12 AREAS RELATED TO CIRCLES CHAPTER

CHAPTER
12
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AREAS RELATED TO CIRCLES
Points to Remember :
1. A circle is a collection of points which moves in a plane in such a way that its distance from a fixed point
always remains the same. The fixed point is called the centre and the fixed distance is known as radius
of the circle.
2. Area and circumference of a circle : If r is the radius of the circle, then
(i) circumference = 2r = d where d = 2r (diameter)
(ii) Area = r 2 
d 2
4
(iii) Area of a semicircle 
1 2
r
2
1 2
(iv) Area of a quadrant  r
4
B
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3. Area of a circular ring : If R and r (R > r) are radii of two concentric circles, then area enclosed by the
two circles.
=  (R2 – r2)
4. Number of revolutions completed by a rotating wheel

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Distance moved
circumference
5. If a sector of a circle of a radius r contains an angle of . Then,
(i) length of the arc of the sector 
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(ii) Perimeter of the sector  2r 
(iii) Area of the sector 

 2 r
360
O

2 r
360

 r 2
360
r

r
A
B
(iv) Area of the segment
= Area of the corresponding segment – Area of the corresponding triangle.


1



r 2  r 2 sin , or
r 2  r 2 sin cos
360
2
360
2
2
6. (i) In a clock, a minute hand rotates through an angle of 6° in one minute.
 1 
(ii) In a clock, an hour hand rotates through an angle of   in one minute.
2
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AREAS RELATED TO CIRCLES
MATHEMATICS–X
ILLUSTRATIVE EXAMPLES
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Example 1. In the given figure, an archery target marked with its five scoring areas from the centre outwards
as Gold, Red, Blue, Black and White. The diameters 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.
[NCERT]
White
Black
Blue
Red
Gold
Solution.
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here,
r
= radius of the region representing Gold score = 10.5 cm.
r1 = radius of the region representing Gold and Red scoring areas = 10.5 cm + 10.5 cm = 21 cm
= 2r cm
r2 = radius of the region representing Gold, Red and Blue scoring areas = 21 cm + 10.5 cm =
31.5 cm = 3 r cm.
r3 = radius of the region representing Gold, Red, Blue and Black scoring areas = 31.5 cm + 10.5
cm = 42 cm = 4r cm.
r4 = radius of the region representing Gold, Red, Blue, Black and white scoring areas
= 42 cm + 10.5 cm = 52.5 cm = 5r cm.
Now,
A 1 = Area of the region representing Gold scoring area
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B
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= r   (10.5)2  346.5 cm 2
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A 2 = Area of the region representing Red Scoring area
=  (2r)2 – r2 = 32 = 3A1 = 3 × 346.5 cm2 = 1039.5 cm2
A 3 = Area of the region representing Blue Scoring area
=  (3r)2 – (2r)2 = 52 = 5A1 = 5 × 346.5 cm2 = 1732.5 cm2
A 4 = Area of the region representing Black Scoring area
=  (4r)2 – (3r)2 = 7r2 = 7A1 = 7 × 346.5 cm2 = 2425.5 cm2
A 5 = Area of the region representing White Scoring area
=  (5r)2 – (4r)2 = 9r2 = 9A1 = 9 × 346.5 cm2 = 3118.5 cm2
Example 2. 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 the speed of 66 km per hour?
[NCERT]
Solution. We have, Speed of the car = 66 km/hr
 Distance travelled by the car in 60 minutes = 66 km
2
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
Distance travelled by the car in 10 minutes  66  10 km
20
= 11 km = 1100000 cm
MATHEMATICS–X
AREAS RELATED TO CIRCLES
189
Also, diameter of car wheels = 80 cm




80
cm=40 cm ( r say)
2
22
Circumference of the wheels  2r  2   40 cm
7
radius of car wheels 
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Distance travelled by the car when its wheels take one complete revolution  2  22  40 cm
7
Number of revolutions made by the wheels in 10 minutes
=
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Distance covered by the car in 10 minutes
Distance covered by the car when its wheels make one complete revolution
1100000
 4375
22
2×  40
7
Hence, each wheel makes 4375 revolutions in 10 minutes.
=
Example 3. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find :
(i) the length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.
[NCERT]
Solution. Let O be the centre of the circle of radius 21 cm such that an arc APB subtends  = 60° at the centre.
(i) Length of the arc APB 
B

