which is generally denoted by R.

If a is the length of each side of a regular

polygon and R is the circum-radius, then we have

the following results:

(a) R =

n a 180ºcosec2

(b ) Area of the polygon =

n na

º180cot

4

1 2

or, Area of the polygon

=

n n n

º180cos

º180sinR2

(c) Area of the circum-circle of an n-sided

regular polygon =

n a

º180cosec

4

22

Particular cases:

l Area of a regular hexagon

=22

2

3330ºcot4

6 a a

l Area of circum-circle of a regular

hexagon = a 2

l Perimeter = 6a . Each angle = 120º.

(E) In-circle of a Regular Polygon: A regular

polygon can also circumscribe a circle. A circle

having centre at the centre of a regular polygon

and touching all sides of it is called the in-circle.

If a is the length of a side of a regular polygon

and r is the radius of the in-circle, then we have

the following results:

(a) r =

n

a 180ºcot

2

(b ) Area of polygon =

n nr º180tan2

(c) Area of the in-circle of an n-sided regular

polygon =

n a

º180cot

4

22

Particular cases:

l Radius of in-circle of a regular hexagon

= a a

2

3º30cot

2

l Area of the in-circle of a regular hexagon

=2

2

4

3

2

3a a

Solved Examples

Ex. 68: F i n d t h e a r e a o f a r e gu l a r h e x a g on

wh o s e s i d e i s 1 0 cm l o n g .

S o l n : Area of a regular hexagon = 2side2

33

Here, side = 10 cm.

Hence, Area of the given regular hexagon

= 22 cm3150102

33

Ex. 69: T h e a r e a o f a r e g u l a r h e xa g o n i s 6 0 0

cm 2 . D et e rm i n e i t s p er i m e t e r .

Soln: We know that the area of a regular

hexagon is equal to 2side2

33.

Area = 23600 cm

or, 3600)side(2

33 2

or,33

31200)side( 2

or, (side)2 = 400

or, side = 20 cm.

So, perimeter = 6 (side) = (6 × 20) cm

= 120 cm.

Ex. 70: F i nd t o t he nea r est m e t r e t he si de o f a

r e g u l a r o c t a g o n a l e n c l o su r e w h o s e

a r e a i s 1 h ec t a r e .

Soln: Area of a regular octagon = 2212 a

2212 a = 1 hectare.

a 2 =)21(2

10000

sq cm

or, a 2 = 2701 sq m approx.

a = 46 metres approx.

Ex. 71: A s qu a r e a n d a r e g u l a r h e x a g o n h a v e

eq u a l p e r i m e t e r s . Compa r e t h e i r a r ea s .

Soln: Let P be the perimeter of both a square

and a regular hexagon.




K KUNDAN

Th en ,

side of the square =4

P and

side of the regular hexagon =6

P.

A1 = Area of the square = (side)2 =

16

P2

A2 = Area of the regular hexagon

= 22

2P

24

3

6

P

2

33sdie

2

33

32

3

243

16

2

2

2

1 P

P

A

A .

Hence, areas of the square and the

hexagon are in the ratio 32:3 .

Ex. 72: T h e s i d e of a r eg u l a r p en t a g on i s 1 0

cm . F i n d i t s a r e a .[Tak e Cot 36°= 1.37 63 ]

Soln: The area of an n -sided regular polygon is

2sidesº180

cot4

n

n .

Here, n = 5 and side = 10 cm

Area of the pentagon =4

5(cot 36°) (10)2

= 125 cot 36° cm2

= 125 × 1.3763

= 172.04 cm2

Ex. 73: F i n d t h e d i f f e r e n c e b e t w e en t h e a r e a

o f a r eg u l a r h e x a g o n ea c h o f w h o s e

s i d e i s 7 2 cm a nd t h e a r e a of t h e ci r c l e

i n s cr i b ed i n i t .

(Tak e7

2 2 ).

Soln: We know that the area of an n -sided

regular polygon is

n na

º180cot

4

1 2 and

area of the incircle is

n a

º180cot

4

1 22;

where a is the side of the polygon.

Here, a = 72 and n = 6.

Required area

=

n a

n na

º180cot

4

1º180cot

4

1 222

= º30cot727

22

4

1º30cot726

4

1 222

= 2cm42.1221937776

= (13468.42 - 12219.42) m2

= 1249 cm2.

Ex. 74: A r e g u l a r h e x a g on i s i n s c r i b e d i n a

c i r c l e of r a d i u s 5 cm . F i n d t h e a r e a o f

t h e c i r c l e wh i c h i s o u t s i d e t h e h ex a g o n .

[Use = 3 .14 an d 3 = 1 . 73 ]

Soln: Area of the circumcircle

= (radius)2 = × 52 = 25 cm2

= 25 × 3.14 cm2 = 78.5 cm2

We know that the area of a regular polygon

of n sides is given by

n n n

180cos

180sinR2

;

where R is the radius of the circumcircle.

Here, n = 6, R = 5 cm.

So, Area of the regular hexagon

= 6 × 52 × sin 30° × cos 30° cm2

= 150 ×2

3

2

1 cm2

= 64.875 cm2

Hence,

required area = Area of the circumcircle

– Area of the hexagon

= 78.5 - 64.875 cm2

= 13.625 cm2

Circle

A circle is a geometrical figure consisting of all

those points in a plane which are at a given

distance from a fixed point in the same plane.

The fixed point is called the centre of the circle

and the constant distance is known as its radius.

(plural radii)

In the given figure, O is the centre and r is the

radius of the circle. A circle with centre O and

radius r is generally denoted by C (O, r ).

The word ci rcle is of ten used for the

circumference.

(i) Some Important Terms

(a) Circular Region: The part of the circle that

consists of the circle and its interior is called the

circular region.

A circular region is also called a circular disc

as shown in the figure given below.




K KUNDAN

(b) Chord of Circle: A line segment joining

any two points on a circle is called a chord of the

circle. It should be noted that a chord is not a part

of the circle.

In the figure given below, PQ is a chord of thecircle.

(c) Diameter: A chord passing through the centre

of a circle is known as its diameter.

Note that a circle has many diameters and a

diameter of a given circle is one of the largest chords

of the circle. Also, all diameters are of the same

length.

(d) Semi-Circle: Clearly, if d is diameter of a

circle of radius = r , then d = 2r.

A diameter of a circle divides the circumference

of a circle into two equal parts each of which is

called a semi-circle.

(e) Quadrant: Two perpendicular diameters of

a circle divide its circumference into four equal

parts each of which is known as a quadrant.

(f) Concentric Circles: Circles having the same

centre but with different radii are said to be

concentric circles. The figure given below shows two concentric

circles.

(g) Congruent Circle: Two circles are said to

be congruent if and only if either of them can be

superpassed on the other so as to cover it exactly.

It follows from the above definition that two circles

are congruent if and only if their radii are equal.

(ii) Circumference of a Circle

The perimeter of a circle is called its circumference. The ratio of the circumference of a circle and

its diameter is always constant.

The ratio

Diameter

nceCircumfere = 3.14 (approximately)

This ratio is denoted by (Pi).

Thus, we have

= 3.14 (approximately) =7

22 (approximately)

Now, Diameter

nceCircumfere

=

or,r 2

C = C = 2r

Thus, circumference C of a circle of radius r is

given by C = 2r

If d denotes the diameter of the circle.

Then, d = 2r

C = d

Note: The number is not a rational number, but

its value upto two decimal places coincides

with7

22. So, we take the value of as

7

22.

In the remaining part of this chapter, unless

stated otherwise, the value of will be taken

as 7

22

.

(iii) Area of a Circle

In this section, we shall first obtain the formula

for the area of a circle and then the same will be

used to solve some simple problems.

To obtain the formula for the area of a circle, let

us consider the following.

Draw a circle of any radius (say, 2 cm) on a thin

card-board. Cut it out, and by folding divide it into

four equal sectors. Cut these four sectors out, and

bisect each of them by folding.

Now you have got eight sectors, arrange these

as shown in the figure below.

Next bisect each sector, as before, and so get 16

equal sectors. Rearrange these as shown in the

figure below.

Now notice that as the number of sectors is

increased, each arc is decreased; so that




K KUNDAN

( i ) the outlines AB, DC tend to become straight

lines, and

( ii) the angles at D and B tend to become right

angles.

Thus, when the number of se ctor s isindefinitely increased the figure ABCD ultimately

becomes a rectangle whose length is the semi-

circumference of the circle , and whose breadth is its

radius .

Hence area of the circle

= )radius()ncecircumfere(2

1

=22

2

1r r r

Hence, area A of a circle of radius r cm is

given by

A = r 2

Also, r =

A

Note:Area of a semi-circle

=2

1 (Area of the circle) =

2

2

1r

Area of a quadrant of a circle

=4

1 (Area of the circle) =

2

4

1r .

(iv) Area Enclosed By Two Concentric

Circles

If R and r are radii of two concentric circles, then

Area enclosed by the two circles

)R)(R()R(R 2222 r r r r

(v) Sector—Its arc and Area

The angle at the centre of a circle contains four

right angles or 360°. Hence, if through the centre

of a circle we draw 360 radii making equal angles

with one another, 360 angles of 1 degree each

would be formed at the centre. Since equal angles

at the centre are subtended by equal arcs, the

whole circumference would be divided into 360equal arcs and the area of the circle would be

divided into 360 equal sectors.

Hence,

(a) Arc of a sector of 1° =360

1× circumference

Arc of a sector of D° =360

D× circumference

Hence, Arc of the sector

= circletheof nceCircumfere

360

angleSector

(b ) Area of a sector of 1° =360

1× area of circle

Area of a sector of D° =360

D× area of circle

Hence, Area of the sector

=360

angleSector × Area of the circle

(c) To show that area of sector

=2

1× radius × length of arc.

Proof. Area of a sector of D°

=360

D× area of circle

=360

D×

2

1 × radius × circumference

=360

D× circumference ×

2

1× radius

= arc ×2

1 × radius

=2

1radius × length of arc

Note: A quadrant is a part of the circle contained

between two perpendicular radii. Hence a quadrant

is a sector of 90°.

(vi) Area of Segment

Any chord of a circle, which is not a diameter,

such as AB, divides the circle into two segments,

one greater and one lesser than a semi-circle.

Greater segment is called major segment and

lesser segment is called minor segment.




K KUNDAN

It will be seen from the figure that area of

segment ACB = sector OACB – OAB.

The area of the segment ADB will be found by

subtracting the area of the segment ACB from the

area of the circle.Note: Area of a minor segment of angle in a

circle of radius r is given by

A =

θ

θ r sin

2

1

360

2 (Always Remember)

(vii) Some Particular Cases

(a) Area of a semi-circle : The sector of a semi-

circle is 180°.

Area of a semi-circle

=22

2

1

360

180r r

(b) Area of a quadrant : The sector angle of a

quadrant of a circle is 90°.

Area of a sector of circle

=22

4

1

360

90r r

(c) Angle described by minute-hand in 60

minutes = 360°.

Angle described by minute-hand in one

minute =

60

360 = 6°.

Thus, minute-hand rotat es th rough an

angle of 6° in one minute.

(d ) Angle described by hour-hand in 12 hours

= 360°

Angle described by hour-hand in one

minute =

2

1

60

30

Thus, hour -han d rot at es thr ough

2

1

6012

360 in one minute.

Solved Examples

Ex. 75: The r a t i o o f t h e r a d i i o f t w o ci r c l e s i s

2 : 5 . W h a t i s t h e r a t i o o f t h e i r

c i r c um f e r e n c e s?

Soln: We have, ratio of radii = 2 : 5. So, let the

radii of two circles be 2 r and 5r

respectively.

Let C1 and C

2 be the circumference of two

circles of radii 2r and 5r respectively.

Th en ,

r r 422C1 and r r 1052C2

5

2

10

4

C

C

2

1

r

r

or, C1 : C

2 = 2 : 5.

