Lesson: Area related to circles

Exercise 11.1 (10)

Choose the correct answer from the given four

options:

Question: 1

If the sum of the areas of two circles with radii

1

R

and is equal to the area of a circle of radius ,

2

R

R

then

a.

1 2

R R R

b.

2 2 2

1 2

R R R

c.

1 2

R +R R<

d.

2 2 2

1 2

R R R

Solution:

(b)

Sum of areas of two circles of radii and = area

1

R

2

R

of circle of radius

R

Hence,

2 2 2

1 2

πR πR πR

Take common from the first expression.

π

R

2 2 2

1 2

π(R R ) π

2 2 2

1 2

R R R

Thus, the correct option is (B).

Question: 2

If the sum of the circumferences of two circles with

radii and is equal to the circumference of a

1

R

2

R

circle of radius , then

R

a.

1 2

R R R

b.

1 2

R R R

c.

1 2

R R R

d. Nothing definite can be said about the

relation among , , and .

R

1

R

2

R

Solution:

(a)

Given: Circumference of circle with radius +

1

R

Circumference of circle of radius =

2

R

Circumference of circle of radius

R

1 2

2πR 2πR 2πR+ =

1 2

2π(R R ) 2πR

1 2

R R R

Thus, the correct option is (A).

Question: 3

If the circumference of a circle and the perimeter of

a square are equal, then

a. Area of the circle = Area of the square

b. Area of the circle > Area of the square

c. Area of the circle < Area of the square

d. Nothing definite can be said about the relation

between the areas of circle and square.

Solution:

(b)

Circumference of the circle with radius, =

R

2πR

Perimeter of square of side, =

a

4a

Circumference Perimeter

of circle of square

2πR 4a

=

2πR

a

4

πR

a

2

Area of circle =

2 2

22

πR R

7

area of square,

a

2

2

πR

2

2

2

22

R

7 2

2

121

R

49

2

2

Area of circle

Area of squ

22

R

7

121

R

4

are

9

22 49

121 7

14

11

1.27

Area of circle = 1.27 Area of square.

So, Area of circle > Area of square.

Thus, the correct option is (B).

Question: 4

Area of the largest triangle that can be inscribed in

a semi-circle of radius units is

r

a. sq. unit

2

r

b. sq. unit

2

1

r

2

c. sq. unit

2

2r

d. sq. unit

2

2r

Solution:

(a)

Let PQR be a triangle inscribed in a semicircle of

area .

r

The base of the triangle QR = Diameter of

semicircle =

2r

Altitude of the triangle OP = radius of the

semicircle =

r

1

Area of trianglePQR ×base×height

2

1

2r r

2

2

r sq.unit

Thus, the area of triangle is .

2

r sq u. nits

Question: 5

If the perimeter of a circle is equal to that of a

square, then the ratio of their areas is

a.

22:7

b.

14:11

c.

7: 22

d.

11:14

Solution:

(b)

Perimeter of circle of radius, =

r

2πr

Perimeter of square of side, =

a

4a

Perimeter Perimeter

of square of circle

4a 2πr

2πr

a

4

πr

2

Area of circle =

2

πr

2

22

r

7

Area of square,

2

2

πr

a

2

2

2

22

r

2 7

2

121

r

49

2

2

Ratio of area of circle

to that

2

of

2

r

7

12

s

1

r

4

u

9

q are

22 49

121 7

14

11

Thus, the ratio of area of circle to that of square is

14:11.

Question: 6

It is proposed to build a single circular park equal in

area to the sum of areas of two circular parks of

diameters 16 m and 12 m in a locality. The radius of

the new park would be

a. 10 m

b. 15 m

c. 20 m

d. 24 m

Solution:

(a)

Let be the radius of new park.

R

Area of the Area of Area of

new park park 1 park 2

2 2 2

1 2

πR πr πr

2 2 2

1 2

R r r

1

1

D 16

r 8m

2 2

2

2

D 12

r 6m

2 2

2 2 2

1 2

R r r

2 2

8 6

64 36

100

So,

2

R 100

R 100

10m

Thus, the radius of the new park is 10 m.

Question: 7

The area of the circle that can be inscribed in a

square of side 6 cm is

a. 36 cm

2

π

b. 18 cm

2

π

c. 12 cm

2

π

d. 9 cm

2

π

Solution:

(d)

Side of given a square,

a 6cm

Radius of circle that can be inscribed inside the

square,

a 6

r 3cm

2 2

Area of the circle inscribed

2 2 2

πr π 3 9π cm

Thus, the area of the square is .