60
22
 2r 
 2   21 cm = 22 cm
360
360
7
cm
60°
cm
21
21
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O
B
P
(ii) Area of sector OAPB    r 2  60  22  21  21 cm 2 = 231 cm2
360
360 7
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(iii) Area of the segment APB   r 2  1 r 2 sin 
360
2
190
1
 

 r2 
  .sin  
2
 360

 60 22 1

 (21)2 . 
  sin 60 cm 2
 360 7 2

 11
3 2
 441  
cm
 21 4 


= 40.05 cm2 Ans.
AREAS RELATED TO CIRCLES
MATHEMATICS–X
Example 4. 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 figure). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in grazing area if the rope were 10 m long instead of 5m. (Use  = 3.14).
[NCERT]
Solution.
Here r = 5 m,  = 90°
(i) The area of the part AEFA of the field in
which horse can graze    r 2
360

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90
 3.14  5  5 m 2
360
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15 m
C
H
F
= 19.625 m2
(ii) Here R = 10 m,  = 90°
A
 Area of the part AGHA of the field in which horse can graze

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5m E
10m
G
B

90
 R 2 
 3.14  10 10 m 2  78.50 m 2
360
360

Increase in area (EGHF) when length of the rope increases from 5 m to 10 m
= 78.50 m2 – 19.625 m2 = 58.875 m2 Ans.
Example 5. 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 the given
figure. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
[NCERT]
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AREAS RELATED TO CIRCLES
191
Solution.
35
mm
2
Total length of wire used in making the silver brooch
= circumference of the brooch + 5 × diameter of the brooch
Here, 2r = 35 mm  r 
(i)
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= (2r + 5 × 2r) mm   22  35  5  35  mm  (110  175) mm = 285 mm
 7

r 2
10
[ The brooch is divided into 10 equal sectors]
(ii) The area of each sector of the brooch 
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22 35 35 1
385

   mm 2 
mm 2 = 96.25 mm 2 Ans.
7 2 2 10
4
Example 6. A round table cover has six equal designs as shown in the given 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 cm2 (use
Solution.
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Here, r = 28 cm and  = 60°
B
[ It is a regular hexagon]
 Required area = 6 × area of the segment ACB
= 6 × {area of the sector OACB – area of the equilateral OAB}
 0 22

3
 6 
  (28)2 
(28)2  cm 2
4

 360 7
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3  1.7).
[NCERT]
28
O
cm
60°
28
cm
B
C
A
= 2464 cm2 – 1999.2 cm2 = 464.8 cm2
 Cost of making the design = Rs. 464.8 × 0.35 = Rs. 162.68 Ans.
Example 7. In a circular table of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the
middle as shown in the figure. Find the area of the design (shaded region).
A
Solution.
B
C
In OBD, we have
OD
BD
 cos 60 and
 sin 60
OB
OB
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AREAS RELATED TO CIRCLES
MATHEMATICS–X

OD 1
DB
3

and

32 2
32
2

OD = 16 cm and BD = 16 3 cm
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O3
 BC = 2 BD = 32 3 cm
Area of the shaded region
= Area of the circle – Area of ABC
2°
c
60° 60° m
B


3
   (32) 2 
 (32 3) 2  cm 2
4


D
C
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 22528


 768 3  cm2 = 1889.65 cm 2 Ans.
 7

Example 8. The given figure depicts a racing track whose left and right ends arc semicircular. The distance
between the 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.
[NCERT]
S
R
10 m
D
60 m
30m
A
Solution.
(i)
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B
106 m
P
Distance around the track along its inner edge
C
30m
B
Q
= 2 × 106 m + Perimeter of two semi-circle of radius
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 212 m  2 
60
m
2
22
1320
2804
 30 m  212 m 
m
m = 400.57 m
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7
7
(ii) Area of the track
= 2 (area of the rectangle SRCD) + 2 (area of the semi-circular track)
1 22


 2 106 10    (402  302 )  m2
2 7


1 22


 2 1060    (40  30)(40  30)  m2
2
7


1 22


 2 1060    70  10  m2
2
7


= 2 (1060 + 1100) m2
= 2 (2160) m2
= 4320 m2 Ans.
MATHEMATICS–X
AREAS RELATED TO CIRCLES
193
Example 9. The area of an equilateral triangle ABC is 17320.5 cm2. 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  = 3.14 and
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–[NCERT]
3  1.73205].
A
60°
B
Solution.
C
We know, area of an equilateral triangle 

3
 (side) 2
4
3
 (side) 2  17320.5
4
4 17320.5
cm 2 
B
4 17320.5 2
cm  40000cm2
1.73205