Ex. 76: A p i e c e o f w i r e i n t h e f o r m o f a

r e ct a n g l e 8 . 9 cm l o n g a n d 5 . 4 cm b r o a d

i s r e s h a p ed a n d b en t i n t o t h e f o r m o f

a c i r c l e . Fi n d t h e r a d i u s o f t h e ci r c l e.

Soln: We have,Length of the wire

= Perimeter of the rectangle

= 2 (l + b )

= 2 × (8.9 + 5.4) cm

= 28.6 cm

Let the wire be bent into the form of a

circle of radius r cm. Then,

Circumference = 28.6 cm

or, 6.282 r

or, 6.287

222 r

or, r =222

76.28

cm

or, r =10222

7286

cm = 4.55 cm.

Ex. 77: A co p p er w i r e, when b en t i n t h e f o rm o f

a squa r e , encl o ses an a r ea o f 48 4 cm 2 .

I f t h e s ame w i r e i s b en t i n t h e f o rm o f

a c i r c l e, f i n d t h e a r e a e n cl o s ed b y i t .

(Use7

2 2 ).

Soln: Area of the square = 484 cm2.

Side of the square

= 484 cm = 22 cm

[ Area = (side)2 side = Area )

So, perimeter of the square = 4 (side)= (4 × 22) cm = 88 cm.

Let r be the radius of the circle. Then,

Circumference of the circle = Perimeter of

the square

or, 882 r

or, 887

222 r

or, r = 14 cm

Area of the circle =

22 14

7

22r cm2

= 616 cm2

Ex. 78: T h e d i am et e r o f t h e w h e el o f a c a r i s

7 7 c m . H ow m a n y r e v ol u t i o n s w i l l i t

m a k e t o t r a v el 1 2 1 k m .Soln: We have,

diameter of the wheel of the car = 77 cm

Circumference of the wheel of the car

= d =

77

7

22cm = 242 cm

Note that in one revolution of the wheel,






K KUNDAN

or, r 7

222 = 220

r = 35 m

Since the track is 7 metres wideeverywhere. Therefore,

R = Outer radius = r + 7 = (35 + 7) m

= 42 m.

Outer circumference = 2R

=

42

7

222 m = 264 m.

Rate of fencing = Rs 2 per metre

Total cost of fencing

= (Circumference × Rate)

= Rs (264 × 2) = Rs 528.

Ex. 84: A c i r c u l a r g r a s s y p l o t o f l a n d , 4 2 m i n

d i a m e t e r , h a s a p a t h 3 . 5 m w i d e

r u n n i n g r o u n d i t o n t h e o u t s i d e. F i n d

t h e c o st o f g r a v el l i n g t h e p a t h a t R s 4

p e r s q u a r e me t r e .

Soln: Radius of the plot =

2

42m = 21m.

Radius of the plot including the path

= (21 + 3.5) m = 24.5 m.

Area of the path

= 22 )21()5.24( m2

= 22 )21()5.24( m2

= 215.24215.24 m2

= 5.35.45 m2

=

5.35.457

22

m2

= 500.5 m2

Hence, cost of gravelling the path

= Rs (500.5 × 4) = Rs 2002.

Ex. 85: A p a p er i s i n t h e f or m o f a r e c t a n gl e

ABCD i n w h i ch AB = 20 cm and BC = 14

cm . A sem i c i r c u l a r p o r t i o n w i t h BC a s

d i a m e t er i s cu t o f f . F i n d t h e a r e a o f

t h e r em a i n i n g p a r t .

S o l n : Length of the rectangle ABCD

= AB = 20 cm

Breadth of the rectangle ABCD= BC = 14 cm

Area of rectangle ABCD

= (20 × 14) cm2 = 280 cm2.

Diameter of the semi-circle

= BC = 14 cm

Radius of the semi-circle = 7 cm.Area of the semi-circular portion cut off

from the rectangle ABCD

=

22 7

7

22

2

1

2

1r cm2 = 77 cm2

Area of the remaining part = Area of

rectangle ABCD - Area of semi-circle

= (280 - 77) cm2 = 203 cm2

Ex. 86: T h e c i r c um f e r e n ce s o f t w o ci r c l e s a r e

i n t h e r a t i o 2 : 3 . Fi n d t h e r a t i o o f t h e i r

a r e a s .

S o l n : Let r 1 and r

2 be the radii of two given circles

and C 1 and C

2 be their circumference. Then,

C1 = 2r

1 and C

2 = 2r

2

Now, C1 : C

2 = 2 : 3

or, 3

2

C

C

2

1 or 3

2

2

2

2

1 r

r or 3

2

2

1 r

r

Let A1 and A

2 be the areas of two circles.

Th en ,

A1 = 2

1r and A2 = 2

2r

9

4

A

A22

21

22

21

2

1

r

r

r

r

9

4

3

222

21

2

1

r

r

r

r

A1 : A

2 = 4 : 9

Hence, the areas of two given circles are

in the ratio 4 : 9.

Ex. 87: T h e a r e a s o f t w o c i r c l es a r e i n t h e

r a t i o 1 6 : 2 5 . Fi n d t h e r a t i o of t h e i r

c i r c um f e r e n c e s .

Soln: Let r 1 and r

2 be the radii of two circles and

let their areas be A1 and A

2 respectively.

Th en ,

211A r and 2

22A r




K KUNDAN

Now, A1 : A

2 = 16 : 25 [Given]

or, 22

21 : r r = 16 : 25

or, 25

1622

2

1

r

r or 2

2

22

2

15

4

r

r

5

4

2

1 r

r ..... (i)

[Taking square root of both sides]

Let C1 and C

2 be the circumferences of

two circles. Then,

C1 = 2r

1 and C

2 = 2r

2.

2

1

2

1

2

1

2

2

C

C

r

r

r

r

or, 5

4

C

C

2

1 .... [Using (i)]

C1 : C2 = 4 : 5Hence, the circumferences of the two

circles are in the ratio 4 : 5.

Ex. 88: A squ a r e p a r k h a s ea c h s i d e of 1 0 0 m .

A t e a ch c or n e r o f t h e p a r k , t h e r e i s a

f l o w e r b ed i n t h e f or m o f a q u a d r a n t

o f r a d i u s 1 4 m a s sh ow n i n t h e f i g u r e

g i v e n b el o w . F i n d t h e a r e a o f t h e

r e m a i n i n g pa r t o f t h e p a r k .

(Tak e7

2 2 ).

Soln:

Area of each quadrant of radius 14 m

= 14147

22

4

1

4

1 2 r [ r = 14]

= 154 m2

Area of 4 quadrants = (4 × 154) m2

= 616 m2.

Area of square park having side 100 m

long = (100 × 100) m2 = 10,000 m2

Hence, area of the remaining part of the

park = (10000 - 616 =) 9384 m2.Ex. 89: Fou r e q u a l c i r c l e s a r e d e s cr i b ed a b o u t

t h e f o u r c or n e r s o f a s q u a r e s o t h a t

e a ch t o u c h e s t w o o f t h e o t h e r s a s

s h ow n i n f i g u r e . F i n d t h e a r ea o f t h e

shaded r eg i on , each si de o f t he squa r e

m e a su r i n g 1 4 c m .

Soln: Let ABCD be the given square each side

of which is 14 cm long. Clearly, the radius

of each circle is 7 cm.

We have:

Area of the square of side 14 cm long

= (14 × 14) cm2 = 196 cm2

Area of each quadrant of a circle of radius

7 cm

=

22 7

7

22

4

1

4

1r cm2

= 38.5 cm

2

Area of 4 quadrants

= 4 × 38.5 cm2 = 154 cm2.

Hence, area of the shaded region

= Area of the square ABCD – Area of 4

quadrants

= (196 – 154) cm2 = 42 cm2.

Ex. 90: An a r c sub t ends an an g l e o f 36º a t t he

cen t r e of a c i r c l e o f r a d i u s 3 . 6 cm , f i n d

t h e l e n g t h o f t h e a r c .

Soln: We know that the length of an arc of a

circle of radius r is given by

ncecircumfere

360

D = r 2

360

D

Here, D = 36° and r = 3.6 cm

Length of the arc

=

6.3

7

222

360

36cm = 2.26 cm.

Ex. 91: A se ct o r i s cu t f r om a c i r c l e o f r a d i u s

21 cm. T he an g le o f t h e sec to r i s 15 0°.

F i n d t h e l en g t h o f i t s a r c a n d a r e a .

Soln: The arc length l and area A of a sector of

angle D° in a circle of radius r are given

by l = r 360

D and

A =2

360

Dr respectively.

Here, r = 21 cm and D = 150°

l =

217

222360

150 cm = 55 cm

And, A =

2

217

22

360

150cm2

=

2

1155 cm2 = 577.5 cm2




K KUNDAN

Ex. 92: The a r ea o f a sect o r o f a c i r c l e i s t h 1 0

1

o f t h e a r ea o f t h e c i r c l e. F i n d t h e a n g l e

o f t h e s ec t o r .S o l n : Let the radius of the circle be r cm and the

sector angle be of x°. Then,

Area of the sector =

2

º360r

x cm2

and, Area of the circle = r 2 cm2

It is given that:

Area of the sector =10

1× Area of the circle

or,22

10

1

º360

ºr r

x

or, x = 2

2 360

10

1

r r

or, x = 36

Hence, the sector angle is of 36°.

Ex. 93: A 36°sec t o r o f a c i r c l e h a s a r e a 3 . 8 5

cm 2 . W h a t i s t h e l e n g t h o f t h e a r c o f

t h e s ec t o r ?

Soln: Let r cm be the radius of the circle. We

have, sector angle = 36° and area of the

sector = 3.85 cm2.

We have,

Area of the sector

=

circletheof Area

360º

angleSector

or, 3.85 =2

7

22

º360

º36r

or,2236

736085.32

r

or, r 2 = 12.25

or, r = 25.12 cm = 3.5 cm

Now,

Area of the sector

=2

1× length of the arc × radius

or, 3.85 = 3.5arctheof length2

1

or, length of the arc

= 5.3

285.3

cm = 2.2 cm

Ex. 94: I n t h e f i g u r e g i v en b e l o w a r e s h o w n

s ec t o r s o f t w o c o n c en t r i c c i r c l e s o f

r a d i i 7 cm a n d 3 . 5 cm . F i n d t h e a r e a

o f t h e sh a d e d r e gi o n .

(Use7

2 2 )

Soln: Let A1 and A

2 be the areas of sectors OAB

and OCD respectively. Then,

A1 = Area of a sector of angle 30° in a

circle of radius 7 cm.

=

27

7

22

360

30 cm2

2

360

D Using A r

=

6

77cm2

Area of a sector of angle 30° in a circle of

radius 3.5 cm.

=

2

5.37

22

360

30cm2

=

2

7

2

7

7

22

12

1cm2 =

24

77cm2

Area of the shaded region = A1 – A

2

=

24

77

6

77cm2 = 14

24

77 cm2

=8

77cm2 = 9.625 cm2

Ex. 95: The pe r im e t er o f a sec t o r o f a c i r c l e o f r a d i u s 5 . 2 cm i s 1 6 . 4 cm . F i n d t h e a r ea

o f t h e s ec t o r .

Soln: Let OAB be the given sector. Then

Perimeter of sector OAB = 16.4 cm

or, OA + OB + arc AB = 16.4 cm

or, 5.2 + 5.2 + arc AB = 16.4or, arc AB = 6 cm

Area of sector OAB

=2

1 × length of the arc × radius

=

2.56

2

1cm2 = 15.6 cm2




K KUNDAN

Ex. 96: T h e a r ea o f a n e qu i l a t e r a l t r i a n g l e i s

3 4 9 cm 2 . T a k i n g e a ch a n g u l a r p o i n t

a s c en t r e , a c i r c l e i s d e s cr i b ed w i t h

r a d i u s eq u a l t o h a l f t h e l en g t h o f t h e

s i d e o f t h e t r i a n g l e a s s h own i n f i g u r e .

