Lesson: Area related to circles

Exercise 12.1 (5)

Question: 1

The radii of two circles are 19 cm and 9 cm

respectively. Find the radius of the circle which has

circumference equal to the sum of the

circumferences of the two circles.

Solution:

circle radius

 

r 19cm

circle radius

 

r 9cm

Assume, the radius of the 3

circle be .

 

Now, find the circumference of the 1

circle.

Circumference 2πr

 

2π 19

38π

Now, find the circumference of the 2

circle.

Circumference 2πr

 

2π 9

18π

Now, find the circumference of 3

circle.

Circumference 2πr

According to the statement,

rd st nd

Circumference Circumference Circumference

of 3 circle of 1 circle of 2 circle

     

 

     

     

2πr 38π 18π 

2πr 56π

56π

2π



r 28

Thus, the radius of the circle is cm.

Question: 2

The radii of two circles are 8 cm and 6 cm

respectively. Find the radius of the circle having

area equal to the sum of the areas of the two

circles.

Solution:

circle Radius

 

r 8cm

circle Radius

 

r 6cm

Let the 3

circle radius be .

 

Now, find the Area of the 1

circle.

Area πr

 

π 8

64π

Now, find the Area of the 2

circle.

Area πr

 

π 6

36π

Now, find the Area of the 3

circle.

Area πr

According to the statement,

rd st nd

Area of Area of Area of

3 circle 1 circle 2 circle

     

 

     

     

2 2 2

1 2

πr πr πr 

πr 64π 36π 

πr 100π

r 100

r 10 

But, radius cannot be negative.

Thus, the radius of the circle is cm.

Question: 3

Given figure depicts an archery target marked with

its five scoring areas from the centre outwards as

Gold, Red, Blue, Black and White. The diameter of

the region representing Gold score is 21 cm and

each of the other bands is 10.5 cm wide. Find the

area of each of the five scoring regions.

[Use ]



Fig. Exc_12.1_3 (Ques.)

Solution:

Fig. Exc_12.1_3 (Ans.)

First, find the 1

circle Radius (gold region)

(r )



10.5cm

According to the statement,

Each circle is 10.5 cm wider than the previous circle.

Thus,

circle radius

(r )

10.5cm 10.5cm 

21cm

circle Radius

 

21cm 10.5cm 

31.5cm

circle Radius

 

31.5cm 10.5cm 

42cm

circle Radius

 

42cm 10.5cm 

52.5cm

Area of gold region Area of 1 circle

πr

 

10.5 

10 5 0 5

.1  

346.5

nd st

Area of Area of

Area of

red region

circle circle

   

 

 

   

 

 

   

2 1

2 2

2 1

πr πr 

2 2

π(21) π(10.5) 

441π 110.25π 

330.75

 

1039.5cm

rd nd

Area of Area of

Area of

blue region

3 circle circle2

   

 

 

   

 

 

   

2 2

3 1

πr πr 

2 2

π(31.5) π(21) 

992.25π 441π 

551.25

 

1732.5cm

th rd

Area of Area of

Area of

black region

4 circle 3 circle

   

 

 

   

 

 

   

2 2

4 3

πr πr 

2 2

π(42) π(31.5) 

1764π 992.25π 

771.75

 

2425.5cm

th th

Area of Area of

Area of

whiteregion

5 circle 4 circle

   

 

 

   

 

 

   

2 2

5 4

πr πr 

2 2

π(52.5) π(42) 

2756.25π 1764π 

992.25

 

3118.5cm

Thus, the areas of gold, red, blue, black and white

regions are

346 5cm ,

1039 5cm ,

1732 5cm ,

and respectively.

2425 5cm

3118 5cm

Question: 4

The wheels of a car are of diameter 80 cm each.

How many complete revolutions does each wheel

make in 10 minutes when the car is travelling at a

speed of 66 km per hour?

Solution:

Diameter of car’s wheel

80cm

Radius of car’s wheel

40cm

Circumference of wheel

2πr

 

2π 40

80π

Speed of car

66 km / hour

66 100000

cm / min





110000 cm / min

Car’s travel distance in 10 minutes

110000 10 

1100000cm

Let be the number of revolutions of the wheel of

the car.