(side) 2 

side = 200 cm

200
radius of each circle drawn 
cm  100 cm
2
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3
For each minor sector :  = 60°
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
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Area of each sector, inside the triangle


 r 2
360
 3.14  (100) 2 
60
cm 2
360
= 5233.33 cm2

Area of the triangle not included in the circles
= Area of triangle – 3 × (area of each sector inside the triangle)
= 17320.5 cm2 – 3 × 5233.33 cm2
= 1620.5 cm2 Ans.
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AREAS RELATED TO CIRCLES
MATHEMATICS–X
Example 10. Calculate the area of the designed region in the given figure common between two quadrants of
circles of radius 8 cm each.
8 cm
D
C
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90°
E
8 cm
8 cm
F
90°
A
Solution.
B
8 cm
Area of the designed region
= 2 (Area of quadrant ABED – Area of ABD)
1
1

 2   (8)2   8  8 cm 2
4
2


256 cm2 = 36.57 cm2 Ans.
 22 16

 2
 32 cm 2 
7
 7

PRACTICE EXERCISE
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Question based on Circumference and Area of a circle :
1. Find the area of the circle whose circumference is 22 cm.
2. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Find the length of
the side of the square.
3. The radius of the circle is 3 m. What is the circumference of another circle, whose area is 49 times that of
the first?
4. Two circles touch externally. The sum of their areas is 130  sq. cm and the distance between their
centres is 14 cm. Find the radii of the circles.
5. 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. [take  = 22/7]
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6. A wire when bent in the form of an equilateral triangle encloses an area of 121 3 cm2 . If the same wire
is bent in the form of a circle, find the area of the circle.
7. A road which is 7 m wide surrounds a circular track whose circumference is 352 m. Find the area of the
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22 

road.  use   
7

8. A race track is in the form of a ring whose inner circumference is 352 m and the outer circumference is 396
m. Find the width of the track.
9. The outer circumference of a circular racing track is 220 m. The track is everywhere 7 m wide. Calculate
22
the cost of levelling the track at the rate of 50 paisa per sq. m.  use   
7

10. The area enclosed between the two concentric circles is 770 cm2. If the radius of the outer circle is 21 cm,
calculate the radius of the inner circle.
11. Find the radius of a circle whose area is equal to the sum of the areas of the three circles whose radii are
3 cm, 4 cm, and 12 cm.
MATHEMATICS–X
AREAS RELATED TO CIRCLES
195
12. A garden roller has a circumference of 3 m. How many revolutions does it make in moving 21 m?
13. A bicycle wheel makes 5,000 revolutions in moving 11 km. Find the diameter of the wheel.
14. The diameter of the driving 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/hr?
15. A wheel of diameter 42 cm, makes 240 revolutions per minute. Find :
(i) the total distance covered by the wheel in one minute.
(ii) the speed of the wheel in km/hr.
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Question based on Area of sector and segment of a circle :
16.
17.
18.
19.
20.
21.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Find the area of the sector of a circle whose radius is 14 cm and angle of sector is 45°.
An arc of length 20  cm subtends an angle of 144° at the centre of the circle. Find the radius of the circle.
A sector of a circle of radius 8 cm contains an angle of 135°. Find the area of the sector.
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute
hand between 9:00 am and 9:35 am.
22. In the given figure, there are shown sectors of two concentric circles of radii 7 cm and 3.5 cm. Find the
22
area of the shaded region.  use   
7

B
B
D
cm
3 .5
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O
30°
7 cm
C
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cm
10
c
90°
10
M
A
m
A
23. The given figure shows a quadrant of a circle of radius 10 cm. Find the area of the shaded region and the
perimeter of the sector.
24. In the given figure, O is the centre of the circle of radius 9 cm and AOB = 150°. Find :
(i) the length of the major arc
22
(ii) the area of the major sector.  use   
7

196
A
O
150° 9 cm
B
AREAS RELATED TO CIRCLES
MATHEMATICS–X
25. An arc of a circle of radius 21 cm has a length of 17.6 cm. Find the angle subtended at the centre of the
circle.
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 7 
26. In the given figure, the length of the minor arc is   of the circumference of the circle. Find :
 24 
(i) AOB
(ii) If it is given that the circumference of the circle is 132 cm, find the length of the minor arc AB and the
radius of the circle.
O
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A
B
27. Find the area of a segment of a circle of radius 7 cm if the arc of the segment has measure 90°.
28. Find the area of the segment of a circle, given that the angle of the sector is 120° and the radius of the
22