F i n d t h e a r e a o f t h e t r i a n g l e n o t

i n c l u d e d i n t h e c i r c l e.

Soln: Let each side of the triangle be a cm. Then,

Area = 349 cm2 or 349

4

3 2 a

2

side4

3 Area

or, a 2 = 49 × 4

a = 14 cm

Thus, radius of each circle is 7 cm.

Now, required area

= Area of ABC – 3 × (Area of a sector of

angle 60° in a circle of radius 7 cm)

=

27

7

22

360

603349 cm2

= ]77349[ cm2

= [49 × 1.73 – 77] cm2 = 7.77 cm2

Ex. 97: The l en g t h o f m i n u t e-h a n d o f a c l o c k i s 1 4 c m . F i n d t h e a r e a s w ep t b y t h e

m i n u t e-h a n d i n o n e m i n u t e.

(Use7

2 2 ).

Soln: Clearly, minute-hand of a clock describes

a circle of radius equal to its length ie 14

cm.

Since the minute-hand rotates through 6°

in one minute, therefore, area swept by

the minute hand in one minute is the area

of a sector of angle 6° in a circle of radius

14 cm.

Hence, required area =2

360

anglesectorr

=

214

7

22

360

6cm2

=

1414

7

22

60

1cm2

=15

154cm2 = 10.26 cm2

Ex. 98: T h e m i n u t e-h a n d o f a c l o ck i s 1 0 c m

l o n g . F i n d t h e a r e a o n t h e f a c e o f t h e

c l o c k d e s cr i b ed b y t h e m i n u t e -h a n d

b et w e en 9 am a n d 9 : 3 5 am .

Soln: Angle described by the minute-hand inone minute = 6º.

So, angle described by the minute-hand

in 35 minutes = (6 × 35)° = 210°

Area swept by the minute-hand in 35

minutes = Area of a sector of angle 210°

in a circle of radius 10 cm

=

2

107

22

360

210cm2

= 183.3 cm2

Ex. 99: T h e s h o r t a n d l o n g h a n d s o f a c l o ck

a r e 4 cm a n d 6 cm l o n g r e s p ec t i v e l y .

F i n d t h e s um o f d i s t a n c es t r a v el l e d b y

t h e i r t i p s i n 2 d a y s .

(Tak e 7

2 2 ) .

Soln: In 2 days, the short hand will complete 4

rounds.

Distance moved by its tip = 4

(circumference of a circle of radius 4 cm)

=

4

7

2224 cm =

7

704cm

In 2 days, the long hand will complete 48

rounds.

Distance moved by its tip

= 48 (circumference of a circle of

radius 6 cm)

=

6

7

22248 cm =

2

12672cm

Hence, the sum of the distances moved

by the tips of two hands of the clock

=

7

12672

7

704cm = 1910.57 cm

Ex.100:F i n d t h e a r e a o f a s egmen t o f a c i r c l e

o f r a d i u s 2 1 c m i f t h e a r c of t h e

segm ent h as m easu r e 60°.

Soln: Let O be the centre of the circle and PXQ

the arc of the segment such that m(PXQ)

= 60º

POQ = 60º

Now,

Area of sector POQ

=

21217

22

º360

º60cm2

=

21322

3

1cm2

= 231 cm2




K KUNDAN

Since OP = OQ and POQ = 60º.

Therefore, OPQ is an equilateral triangle.

Therefore,

Area of OPQ = 2side4

3

=

2121

4

3cm2

=4

44173.1 cm2 = 190.73 cm2

Area of segment PXQ

= Area of sector OPQ – Area of OPQ

= (231 – 190.73) cm2 = 40.27 cm2.

Ex.101: I f t h e a r c o f a s egmen t o f a c i r c l e h a s

m easure 12 0°. I f t h e ra d i us o f t he c i r c l e

i s 6 cm , f i n d t h e a r e a o f t h e se gmen t .

Soln: Let O be the centre of the circle and PXQ

the arc of the segment such that m (PXQ)

= 120°.

POQ = 120°

Now, Area of sector POQ =2

º360

º120r

=

66

7

22

º360

º120cm2

=

667

22

3

1

cm2

=7

264cm2

Let ONPQ. Then OPN is right angled

triangle, right angled at N and OPN =

30°.

In PNO,

sin 30° =OP

ON

ON = sin 30° × OP

ON =2

1(Hypotenuse)

[since, sin 30° =2

1]

=

6

2

1OP

2

1cm = 3 cm

In OPN, we have

222 PNONOP

222 ONOPPN

or, PN = 22 PNOP

or, PN = 936 = 27 cm = 33 cm

PQ = 2PN = 332 cm = 36 cm

Area of OPQ =2

1× PQ × ON

=22 cm39cm336

2

1

Hence,

Area of segment PXQ = Area of sector OPQ

– Area of OPQ

=

39

7

264cm2

= (38.142 – 15.588) cm2

= 22.554 cm2 ]732.13[

Ex.102:A ho r s e i s p l a c e d f o r g r a z i n g i n s i d e a

r e ct a n g u l a r f i e l d 7 0 m b y 5 2 m a n d i s

t e t h e r e d t o o n e co r n e r b y a r o p e 2 1 m

l o n g . On how m uch a r e a ca n i t g r a z e ?

Soln: Shaded portion indicates the area which

the horse can graze. Clearly, shaded area

is the area of a quadrant of a circle of

radius r = 21 m.

Hence, required area =2

4

1r

=

2

217

22

4

1cm2

=2

693cm2 = 346.5 cm2




K KUNDAN

Ex.103: PQRS i s a d i am e t er o f a c i r c l e o f r a d i u s

6 cm . T h e l e n g t h s PQ , QR and RS a r e

equa l . Sem i -c i r c l es a r e d r aw n on PQ and

QS a s d i am et e r s a s s h own i n t h e f i g u r e

g i v e n be l ow . Fi n d t h e p er i m e t e r o f t h e s h a d e d r eg i o n .

Soln: We have PS = diameter of a circle of radius

6 cm = 12 cm

cm43

12RSQRPQ

QS = QR + RS = (4 4) cm = 8 cm

Hence, required perimeter

= Arc of semi-circle of radius 6 cm + Arc

of semi-circle of radius 4 cm + Arc of semi-circle of radius 2 cm

cm12cm)246(

Ex.104: I n t h e f i g u r e g i v en b e l o w , A O B CA

r e p r e sen t s a q u a d r a n t o f a c i r c l e o f

r a d i u s 3 . 5 cm w i t h c en t r e O . Ca l c u l a t e

t h e a r e a o f t h e s h a d e d p o r t i o n .

(Take7

2 2 ).

Soln: Area of quadrant AOBCA =2

4

1r

= 25.37

22

4

1

=8

77

2

7

2

7

7

22

4

1 cm2 = 9.625 cm2

Area of AOD = HeightBase2

1

= OBOA2

1

= 2cm25.32

1 = 3.5 cm2

Hence, area of the shaded portion

= Area of quadrant – Area of AOD

= (9.625 – 3.5) cm2 = 6.125 cm2

Ex.105: F i n d t h e a r e a o f t h e s h a d e d r eg i o n i n

t h e f i g u r e g i v e n b el ow .

Soln: Clearly, radius of the bigger semi-circle

= 14 cm

Area of the bigger semi-circles

= 222 cm147

22

2

1

2

1r = 308 cm2

Radius of each of the smaller circle

= 7 cm

Area of 2 smaller semi-circles

=222 cm154cm7

7

22

2

12

Hence, required area

= (308 + 154) cm2 = 462 cm2

Ex.106: ABCD i s a f l owe r bed . I f OA = 21 m and

OC = 14 m , f i n d t h e a r ea o f t he bed .

(Take7

2 2 ).

S o l n : We have OA = R = 21 m and

OC = r = 14 m.

Area of the flower bed = area of a

quadrant of a circle of radius R – Area of a

quadrant of a circle of radius r

2222

44

1

4

1r R r R

222 cm14217

22

4

1

m14andm21 r R

2m142114217

22

4

1

22 m5.192m7357

22

4

1




K KUNDAN

Ex.107: ABCP i s a q u a d r a n t o f a c i r c l e o f r a d i u s

1 4 c m . W i t h A C a s d i a m et e r , a s em i -

c i r c l e i s d r a w n . F i n d t h e a r e a o f t h e

s h a d e d p o r t i o n .

Soln: In the right-angled triangle ABC,

We have AC2 = AB2 + BC2

or, AC2 = 142 + 142

or, 214142AC 2 cm

or, 272

214

2

AC

cm

Now, required area = Area APCQA

= area ACQA – Area ACPA

= Area ACQA – (Area ABCPA –

Area of ABC)

= (Area of semi-circle with AC as

diameter) – [Area of a quadrant of

a circle with AB as radius – Area of

ABC]

14142

114

7

22

4

1

277

22

2

1

2

2

cm2

1414

2

1

1414

7

22

4

1249

7

22

2

1

cm2

22 cm98cm 98154154

Ex.108: I n a n e q u i l a t e r a l t r i a n g l e of s i d e 2 4 cm ,

a c i r c l e i s i n s c r i b e d t o u c h i n g i t s s i d e s.

F i n d t h e a r e a o f t h e r ema i n i n g p o r t i o n

o f t h e t r i a n g l e.

(Tak e 7 3 2 .1 3 ).

Soln: Let ABC be an equilateral triangle of side

24 cm, and let AD be perpendicular from

A on BC. Since the triangle is equilateral,

so D bisects BC.

BD = CD = 12 cm

The centr e of the inscr ibed ci rcle wi ll

coincide with the centroid of ABC

Therefore, AD31OD

In ABD, we have

222 BDADAB [Using Pythagoras Theorem]

or, 222 12AD24

or, 22 1224AD

12241224

3121236 cm

OD = AD3

1 cm

= 343123

1

Area of the incircle

= 222 cm347

22OD

22 cm85.150cm487

22

Area of the triangle ABC

= 2224

4

3

4

3 side

= 279.4 cm2

Area of the remaining portion of thetriangle

2cm85.1504.249

2cm55.98

Ex.109: Two c i r c l e s t o u c h e x t e r n a l l y . T h e s um

o f t h e i r a r ea s i s 1 3 0 s q cm a n d t h e

d i s t ance be tween t h e i r cen t r e s i s 14 cm .

F i n d t h e r a d i i o f t h e c i r c l es .

Soln: Note that if two circles touch externally,

the distance between their centres is equal

to the sum of their radii.

Let the radii of the two circles be r 1 cm

and r 2 cm respectively.

Let C1 and C

2 be the centres of the given

circles. Then,2121CC r r

or, 2114 r r (given)cm14CC 21

or, 1421 r r .... (i)

It is given that the sum of the areas of two

circles is equal to 130 cm2.




K KUNDAN

13022

21 r r

or, 13022

21 r r .... (ii)

Now, 212

22

12

21 2 r r r r r r

or, 212 213014 r r [Using (i) and (ii)]

or, 212130196 r r

3321 r r ... (iii)

Now, 212

22

12

21 2 r r r r r r

or, 332130221 r r

[Using (ii) and (iii)]

or, 64221 r r

821 r r ... (iv)

Solving (i) and (iv), we get

r 1 =11 cm and r

2 = 3 cm

Hence, radii of the two circles are 11 cm

and 3 cm.

Ex.110: I n t h e g i v e n f i g u r e , a s em i c i r c l e i s

d r a w n w i t h s egm e n t P R a s d i a m et e r .

Q i s t h e m i d -p o i n t o f s egmen t PR , t w o

sem i c i r c l e s w i t h segmen t PQ and QR as

d i am et e r s a r e d r awn . A ci r c l e i s d r awn

wh i c h t o u c h e s t h e t h r ee sem i c i r c l e s . If PR = 28 cm , f i n d t he a r ea o f t h e shad ed

r e g i o n .