2

9π cm

Question: 8

The area of the square that can be inscribed in a

circle of radius 8 cm is

a. 256 cm

2

b. 128 cm

2

c. 64 cm

2

2

d. 64 cm

2

Solution:

(b)

Diagonal of a square, SQ =

2r 2 8 16cm

Applying Pythagoras theorem,

2 2 2

SR RQ SQ

2 2 2

a a 16

2

2a 256

2

a 128

Thus, area of square that can be inscribed in the

circle is .

2

128cm

Question: 9

The radius of a circle whose circumference is equal

to the sum of the circumferences of the two circles

of diameters 36 cm and 20 cm is

a. 56 cm

b. 42 cm

c. 28 cm

d. 16 cm

Solution:

(c)

Diameters of two circles are 36 cm and 20 cm.

Let and be the radii of two circles.

1

r

2

r

1

Circumference of first circle 2πr

1

D

2π

2

36

2π

2

36π

2

Circumference of second circle 2πr

2

D

2π

2

20

2π

2

20π

Circumference of Circumference Circumference

the new circle of circle 1 of circle 2

2πr 36π 20π

2πr 56π

56

r

2

28cm

Thus, radius of the circle is 28 cm.

Question: 10

The diameter of a circle whose area is equal to the

sum of the areas of the two circles of radii 24 cm

and 7 cm is

a. 31 cm

b. 25 cm

c. 62 cm

d. 50 cm

Solution:

(d)

Radius of two circles = 24 cm and 7 cm

Area of Area of Area of

the circle circle 1 circle 2

2 2 2

1 2

πr πr πr

2 2 2

r 24 7

2

r 625

r 625

r 25cm

Diameter of the circle

2r 2 25 50cm

Thus, diameter of the circle is 50 cm.

Exercise 11.2 (14)

Question: 1

Is the area of the circle inscribed in a square of side

cm, ? Give reasons for your answer.

a

2 2

πa cm

Solution:

Radius of the circle,

a

r cm

2

Area of the circle

2

πr

2

a

π

2

2

2

πa

cm

4

So,

2

2 2 2

πa

cm πa cm

4

Hence, the above statement is false.

Question: 2

Will it be true to say that the perimeter of a square

circumscribing a circle of radius cm is 8a cm?

a

Give reasons for your answer.

Solution:

True.

Side of the square =

2a cm

Perimeter of the square

4 sideof square

4 2a

8a cm

Hence, the above statement is true.

Question: 3

In the adjoining figure, a square is inscribed in a

circle of diameter and another square is

d

circumscribing the circle. Is the area of the outer

square four times the area of the inner square? Give

reasons for your answer.

Solution:

No, the outer square area is not four times the area

of inner square.

Diagonal of inner square = diameter of the circle =

d.

Apply Pythagoras theorem,

2 2 2

EF EH FH

2 2 2

a a d

2

2

d

a

2

Area of inner square

2

2

d

a

2

Area of outer square =

2

2

side d

So, area of larger square is two times the area of

smaller square.

Thus, the given statement is false.

Question: 4

Is it true to say that area of a segment of a circle is

less than the area of its corresponding sector? Why?

Solution:

Area of major segment PQR > Area of

corresponding sector OPQR

Whereas Area of minor segment ACB < Area of

corresponding sector OACB.

Thus, the above statement is wrong.

Question: 5

Is it true that the distance travelled by a circular

wheel of diameter d cm in one revolution is

2πd

cm? Why?

Solution:

False

Distance travelled by wheel in one revolution =

Circumference of the circular wheel

Circumference of the circle

2πr

d

2π

2

πd=

Hence, the distance travelled by wheel in one

revolution =

πd

Thus, the above statement is false.

Question: 6

In covering a distance m, a circular wheel of

s

radius m makes revolutions. Is this statement

r

s

2πr

true? Why?

Solution:

True.

Let n be the number of revolutions.

Total distance covered by wheel in n revolutions,

s 2πrn

Number of revolutions,

s

n

2πr

Thus, the above statement is true.

Question: 7

The numerical value of the area of a circle is greater

than the numerical value of its circumference. Is

this statement true? Why?

Solution:

False.

Let radius of circle be 1 cm.

Area of circle

2 2 2

πr π 1 π cm

Circumference of the circle =

2πr 2π 1 2π cm.

Thus, area of circle is not always greater than its

circumference.

Question: 8

If the length of an arc of a circle of radius is equal

r

to that of an arc of a circle of radius , then the

2r

angle of the corresponding sector of the first circle

is double the angle of the corresponding sector of

the other circle. Is this statement false? Why?