circle is 21 cm.  Take  
and 3  1.73
7


B
29. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find :
(i) Area of the minor sector
(ii) Area of the minor segment
(iii) Area of major sector
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(iv) Area of major segment
 use   3.14
30. 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. [use  = 3.14,
3 = 1.73]
31. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an
angle of 115°. Find the total area cleaned at each sweep of the blades.
–[NCERT]
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32. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80°
to a distance of 16.5 km. Find the area of the sea over which the ships are warned. [use  = 3.14]
–[NCERT]
33. 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.
–[NCERT]
MATHEMATICS–X
AREAS RELATED TO CIRCLES
197
Questions based on combinations of Plane figures :
34. A square is inscribed in a circle of radius 10 cm. Find the area of the circle, not included in the square [use
 = 3.14]
35. In the given figure, ABC is an equilateral triangle inscribed in a circle of radius 4 cm. Find the area of the
shaded region.
A
O
C
B
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A
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36. The side of a square is 42 cm. Find the area of (i) the inscribed circle (ii) the circumscribed circle.
37. The following figure shows a rectangle ABCD inscribed in a circle.
D
C
O
B
B
A
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(i) If AB = 8 cm and BC = 6 cm, find the area of the circle not included in the rectangle.
(ii) If diameter of the circle is 25 cm and BC = 15 cm, find the area of the circle not included in the given
rectangle.
38. The given figure shows a square OABC, inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the
shaded region.
[NCERT]
P
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A
B
Q
O
C
39. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal, semi-circles are drawn
on PQ and QS as diameters as shown in figure. Find the perimeter and area of the shaded region.
P
Q
R
S
40. A paper is in the form of a rectangle ABCD in which AB = 20 cm and BC = 14 cm. A semi-circular portion
with BC as diameter is cut off. Find the area of the remaining part.
198
AREAS RELATED TO CIRCLES
MATHEMATICS–X
41. A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a
22
quadrant of radius 14 m as shown. Find the area of the remaining part of the park.  use   
7