[ = 3.14]

Soln: Let A be the centre of the circle touching

the three semicircles at points D, E and F

respectively.

Let r be the radius of the circle

Then PR6

1

r [From Geometry]

286

1r cm =

3

14cm

Now area of shaded region

= Area of semi-circle PDR – Area of semi-

circle PEQ – Area of QFR – Area of circle

with centre A

= 2

222

3

147

2

17

2

114

2

1

=

2

196

2

49

2

4998

=

9

19649

=9

770

9

3522

9

245

7

22

= 85.56 cm2.

Ex.111: F i n d t h e a r e a o f a r i g h t -a n g l e d t r i a n g l e,

i f t h e r a d i u s o f i t s c i r c u m -c i r l c e i s 5

c m a n d t h e a l t i t u d e d r a w n t o t h e

h y p o t e n u es i s 4 cm .

Soln: We know that the circumcentre of a right-

angled triangle is the mid-point of its

hypotenues and the circum-radius is half

of the hypotenues.

Let ABC be the given triangle with right-

angle at B. Let O be the mid-point of

hypotenues AC. Let BD be the

perpendicular from B on AC. Then,

AC = 2(OA) = 2 × 5 = 10 cm

[OA=radius of the circumcirlce=5 cm]

and, BD = 4 cm (given)

Area of ABC =2

1(Base × Height)

= BDAC2

1

= )410(2

1 cm2

= 20 cm2.




K KUNDAN

1. Calculate the area of a rectangle 23 metres 7

decimetres long and 14 metres 4 decimetres

8 centimetres wide.

2. Find the diagonal of a rectangle whose sides

are 12 metres and 5 metres.

3. How many metres of carpet 75 cm wide will

be required to cover the floor of a room which

is 20 metres long and 12 metres broad?

4. How many paving stones, each measuring2

12

metres by 2 metres, are required to pave a

rectangular courtyard 30 metres long and

2

116 metres broad?

5. 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 which will exactly

fit and the number required.

6. A wire is in the shape of a square of side 10

cm. If the wire is rebent into a rectangle of

length 12 cm, find the breadth. Which

encloses more area, the square or the

rectangle?

7. The area of a square and a rectangle are equal.

If the side of the square is 40 cm and the

breadth of the rectangle is 25 cm, find the

length of the rectangle. Also find the perimeter

of the rectangle.8. A map is drawn to a scale of 120 cm to the

km. How many square cm on the map will

represent a hectare of ground?

9. Find the width of a roller which traverses

128 sq km while cutting 6.4 hectares of grass.

10. The diagonal of a rectangular field is 15 m

and its area is 108 m2. What will be the cost

of fencing this field if the cost of fencing for

one metre is Rs 5.

11. A strip of paper 2.2 km long and .075 mm

thick is rolled up into a solid cylinder.

Assuming the area of a circle to be7

13 times

the square of its radius, find approximately

the radius of the circular ends of the cylinder.

12. A square field containing 31684 square metresis to be enclosed with wire placed at heights

1, 2, 3 and 4 metres above the ground. What

length of the wire will be needed, if the length

required for each circuit is 5% greater than

the perimeter of the field?

13. The area of a rectangular field is 27000 square

metres and the ratio between its length and

breadth is 6 : 5. Find the cost of the wire

required to go four times round the field at Rs

740 per kilometre of length of the wire.

14. A rectangular park is 100 metres long and 80

metres wide. There are two paths, each 5

metres wide, in the middle of the park

running one parallel to the length and the

other parallel to the width of the park. Find,

(i ) the area of the paths,

( ii) the expenditure involved in constructing

the paths at 25 paise per square metre,

and

( iii) the expenditure involved in laying grass

in the remaining portion of the park at5 paise per square metre.

15. A rectangular field 150 metres long and 100

metres wide, has within it a 10 metres wide

uniform path running round it. Find,

(i) the area of the path, and

( ii) the cost of cultivating the remaining part

of the field at Rs 1.50 per square metre.

16. A school hall 20 m long and 15 m broad is

surrounded by a verandah of uniform width

of2

12 m. Find the cost of flooring the

verandah at Rs 2.50 per square metre.

17. A room is 8 m long and 6 m wide. It is

surrounded by a verandah. Find the width of

the verandah if it occupies 72 sq metres.18. A path 2 m wide, running all around outside

a square garden occupies 204 sq metres. Find,

(i) the length of the square garden.

( ii) the area of the part of the garden enclosed

by the path.

19. A square carpet is spread in the centre of a

hall 9 m square leaving some margin of equal

width all around. The total cost of carpeting

at Rs 2.50 per sq m and decorating the margin

at 20 paise per sq m is Rs 163.40. Find the

width of the margin.

20. A rectangular field is 200 metres long and

121 metres broad. It is planted with trees in

rows perpendicular to the length, one metre

from row-to-row, and one metre from tree-to-

tree in the same row. If the width of a metreall-round the field remains unplanted, find

the number of trees.

21. A path 2 metres wide running all-round a

square garden has an area of 9680 sq metres.

Find the area of the part of the garden enclosed

by the path.

22. A marginal walk all-round the inside of a

Practice Exercise

Exercise–1

(Rectangle, Square and Area of Path)




K KUNDAN

rectangular space 37 m by 30 m occupies 570

sq metres. Find the width of the walk.

23. A garden, whose length is 22 metres, has a

path 1.5 metres wide on the two sides and at

one end. If it costs Rs 2460 to turf theremainder at Rs 15 a sq metre, what is the

width of the garden?

24. In the centre of a room 10 metres square, there

is a square of turkey carpet, and the rest of

the floor is covered with oilcloth. The carpet

and the oilcloth cost respectively Rs 150 and

Rs 65 per square metre, and the total cost of the carpet and the oilcloth is Rs 13385. Find

the width of the oilcloth border.

Exercise–2(Triangle)

7. Find the area of an isosceles right triangle,

the length of whose each side containing the

right angle is 15 cm.

8. The sides of an equilateral triangle is 8 cm.

Find its area and the height.

9. From a point in the interior of an equilateral

triangle, perpendiculars drawn to the three

sides, are 16 cm, 20 cm and 22 cm

respectively. Find the area of the triangle.10. A triangular park ABC has sides 120 m, 80 m

and 50 m (see figure). A gardener Dhania has

to put a fence all round it and also plant grass

inside. How much area does he need to plant?

Find the cost of fencing it with barbed wire

at the rate of Rs 20 per metre leaving a space

3 m wide for a gate on one side.

1. The base of triangular field is 880 metres and

its height 550 metres. Find the area of the

field. Also calculate the charges for supplying

water to the field at the rate of Rs 242.50 per

sq hectometre.

2. The base of a triangular field is three times

its height. If the cost of cultivating the field

at Rs 1505.52 per hectare is Rs 20324.52,

find its base and height.3. Find the area of 1 triangular field whose sides

are 50 metres, 78 metres, 112 metres

respectively and also find the perpendicular

from the opposite angle on the side 112

metres. If it is lent at Rs 10000 per hectare,

find the rent of the field.

4. X is a point on side CD of a square ABCD

such that CX = 5 cm. If area of the triangle

ADX is 43 cm2, find the length of the side of

the square.

5. Find the area of a triangle, one of whose angles

is 90°, hypotenuse is 12.5 cm and the base is

7.5 cm.

6. The area of a tri angle equals the area of a

square whose side is 45 m. Find the length

of the side of the triangle which is 75 m from

the opposite vertex.

3. Find the area of a quadrilateral piece of ground

ABCD in which AB = 85 metres, BC = 143

metres, CD = 165 metres, DA = 85 metres

and DB = 154 metres.

Exercise–3(Quadrilateral, Parallelogram, Rhombus, Trapezium and Regular Polygon)

1. Find the area of the quadrilateral ABCD in

which the diagonal DB = 10 m and the

perpendiculars AL and CM drawn on it from

A and C are respectively 4 m and 6 m. (These

perpendiculars are called offsets).

2. The sides of a parallelogram are 40 m and 30

m respectively and its diagonal is 50 m. Find

its area.




K KUNDAN

Exercise–4

(Circle)

10. Find the ratio of area of a square inscribed in

a semi-circle of radius r to the area of another

square inscribed in the entire circle of radius

r .

11. Find to the three places of decimals the radius

of the circle whose area is the sum of the

areas of two triangles whose sides are 35,

53, 66 and 33, 56, 65 measured in

centimetres.

(Take = 22/7).

12. PQRS is a diameter of a circle of radius 6 cm.

The length PQ, QR and RS are equal. Semi-

circles are drawn on PQ and QS as diameters

as shown in the figure below. Find the area

of the shaded region.

13. A square water tank has its side equal to 40

m. There are four semi-circular grassy plots

all round it. Find the cost of turfing the plots

at Rs 1.25 per sq cm.

(Use = 3.14)

14. A rectangular park is 100 m by 50 m. It is

surrounded by semi-circular flower beds all

around. Find the cost of levelling the semi-

circular flower beds at 60 per sq metre.

(Use = 3.14)

15. A park is in the form of a rectangle 120 × 100

m. At the centre of the park there is a circular

lawn. The area of the park excluding the lawn

is 8700 m2. Find the radius of the circular

lawn.

(Use = 22/7)

1. The radius of a circular wheel is4

31 m. How

many revolutions will it make in travelling

11 km?

2. The circumference of a circular garden is 1012

metres. Find the area. Outside the garden a

road of 3.5 metres width runs round it.

Calculate the area of this road and find the

cost of gravelling the road at Rs 32 per 100 sq

metres.

3. A bicycle wheel makes 5000 revolutions in

moving 11 km. Find the diameter of the wheel.

4. A boy is cycling such that the wheels of the

cycle are making 140 revolutions per minute.

If the diameter of the wheel is 60 cm, calculate

the speed per hour with which the boy is

cycling.

5. The diameter of the wheel of a bus is 140 cm.

How many revolutions per minute must the

wheel make in order to keep a speed of 66

km per hour?

6. A copper wire, when bent in the form of a

square, encloses an area of 484 cm2. If the

same wire is bent in the form of a circle, find

the area enclosed by it.

(Use = 22/7)

7. 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.

8. A bucket is raised from a well by means of a

rope which revolves round a wheel of diametre

75 cm. If the bucket ascends in 1 minute 12

seconds with a uniform speed of 1.3 m per

second, find the number of complete

revolutions made by the wheel in raising the

bucket.

9. The radius of circle is 20 cm. Three more

concentric circles are drawn inside it in such

a manner that it is divided into 4 equal parts.

Find the radius of the smallest circles?

4. The perimeter of a rhombus is 146 cm and

one of its diagonals is 55 cm. Find the other

diagonal and the area of the rhombus.

5. The parallel sides of a field, which is in the

shape of a trapezium, are 20 m and 41 m andthe remaining two sides are 10 m and 17 m.

Find the cost of levelling the field at the rate

of Rs 30 per sq metre.

6. Ratio between the parallel sides of the

trapezium is 1:3, while ratio between

unparallel sides of the trapezium is 2:3. Ratio

between bigger parallel and unparallel sides

is 2:1. If height of the trapezium is4

1515,

then find the area of the trapezium?

7. A regular hexagon of side 6 cm is inscribedin a circle. Find the area of the region in the

circle which is outside the hexagon.

[Use = 3.14, 732.13 ]

8. The diagonals of a rhombus are 8 cm and 6

cm, find the sides and the area.




K KUNDAN

in the figure. Find the area of the shaded

region.

[ = 22/7, 3 = 1.73]

21. In a circle of radius 21 cm an arc subtends

an angle of 60º at the centre. Find

(i) the length of the arc,

(ii) the area of the sector bounded by the arc

and

( iii) the area of the segment made by this arc.