Solution:

Length of arc of a circle

θ

2πr

360

Length of arc of first circle,

1

1

θ

l 2πr

360

Length of arc of second circle,

2

2

θ

l 2π 2r

360

2

θ

4πr

360

Both the lengths are equal.

1 2

l l

1 2

θ θ

2πr 4πr

360 360

1

2

θ

2

θ

Thus, the given statement is true.

Question: 9

The areas of two sectors of two different circles

with equal corresponding arc lengths are equal. Is

this statement true? Why?

Solution:

False.

Area of sector

2

θ

πr

360

Area of sector of circle 1,

2

1 1

1

πr θ

A

360

Area of sector of circle 2,

2

2 2

2

πr θ

A

360

Length of arc of the two circles are equal.

1 1 2 2

2πr θ 2πr θ

360 360

1 1 2 2

r θ r θ

Let

1 1 2 2

r θ r θ a

Now, find the ratio of area of the two sectors.

2

1 1

1

2

2 2

2

πr θ

A

360

πr θ

A

360

2

1 1

2

2 2

r θ

r θ

1

2

r a

r a

1

2

r

r

So, the areas will only be equal, when the two

circles are of same radius.

Thus, the above statement is False.

Question: 10

The areas of two sectors of two different circles are

equal. Is it necessary that their corresponding arc

lengths are equal? Why?

Solution:

False.

Ratio of areas of the two circles

2

1 1 1

2

2 2 2

A r θ

A r θ

1 2

A A

2 2

1 1 2 2

r θ r θ

2

1 2

2

2 1

θ r

θ r

1

2

Ratio of length of arcs

of the two circles

l

l

1 1

2 2

2πr θ

2πr θ

2

2 1

2

1 2

r r

r r

2

1

r

r

Length of arc will be equal when the two circle will

have the same radius.

Thus, the above statement is false.

Question: 11

Is the area of the largest circle that can be drawn

inside a rectangle of length a cm and breadth b cm

(a > b) is ? Why?

2 2

πb cm

Solution:

False.

Diameter of largest circle = breadth of the

rectangle.

Length of rectangle = cm

a

Breadth of rectangle = cm

b

Therefore, diameter of circle = cm

b

Area of the circle

2

2

πd

πr

4

2

2

πb

cm

4

Thus, the above statement is false.

Question: 12

Circumferences of two circles are equal. Is it

necessary that their areas be equal? Why?

Solution:

True.

Circumference of two circles are equal.

1 2

2πr 2πr

1 2

r r

Circle having same radius will have similar areas.

Thus, the above statement is true.

Question: 13

Areas of two circles are equal. Is it necessary that

their circumferences are equal? Why?

Solution:

True.

Area of the two circles are equal.

2 2

1 2

πr πr

2 2

1 2

r r

1 2

r r

Circles with same radius will have same

circumference.

Thus, the above statement is true.

Question: 14

Is it true to say that area of a square inscribed in a

circle of diameter p cm is ? Why?

2 2

p cm

Solution:

Diagonal of the square inscribed = Diameter of the

circle = p cm

2 2 2

PS PQ QS

2 2 2

a a p

2

2

p

a

2

Area of the square

2

2 2

p

a cm

2

Thus, the given statement is false.

Exercise 11.3 (16)

Question: 1

Find the radius of a circle whose circumference is

equal to the sum of the circumferences of two

circles of radii 15 cm and 18 cm.

Solution:

Radius of two circles = 15 cm and 18 cm

Circumference Circumference Circumference

of the circle of circle1 of circle2

1 1

2πr 2πr 2πr

1 2

r r r

15 18

33cm

Thus, the radius of the circle is 33 cm.

Question: 2

In the adjoining figure, a square of diagonal 8 cm is

inscribed in a circle. Find the area of the shaded

region.

Solution:

Let a be the side of the square.

Diagonal of the inscribed square = Diameter of the

circle = 8 cm

2 2 2

PS PQ QS

2 2 2

a a 8

2 2

64

a 32cm

2

Therefore, area of the square

2 2

a 32cm

Area of the circle

2

2

πd

πr

4

2

22 8

7 4

2

22 8

7 4

2

352

cm

7

Area of the Area of Area of

–

shaded region circle square

2 2

πr a

352

32

7

352 224

7

2

128

cm

7

Thus, area of shaded region is .

2

128

cm

7

Question: 3

Find the area of a sector of a circle of radius 28 cm

and central angle .

45

Solution:

Radius of the circle,

r 28cm

Central angle,

θ 45

Area of the sector

2

πr θ

360

2

π 28 45

360

2

308cm

Thus, area of the sector is 308 cm

2

.