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100 m
14 m
100 m
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42. Four equal circles are described about the four corners of a square so that each touches two of the
others as shown in figure. Find the area of the shaded region, if each side of the square is 14 cm.
D
C
A
B
B
43. ABCP is a quadrant of a circle of radius 14 cm. With AC as diameter, a semi-circle is drawn as shown in
figure. Find the area of the shaded portion.
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Q
A
14 cm
P
B
M
A
C
14 cm
44. In the following figure, ABC is an equilateral triangle. Circles are drawn with vertices of the triangle ABC
as centres so that every circle touches the remaining two. If the perimeter of the triangle ABC is 84 cm.
A
B
C
Find
(i) area of sector, inside the triangle, of each circle.
(ii) area of the triangle which is not included in the circle.
MATHEMATICS–X
[use  = 3.14, 3 = 1.73]
AREAS RELATED TO CIRCLES
199
45. 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.
B
D
J
A
C
O
A
46. In the given figure, find the area of the shaded region, where ABCD is a square of side 10 cm.
[use  = 3.14].
D
C
A
B
10 cm
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47. Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one
another (see figure). What area will be left ungrazed?
B
C
D
T
I
A
50 m
50 m
B
48. In the given figure, diameter of the biggest semi-circle is 108 cm, and diameter of the smallest circle is
36 cm, calculate the area of the shaded region.
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49. Four equal circles are described at the four corners of a square so that each touches two of the others.
The shaded area, enclosed between the circles is
24 2
cm . Find the radius of the each circle.
7
D
C
A
B
50. Three horses are tethered with 7 m long ropes at the three corners of a triangular field having sides 20 m,
34 m and 42 m. Find the area of the plot which can be grazed by the horses. Also, find the area of the plot
which remains ungrazed.
[CBSE 2001]
200
AREAS RELATED TO CIRCLES
MATHEMATICS–X
51. In an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion
of the triangle not included in the circle. [use 3  1.73 and   3.14]
52. In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as
diameter is drawn. If AC = 54 cm and BC = 10 cm, find the area of the shaded region.
[CBSE 2000]
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D
F
A
O
C
B
G
E
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53. In the given figure, ABCD is a rectangle with sides 4 cm and 8 cm. Taking 8 cm as the diameter, two semicircles are drawn. Find the area overlapped by the two semi-circles.
C
D
B
4 cm
A
8 cm
B
54. A circle has been inscribed in a square of side 4 cm. Determine the left out area. What will be the left out
area of the circle if a square is inscribed in the circle? [use  = 3.14]
T
I
55. Find the area of the shaded region shown in the 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.
M
A
O
B
A
56. In the given figure, ABCD is a square. AEB and DFC are arcs of a circle whose centre is the centre of the
square. If the area of the square is 49 cm2, find the area of the shaded region.
E
MATHEMATICS–X
A
B
O
C
D
F
AREAS RELATED TO CIRCLES
201
57. In the given figure, a piece of cardboard is given in the shape of a trapezium ABCD where AB || DC and
BCD = 90°. A quarter circle BFEC is removed from the cardboard. Given AB = BC = 3.5 cm and DE = 1.5
cm. Calculate the area of the remaining piece of the cardboard.
3.5 cm
A
D
1.5 cm
J
A
B
F
3.5 cm
E
C
J
A
58. In the given figure, there are three semi-circles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm. Calculate :
(i) the area of the shaded region.
(ii) the cost of painting the shaded region at the rate of 50 paisa per cm2, to the nearest Rs.
E
4.5 cm
D
A
B
C
3 cm
B
3 cm
T
I
3 cm
59. Find the perimeter and area of the shaded region whose particulars are given below in each of the
following :
14 cm
28 cm
O 28 cm
(iii)
(ii)
14 cm
3.5 cm
10 cm
M
A
(i)
(iv)
28 cm
202
10 cm
AREAS RELATED TO CIRCLES
MATHEMATICS–X
(vi)
10 cm
8 cm
10 cm
J
A
12 cm
(viii)
T
I
7 cm
6 cm
(ix)
14 cm
M
A
B
(x)
6.5 cm
4 cm
14 cm
33 cm
7 cm
4 cm
(vii)
18 cm
33 cm
18 cm
J
A
10 cm
8 cm
(v)
8 cm
3.5 cm
(i)
56
60. Find the area of the shaded parts in the following figures, all dimensions being in cm.
3.5
3.5
3.5
(ii)
3.5
56
MATHEMATICS–X
AREAS RELATED TO CIRCLES
203
(iii)
J
A
14
(iv)
14
14
14
48
A
5
(v)
(vi)
B
D
given, area of ABC = 25 cm2
(vii) A
C
D
C
B
T
I
14
M
A
14
B
(viii)
given, AC = CD = DB = 7 cm
(ix)
21
14
J
A
28
56
(x)
here, radius of each circle = 7 cm
HINTS TO SELECTED QUESTIONS
7. Circumference of circular track = 2 r
7 cm
Given, 2 r = 353 m  r = 56 m
Area of road  [  (56  7) 2  (56) 2 ]m 2
2
2
2
 [(63)  (56) ]m  2618 m
204
r
2
AREAS RELATED TO CIRCLES
MATHEMATICS–X
22
 21 cm = 132 cm = 1.32 m
7
Now, no. of rounds made in 1 minute = 240
 The total distance covered by wheel in 1 minute = 1.32 × 240 m = 316.8 m
15. Distance covered in 1 round  2r  2 
J
A
distance 316.8

m/s = 5.28 m/s = 19.008 km/hr
time
60
21. Angle destriced by the minute hand in 1 minute = 6°
 Angle described by the minute hand in 35 minutes = (6 × 35)° = 210°
 Area swept by the minute hand in 35 minutes
Now, speed of the wheel 

210
 (10) 2 cm 2  183.3 cm 2
360
J
A

7
7
 2r 
 2 r   
 360  105
360
24
24
Now, 2r = 132  r = 21 cm
26. Let AOB = , then
Also, length of minor arc 
7
 132 cm = 38.5 cm
24
B
33. Since ribs are equally spaced.  Angle made by two consecutive ribs at the centre  360  45
8
Thus, area between two consecutive ribs
= Area of a sector of a circle of radius 45 cm and sector angle 45°

45 22
  45  45 cm 2  795.53 cm 2
360 7
T
I
35. Clearly AD  BC. Also, AD is median of ABC, and O is the centroid.
 AO : AD = 2 : 1  4 : OD = 2 : 1
 OD = 2 cm.
A
Now, In ODB, BD  OB 2  OD 2  4 2  2 2 cm  12 cm