22. A circular swimming pool is surrounded by a

circular path which is 4 m wide. If the area

of the path is25

11th part of the area of the

swimming pool, then find the radius of the

swimming pool (in metres).

23. If the circumference of a circle is 80 cm, then

find the side of a square inscribed in the circle.

24. In the figure given below, square OABC is

inscribed in a sector OPBQ. If OC = 20 cm,

find the area of the shaded region.

(use = 3.14)

16. In the figure given below, ABCD is a rectangle.

The radius of the semic ircles drawn on AD

and BC as diameters and radius of circle drawn

in between is the same. If BC = 7 cm, find

the area of the shaded region.

17. An athletic track 14 m wide consists of two

straight sections 120 m long joining semi-

circular ends whose inner radius is 35 m.

Calculate the area of the shaded region.

18. In the given figure, the centre of the circle is

A and ABCDEF is a regular hexagon of side 6

cm. Find the following:

(i) Area of segment BPF

(ii) Area of the shaded portion.

19. Quadrilateral ABCD is a rectangle. Sectors

with centre C and D are drawn as shown in

the figure. If AB = 21 cm, CB = 14 cm, find

the area of the shaded portion.

20. An equilateral ΔABC has each of its sides

14 cm with each of its vertices as centres,

and radius as 7 cm, arcs are drawn as shown




K KUNDAN

25. In the given figure, cresent is formed by two

circles which touch at the point A. O is the

centre of the bigger circle. If CB = 9 cm and

DE = 5 cm, find the area of the shaded

portion.

26. Two circles touch internally. The sum of their

areas is 116 sq cm and the distance between

their centres is 6 cm. Find the radii of the

circle.

27. Find the area of the shaded portion of the

given diagram. Give your answer correct to

three significant figures.

28. Find the area of the shaded portion in the

given figure, where the arcs are quadrants of

a circle.

29. Find the area of the shaded portion in the

figure given below:

[use = 22/7]

30. In a right ABC, A = 90°, AB = 4 cm, AC = 3

cm. On its three sides as diameters, three

semi-circles are drawn as shown in the figure

given below. Find area of shaded parts.

31. The area of the shaded circular ring is 770

sq cm and the difference between the radii of

the two circles is 7 cm. Find the area of the

unshaded region.

32. In the figure given below, ABC is an

equilateral triangle of side 7 cm and segment

BC is diameter of the semicircle. Find the

area of the shaded region.

( 3 = 1.732)

33. In the given diagram AC is a diameter of a

circle with radius 5 cm. If AB = BC and CD =

8 cm, calculate area of the shaded region.

34. In the figure as mentioned below, POQ and

ROS are diameters of a circle with centre O




K KUNDAN

42. Find the area of the shaded region in the

given figure, where ABCD is a square of side

14 cm.

43. Find the area of the shaded region in the

given figure, where ABCD is a square of side

10 cm and semi-circle are drawn with each

side of the square as diameter.

(use = 3.14)

44. The circumference of a circle is 16.8 cm more

than its diameter. What will be the radius of

the circle?

(use = 3.14)

45. A square has been inscribed in a circle. What

will be the ratio of the areas of circle and the

square?

46. The length of sides AB, BC and CA of a ABC

are 4 cm, 6 cm and 8 cm respectively. Three

sectors of circles drawn with centres A, B, Cand each with one centimetre of radius,

starting and terminating with the sides of

the triangle are cut off. Find the area of the

remaining position of the triangle.

47. In a circle of radius 28 cm, an arc subtends

an angle of 72° at the centre. Find the length

of an arc and the area of the sector so formed.

48. The radii of three concentric circles are in

the ratio 1 : 2 : 3. Find the ratio of the area

between the two inner circles to that between

the two outer circles.

and radius 14 cm. Find the area of shaded

region.

(use = 22/7).

35. A rectangular field is surrounded by four

semicircular flower-beds. If the length and

the breadth of the field are 6 m and 4 m

respectively, find the cost of raising the

flower-beds at the rate of Rs 8 per m2.

(Take = 3.14)

36. The length of the side of a square is 14 cm.

Taking ver tices of the square as centres, 4circles are drawn each with a radius of 7 cm.

Find the area of the region of the square that

remains outside the region of any of the

circles.

37. A brick, 5 cm thick (high), is placed against a

wheel to act for a stop. The horizontal distance

of the face of the brick stopping the wheel

from the point where the wheel touches the

ground is 15 cm. What is the radius of the

wheel?

38. The length of the minute-hand of a clock is

10 cm. What is the area swept by the minute-

hand in one minute?

(Use = 3.14)

39. A chord AB of a circle of radius 10 cm makes

a right angle at the centre of the circle. Findthe area of the major and minor segments.

(Take = 3.14).

40. A chord AB of a circle of radius 15 cm makes

an angle of 60° at the centre of the circle.

Find the area of the major and minor segment.

(Take = 3.14, 3 = 1.732).

41. In the given figure, two circular flower beds

have been shown in two sides of a square

lawn ABCD of side 56 m. If the centre of each

circular flower bed is the point of intersection

O of the diagonals of the square lawn, find

the sum of the areas of the lawn and the

flower beds.




K KUNDAN

1. Length = 23.70 metres

Breadth = 14.48 metres

Area = (23.70 × 14.48) square metres

= 343.176 square metres

2. Length of the diagonal

= 22 512 metres = 169 metres

= 13 metres

3. Area of carpet = Area of the floor

= (20 × 12) sq metres

Width of carpet = 75 cm =4

3metre

Length of carpet =

4

3

1220 = 320 metres

4. Area of courtyard =2

11630 sq m

= 495 sq metres

Area of each paving stone =

2

2

12 sq m

Number of stones required =5

495 = 99

5. 39 m 10 cm = 3910 cm

35 m 70 cm = 3570 cm

The side of the largest square tile

= HCF of 3910 and 3570= 170 cm

= 1 m 70 cm

Number of tiles =170170

35703910

= 483

6. Side of the square = 10 cm

Length of the wire = Perimeter of the square

= 4 × side = (4 × 10 =) 40 cm

Length of the rectangle = l = 12 cm. Let b be

the breadth of the rectangle.

Perimeter of the rectangle = Length of wire

= 40 cm

Perimeter of rectangle = 2 (12 + b )

Thus, 40 = 2 (12 + b )

or, 20 = 12 + b

b = (20 – 12 =) 8 cm The breadth of the rectangle = 8 cm.

Area of the square = (Side)2

= 10 cm × 10 cm = 100 cm2

Area of the rectangle = l × b = 12 cm × 8 cm

= 96 cm2

So, the square encloses more area even though

its perimeter is the same as that of the rectangle.

7. Area of the square = (Side)2

= 40 cm × 40 cm = 1600 cm2

It is given that, the area of the rectangle

= The area of the square

Area of the rectangle = 1600 cm2

Breadth of the rectangle = 25 cm

Area of the the rectangle = l × b

or, 1600 = l × 25

l =25

1600 = 64 cm

Hence, the length of rectangle is 64 cm

Perimeter of the rectangle = 2 (l + b )

= 2 (64 + 25) cm

= 178 cm

Hence, the perimeter of the rectangle is 178

cm.

8. 1 hectare = 10000 sq m

1000 metres are represented by 120 cm.

(1000 × 1000) sq m are represented by

(120 × 120) sq m.

1 sq m is represented by10001000

120120

sq cm

10000 sq m are represented by

10001000

10000120120sq cm = 144 sq cm

9. 128 km = 128000 m

6.4 hectare = (6.4 × 10000) sq m

Imagining the grass area to be 128000 m long,

and as wide as the roller, we have

Width required =128000

100004.6 m =

2

1m = 50 cm

10. Let the length of the rectangle be x metres

and breadth be y metres.

Area of rectangular field = x × y = 108 m2 ... (i)

Area of rectangle = length × breadth

And 15

2

= x

2

+ y

2

or, 225 = x 2 + y 2 .... (ii)

In a right-angled triangle

Hypotenuse2 = Base2 + Height2

From equations (i) and (ii)

(x + y )2 = x 2 + y 2 + 2xy

= 225 + 2 × 108 = 441 = (21)2

or, x + y = 21

Answers and explanations

Exercise–1




K KUNDAN

Now, the perimeter of the field

= 2(length + breadth)

Perimeter of field = 2(x + y)

= 2 × 21 = 42 metres

Cost of fencing the field = (42 × 5 =) Rs 21011. By rolling up a rectangular piece of paper it

will be seen that the area of the longer edge

of the paper rolled into solid cylinder is equal

to that of the circular end of the cylinder.

2.2 km = 2.2 × 1000 × 100 cm

.075 mm = .0075 cm

(radius)2 ×7

13

= 2.2 × 1000 × 100 × 0.0075 sq cm

(radius)2 =

22

7

1000010

75100100022 sq cm

radius = 22.91 cm (Approx.)

12. Area of the field = 31684 sq m

perimeter = 431684 metres

= 178 × 4 metres

length of each circuit = 178 × 4 ×100

105 m

Since the wire goes round 4 times.

total length of wire required

= 178 × 4 ×100

105× 4 m = 2990.4 m

13. Let the length be 6x m and the breadth be 5x m.

area = )56( x x sq m = 27000 sq m

or, 30x 2 = 27000

or, x 2 = 900

x = 30

Hence length = 180 m and breadth = 150 mLength of wire required to go round the field

four times = [4× 2(180 + 150)] m = 2.64 km

required cost = Rs (2.64 × 740)

= Rs 1953.60

14. (i ) The area of the path HG

= (100 × 5 =) 500 sq m

The area of the path EF

= (80 × 5 =) 400 sq m

Area of the shaded portion

= (5 × 5 =) 25 sq m

The shaded port ion is common to both

the roads, so while finding the actual area

of the paths, we should subtract this

common area from the sum of the areas of

the two paths.

Area of the paths

= (500 + 400 – 25 =) 875 sq units.

( ii) The expenditure involved in constructingthe paths

= Rs 875100

25 = Rs

4

875 = Rs 218.75

(iii) Area of the park = 100 × 80 = 8000 sq m

Area of the remaining portion of the park

= (8000 – 875 =) 7125 sq m

The expenditure of laying grass in the

remaining portion of the park

= Rs 7125100

5 = Rs 356.25

15. (i ) Area of the field = 150 × 100 = 15000 sq m

Length of the inner rectangle

= {150 – (10 + 10) =} 130 m

Width of the inner rectangle

= {100 – (10 + 10) =} 80 m

Area of the inner rectangle

= (130 × 80 =) 10400 sq m

Area of the path

= (15000 – 10400 =) 4600 sq m( ii) The area of the remaining part of the field

to be cultivated = 10400 sq m

Cost of cultivating the remaining part of

the field = Rs 104002

3 = Rs 15600

16. The area of the school hall ABCD

= (15 × 20 =) 300 sq m

The length of the rectangular region PQRS

=

2

12

2

1220 25 m




K KUNDAN

The width of the rectangular region PQRS

=

2

12

2

1215 m = 20 m

Area of the rectangular region PQRS

= (25 × 20 =) 500 sq m Area of the verandah

= (500 – 300) sq m = 200 sq m

Cost of flooring the verandah

= Rs 200 ×2

5 = Rs 500

17. Let the width of the verandah be x metres.

Area of the room ABCD = (8 × 6 =) 48 sq m

Length of PQRS = (8 + x + x ) m = (8 + 2x ) m

Width of PQRS = (6 + x + x ) m = (6 + 2x ) m

Area of PQRS = (8 + 2x ) × (6 + 2x ) sq m

= (48 + 28x + 4x 2) sq m

Area of the room + area of the verandah

= (48 + 28x + 4x 2) sq m

2428487248 x x

or, 72284 2 x x

or, 1872 x x

or, 01872 x x

or, 018292 x x x

or, 0)9(2)9( x x x

or, 0)2)(9( x x

Either x = –9 or x = 2

x cannot be –9 because width of the verandah

cannot be negative

x = 2

Width of the verandah = 2 m

18. (i ) Let the length of the square garden ABCD

be x metres.