Question: 4

The wheel of a motor cycle is of radius 35 cm. How

many revolutions per minute must the wheel make

so as to keep a speed of 66 km/h?

Solution:

Distance travelled by the wheel in one revolution

2πr 2π 0.35

0.7π cm

Speed of the wheel

5

66km / h m / s

18

55

m/s

3

Let, total number of revolution in 1 min be .

n

Distance

Speed

Time

Total distance travelled in revolutions = 0.7

n

π n

Total time taken = 1 min = 60 sec.

55 0.7π n

3 60

55 60 7

n

3 22 0.7

500rev/min

The, the total number of revolutions in 1 min are

500.

Question: 5

A cow is tied with a rope of length 14 m at the

corner of a rectangular field of dimensions

. Find the area of the field in which the

20m 16m

cow can graze.

Solution:

Cow will graze in the area corresponding to sector

STU of the rectangle PQRS.

Radius of the area grazed = 14 m

Central angle =

90

2

Area grazed Area of

by cow the sector

πr θ

360

2

22 14 90

7 360

22 14 14

7 4

2

154m

Thus, the grazed area is 154 m

2

.

Question: 6

Find the area of the flower bed (with semi-circular

ends) shown in figure.

Solution:

Radius of semi-circle,

r

10

cm 5cm

2

Length of the rectangle, = 38 cm

l

Breadth of the rectangle, = 10 cm

b

Now, find the area of the flower bed.

Area of the Area of the Area of the

flower bed two semicircles rectangle PQRS

2

πr

l b

2

2

2

πr 38 10

2

25π 380 cm

Thus, area of the flower bed is .

2

25π 380 cm

Question: 7

In the adjoining figure, AB is a diameter of the

circle, AC = 6 cm and BC = 8 cm. Find the area of

the shaded region (Use π = 3.14).

Solution:

In ,

ABC

Base, AC =

6cm

Height, BC =

8cm

Now, find the area of triangle.

2

25π 380 cm

1

6 8

2

2

24 cm

Apply Pythagoras theorem,

2 2 2

AC BC AB

2 2 2

AB 6 8

AB 36 64

100

10cm

So, diameter of the circle is .

10cm

Therefore, radius of the circle = 5 cm

2

Area of

the cir le

πr

c

2

3.14 5

314

5 5

100

7850

100

2

78.5cm

Area of the Area of Area of

–

shaded region circle triangle ABC

78.5 24

54.5cm

Thus, area of the shaded region is 54.5 cm.

Question: 8

Find the area of the shaded field shown in the given

figure.

Solution:

Length of rectangle = 8 m

Breadth of rectangle = 6 m

Radius of the semicircle = 6 – 4 = 2 m

Area of the Area of the Area of the

shaded field semicircle rectangle

2

πr

l b

2

2

π (2)

4 8

2

2

2π 32 m

Thus, area of the shaded region is .

2

32 2π m

Question: 9

Find the area of the shaded region in the given

figure.

Solution:

Length of triangle PQRS,

1

l 12m

Breadth of triangle PQRS,

2

b 26cm

Diameter of semicircle

12 4 4

4m

Radius,

4

r 2m

2

Length of the rectangle TUVW,

2

l 26 (3 2 2 2)

16m

Breadth of the rectangle TUVW,

2

b 12 8

4m

Area of

rectangle PQRS

Area of the Area of two

shaded region semicircles

Area of

rectangle TUVW

2

1 1 2 2

πr

l b 2 l b

2

π 2 2

12 26 2 4 16

2

2

312 2 2π 64 m

2

248 4π m

Thus, area of the required region is .

2

248 4π m

Question: 10

Find the area of the minor segment of a circle of

radius 14 cm, when the angle of the corresponding

sector is .

60

Solution:

Radius of the circle,

r 14cm

Angle of corresponding sector,

θ 60

In triangle PQR,

PR = PQ = (radii of the circle)

r

Let, (angles opposite to equal sides

R Q x

are equal)

By the Angle P Q R 18 sum property0

x x 60 180

2x 180 60

x 60

So, the triangle PQR is an equilateral triangle.

Side of the triangle PQR,

a r 14cm

Area of minor Area of Area of

segment sector PQR

2

2

πr θ 3

r

360 4

2

2

π 14 60 3

14

360 4

22 14 14 60 3

14 14

7 360 4

2

102.67 84.87 cm

2

17.8cm

Thus, area of the minor segment is .