4 cm
O
BC  2BD  2 12 cm
M
A
B
3
(2 12)2 cm 2  12 3 cm 2
 Area of ABC 
4
Required Area = Area of circle – Area of triangle
D
C
=  (4)2 cm2 – 12 3 cm2 = (16  – 12 3 ) cm2
37. (i) Join A to C. Here AC  AB2  BC 2  82  6 2 cm  10 cm
Clearly, AC is diameter of the circle.  Radius of the circle 
10
cm  5 cm
2
 Required Area = Area of circle – Area of rectangle
 (5) 2 cm 2  8  6 cm 2  (25   48) cm 2 = 30.55 cm2
39. here, PQ = QR = RS  12 cm  4 cm.
3
QS = QR + RS = 4 cm + 4 cm= 8 cm.
MATHEMATICS–X
AREAS RELATED TO CIRCLES
205
 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
=  × (6 + 4 + 2) cm = 12  cm.
Also, Required area = Area of semi-circle with PS as diameter + Area of semi-circle with PQ as diameter
– Area of semi-circle with QS as diameter
J
A
1 22
   [62  2 2  4 2 ] cm 2  37.71 cm 2
2 7
43. In right ABC, AC  142  142 cm  14 2 cm 
J
A
AC 14 2

cm  7 2 cm
2
2
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 circle with
AB as radius – Area of ABC)
B
1
1
1

 (7 2) 2 cm 2   (14)2 cm 2   14  14 cm 2 
2
2
4

= 98 cm2
46. Let us mark the four unshaded regions as I, II, III and IV.
Now, Area of I + Area of III
= Area of ABCD – Area of two semi-circles of radius 5 cm each.
T
I
1


 10 10  2    (5)2  cm 2  21.5 cm 2
2


D
I
II
IV
III
A
Similarly, Area of II + Area of IV = 21.5 cm2
10 cm
So, Area of shaded portion = Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm2 = 57 cm2
48. Clearly, Required Area
= Area of semi-circle of radius 54 cm –
[2× Area of semi-circle of radius 27 cm + Area of circle of radius 18 cm]
M
A

C
B
1
 1

(54)2 cm2   2  (27)2  (18) 2  cm2  1272.86 cm2
2
 2

49. Clearly, Area of square – Area of 4 sectors  24 cm 2
7


206
(2r ) 2  4 
4r 2 
90 2 24
r 
360
7
22 2 24
r 
 r 2  4  r  2 cm .
7
7
AREAS RELATED TO CIRCLES
MATHEMATICS–X
50. Area which can be grazed = sum of areas of 3 sectors
A
7
r 2  r 2 . r 2 .



, where r = 7 m
360 360 360

m
r 2
r 2
1
(     ) 
180  r 2
360
360
2

1 22
  (7) 2 m 2  77 m 2
2 7
Now, Area of plot = Area of ABC = 336 m2
 Area ungrazed = 336 m2 – 77 m2 = 259 m2.
53. In RMQ,
RM 2 1
     60
RQ 4 2
 PRQ = 120°
Area overlapped by two semi-circles
= 2 (Area of segment PSQ)
= 2 (Area of sector RPSQ – Area of RPQ)
B
7m
16  1

 2
 (2  RQ sin 60)  2 cm 2
2
 3

T
I
J
A
[ using heron’s formula]
cos  
1
 120

 2
 (4)2   PQ  RM  cm 2
360

2


C
7m


J
A

R
D

P
B
A
M
S
8 cm
Q
C
4 cm
B
16 22
3
2
2
 2    2 4
 cm  19.66 cm
3
7
2


55. Clearly AOB = 60°.
Area of shaded region = Area of major sector of circle + Area of AOB
360  60 22
3
  (6)2 cm 2 
(12) 2 cm 2
360
7
4
M
A

 660


 36 3  cm2
 7

56. Clearly AOB = 90° here, side of square  49 cm  7 cm .
 diagonal BD = 7
2 cm
7 2
7
cm 
cm.
2
2
Now, required area = 2 (Area of segment AEB)

OB 
2
2
 90

 7  1 7 
2
 2

  
 .sin 90  cm
 2  2 2 
 360

MATHEMATICS–X
AREAS RELATED TO CIRCLES
207
 1 22 49 1 49 
 2      1 cm 2  14 cm 2
4 7 2 2 2 
57. Required Area = Area of trapezium – Area of quadrant

J
A
1
1 22
(3.5  5)  3.5 cm 2    (3.5) 2 cm 2
2
4 7
 14.875 cm 2  9.625 cm 2  5.25 cm 2
58. Required Area
= Area of semi-circle with radius 4.5 cm – [Area of circle with diameter 4.5 cm +
2 (Area of semi-circle with diameter 3 cm)] + Area of semi-circle with diameter 3 cm
J
A
  4.5 2