 

n Distance travelled

Distance travelled

in 1 revolution

in 10 minutes

that is, circumference

 



 

 



 

 

 

 

n 80π 1100000 

1100000 7

80 22







4375

Thus, each wheel of the car will make 4375

revolutions.

Question: 5

Tick the correct answer in the following and justify

your choice: If the perimeter and the area of a circle

are numerically equal, then the radius of the circle

is:

(A) 2 units

(B) units

(D) 7 units

Solution:

(A)

Let be the radius of the circle.

We know that,

Circumference of circle

2πr

Area of circle

πr

Given: The area of circle & the circumference of

circle are equal.

2πr πr

2 r

Thus, the radius of the circle is 2 units.

Exercise 12.2 (14)

Question: 1

Find the area of a sector of a circle with radius 6 cm

if angle of the sector is .

60

Solution:

Fig. Exc_12.2_1

Assume OPQR be a sector of the circle making

angle at point O in the circle.

60

Area of sector of angle θ πr

360





Now, calculate the area of the sector OPQR.

60 22

Area of sector OPQR (6)

360 7



  



1 22

6 6

6 7

   

132



Thus, the area of the sector of the circle is .

132

Question: 2

Find the area of a quadrant of a circle whose

circumference is 22 cm.

Solution:

Fig. Exc_12.2_2

Let be the radius of the circle.

Circumferenceof circle 22cm

2πr 22

2π



Circle quadrant will subtend angle at the centre

90

of the circle.

Area of such circle quadrant= πr

360



1 11

4 π

 

  

 

 

121





121 7

4 22









Thus, the area of the quadrant is .

Question: 3

The length of the minute hand of a clock is 14 cm.

Find the area swept by the minute hand in 5

minutes.

Solution:

Fig. Exc_12.2_3

In 1 hour (60 minutes), the minute hand rotates

360

In 5 minutes, minute hand will rotate

360



 

30 

Thus, in 5 minutes, area covered by the minute

hand, will be the equal to the area of a sector of

30

in a circle whose radius is 14 cm.

Area of sector of angle θ r

360

  



30 22

Area of sector of 30 14 14

360 7



    



2 14

  

11 14





154



Thus, the area is .

154

Question: 4

A chord of a circle of radius 10 cm subtends a right

angle at the centre. Find the area of the

corresponding:

(i) Minor segment

(ii) Major sector

[Use ]

π 3 4.1

Solution:

Fig. Exc_12.2_4

Let PR be the chord of the circle subtending at

90

the centre O of the circle.

Firstly, we will calculate the major sector.

(ii) Find the area of the major sector OPSR.

Area of major

sector OPSR

360 90

π r

360

 

  

 

  

 

 



 

 

270

π r

360



 

  

 



 

3 14 10 10

.   

235.5cm

Now, calculate the area of the minor sector OPQR.

Area of minor

secto

π r

OP R

 





 

  

 



 



 

3 14 10 10

.   

78.5cm

Now, find the area of .

OPR

Area of OPR OP OR

   

10 10

  

50cm

(i) Calculate the area of the minor segment PQR.

Area of minor Area of minor Area of

segment PQR sector OPQR OPR

     

 

     



     

78.5 50 

28.5cm

Question: 5

In a circle of radius 21 cm, an arc subtends an angle

of at the centre. Find:

60

(i) The length of the arc

(ii) Area of the sector formed by the arc

(iii) Area of the segment forced by the

corresponding chord

Solution:

Radius of circle

 

21cm

Angle subtends by the given arc

60 

Fig. Exc_12.2_5

(i) Find the length of the arc EFG.

Length of the

arc E

2πr

360

 



 

 

 

60 22

2 21

360 7



   



2 22 3

   

22cm

(ii) Find the area of the arc IEFG.

Area of the

arc IEF

πr

 





 



 

1 22

21 21

6 7

   

231cm

In ,

IEG

IEG IGE (As IE IG)   

IEG EIG IGE 180      

2 IEG 60 180    

IEG 60  

Thus, is an equilateral triangle.

IEG

So,

Area of IEG (Side)

  

(21)

 

441 3



(iii) Now, find the area of segment EFG.