2

17.8cm

Question: 11

Find the area of the shaded region in the adjoining

figure, where arcs drawn with centres P, Q, R and S

intersect in pairs at mid-points W, T, U and V of the

sides PQ, PS, RS and QR, respectively of a square

PQRS (Use π = 3.14).

Solution:

Side of the square,

a 12cm

Radius of the sectors

a 12

6cm

2 2

Central angle of the sector =

90

Area of the Area of Area of the

shaded part the square 4 sectors

2

2

πr θ

a 4

360

2

2

3.14 6 90

12 4

360

2

3.14 6 6 90

12 4

360

2

144 4 28.26 cm

2

144 113.04cm

2

30.96cm

Thus, area of shaded region is .

2

30.96cm

Question: 12

In the adjoining figure, arcs are drawn by taking

vertices A, B and C of an equilateral triangle of side

10 cm. to intersect the sides BC, CA and AB at their

respective mid-points D, E and F. Find the area of

the shaded region (Use π = 3.14).

Solution:

The sides of an equilateral triangle are equal.

And, all the angles of an equilateral triangle,

θ 60

Now, arcs are drawn by taking the vertices P, Q and

R of the equilateral triangle.

Each side of equilateral triangle = 10 cm

According to the question,

1

Radiusof thearc,r Sideof equailateral triangle

2

1

10

2

5cm

Area of shaded

3 Area of sector madeby arc

region

2

πr θ

3

360

2

3.14 5 60

3

360

2

3 13.08 cm

2

39.25cm

Thus, area of the shaded region is 39.25 .

2

cm

Question: 13

In the given figure, arcs have been drawn with radii

14 cm each and with centres P, Q and R. Find the

area of the shaded region.

Solution:

It is given that,

Radius of each arc, = 14 cm

r

Now, from the above figure,

(Interior angles of a

1 2 2

θ θ θ 180

triangle)

Area of Sum of areasof

shaded region thethreesectors

2 2 2

1 2 3

πr θ πr θ πr θ

360 360 360

2

1 2 3

22

14

7

θ θ θ

360

22 14 2

180

360

2

308cm

Thus, area of shaded region is .

2

308cm

Question: 14

A circular park is surrounded by a road 21 m wide.

If the radius of the park is 105 m, find the area of

the road.

Solution:

Radius of the park,

1

r 105m

Width of the road

21m

Radius of road and park combined,

2

r 105 21

126m

Now, find the area of the road.

Area of the Area of road Area of

–

road and park the park

2 2

2 1

πr -πr

2 2

π 126 - π 105

2 2

π 126 105

22

4851

7

2

15246m

Thus, area of the road is 15246 .

2

m

Question: 15

In the given figure, arcs have been drawn of radius

21 cm each with vertices P, Q, R and S of

quadrilateral PQRS as centres. Find the area of the

shaded region.

Solution:

Radius of each sector, =

r

21cm

Sum of angles of quadrilateral

1 2 3 4

θ θ θ θ

360

Now, find the area of the shaded region.

Area of shaded Sum of areas of

region all the 4 arcs.

2 2 2 2

1 2 3 4

πr θ πr θ πr θ πr θ

360 360 360 360

2

1 2 3 4

22

21

7

θ θ θ θ

360

22 21 3

360

360

2

3186cm

Thus, area of the shaded region is .

2

1386cm

Question: 16

A piece of wire 20 cm long is bent into the form of

an arc of a circle subtending an angle of at its

60

centre. Find the radius of the circle.

Solution:

Length of the arc = length of the wire =

20cm

Angle made by the arc, =

θ

60

Now, find the radius of the arc.

Length of

the arc

2πrθ

360

2πrθ

20

360

20 360

r

2π 60

60

r cm

π

Thus, radius of the circle = .

60

cm

π

Exercise 11.4 (20)

Question: 1

The area of a circular playground is 22176 . Find

2

m

the cost of fencing this ground at the rate of Rs 50

per meter.

Solution:

Area of the circular playground =

2

22176m

Now, find the radius of the playground,

Area 22176

2

πr 22176

2

22176 7

r

22

2

155232

r

22

r 7056

84m

Now, find the length of fencing.

Length of Circumference of

fencing the playground

2πr

22

2 84

7

2 22 12

528m

Cost of fencing per meter

Rs.50

Cost of fencing 528 meter =

528 50 26400Rs.

Thus, the cost of fencing is Rs. 26400.

Question: 2

The diameters of front and rear wheels of a tractor

are 80 cm and 2 m respectively. Find the number of

revolutions that rear wheel will make in covering a

distance in which the front wheel makes 1400

revolutions.