1
1
1
 (4.5)2 cm 2    
 2  (1.5)2  cm 2  (1.5)2 cm 2

2
2
2
  2 

= 10.125  cm2 – 7.3125  cm2 + 1.125  cm2 = 12.375 cm2
MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
B
1. A circular track has an inside circumference of 440 m. If the width of the track is 7 m, the outside
circumference is :
(a) 441 m
(b) 484 m
(c) 625 m
(d) none of these
2. The radius of a car wheel is 49 cm. The number of times the wheel of a car rotate in a journey of 1925 m
is:
(a) 525
(b) 600
(c) 625
(d) 725
3. The given figure consists of a rectangle and a semicircle. The area of the figure is :
(a) 432 cm2
(b) 450 cm2
(c) 532 cm2
(d) 550 cm2
T
I
28 cm
M
A
8 cm
4. A student takes a rectangular piece of a paper 30 cm long and 21 cm wide. The area of the greatest circle
he can cut from the paper is :
(a) 246.5 cm2
(b) 346.5 cm2
(c) 446.5 cm2
(d) none of these
5. In the given figure, the larger circle has the radius 8 cm with O as its centre. The area of the shaded region
is :
(a) 150.85 cm2
208
O
(b) 200.75 cm2
(c) 250.85 cm2
AREAS RELATED TO CIRCLES
(d) none of these
MATHEMATICS–X
6. The minute hand of a clock is 7 cm long. The area traced out by the minute hand of the clock between
5 : 15 pm to 5 : 35 pm on a day is :
1 2
(a) 50 cm
3
1 2
(b) 51 cm
3
1 2
(c) 52 cm
3
J
A
(d) none of these
7. In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining
portion of the triangle is:
(a) 88.55 cm2
(b) 90.55 cm2
(c) 95.55 cm2
(d) 98.55 cm2
8. In the given figure, the area of the shaded segment of the circle with centre O is
4c
m
O
60°
B
A
(a) 1.45 cm2
(b) 2.15 cm2
(c) 3.25 cm2
B
J
A
(d) 4.75 cm2
9. The perimeter of a sector of a circle of radius 9 cm is 33 cm. The area of this sector is :
(a) 57.6 cm2
(b) 61.25 cm2
(c) 67.5 cm2
(d) none of these
10. OABC is a quadrant with radius 14 cm and a semi-circle with OA as diameter. The area of the shaded
portion is :
A
T
I
O
(a) 55 cm2
(b) 66 cm2
M
A
B
C
(c) 77 cm2
(d) 88 cm2
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. If the perimeter of a protractor is 72 cm, what is its radius?
2. The length of a minute hand of a wall clock is 7 cm. What is the area swept by it in half an hour?
3. What is the area of a sector of a circle with diameter 21 cm and central angle 120°?
4. In figure, OACB is a quadrant of a circle of radius 14 cm. What is its perimeter?
MATHEMATICS–X
B
C
O
A
AREAS RELATED TO CIRCLES
209
5. What is the radius of a circle whose circumference and area are numerically equal?
6. In figure, what is the area of shaded region when ABCD is a square of side 14 cm.
D
C
J
A
A
B
7. If an arc forms 120° at the centre O of the circle. What is the ratio of its length to the circumference of the
circle?
8. The archery target has three concentric circular regions. The diameter of the regions are in the ratio
1 : 2 : 3. What is the ratio of their areas?
9. A bicycle wheel makes 10 revolutions in moving 880 cm. What is the radius of the wheel?
10. A wire is in the form of a circle of radius 14 cm. It is bent into a square. Determine the side of the square.
11. In figure, given a sector of a circle of radius 10.5 cm. What is the perimeter of the sector?
O
60°
B
B
A
12. In figure, find the area of the shaded region.
T
I
14 cm
J
A
13. What is the area of the largest triangle that can be inscribed in a semicircle of radius a cm?
14. In figure, ABCD is a square of side 7 cm. What is the area of the shaded region?
M
A
D
C
A
B
7 cm
15. In figure, what is the area of the shaded region included between two concentric circles?
210
14 cm
10.5 cm
AREAS RELATED TO CIRCLES
MATHEMATICS–X
PRACTICE TEST
M.M : 30
Time : 1 hour
J
A
14 cm
General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. If the area of a semi-circular region is 1,232 cm2, find its perimeter.
2. The length of a minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
3. Find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
A
B
P
J
A
D
C
14 cm
4. A wheel has diameter 84 cm. Find how many complete revolutions must it take to cover 1584 meters.
5. A horse is placed for grazing inside a rectangular field 80 m by 55 m and is tethered to one corner by a
rope 21 m long. On how much area can it graze?
6. The area of an equilateral triangle is 49 3 cm2. Taking each angular point as centre, a circle is described
with radius equal to half the length of the side of the triangle as shown. Find the area of the triangle not
included in the circle.
T
I
B
60°
cm
6
O
6
M
A
cm
7. A chord AB of a circle of radius 6 cm subtends an angle of 60° at the centre O. Find the area of the minor
segment ACB.
B
A
C
8. Find the area of the shaded region, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Q
MATHEMATICS–X
O
R
P
AREAS RELATED TO CIRCLES
211
9. An athletic track 14 m wide consists of two straight sections 120 m long joining semi-circulars ends
whose inner radius is 35 m. Calculate the area of the shaded region.
J
A
120 m
14m
14m
35m
10. Find the area of the shaded portion :
J
A
5
3.
35 cm
B
60 cm
(i)
T
I
O
cm
8.75 cm
(ii)
ANSWERS OF PRACTICE EXERCISE
2
1. 38.5 cm
2. 44 cm
6. 346.5 cm2 7. 2618 m2
11. 13 cm
12. 7
M
A
16. 9.625 cm2
4. 11 cm and 3 cm 5. 616 cm2
8. 7 m
9. Rs. 693
10. 14 cm
13. 70 cm
14. 250
15. (i) 316.8 m (ii) 19.008 km/hr
17. 77 cm2
21. 183.3 cm2 22. 9.625 cm2
25. 48°
3. 132 m
18. 25 cm
23. 28.57 cm2, 35.71 cm
26. (i) 105° (ii) 38.5 cm, 21 cm
33. 795.54 cm2
34. 114 cm2
20. 45.03 m2
24. (i) 33 cm (ii) 148.5 cm2
27. 14 cm2
29. (i) 78.5 cm2 (ii) 28.5 cm2 (iii) 235.5 cm2 (iv) 285.5 cm2
32. 189.97 km2
19. 24  cm2
28. 271.2675 cm2
30. 88.44 cm2
31. 1254.96 cm2
35. (16  – 12 3 ) cm2
36. (i) 1386 cm2 (ii) 2772 cm2
37. (i) 30.55 cm2 (ii) 190.635 cm2
38. 228.57 cm2
39. 12  cm, 37.71 cm2 40. 203 cm2
41. 9384 m2
42. 42 cm2
43. 98 cm2
44. (i) 102.57 cm2 (ii) 31.37 cm2
45. 66.5 cm2
46. 57 cm2
47. 535.71 m2
48. 1272.86 cm2
212
AREAS RELATED TO CIRCLES
MATHEMATICS–X
49. 2 cm
50. 77 m2, 259 m2
51. 24.6 cm2
53. 19.66 cm2
54. 3.44 cm2; 4.56 cm2
 660