Area of Area of Area of

segment EFG segment IEFG IEG

     

 

     



     

441 3

231 cm

 

 

 

 

Question: 6

A chord of a circle of radius cm subtends an

angle of at the centre. Find the areas of the

60

corresponding minor and major segments of the

circle.

Useπ 3 14and 3 1 73. .

 

 

 

Solution:

Fig. Exc_12.2_6

Let be the radius of circle.

Radius

15cm

Area of

segment HAC

π r

360

 



 





 





3.14 (15)

  

117.75cm

In ,

HAC

HAC HCA (As HA HC)   

HAC HCA AHC 180      

HAC HAC 60 180      

2 HAC 120  

HAC 60  

Thus, is an equilateral triangle.

HAC

We know that,

Area of HAC (side)

  

(15)

 

225 3



56.25 3 cm

97.3125cm

Area of Area of Area of

sector ACB sector HABC sector HAC

     

 

     



     

117.75 97.3125 

20.4375cm

Area of major Area of Area of

sector ADC circle sector ABC

     

 

     

     

π(15) 20.4375 

3.14 225 20.4375  

706.5 20.4375 

686.0625cm

Question: 7

A chord of a circle of radius 12 cm subtends an

angle of 120° at the centre. Find the area of the

corresponding segment of the circle.

Useπ 3 14and 3 1 73. .

 

 

 

Solution:

Fig. Exc_12.2_7

Construction: Draw a perpendicular DF on chord

AC.

This bisects the chord AC in two parts.

AF FC

DFA,

cos60

 

DF 1

12 2



DF 6cm

sin60





Given: and

AD 12cm

sin60

 

AF 3

12 2



AF 6 3 cm

AC 2AF

2 6 3 

12 3 cm

Now, find the area of .

DAC

Area of DAC AC DF

   

12 3 6

  

36 3

36 1.73 

62.28cm

Now, find the area of sector DABC.

120

Area of sector DABC π r

360



  



120

3.14 (12)

360



  



3.14 144

  

150.72cm

Area of Area of Area of

sector ABC sector DAFC DAC

     

 

     



     

150.72 62.28 

88.44 cm

Hence, the area of the corresponding segment of

the circle is .

88.44 cm

Question: 8

A horse is tied to a peg at one corner of a square

shaped grass field of side 15 m by means of a 5 m

long rope (see the given figure). Find:

(i) The area of that part of the field in which the

horse can graze.

(ii) The increase in the grazing area of the rope

were 10 m long instead of 5 m.

[Use ]

π 3 14 

Fig. Exc_12.2_8 (Ques.)

Solution:

Fig. Exc_12.2_8 (Ans.)

From the figure, observe that the horse can graze a

sector of in a circle of 5 m radius.

90

(i) Area of that part of the field, in which the

horse can graze area of sector OACB



Area of sector OACB

πr

360







3.14 (5)

  

3.14

 

19.625m

Area of that part of the field, in which the

horse can graze, if the length of the rope is 10

πr

360







3.14 (10)

  

78.5m

(ii) Now, calculate the increase in the grazing area

of the rope.

Increase in the

grazing

19.625

 

 

 

 

58.875m

Question: 9

A brooch is made with silver wire in the form of a

circle with diameter 35 mm. The wire is also used in

making 5 diameters which divide the circle into 10

equal sectors as shown in figure. Find.

(i) The total length of the silver wire required.

(ii) The area of each sector of the brooch.

Fig. Exc_12.2_9

Solution:

Radius of circle



17.5mm

Circumference of brooch

2πr

22 35

7 2

 

  

 

 

110mm

(i) Length of the silver wire required

110 5 35  

285mm

Observe, that each 10 sectors of the circle are

subtending at the center of the circle.

36

(ii) Now, find the area of each sector of the

brooch.

Area of each sector πr

360



 



1 22 35

10 7 2

 

  

 

 

 

1 22

17.5

10 7

  

96.25mm

Thus, the area of each sector of the brooch is

96.25mm

Question: 10

An umbrella has 8 ribs which are equally spaced

(see figure). Assuming umbrella to be a flat circle of

radius 45 cm, find the area between the two

consecutive ribs of the umbrella.