Solution:

Diameter of front wheel

80cm 0.8m

So, radius of the front wheel,

1

0.8

r m

2

Number of revolutions made by front wheel,

1

n 1400

Diameter of rear wheel

2m

So, radius of the rear wheel,

2

2

r m

2

Let, number of revolutions made by rear wheel be

.

2

n

Total distance Total distance

covered by front wheel covered by rear wheel

1 1 2 2

2πr n 2πr n

2

22 0.8 22 2

2 1400 2 n

7 2 7 2

2

7040 7

n

2 22 2

560rev.

Thus, the number of revolutions made by rear

wheel is 560.

Question: 3

Sides of a triangular field are 15 m, 16 m and 17 m.

With the three corners of the field a cow, a buffalo

and a horse are tied separately with ropes of length

7 m each to graze in the field. Find the area of the

field which cannot be grazed by the three animals.

Solution:

Radius of the sector = length of ropes

1 2 3

r r r 7m

Let, be the semi-perimeter and , and be the

s

a

b

c

sides of the triangle PQR.

Here,

a 15m

b 16m

c 17m

a b c

Semiperimeter, s

2

17 16 15

2

48

2

24m

1 2 3

Sum of angles made

by the trian

θ θ

gle

θ 180

Now, find the area not grazed by the animals.

Area not grazed by Area of triangle Area of the

–

the three animals PQR three sectors.

2 2 2

1 1 2 2 3 3

πr θ πr θ πr θ

s s a s b s c

360 360 360

2

1 2 3

22

7

7

24 24 17 24 16 24 15 θ θ θ

360

22 7

24 7 8 9 180

360

2 2

24 21m 77m

2

24 21 77 m

Thus, area of region not grazed by animals is

.

2

24 21 77 m

Question: 4

Find the area of the segment of a circle of radius 12

cm whose corresponding sector has a central angle

of (Use π = 3.14).

60

Solution:

Radius of the circle,

r 12cm

Central angle,

θ 60

Now, from the figure,

In triangle PQR,

PR = PD = (radius of the circle)

r

Let, (angle opposite to equal sides

R Q x

are equal)

(Angle sum property)

P Q R 180

x x 60 180

2x 180 60

x 60

So, triangle PQR is an equilateral triangle.

Now, find the area of the minor segment.

Area of minor Area of Area of

segment sector triangle PQR

2

2

πr θ 3

r

360 4

2

2

3.14 12 60 3

12

360 4

2

75.36 36 3 cm

Thus, area of the minor segment is

.

2

75.36 36 3 cm

Question: 5

A circular pond is 17.5 m in diameter. It is

surrounded by a 2 m wide path. Find the cost of

constructing the path at the rate of Rs 25 per

2

m

Solution:

Radius of the pond,

1

17.5

r m 8.75m

2

Width of the path

2m

Radius of pond and path combined,

2

r 8.75 2

10.75m

Area of the Area of pond Area of the

–

path and path pond

2 2

2 1

πr πr

2 2

π 10.75 π 8.75

2 2

π 10.75 8.75

22

39

7

2

122.57m

Cost of constructing 1 of path

2

m

25Rs.

Cost of constructing path

2

122.57m

25 122.57

3064.25Rs.

Thus, cost of constructing the path is

3064.29Rs.

Question: 6

In the adjoining figure, ABCD is a trapezium with

AB || DC, AB = 18 cm, DC = 32 cm and distance

between AB and DC = 14 cm. If arcs of equal radii 7

cm with centres A, B, C and D have been drawn,

then find the area of the shaded region of the

figure.

Solution:

Radius of each sector =

7cm

Length of the two parallel sides of the trapezium

are,

AB 18 cm

DC 32 cm

Distance between the parallel sides of trapezium

14cm

1 2 3 4

Sum of angles of

quadrilate

θ θ θ θ

r l

3 0

a

6

Now, find the area of the shaded region.

Area of shaded Area of Sum of areas of

region trapezium all the 4 arcs

2

1

2

2

2

3

2

4

πr θ

360

πr θ

1

360

sum of parallel sides height

2

πr θ

360

πr θ

360

2

1 2 3 4

22

7

1

7

18 32 14 θ θ θ θ

2 360

22 7

350 360

360

2

350 154 cm

2

196cm

Thus, the area of shaded region is .

2

196cm

Question: 7

Three circles each of radius 3.5 cm are drawn in

such a way that each of them touches the other

two. Find the area enclosed between these circles.