 36 3  cm2
55. 
7


56. 14 cm2
57. 5.25 cm2
58. (i) 12.375 cm2 (ii) Rs. 6
59. (i) 88 cm, 154 cm2
52. 770 cm2
J
A
(ii) 44 cm, 42 cm2
(iii) 176 cm, 1232 cm2
(iv) 34 cm, 61.5 cm2
(v) 32.6 cm, 29.7 cm2
(vi) 31.4 cm, 57.1 cm2
(vii) 58.9 cm, 225 cm2
(viii) 189 cm, 1697 cm2
(ix) 50.3 cm, 116 cm2
(x) 37.8 cm, 42.6 cm2
60. (i) 672 cm2
(ii) 19.3 cm2
(iii) 462 cm2
(v) 14.3 cm2
(vi) 192.5 cm2
(vii) 231 cm2
(ix) 231 cm2
(x) 378 cm2
B
J
A
(iv) 2328
2 2
cm
7
(viii) 336 cm2
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (b)
6. (b)
2. (c)
7. (d)
3. (c)
8. (a)
T
I
4. (b)
9. (c)
5. (a)
10. (c)
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
1. 14 cm
2. 77 cm2
6. 42 cm2
7. 1 : 3
11. 32 cm
12. 42 cm2
M
A
3. 115.5 cm2
4. 50 cm
5. 2 units
8. 1 : 4 : 9
9. 14 cm
10. 22 cm
13. a2 cm2
14. 10.5 cm2
15. 269.5 cm2
ANSWERS OF PRACTICE TEST
1. 144 cm
6. 7.77 cm2
2. 30.8 cm2
7. 3.27 cm2
MATHEMATICS–X
3. 42 cm2
8. 161.54 cm2
4. 600
9. 7056 m2
5. 346.5 m2
10. (i) 1137.5 cm2 (ii) 231 cm2
AREAS RELATED TO CIRCLES
213