Fig. Exc_12.2_10

Solution:

Ribs in Umbrella

8

Angle between two consecutive ribs

360





45 

Now, calculate the area.

Area between

two consecutive

ribs o

π r

f cir

360

cle

 



 

  

 



 

 

 

1 22

8 7

  

2025

 

2025

 

22275



795.54 cm

Thus, the area between two consecutive ribs of the

circle is .

795.54 cm

Question: 11

A car has two wipers which do not overlap. Each

wiper has blade of length 25 cm sweeping through

an angle of . Find the total area cleaned at each

115

sweep of the blades.

Solution:

Fig. Exc_12.2_11

From the figure, each blade of wiper will sweep an

area of in a circle of 25 cm radius.

115

115

Area of such sector π r

360



  



 

115 22

360 7



  



23 22

25 25

72 7

   

23 22

25 25

72 7

   

627.48cm

Area swept by 2 bl

158

ade

2 2

 

158125

126



1254.9603cm

Thus, the area swept by 2 blades is .

1254.9603cm

Question: 12

To warn ships for underwater rocks, a lighthouse

spreads a red coloured light over a sector of angle

to a distance of 16.5 km. Find the area of the

80

sea over which the ships warned. [Use ]

π 3 14 

Solution:

Fig. Exc_12.2_12

The above figure shows that the lighthouse spreads

light across a sector of in a circle of 16.5 km

80

radius.

Area of sector OPQR r

360

  



 

3.14 16.5

360



  



 

3.14 16.5

  

189.97km

Thus, the area of the sector is .

189.97km

Question: 13

A round table cover has six equal designs as shown

in figure. If the radius of the cover is 28 cm, find the

cost of making the designs at the rate of Rs. 0.35 per

. [Use ]

3 1 3.7

Fig. Exc_12.2_13 (Ques.)

Solution:

Fig. Exc_12.2_13 (Ans.)

Observe that these designs are the segments of the

circle.

Take segment PQR.

Chord PR is one side of the hexagon.

Substitute every chord at the center of

360



 

the circle.

OPR,

OPR ORP (As OP OR)   

POR 60  

OPR ORP POR 180      

2 OPR 180 60 120      

OPR 60  

Thus,

OPR is an equilateral triangle.

 

Area of OPR side

  

 

 

784

 

196 3

196 1.7 

333.2cm

Area of sector OPQR r

360



  



 

60 22

360 7



  



1 22

28 28

6 7

   

410 67cm 

Area of Area of Area of

sector PQR sector OPQR OPR

     

 

     



     

 

410 667 333.2  

77.467cm

Area of designs

77 76 .46 

464 8cm 

Making cost of designs

1cm

Rs 0.35

Making cost of designs

464.8cm

464.8 0.35 

Rs.162.68

Thus, the making cost of such designs is Rs. .

162.68

Question: 14

Tick the correct answer in the following: Area of a

sector of angle (in degrees) of a circle with radius

is:

(A)

2πR

180



(B)

2πR

180



(C)

2πR

360



(D)

2πR

720



Solution:

(D)

Fig. Exc_12.2_14

The area of the sector of angle

θ πR

360

 



 

Area of sector of angle p πR

360

 



 



 

 

2πR

720

 



 

 

Thus, the correct option is (D).

Exercise 12.3 (16)

Question: 1

Find the area of the shaded region in the given

figure, if PQ = 24 cm, PR = 7 cm and O is the centre

of the circle.

Fig. Exc_12.3_1

Solution:

From the figure, observe that diameter of the circle

is RQ.

By applying Pythagoras theorem in PQR,

2 2 2

RQ RP PQ 

   

2 2

RQ 7 24 

RQ 49 576 

RQ 625

RQ 25

Radius of circle , OR .

 

RQ 25

12.5

2 2

  

We know that, diameter of a circle divides circle in

two equal parts, so RQ is the diameter of the circle

that divides it in two equal parts.

Area of semi-circle RPQOR π r

  

1 22 25

2 7 2

 

  

 

 

1 22 625

2 7 4

  

6875



245.54 cm

Area of PQR PQ PR

   

24 7

  

84 cm

Area of Area of semi-Circle Area of

shadedregion RPQOR PQR

     

 

     



     

6875

 

6875 2352





4523



161.54 cm

Thus, the area of the shaded region is .