Area of the square garden ABCD

= (x × x =) x 2 sq m

Length of square PQRS

= )4(m)22( x x m

Area of square PQRS = (x + 4)2 sq m

(x + 4)2 = x 2 + 204

or, x 2 + 8x + 16 = x 2 + 204

or, 8x = 204 – 16

or,2

123

2

47

8

188 x m

The length of the square garden ABCD

=2

123 m

( ii) Area of the part of the garden enclosed by

the path, ie, of ABCD

=

4

2209

2

47

2

47 = 552.25 sq m

19. Let the width of the margin be x m.

Area of the square hall ABCD = (9 × 9 =) 81 sq m

Area of the square hall PQRS = (9 – 2x )2 sq m

= 81 – 36x + 4x 2 sq m

Area of the margin = 81 – (81 – 36x + 4x 2) sq m

= 81 – 81 + 36x – 4x 2 sq m

= 36x – 4x 2 sq m

Cost of the carpet = Rs 2

543681 2 x x

= Rs2

20180405 2x x

Cost of decorating the margin

= Rs 5

1436 2 x x = Rs

5

436 2x x

It is given that total cost of carpeting at Rs

2.50 per sq m and decorating the margin at

20 paise per sq m is Rs 163.40.

4.1635

436

2

20180405 22

x x x x

or, 10

1634

10

8721009002025 22

x x x x

or, 16348721009002025 22 x x x x

or, 039182892 2 x x

or, 03917824692 2 x x x

or, 46x (2x – 1) – 391(2x – 1) = 0

or, (46x – 391) (2x – 1) = 0






K KUNDAN

Solving equations (i) and (ii), we have

x = 81

The area of the square carpet is 81 sq

metres.

Therefore, the carpet is 9 metres in lengthand breadth. But the room is 10 metres in

length and breadth.

Hence double the width of the border is

(10 – 9 =) 1 metre

the width of the border =2

1metre = 5 dm

Alternative Method:

The area of the square room = 100 sq metres

The mean cost per sq metre = Rs

100

13385

= Rs 13.385

= 81 : 19

By the Alligation Rule, the area of the square

carpet is 81 sq metres. Therefore, the carpet

is 9 metres in length and breadth.

But room is 10 metres in length and breadth.

Hence double the width of the border is

(10 – 9 =) 1 metre

the width of the border =2

1metre = 5 dm

Exercise–2

(s – a ) = (120 – 50 =) 70 metres

(s – b ) = (120 – 78 =) 42 metres

(s – c ) = (120 – 112 =) 8 metres

Area = 84270120 = 1680 sq m

Perpendicular =Base

Area2 =

112

21680 metres

= 30 metres

Rent per hectare = Rs 10000

required rent = Rs

10000

168010000

= Rs 1680

4. Let ABCD be the given square and X is pointon side CD.

CX = 5 cm

Let the length of the side of square be x cm.

XD = (x – 5) cm

Now, ADX is a right-angled triangle

2

1 × (DX) × (AD) = 42

or, )()5(2

1x x = 42

or, x 2 – 5x – 84 = 0

or, x 2 – 12x + 7x – 84 = 0

1. Area of the field =2

HeightBase

=

2

550880 sq metres

=

100100

550440 sq hectometres

= 24.20 sq hectometres

Cost of supplying water to 1 sq hectometre

= Rs 242.50

Cost of supplying water to the whole field

= Rs 24.20 × 242.5 = Rs 5868.5

2. Area of the field =2

2752.150552.20324

hectares

Also, area of the field =2

1 × Base × Height

=2

1 × 3 × Height × Height =

2

3(Height)2

2

3(Height)2 =

2

27hectares

(Height)2 =

3

2

2

27 9 hectares

= 90000 sq metres

Height = 90000 m = 300 m

Also, Base = 3 × Height = 900 m3. Here, a = 50 metres, b = 78 metres, c = 112

metres

s = )1127850(2

1 metres

=

240

2

1metres = 120 metres




K KUNDAN

or, x (x – 12) + 7(x – 12) = 0

or, (x – 12) (x + 7) = 0

or, x = 12 because x –7

The side of square = 12 cm

5.

Let ABC be a right-angled triangle,

whose ABC = 90°

Hypotenuse, AC = 12.5 cm

Base, BC = 7.5 cm

Perpendicular (AB) = 22 BCAC

= 22 )5.7()5.12(

= )5.75.12()5.75.12(

= 520 = 100 = 10 cm

required area =2

1 × Base × Height

= 105.72

1 = 375 sq cm

6. Area of the square = (45 × 45 =) 2025 sq m

Area of the triangle = 2025 sq m

Height of the triangle = 75 m

Required side of the triangle

=75

22025 = 54 m

7. Area of an isosceles right triangle

=2

1 × (length of one of its two equal sides)2

= 2152

1 sq cm =

2

225= 112.5 sq cm

8. Area of an equilateral triangle = 2side4

3

= 884

3 sq cm

= 1.732 × 16 = 27.712 sq cm

Height of an equilateral triangle = side2

3

= 82

3 = 1.732 × 4 = 6.928 cm

9. Let each side of ABC be x cm.

Area of BOC =2

122x = 11x sq cm

Area of AOC =2

116x = 8x sq cm

Area of AOB = x x 102

120 sq cm

Area of ABC

= area of BOC + area of AOC + area of AOB

= )10811( x x x sq cm = 29x sq cm

But area of equilateral ABC=2

4

3x sq cm

x x 294

3 2 or, 294

3x

or, 97.663

116

3

429

x cm

Area of ABC =2

4

3x

= 297.664

732.1 = 1942.0 sq cm

10. For finding area of the park, we have2s = 50 m + 80 m + 120 m = 250 m

ie s =

2

250125 m

Now, (s – a ) = (125 – 120 =) 5 m

(s – b ) = (125 – 80 =) 45 m

(s – c ) = (125 – 50 =) 75 m

Therefore, area of the park

= )()()( c s b s a s s

= 75455125 m2

= 15375 m2

Also perimeter of the park = AB + BC + CA

= 250 m

Ther efor e, length of th e wi re needed for

fencing

= 250 m – 3 m (to be left for gate)

= 247 m

And so the cost of fencing = Rs (20 × 247)

= Rs 4940




K KUNDAN

1. Here the area of the quadrilateral ABCD = the

area of the triangle ABD + the area of the

triangle BCD =2

1 × BD × AL +

2

1 BD × CM

=2

1 × BD (AL + CM)

ie, The area of a quadrilateral

=2

1 × diagonal × sum of offsets.

Ther ef ore, the requi re d ar ea of the

quadrilateral

= 64102

1 sq m

=

1010

2

1 50 sq m

2. Area of parallelogram ABCD

= area of ABC + area of ACD

= 2 area of ABC

( Each diagonal of a parallelogram bisects it)

Semi-perimeter (s) of ABC

=

2

304050 60 m

Area of ABC = )()()( c s b s a s s

= )5060()3060()4060(60

= 10302060

= 360000 = 600 sq m

Area of parallelogram ABCD = 2 × 600 sq m

= 1200 sq m

3. Area of quadrilateral ABCD

= Area of triangle ADB + Area of triangle DBC.

In triangle ADB,

2

1 × perimeter =

21548585 162 m

Area of the triangle ADB

= )154162()85162()85162(162 sq m

= 87777162

= 2772 sq m

In triangle DBC,

2

1× perimeter =

2

154143165231 m

Area of the triangle DBC

= )154231()143231()165231(231 sq m

= 778866231 = 10164 sq m Area of the quadrilateral ABCD

= (2772 + 10164) sq m

= 12936 sq m

4.

Let ABCD be the rhombus in which AC

= 55 cm Perimeter of the rhombus = 146 cm

AB =4

146= 36.5 cm

and AO =2

55= 27.5 cm

Exercise–3




K KUNDAN

BO = 22 )5.27()5.36( cm = 24 cm

Hence the other diagonal BD = 48 cm

Area of the rhombus = 2

1

× AC × BC

=

4855

2

1

= 1320 sq cm

5.

Let ABCD be the trapezium such that AB||CD,AB = 41 m, DC = 20 m, AD = 10 m and

BC = 17 m

Draw CE||DA and CF AB.

Clearly, AECD is a parallelogram

Now, EB = AB – AE = AB – DC [AE = DC]

= (41 – 20 =) 21 m

Also, EC = AD = 10m

Thus, in ECB, we have

EB = 21 m, EC = 10 m and BC = 17 m

Let s be the semi-perimeter of the ECB. Then

s =

2

171021 24 m

Area of ECB = )()()( c s b s a s s

= )1724()1024()2124(24 sq m

= 714324 = 84 sq m ....(i)

Also, area of ECB =2

1× Base × Height

=2

1× 21 × (CF) ....(ii )

From equations (i) and (ii), we get

84CF2

21

CF =21

284 = 8 m

Area of parallelogram AECD = Base × Height

= AE × CF = (20 × 8 =) 160 sq m

Now, area of trapezium ABCD

= (Area of parallelogram AECD) + (Area of ECB)

= (160 + 84) sq m = 244 sq m

Cost of levelling the field at the rate of Rs

30 per sq metre = Rs (30 × 244) = Rs 7320

6. Let x be present in the parallel sides of the

trapezium.

parallel sides are x and 3x .

Let k be present in the non-parallel sides of

the trapezium. Non-parallel sides are 2k and 3k .

According to the question,

1

2

3

3

k

x

or, k x 2

Combining triangle AEC and BFD, we get a

triangle of base 6k – 2k = 4k and two other

sides are 2k and 3k.

Semi-perimetre of combined triangle

= x =2

324 k k k =

2

9k

Area of the triangle

= )4()3()2( k s k s k s s

=

k

k k

k k

k k 4

2

93

2

92

2

9

2

9

=

22

3

2

5

2

9 k k k k

= 3594

2

k

= 154

3 2k

Also area of the triangle

= 151524

15154

2

1 k k

or, 154

3 2k = 1515

2

k

k = 10 units

Parallel sides are 20 units and 60 units.

Area of trapezium = 4

15156020

2

1 units2

= 2units1575 .

7. Each side of the hexagon inscribed in the

circle is 6 cm, the radius of the circle is 6 m.

Area of the circle =2r

= 3.14 × (6)2 = 3.14 × 36 = 113.04 cm2




K KUNDAN

1. Distance to be travelled = 11 km = 11000 m

Radius of the wheel =4

31 m

Circumference of the wheel

=

4

31

7

222 m = 11 m

In travelling 11 m the wheel makes 1

revolution.

In travelling 11000 m the wheel makes

11

110001000 revolutions.

2. Area = r 2, circumference = 2r

2r = 1012 metres.

r = 1012 ×22

7

2

1 m = 161 m

area of the garden = r 2

= 1611617

22 sq m

= 81466 sq m

Area of the road

= area of bigger circle – area of the garden.

Radius of the bigger circle

=

2

13161 m =

2

329 m

area of bigger circle =2

329

2

329

7

22 sq m

= 2

185046 sq m

area of road = 814662

185046

=2

13580 sq m

required cost = Rs100

32

2

7161 = Rs 1145.76

3. Distance covered by the wheel in one

revolution

=srevolutionof Number

movedDistance

=

1001000

500011km

500011

cm

= 220 cm

Circumference of the wheel = 220 cm

Let the radius of the wheel be r cm. Then,

Circumference = 220 cm

or, 2r = 220 cm

Exercise–4

Area of the hexagon =2

33(side)2

=22 cm354)6(

2

33

= 54 × 1.732 = 93.53 cm2

Hence the area of the region of the circle

which is outside the hexagon

= 113.04 cm2 – 93.53 cm2 = 19.51 cm2.

