Chapter 12 : Areas Related to Circles — Class 10 Mathematics NCERT Solutions | CBSE
Chapter 12 : Areas Related to Circles – NCERT Class 10 Mathematics Solutions | CBSE
Chapter 12. AREAS RELATED TO CIRCLES
Chapter 12 : Areas Related to Circles |
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Exercise 12.1 solution complete Exercise 12.2 solution complete Exercise 12.3 solution complete |
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1. Circumference of a circle 3. The sector of a circle : Here, OAPB is the minor sector of the circle and OAQB is the major sector of the circle . 4. The segment of a circle : Here, APB is the minor segment of the circle and AQB is the major segment of the circle . OR Area of the segment APB = Area of the sector OAPB – Area of ∆ OAB = 8. In figure : Area of the major sector OAQB = In figure : Area of major segment AQB = |
.
.
⋅
– Area of the minor sector OAPB and
EXERCISE 12.1
Unless stated otherwise , use 
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: Let 
be the radius of new circle .
Here, 
cm and
cm .
A/Q , 
cm 
cm
Therefore , the radius of the new circle is 28 cm .
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: Let be the radius of new circle .
Here, 
cm and 
cm .
A/Q, 
cm
Therefore, the radius of the new circle is 10 cm .
3. Fig. 12.3 depicts an archery target marked with its five scoring regions 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 .
Solution: For Gold region : Here , 
cm , 
cm
Area of the gold region 
For Red region : Here , 
and 
Area of the red region 
For blue region : Here , 
and 
Area of the red region 
For Black region : Here , 
Area of the black region 
For White region : Here , 
and 
Area of the white region 
4. The wheels of a car are of diameter 80 cm each . How many complete revolution does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour ?
Solution: Here, 
and 
The perimeter of the wheels of a car 
The speed of the car 
The distances travel by the car 
Speed of the car × time [ Time 
min.]
The number of complete revolution of the wheels of the car 
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 (C) 4 units (D) 7 units
Solution: (A) 2 units
[ Let 
be the radius of the circle .
A/Q , 
or 
units ]
EXERCISE 12.2
Unless stated otherwise , use .
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60° .
Solution : Here, 
and 
The area of a sector of a circl 
2. Find the area of a quadrant of a circle whose circumference is 22 cm .
Solution : We have, 
The area of a quadrant of a circle 
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 : Here, 
The angle of swept in 5 minutes 
Therefore, the area swept by the minute hand in 5 minutes
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 
)
Solution : Here, 
The area of the minor segment of a circle 
(ii) Here, 
, 
Area of major sector of a circle 
5. 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 .
Solution: Here, Radius 
cm and
(i) the length of an arc of a sector of a circle 
(ii) The area of the sector formed by the arc 
(iii) Here, OA = OB = 21 cm and 
So, OAB be an equilateral triangle .
Area of triangle OAB 
The area of segment formed by chord =
Area of sector – Area of triangle
(6) A chord of a circle of radius 15 cm subtends an angle of 60° at the centre . Find the areas of the corresponding minor and major segments of the circle . (Use 
and 
)
Solution: Here , radius 
cm , 
Area of the minor segment = Area of the sector – Area of the triangle formed by radius and chord
Area of major segment = Area of the circle – Area of the minor segment
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 . (Use and )
Solution : Here, 
and 
The area of the corresponding segment of the circle 
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 Fig. 12.11) . Find : (i) the area of that part of the field in which the horse can graze . (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m . (Use )
Solution : Since , ABCD is a square .
So, AB = BC = CD = AD = 15 cm
(i) Here, 
, 
Therefore, the area of the field in which the horse can graze 
(ii) Here, 
, 
and 
The increasing area of the field 
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 Fig. 12.12 . Find :
(i) the total length of the silver wire required . (ii) the area of each sector of the brooch .