161.54 cm

Question: 2

Find the area of the shaded region in given figure, if

radii of the two concentric circles with centre O are

7 cm and 14 cm respectively and .

AOC 40  

Fig. Exc_12.3_2 (Ques.)

Solution:

Inner circle radius = 7 cm

Outer circle radius = 14 cm

Fig. Exc_12.3_2 (Ans.)

Area of Area of Area of

shadedregion sector OAQC sector OBPD

     

 

     

     

   

2 2

40 40

π 14 π 7

360 360



     

 

1 22 1 22

14 14 7 7

9 7 9 7

       

616 154

9 9

 

462



51.33cm

Thus, the area of the shaded region is .

51.33cm

Question: 3

Find the area of the shaded region in the given

figure, if ABCD is a square of side 14 cm and APD

and BPC are semicircles.

Fig. Exc_12.3_3

Solution:

From the given figure, observe that the radius of

every semi-circle is 7 cm.

Area of each semi circle πr



 

1 22

2 7

  

11 7 

77cm

Now, find the area of square.

 

Area of square ABCD 14

196cm

Area of Area of Area of Area of

shadedregion square ABCD semi-circleAPD semi-circleBPC

       

  

       

       

196 77 77  

42cm

Thus, the area of the shaded region is .

42cm

Question: 4

Find the area of the shaded region in the given

figure, where a circular arc of radius 6 cm has been

drawn with vertex O of an equilateral triangle OAB

of side 12 cm as centre.

Fig. Exc_12.3_4 (Ques.)

Solution:

Each interior angle of an equilateral triangle is of

60

Fig. Exc_12.3_4 (Ans.)

Area of segment OCDE r

360



 



 

60 22

360 7



  



1 22

6 7

  

132



 

Area of AOB 12

 

3 12 12

 



36 3

Area of circle πr

 



792



Area of Area of Area of Area of

shadedregion OAB Circle sector OCDE

       

  

       



       

792 132

36 3

7 7

  

660

36 3 cm

 

 

 

 

Thus, the area of the shaded region is

660

36 3 cm

 



 

 

Question: 5

From each corner of a square of side 4 cm a

quadrant of a circle of radius 1 cm is cut and also a

circle of diameter 2 cm is cut as shown in the given

figure. Find the area of the remaining portion of the

square.

Fig. Exc_12.3_5 (Ques.)

Solution:

Fig. Exc_12.3_5 (Ans.)

Each quadrant is a sector of in a circle having

90

radius of 1 cm.

Area of each quadrant

πr

360







 

1 22

4 7

  



 

Area of square PQRS 4

16cm

Area of Area of Area of Area of

shadedregion square circle quadrant

       

   

       

       

22 22

16 4

7 28

   

22 22

7 7

  

 

112 44





Thus, the area of the shaded portion is .

Question: 6

In a circular table cover of radius 32 cm, a design is

formed leaving an equilateral triangle ABC in the

middle as shown in the given figure. Find the area

of the design (shaded region).

Fig. Exc_12.3_6 (Ques.)

Solution:

Fig. Exc_12.3_6 (Ans.)

Radius of circle

 

32cm

In , AP is the median.

ABC

AP 32

 

48cm

In ,

ABP

   

2 2

AB AP BP 

 

 

 

 

 

 

3AB



That is,

 

3AB



48 2





32 3 cm

Now, calculate the area of the equilateral triangle.

 

Area of equilateral

triangle ABC

 



 



 

 

32 3



32 32 3

   

96 8 3  

768 3 cm

Area of circle πr

 

 

22528



Area of Area of Area of

–

design circle ABC

     



     



     

22528

768 3 cm

 

 

 

 

Thus, the area of the design is

22528

768 3 cm

 



 

 

Question: 7

In the adjoining figure, ABCD is a square of side 14

cm. With centres A, B, C and D, four circles are

drawn such that each circle touch externally two of

the remaining three circles. Find the area of the

shaded region.

Fig. Exc_12.3_7 (Ques.)