Solution:

From the figure,

(Equilateral triangle)

θ 60

Radius of each sector

3.5cm

So, side of equilateral triangle,

a 7cm

Area of shaded Area of equilateral Area of

region triangle STU 3 sectors

2

2

3 πr θ

a 3

4 360

2

2

3 3.14 3.5 60

7 3

4 360

2

21.22 19.25 cm

2

1.97cm

Thus, the required area is .

2

1.97cm

Question: 8

Find the area of the sector of a circle of radius 5 cm,

if the corresponding arc length is 3.5 cm.

Solution:

Length of the arc =

3.5cm

Radius of the circle, =

r

5cm

Now, find the angle subtended by the arc.

θ

Length of

the ar

2πrθ

3.

6

c

5

3 0

3.5 360

θ

2π 5

3.5 36

θ

π

Now, find the area of the sector.

2

Area of the

secto

πr θ

36

r

0

2

π 5 3.5 36

360 π

2

8.75cm

Thus, the required area of the circle is .

2

8.75cm

Question: 9

Four circular cardboard pieces of radii 7 cm are

placed on a paper in such a way that each piece

touches other two pieces. Find the area of the

portion enclosed between these pieces.

Solution:

According to the above figure,

At the contact point P, UP and PV are radii.

So, if a tangent pass through point P, UV will be

perpendicular to that tangent.

So, the interior angles of quadrilateral UVWT are

each.

90

Radius of each sector

7cm

Side of the square,

a 14 cm

Now, find the area of the shaded region.

Area of shaded Area of square Area of the4

region UTVW sectors

2

2

πr θ

4

360

a

2

2

3.14 7 90

14 4

360

2

196 154cm

2

42cm

Thus, the required area is .

2

42cm

Question: 10

On a square cardboard sheet of area 784 four

2

cm

congruent circular plates of maximum size are

placed such that each circular plate touches the

other two plates and each side of the square sheet

is tangent to two circular plates. Find the area of

the square sheet not covered by the circular plates.

Solution:

Let us consider each side of the square be .

a

2

Area of thesquarePQRS 784 cm

2

a 784

a 784

a 28cm

Now, according to the above figure.

4r RQ 28

28

r

4

r 7cm

Therefore, radius of the circle is 7 cm.

Now, find the area of the region not covered by

circular plates.

Area of region not covered Area of Area of the

–

by circular plates the square 4 circles

2

784 4πr

2

784 4π 7

784 616

2

168cm

Thus, the required area is .

2

168cm

Question: 11

Floor of a room is of dimensions 5 m × 4 m and it is

covered with circular tiles of diameters 50 cm each

as shown in the given figure. Find the area of floor

that remains uncovered with tiles. (Use π = 3.14).

Solution:

Diameter of each tile = 50 cm = 0.5 m

So, radius of each tile,

0.5

r 0.25m

2

Number of tiles along the length =

5

10

0.5

Number of tiles along the width =

4

8

0.5

Total number of tiles =

10 8 80tiles

Total area covered by tiles

2

πr 80

2

3.14 0.25 80

2

15.7m

Now, area of the rectangular floor

5 4

2

20m

So, area not covered by circular tiles

20 15.7

2

4.3m

Thus, the area not covered by circular tiles is .

2

4.3m

Question: 12

All the vertices of a rhombus lie on a circle. Find the

area of the rhombus, if area of the circle is 1256

2

cm

. (Use π = 3.14).

Solution:

Area of circle =

2

1256cm

Now, find the radius of the circle.

2

πr 1256

1256

r

3.14

20cm

Now, length of both the diagonals =

2r

Therefore, length of each diagonal of the rhombus =

2 20 40cm

Now, find the area of the rhombus.

Area of the rhombus

1 2

1

d d

2

1

40 40

2

2

800cm

Thus, the required area is .

2

800cm

Question: 13

An archery target has three regions formed by three

concentric circles as shown in the given figure. If

the diameters of the concentric circles are in the

ratio , then find the ratio of the areas of three

1:2:3

regions.

Solution:

Given: Ratio of diameters of three circles =

1:2:3

Hence, ratio of their radii will be .

1:2:3

Let,

1

r r,

2

r 2r,

3

r 3r

Now, find the areas of the three circles.

2 2

1 1

A πr πr

2 2

2 2

A πr 4πr

2 2

3 3

A πr 9πr

Now, according to the question,

2 2

Area of region between

first and second circle

4πr πr

2

3πr

2 2

Area of region between

second and third circle

9πr 4πr

2

5πr

Required ratio of areas is .

2 2 2

πr :3πr :5πr 1:3:5

Thus, the ratio is .

1:3:5

Question: 14

The length of the minute hand of a clock is 5 cm.

Find the area swept by the minute hand during the

time-period 6:05 am and 6:40 am.