8. We know that the diagonals of a rhombus

bisect one another at right angles. Therefore

from the given figure the area of the rhombus

ABCD = area of the triangle ABC + area of the

triangle ADC = 2 × area of the triangle ABC.

=2

12 × AC × OB = AC × OB

= 2

1 × AC × BD

BD2

1

OB

ie, the area of a rhombus

=2

1× product of its two diagonals.

Here, AC = 8 cm and BD = 6 cm

AO = 4 cm and BO = 3 cm

AB = 22 34 = 5 cm

Each side of the rhombus is 5 cm.

The required area of the rhombus

= 682

1 sq cm = 24 sq cm




K KUNDAN

or, 887

222 r

or, r = 14 cm

Area of the circle

= 2222 cm616cm147

22

r

7. Length of the wire = circumference of the

circle

=

28

7

222 cm [Using C = r 2 ]

= 176 cm ... (i)

Let the side of the square be x cm. Then,

perimeter of the square = length of the wire

or, 4x = 176 [Using (i)]

or, x = 44 cm

Hence, the length of the side of the square is

44 cm.

8. Radius of the wheel =2

75cm

Now, perimeter of the wheel = r 2

=

7

7522

2

75

7

222 cm

Total time taken to pull up bucket is 1 minute

12 seconds = (60 + 12 =) 72 seconds

Distance travelled in one second = 1.3 m

Distance travelled in 72 seconds

= (1.3 × 72 =) 93.6 m or 9360 cm

Let number of revolutions be N.

According to the question,

Distance covered by bucket = perimeter of

wheel × number of revolutions made by the

wheelN × (perimeter of the wheel) = 9360 cm

or, N × 93607

7522

or, N = 70.397522

79360

Number of complete revolutions = 39.

9. Area of circle = r 2, where r is the radius.

Area of the biggest circle = 22 cm40020

Area of the smallest circle =2cm400

4

1

= 100 cm2

or, 2207

222 r

or, r = 35 cm

diameter = 2r cm = (2 × 35) cm = 70 cm

Hence, the diameter of the wheel is 70 cm.

4. Radius of the wheel = r =2

60cm = 30 cm


=

30

7

2222 r cm =

7

1320 cm

Distance covered in one revolution

= circumference =7

1320 cm

Distance covered in 140 revolutions

=

140

7

1320cm = (1320 × 20) cm

= 26400 cm =

100

26400

m

= 264 m =

1000

264km

It is given that the wheels are making 140

revolutions per minute. So, distance covered

in one minute = Distance covered in 140

revolutions

=

1000

264km

Distance covered in one hour

=

60

1000

264km = 15.84 km

Hence, the speed with which the boy is

cycling= 15.84 km/hr

5. Distance covered by the wheel in one minute

=

60

100100066 = 110000 cm


=

70

7

222 = 440 cm

Number of revolutions in one minute

=

440

110000 = 250

6. Area of the square = 484 cm2

side of the square = 484 cm = 22 cm

]Areaside)side(Area[ 2

So, Perimeter of the square

= 4(side) = (4 × 22) cm = 88 cm


circumference of the circle

= Perimeter of the square

2r = 88




K KUNDAN

Now, if the radius of the smallest circle be r .

Now, according to the question,

1002r

or, r 2 = 100

or, r = 10

radius of the smallest circle = 10 cm

10. Each side of the square inscribed in semicircle

= BC = OA = a

In right angle triangle OAB

OA2 + AB2 = OB2

or,2

22

4

r a

a

or,22

4

5r a

5

4 22 r

a

Area of the square inscribed in semicircle

=5

4 22 r

a

Diagonal of the square inscribed in a circle

= 2r

Area of this square =22 2)2(

2

1r r

Required ratio = 1:5

2

2:5

4 22

r

r

or 2 : 5

11. For the first triangle, we have

a = 35, b = 53 and c = 66

s =

2

665335

2

c b a 77 cm

1

= Area of the first triangle

= )()()( c s b s a s s

= )6677()5377()3577(77

= 11244277

= 114667117

= 2222 26117

= 7 × 11 × 6 × 2 = 924 cm2 ....(i)

For the second triangle, we have

a = 33, b = 56, c = 65

s =

2

655633

2

c b a 77 cm

2

= Area of the second triangle

= )()()( c s b s a s s

= )6577()5677()3377(77

= 12214477

= 4373114117

= 2222 34117

= 7 × 11 × 4 × 3 = 924 cm2 ....(ii)


Area of the circle = Sum of the areas of two

triangles

or, 212 r

or, 9249242 r

or, 18487

22 2 r

or, 58878422718482 r

or, 249.24588 r cm

12. For semi-circle with diameter PS,

Radius = 6 cm

Area of such semi-circle = 2

2

1r

=7

396)6(

7

22

2

1 2 cm2

Diameter PS = 6 + 6 = 12 cm

PQ = QR = RS =

3

124 cm

QS = QR + RS = 8 cm

For semi-circle with diameter QS, radius= 4 m

Its area =7

176)4(

7

22

2

1

2

1 22 r cm2

For semi-circle with diameter PQ, radius

= 2 cm

Its area = 7

442

7

22

2

1 2 cm2

Area of shaded region

=7

264

7

44

7

176

7

396 = 37.71 cm2.

13. Diameter of each of the semi-circle = 40 m.

Radius, r =2

40




K KUNDAN

Area of each semi-circle =2

2

1r

Area of semi-circle grassy plots

=22 2

2

14 r r

= 2 × 3.14 × 20 × 20 = 2512 sq m

Cost of turfing the plots at the rate of Rs

1.25 per sq m = Rs (1.25 × 2512 =) Rs 3140.

14. Radius of flower beds I and II =

2

10050 m.

Area of each of flower beds I and II = 221 r

= 505014.32

1 sq m = 3925 sq m

Total area of flower beds I and II

= 2 × 3925 = 7850 sq m

Radius of each of flower beds III and IV

=

2

5025 m

Area of each of flower beds III and IV =2

2

1r

= 252514.3

2

1 sq m = 981.25 sq m

Total area of flower beds III and IV

= 2 × 981.25 sq m = 1962.62 sq m

Total area of 4 semi-circular beds

= (7850 + 1962.50) sq m = 9812.50 sq m

Cost of levelling the flower beds at the rate

of 60 p per sq m

= Rs 50.9812100

60 = Rs 5887.50

15. Area of rectangular park

= 120 m × 100 m = 12000 m2

Area of the park excluding the circular lawn

= 8700 m2

Area of circular lawn

= (12000 – 8700) m2 = 3300 m2

Let r be the radius of circular lawn

33002 r or 33007

22 2 r

or, r 2 =22

73300 = 150 × 7 = 1050

r = 2/1)1050(1050

Radius of the circular lawn = 32.40 m.

16. Area of rectangle ABCD

= Length × Breadth = AB × BC

= (3.5 + 7 + 3.5) × 7 = 98 cm 2

Area of unshaded region

=2222 2

2

1

2

1r r r r

= 5.35.37

44)5.3(

7

222 2 = 77 cm2

Area of the shaded region

= Area of rectangle ABCD – Area of unshaded

region

= 98 cm2 – 77 cm2 = 21 cm2.

17. Area of the shaded region

= 2(Area of rectangle with sides 120m and 14m)

+ 2[Area of the semi-circle with radius (35+14)

cm

ie 49 cm - Area of the semi-circle with radius

35 m]

= 2 (120 × 14) +

)3549(

7

22

2

12 22

= )3549()3549(7

22)1680(2

=

1484

7223360 m2

= (3360 + 22 × 84 × 2) m2

= (3360 + 3696) m2 = 7056 m2

18. (i ) Area of segment BPF

= Area of sector ABPF – Area of ABF

Draw AK BF

BAF is an angle of regular hexagon

BAF =120° . .. .(1)

The perpendicular from the centre of a

circle to a chord disects the chord

BK = KF ....(2)

AB = AF (Side of a regular hexagon)

AK = AK (common side)




K KUNDAN

AFKABK (SSS congruence rule)

FAKBAK

Also 120ºBAFFAKBAK

BAK = 60º ABK is a 30º – 60º – 90º triangle

AK=2

1hypotenuse (AB) = 36

2

1

and 3362

3)AB(

2

3BK cm

BF = 2BK = 2 × 3 363 cm.

Now Area of sector ABPF

=2

360

anglesectorr

= 68.3714.312614.3

360

120 2 cm2

Area of ABF = AKBF2

1

= 393362

1

= 9 × 1.73 = 15.57 cm2

Area of segment BPF

= Area of sector ABPF – Area of ABF

= 37.68 m2 – 15.57 cm2 = 22.11 cm2.

( ii) Area of shaded portion

= Area of hexagon ABCDEF – area of ABF)

= 57.1562

33 2

= 57.15354 = 54 × 1.73 – 15.57

= 93.42 cm2 – 15.57 cm2 = 77.85 cm2.

19. Radius of sector CXB = 14 cm

Radius of sector DXY = 21 cm – 14 cm = 7 cm

Area of sector CXB = 2147

22

360

90 = 154 cm2

Area of sector DXY = 277

22

360

90

= 38.5 cm2

Area of rectangle = ABCD = AB × BC

= 21 cm × 14 cm = 294 cm2

Area of the shaded portion

= Area of rectangle – Area of two sectors

= 294 – (154 + 38.5) = 294 – 192.5

= 101.5 cm2.

20. Area of equilateral4

3 (side)2

= 1964

732.114

4

3 2

= 1.732 × 49 = 84.868 cm2

Area of three sectors

= 22 77

22

120

60

360

anglesector3

r

= 11 × 7 = 77 cm2

Area of shaded region = Area of equilateral – Area of three sectors

= (84.868 – 77 =) 7.868 cm2.

21. The radius of circle = 21 cm

An arc ABC subtends an angle of 60º at the

centre = OA = OB = 21 cm

6060 – 1802

1OBAOAB

OAB is equilateral.

A

B

O60°

21cm

(i ) Length of the arc =

360

60 × circumference

= 217

222

360

60 cm = 22 cm

( ii) Area of the sector =360

60 × area of the circle

= 21217

22

360

60 sq cm = 231 sq cm

(iii) Area of segment

= Area of sector – Area of equilateral AOB

of side 21 cm

=

2121

4

3231 sq cm

=

4

441732.1231 sq cm

= (231 – 190.953) sq cm = 40.047 sq cm.




K KUNDAN

22. Let the radius of the swimming pool be r m.

Width of the path = 4 m

Then area of the swimming pool = r 2 sq m

and area of the swimming pool including the

path

= msq )4( 2 r

222

25

11)4( r r r

It is given that the area of the path is2511 th

part of the area of the swimming pool.

or, 222

25

114 r r r

or, 222 1125)168(25 r r r r

or, 222 112540020025 r r r r

or, 040020011 2 r r

22

1760040000200 r

2257600200 r

= ,22

440

22

240200

22

40,20

20

40

Since radius of the swimming pool cannot be

negative. Hence radius = 20 m.

23. Circumference of circle = 80 cm

or, r 2 = 80 cm or r =

40cm

Let the side of square be 2a cm.

Since from the centre of a circle to a chord

bisects the chord,

OM = a and AM = a

Now in right-angled triangle OAM, we have

222 AMOMOA or 222 a a r

or, ,2

2

240

a

or

22 40

2

1

a

or,

40

2

1a

Hence side of square

=

402

40

2

122a cm

24. OC = CB = 20 cm

In right-angled OCB,

OB = cm 220202022

Radius of sector OPBQ = 220 cm.

Area of sector OPBQ

= 22 22014.3360

º90

360

anglesector r

= 80014.34

1 = 628 cm2

Area of square OABC = {(20)2 =} 400 cm2.