Solution:

Fig. Exc_12.3_7 (Ans.)

In the figure, area of each sector is equal & in a

circle the sector is of having radius of 7 cm.

90

Area of each sector πr

360



 



 

90 22

360 7



  



1 22

4 7

  

38.5cm

 

Area of squareABCD Side

 

14

196cm

Area of Area of Area of

shadedregion square ABCD each sector

     

  

     

     

196 4

  

196 154 

42cm

Thus, the area of the shaded portion is .

42cm

Question: 8

The given figure depicts a racing track whose left

and right ends are semi-circular.

The distance between the two inner parallel line

segments is 60 m and they are each 106 m long. If

the track is 10 m wide, find:

(i) the distance around the track along its inner

edge.

(ii) the area of the track.

Fig. Exc_12.3_8 (Ques.)

Solution:

Fig. Exc_12.3_8 (Ans.)

Distance around the track

AB arcBEC CD arcDFA

along its inner edge

 

   

 

 

1 1

106 2πr 106 2πr

2 2

     

1 22 1 22

212 2 30 2 30

2 7 2 7

        

212 2 30

   

212 188.57 

400.57m

Area of the track = (Area of GHIJ − Area of ABCD)

+ (Area of semi-circle HKI − Area of semi-circle

BEC) + (Area of semi-circle GLJ − Area of semi-

circle AFD)

   

2 2

106 80 106 60

1 22 1 22

40 30

2 7 2 7

1 22 1 22

40 30

2 7 2 7

 

 

  

 

      

 

     

 

 

     

2 2

22 22

106 80 60 40 30

7 7

     

     

2 2

106 20 40 30

 

  

 

  

2120 10 70

 

 

 

 

2120 2200 

4320m

Thus, the area of the track is 4320 m

Question: 9

In the given figure, AB and CD are two diameters of

a circle (with centre O) perpendicular to each other

and OD is the diameter of the smaller circle. If OA =

7 cm, find the area of the shaded region.

Fig. Exc_12.3_9 (Ques.)

Solution:

Fig. Exc_12.3_9 (Ans.)

From the figure, observe that,

Radius of larger circle = 7 cm

 

Radius of smaller circle =

 

Area of smaller circle πr

22 7 7

7 2 2

  



38.5cm

Area of semi-circle AECFB πr



 

1 22

2 7

  

 

77cm

Area of ABC AB OC

   

14 7

  

49cm

Area of

smaller Circle

Area of Area of

shadedregion semi-circleAECFB

Area of

ABC

 

 

 

 

   

 

 

   

 

   

 

 

 



 



 

77 49

  

 

38.5 28 

66.5cm

Thus, the area of the shaded region is 66.5 cm

Question: 10

The area of an equilateral triangle ABC is 17320.5

. With each vertex of the triangle as centre, a

circle is drawn with radius equal to half the length

of the side of the triangle. Find the area of the

shaded region. (Use and )

π 3 4.1

3 1 73205 

Fig. Exc_12.3_10 (Ques.)

Solution:

Let be the side of an equilateral triangle.

Area of equilateral triangle is 17320.5

Area of

equilateral triangl

 



 

 

17320.5 a



1.73205

17320.5 a



a 4 10000 

a 200cm

Fig. Exc_12.3_10 (Ans.)

Each segment is a measure of 60°.

Area of

π r

segment ADEF

360

 



  

 



 

 

= π 100

 

3.14 10000





5233.33cm 

Area of Area of Equilateral Area of each

shadedregion triangle segment

     

 

     

     

17320.5 3 5233.33  

17320.5 15700 

1620.5cm

Thus, the area of the shaded region is 1620.5 cm

Question: 11

On a square handkerchief, nine circular designs

each of radius 7 cm are made. Find the area of the

remaining portion of the handkerchief.

Fig. Exc_12.3_11 (Ques.)

Solution:

Fig. Exc_12.3_11 (Ans.)

From the figure, observe that the side of the square

is 42 cm.