Solution:

Length of minute hand = radius =

5cm

Area swept by minute hand in 60 minutes =

2

πr

Area swept in 1 minute

2

πr

60

Time difference =

35min

2

Area swept by minute

hand in 35 minutes

πr

35

60

22 5 5

35

7 60

2

5

45 cm

6

Thus, the required area is .

2

5

45 cm

6

Question: 15

Area of a sector of central angle of a circle is

200

770 Find the length of the corresponding arc of

2

cm

this sector.

Solution:

Central angle,

θ 200

Area of sector

2

2

πr θ

770cm

360

Now, find the radius of the circle.

Radius of the circ

770

l

360

r

π 200

e,

r 441

r 21cm

Length of the corresponding arc

2πrθ

360

2 3.14 21 200

360

1

73 cm.

3

Thus, length of the required arc is .

1

73 cm

3

Question: 16

The central angles of two sectors of circles of radii 7

cm and 21 cm are respectively

and

. Find the

120

40

areas of the two sectors as well as the lengths of the

corresponding arcs. What do you observe?

Solution:

For the first circle,

Radius of circle =

7cm

Central angle =

120

Length of arc

2πrθ

360

2 3.14 7 120

360

44

cm

3

Area of the sector

2

πr θ

360

22 7 7 120

7 360

2

154

cm

3

For the second circle,

Radius of the circle =

21cm

Central angle

40

Length of arc

2πrθ

360

2 3.14 21 40

360

44

cm

3

Area of the sector

2

πr θ

360

22 21 21 40

7 360

2

154 cm

Thus, it can be observed that, the lengths of arcs of

the two circles are equal, whereas areas of the

corresponding sectors are not equal.

Question: 17

Find the area of the shaded region given in the

given figure.

Solution:

Area of square PQRS

2

2

14

196cm

Inner figure consists of 4 semicircles of radius and

r

square JKLM of side .

2r

So, find the radius of semicircle.

According to the question,

2r 2r 6 14

4r 8

r 2cm

Area covered by Area of square Area of the4

inner figure JKLM semicircles

2

2

πr

2r 4

2

2

2

2 2 2π 2

2

16 8π cm

Area of Area of Area covered

shaded region squarePQRS by inner figure

196 16 8π

2

180 8π cm

Thus, the area of shaded region is .

2

180 8π cm

Question: 18

Find the number of revolutions made by a circular

wheel of area 1.54 in rolling a distance of 176 m.

2

m

Solution:

Distance covered by the wheel

2πrn 176m

Here, is the number of revolutions made.

n

Area of wheel =

2 2

πr 1.54m

Now, find the radius of the wheel.

Radius of the

1.54

rw ,

π

heel

1.54

r

3.14

0.7m

Now, find the number of revolutions made by the

wheel.

176 7

n

2 22 0.7

40revolutions.

Thus, the circular wheel made 40 revolutions in

covering a distance of 176 m.

Question: 19

Find the difference of the areas of two segments of

a circle formed by a chord of length 5 cm

subtending an angle of at the centre.

90

Solution:

Let us consider, chord SR = 5 cm

It divides the circle in minor segment SPR and

major segment SQR.

Also, let the radius of the circle be .

r

Now, in triangle SOR,

Apply Pythagoras theorem.

2 2 2

OS OR SR

2 2 2

r r 5

2

25

r

2

Now, find the area of the minor segment.

Area of minor Area of sector Area of triangle

–

segment ORSP SOR

2

πr θ 1

r r

360 2

2 2

πr r

4 2

2

π 25 25

cm

8 4

Now, find the area of the major segment.

Area of major Area of Area of minor

–

segment circle segment

2 2

π 25 25

πr cm

8 4

2

π 25 π 25 25

cm

2 8 4

Area of major

segment

Difference in areas of

major and minor segments

Area of minor

segment

2

π 25 25 π 25 25

πr

8 4 8 4

25π 25π 25 25π 25

2 8 4 8 4

2

25π 25

cm

4 2

Thus, the difference in area of major segment and

minor segment is .

2

25π 25

cm

4 2

Question: 20

Find the difference of the areas of a sector of angle

and its corresponding major sector of a circle

120

of radius 21 cm.

Solution:

Let, be the angle made by major segment and

2

θ

1

θ

be the angle made by minor segment.

(Given)

1

θ 120

2

θ 360 120

240

Radius of the circle =

21cm

2

2 1

Difference in area of major

and mi

πr

θ

nor segment

θ

360

2

π 21

240 120

360

2

462cm

Thus, the required area is .

2

462cm