Hence the area of the shaded region

= Area of sector OPBQ – Area of square OABC

= 628 cm2 – 400 cm2 = 228 cm2.25. Let r be the radius of inner circle and R be

the radius of outer circle

Then, 9R2 r ...(i)

ΔDOC~ΔAOD

ODOC

OAOD

or, OCOAOD2 or 9RR5R 2

or, 9RR2510RR 22 or R = 25

2(25 – r ) = 9 [using (i)]

or, 2r = 50 – 9

or, 2r = 41




K KUNDAN

r =2

41 = 20.5

Area of shaded portion =22R r

= [(25)2 – (20.5)2] = 25.4206257

22

= 75.2047

22 = 643.5 cm2.

26. Let the radii of the given circles be R and r

respectively.

Sum of their areas = 116 cm2

116R 22 r

or, 116)R( 22 r

or, 116R2 r ....(i)

If O and O be the centres of the given circles,

then r ROO

6R r (Given) ....(ii)

Now, 2222 R2RR r r r

or, 22 6R r = 2 × 116 [Using (i) and (ii)]

or, 196362322 r R

14196R r ....(iii)

Solving (ii) and (iii), we get

2R = 20

R =2

20 = 10

From (iii), 10 + r = 14 or r = 14 – 10 = 4

Hence radii of the given circles are 10 cm

and 4 cm respectively.

27. Area of rectangle = (28 × 26 =) 728 m2

Area of one corner (unshaded)

= 2)10(4

1 = 10014.3

4

1 =

4

314= 78.5 cm2

Total area of 4 unshaded corners

= 78.5 × 4 = 314 m2

Area of shaded portion

= Area of rectangle – Area of 4 corners

= (728 – 314 =) 414 m2

.28. Required area

= Area of square – 4 × Area of one sector

=

27

4

141414

= 497

22

4

14196

= (196 – 154 =) 42 cm2.

29. Area of shaded region

= Area of a square

– 2(Area of a semi-circle of radius 7 cm)

=

77

7

22

2

1214

2

= (196 – 154 =) 42 cm2

.30. In ABC, A = 90°

By pythagoras theorem, we get

222 ACABBC = 25)3()4( 22

25BC cm = 5 cm

Area of shaded portion

= 222 5.22

15.1

2

12

2

1

= 25.625.247

22

2

1

= 64.197

5.1375.12

7

11 cm2.

31. Let r cm be the radius of the inner circle, then

The radius of the outer circle = (r + 7) cm.

Area of the shaded region = 770 cm2

or, 770)7( 22 r r

or, 770)4914( 22 r r r

or, 77049147

22r

or, 3572 r

or, 282 r

r = 14

Radius of outer circle = (14 + 7) cm = 21 cm.




K KUNDAN

Side of the square = 2 × 21 cm = 42 cm

The area of the unshaded region of the

square

= (42)2 – 770 = (1764 – 770) cm2 = 994 cm2.

32.

Area of the shaded region

= Area of the equilateral + Area of the

semicircle

= 22

2

1side

4

3r =

22

2

7

7

22

2

17

4

3

[ BC = 7 cm is the diameter of semicircle

cm r 2

7

]

=4

7749

4

732.1

4

49

7

1149

4

3

= 12.25 × 1.732 + 19.25

= 21.22 + 19.25 = 40.47 cm2.

33. AC = 2 × radius = 2 × 5 cm = 10 cm.

As angle in a semi-circle is 90º.

So ADC = 90° and ABC = 90°

By pythagoras theorem;

AD2 = AC2 – CD2

= 102 – 82 = 100 – 64 = 36

AD = 36 = 6 cm

and AB2 + BC2 = AC2

or, AB2 + AB2 =100

or, 2AB2 =100

or, AB2 = 50

or, AB = 2550 cm

The area of the shaded region

= area of the circle – area of ADC – area of

ABC

= BCAB2

1CDAD

2

12 r

= 25252

186

2

15

7

22 2

[ AD = 5 cm, CD = 8 cm

AD = 25 , BC = 25 ]

= 25247

550

= 78.59 – 49 = 29.57 cm2.

34. From the given figure it can be seen that

Radius = OP = OR = OQ = OS = 14 cm

Taking the semi-c ircle POQS first

Area of shaded small circle =

2

2

OS

=4

(OS)2 =

47

141422

= 154 cm2

Now taking the semi-circle OPRQ

Area of the semi-circle OPRQ =2

(OR)2

=

27

141422

= 308 cm2

Area of triangle PRQ=2

1 × base × height

=2

1 × PQ × OR

=2

1 × 14 × 28

= 196 cm2

Area of shaded region in semi-circle OPRQ

= (308 – 196) cm2 = 112 cm2

The area of the shaded region = Area of shaded

region in semi-circle OPQR + Area of shaded

region in POQS = (154 + 112) cm2 = 266 cm2

35.

In the given figure PQRS is a rectangular field

in which SR = 6 m and PS = 4 m. We have to

find the cost of shaded portion in the given

figure. The semicircle PAS and QBR makes a

complete circle of radius

=2

4

2

PS

2

QR = 2 m

area of circle = r 2 = 3.14 × 4 m2

Similarly, PCQ and SDR also makes a complete

circle of radius

=2

6

2

SR

2

PQ = 3 m

area of circle = r 2 = 3.14 × 9 m2




K KUNDAN

Total area in which flower is raised

= 3.14 × 4 + 3.14 × 9 = 3.14 × 13 m2

total cost = (3.14 × 13 × 8 =) Rs 326.56

36.

Area of the square = (14 cm)2 = 196 cm2.

Area of circular part at one vertex

= 777

22

º360

º90 =

4

154cm2

Total area of circular parts = 44

154

= 154 cm2.

Area of the region of the square that remains

outside the region of any of the circle

= (196 – 154) cm2 = 42 cm2.

37. In the following figure, BDEF is the brick put

to stop the wheel.

Here BD = AC = EF = 5 cmAB = CD = 15 cm

OA = OD = r (say) [Radius of the wheel]

In the OCD; OC = OA – CA = (r – 5) cm

CD = 15 cm

Here, OD2 = OC2 + CD2

or, 222 15)5( r r

or, 225102522 r r r

or, 10r = 250

r = 25 cm

Therefore, the required radius of the wheel

= 25 cm.

38. Angle described by minute-hand in 60 minutes

= 360°

Angle described by minute-hand in one

minute =60

360 = 6°

We know that the area A of a sector of angle

D, in a circle of radius r is given.

A =2

360

Dr

Here, r = 10 cm and D = 6°

A =30

157101014.3

360

6

sq cm

=3075 sq cm

39. Area of the minor segment

= Area of sector OAB – Area of the right-

angled triangle OAB

=

1010

2

1101014.3

360

90cm2

= (78.5 – 50) cm2 = 28.5 cm2

Area of the major segment

= Area of the circle – Area of the minor segment

= [3.14 × 102 – 28.5] cm2

= [314 – 28.5] cm2 = 285.5 cm2

Note: We know that the area of a minor

segment of angle in a circle of radius r is

given by

sin

2

1

360A 2r [Always Remember]

Here, r = 10, = 90°

A =

90sin

2

1

360

9014.3)10( 2

cm2

=

2

1

4

14.3)10( 2

cm2

= [3.14 × 25 – 50] cm2

= (78.5 – 50)cm2 = 28.5 cm2

40. We know that the area of a minor segment of

angle in a circle of radius r is given by

A =

sin

2

1

360

2r

=

60sin

21

3606014.3)15( 2

cm2

=

4

3

6

14.3225 cm2

= 225 [0.5233 – 0.4330] cm2

= 225 × 0.0903 cm2 = 20.317 cm2




K KUNDAN

Area of the major segment

= Area of the circle – Area of the minor segment

= [3.14 × (15)2 – 20.317] cm2

= [706.5 – 20.317] cm2 = 686.183 cm2

Note: We can have another method for findingarea of the minor segment. Clearly, from the

figure, triangle OAB is an equilateral triangle

with the side 15 cm.

Now, area of the minor segment

= Area of the sector OAB – Area of the

triangle OAB

=

22 )15(

4

3)15(

360

60cm2

=

4

3

6

14.3152

cm2

= 20.317 cm2

41. Area of the square lawn ABCD = (56 × 56) sq m

Let OA = OB = x metresso, x 2 + x 2 = 562

or, 2x 2 = 56 × 56

or, x 2 = 28 × 56

Now, area of sector OAB =22

4

1

360

90x x

=

5628

7

22

4

1sq m

[Putting the value of x 2 = 28 × 56]

Also, area of OAB =

5656

4

1sq m

(AOB = 90°)

So, area of flower bed AB

=

5656

4

15628

7

22

4

1sq m

=

2

7

225628

4

1sq m

=

7

85628

4

1sq m

Similarly, area of the other flower bed

=

7

85628

4

1sq m

Therefore, total area

=

7

85628

4

1

7

85628

4

15656 sq m

=

7

2

7

225628 sq m

=

7

185628 sq m = 4032 sq m

Alternative Method:

Total area

= Area of sector OAB + Area of sector ODC +

Area of OAD + Area of OBC

=

5628

7

22

360

905628

7

22

360

90

5656

4

15656

4

1sq m

= 141422227

567

sq m

= (56 × 72 =) 4302 sq m

42. Area of square ABCD

= (14 × 14) sq cm = 196 sq cm

Diameter of each circle =

2

14 7 m

Radius of each circle = 2

7cm

Area of one circle =2r =

2

7

2

7

7

22sq cm

=2

77sq cm

Therefore, area of the four circles

=

2

774 154 sq cm

Hence, area of the shaded region

= (196 – 154 =) 42 sq cm

43.

Let us mark the four unshaded regions as I,

II, III and IV as in the figure.

Area of I + Area of III

= Area of ABCD – Areas of two semicircles

of each of radius 5 cm

=

25

2

121010 sq cm

= (100 – 3.14 × 25) sq cm

= 21.5 sq cmSimilarly, Area of II + Area of IV = 21.5 sq cm

So, area of the shaded region

= Area of ABCD – Area of (I + II + III + IV)

= (100 – 2 × 21.5) sq cm

= (100 – 43 =) 57 sq cm




K KUNDAN

44. Let the radius of the circle be r cm.

As per given information,

r r 22 = 16.8

or, )1(2 r = 16.8

or,

1

7

222r = 16.8

or,7

152 r = 16.8

or, r =152

78.16

= 3.92

Hence, radius of the circle = 3.92 cm

45.

Let the radius of circle be x .

Area of the circle =22 x r

A square has been inscribed in the circle.

Diagonal of square = Diameter of circle = 2x

Side of square =2

2x

Its area = 22

22

4

2

2

2

2x

x x x

Required ratio =22 2: x x

= 2:

7

22

= 11 : 7

46.

Let a = 6 cm, b = 8 cm and c = 4 cm

Semi-perimeter(s) =2

486

2

c b a = 9 cm

Area of the triangle = )()()( c s b s a s s

= )49()89()69(9

= 5139 = 153 sq m

Area of sectors = )QQQ( 321

2

r

= 180360

12

( Sum of all the angles of a

triangle is equal to 180°)

=7

11

27

22

sq m

Area of the remaining portion of the triangle

=

7

11153 sq cm

= (3 × 3.9 – 1.6) sq cm

= (11.7 – 1.6) sq cm

= 10.1 sq cm

47.

Area of the sector AOB

=360

AngleSector × Area of the circle

=

2828

7

22

360

72cm2

=5

2464 cm2 = 492.8 cm2

Length of the arc AB

=360

AngleSector× Circumference of the circle

=

287

22

2360

72

cm

=5

176 cm = 35.2 cm

48. Let the radii of three concentric circles be x ,

2x and 3x respectively.

required ratio

=circlesoutertwothebetweenArea

circlesinnertwothebetweenArea

= 22

22

)2()3(

)()2(

x x

x x

= 2

2

)49(

)14(

x

x

=5

3 = 3 : 5

Date post:	06-Jul-2018
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Area and Perimeter for Ssc Exams

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