 

Area of square Side

 

42

1764 cm

Area of each circle πr

 

 

154 cm

Area of 9 circles 9 154 

1386cm

Handkerchief remaining portion area:

Area of remaining portion

1764 1386

of handkerchief

 

 

 

 

378cm

Thus, the area of the shaded region is 378 cm

Question: 12

In the given figure, OACB is a quadrant of a circle

with centre O and radius 3.5 cm. If OD = 2 cm, find

the area of the

(i) quadrant OACB

(ii) shaded region

Fig. Exc_12.3_12 (Ques.)

Solution:

Fig. Exc_12.3_12 (Ans.)

(i) Since OACB is a quadrant, it will subtend 90°

angle.

area:

quadrant OACB

Area of

πr

360

 





 



 

 

1 22

3.5

4 7

  

1 22 7

4 7 2

 

  

 

 

11 7 7

2 7 2 2

 



  



(ii)

Area of OBD OB OD

   

3.5 2

  



Area of Area of Area of

shadedregion quadrant OACB OBD

     

 

     



     

77 7

8 2

 

77 28





Thus, the area of the shaded region is .

Question: 13

In the given figure, a square OABC is inscribed in a

quadrant OPBQ. If OA = 20 cm, find the area of the

shaded region. (Use )

π 3 4.1

Fig. Exc_12.3_13 (Ques.)

Solution:

Fig. Exc_12.3_13 (Ans.)

In ,

OAB

2 2 2

OB OA AB 

   

2 2

OB 20 20 

OB 400 400 

OB 20 2

Radius of circle

20 2 cm

Area of quadrant OPBQ π r

360



  



 

3.14 20 2

360



  



3.14 800

  

628cm

Area of Area of Area of

shadedregion quadrant OPBQ OABC

     

 

     

     

 

628 400 

288cm

Thus, the area of the shaded region is 288 cm

Question: 14

AB and CD are respectively arcs of two concentric

circles of radii 21 cm and 7 cm and centre O. If



AOB = 30°, find the area of the shaded region.

Fig. Exc_12.3_14 (Ques.)

Solution:

Fig. Exc_12.3_14 (Ans.)

Area of Area of Area of

shadedregion segment OAPB segment OCQD

     

 

     

     

   

2 2

30 30

π 21 π 7

360 360

 

     

 

   

2 2

π 21 7

360



 

   

 



  

1 22

21 7 21 7

12 7

   

 

 

22 14 28

12 7

 





308



Thus, the area of the shaded region is .

308

Question: 15

In the given figure, ABC is a quadrant of a circle of

radius 14 cm and a semicircle is drawn with BC as

diameter. Find the area of the shaded region.

Fig. Exc_12.3_15 (Ques.)

Solution:

Fig. Exc_12.3_15 (Ans.)

From the figure, observe that ABC is the quadrant

of the circle, is right-angle.

BAC

In ,

ABC

2 2 2

BC AC AB 

   

2 2

BC 14 14 

BC 14 2

 

14 2

Radius r of semi-circle drawn on BC



7 2 cm

Area of ABC AB AC

   

14 14

  

98cm

Area of sector ABCD π r

360



  



90 22

360 7



  



1 22

196

4 7

  

154 cm

Area of semi-circle

drawn

on B

 

  

 

 

 

1 22

7 2

2 7

  

1 22

2 7

  

154 cm

Area of

Semi-circle

Area of Area of

shadedregion segment ABCD

Area of

ABC

 

 

 

 

 

   

 



 

 

 

   

 



 

 

 



 

 



 

 

 

 

154 154 98  

98cm

Thus, the area of the shaded region is 98 cm

Question: 16

Calculate the area of the designed region in the

given figure common between the two quadrants of

circles of radius 8 cm each.

Fig. Exc_12.3_16 (Ques.)

Solution:

Fig. Exc_12.3_16 (Ans.)

Find the designed area between two segments

BAEC & DAFC.

Segment BAEC area:

Area of

π r

segment BAEC

360

 



  

 



 

90 22

360 7



  



1 22

4 7

  

352



Area of BCA BA BC

   

8 8

  

32cm

Designed region area:

Area of Area of

designedregion segment AEC

   

 

   

   

Area of Area of

segment BAEC BAC

 

   

 

 

   



   

 

352

2 32

 

  

 

 

352 224



 

 

 

 

2 128





256



Thus, the area of the designed region is .

256