Solution: (i) Here, diameter 
and Radius 
The circumference =2πr 
The length of the wire required to make 5 diameters 
Therefore, the total length of the silver wire required 
(ii) The angle of each brooch 
So, the area of each sector of the brooch 
10 . An umbrella has 8 ribs which are equally spaced (see Fig. 12.13) . Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella .
Solution : Here , 
cm
The area between the two consecutive ribs of the umbrella 
11. 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 .
Solution : Here , 
The area cleaned at each sweep of the blades 
Therefore, the total area cleaned at each sweep of the blades 
12. 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 ) .
Solution: Here , 
The area of the sea over which the ships are warn
13. A round table cover has six equal designs as shown in Fig. 12.14 . 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 )
Solution : Here, 
and 
Area of a design part of the round table cover 
Area of six design part of the round table cover 
Therefore, the cost of making the designs 
14.Tick the correct answer in the following :
Area of a sector of angle 
(in degrees) of a circle with radius R is :
(A) 
(B) 
(C) 
(D) 
Solution : (D)
EXERCISE 12.3
Unless stated otherwise , use .
1. Find the area of the shaded region in Fig.12.19, if 
,
and O is the centre of the circle .
Solution: Here , 
and
In 
we have ,
Radius 
Area of 
Area of semicircle 
The area of the shaded region 
Area of semicircle QPR 
Area of 
2. Find the area of the shaded region in Fig. 12.20 , if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and 
.
Fig. 12.20
Solution: Here, 
, 
, 
The area of the shaded region 
3. Find the area of the shaded region in Fig. 12.21 , if ABCD is a square of side 14 cm and APD and BPC are semicircles .
Solution: For square : Since, ABCD is a square .
So, AB = BC = CD = AD = 14 cm
Area of the square ABCD 
For semicircle : Here, 
, 
Area of the two semicircle 
The area of the shaded region 
Area of the square ABCD – Area of 2 semicircle
4. Find the area of the shaded region in Fig. 12.22 , where a circular arc of radius 6 cm has been draw with vertex O of an equilateral triangle OAB of side 12 cm as centre .
Fig. 12.22
Solution: For equilateral triangle OAB : Here, OA = AB = OB = 12 cm and 
Area of an equilateral triangle OAB 
For circle : Here, 
The major angle of the sector 
The area of the major sector 
The area of the shaded region area of major sector + area of an equilateral triangle OAB
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 Fig. 12.23 . Find the area of the remaining portion of the square .
Solution: For square ABCD : Here, AB = BC = CD = AD = 4 cm , 
and 
The area of the remaining portion of the square = Area of the square ABCD – Area of the circle – Area of 4 quadrant
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 Fig. 12.24 . Find the area of the design .
Solution: Here, 
and 
In 
, we have
Since, 
Therefore, the area of the design = Area of the circular cover – Area of an equilateral triangle ABC
7. In Fig. 12.25, 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 .
Solution: Since , AB = BC = CD = AD = 14 cm , 
, 
Therefore, the area of the shaded region = Area of the square ABCD – Area of four quadrant
8.In Fig. 12.36 depicts a racing track whose left and right ends are semicircular . 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 .
Solution: In given figure :
(i) Here, AD = BC = EH = FG = 106 m , AH = DE = 60 m
Diameter 
and Radius 
(i) The distance around the track along its inner edge = AD + HE + 2 × circumference of the semicircle
(ii) Here, AB = CD = EF = HG = 10 m , 
and 
The area of the track = 2 × area of rectangle ABCD (= EFGH) + 2 × Area of the semicircular part
9. In Fig. 12.27, 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 
, find the area of the shaded region .
Fig. 12.27
Solution: Here, 
, 
, 
and 
Therefore. the area of the shaded region = Area of circle with radius 
+ Area of semicircle with radius 7 cm – Area of triangle ACB
10. 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 (see Fig. 12.28) . Find the area of the shaded region . (Use 
)
Solution: Since, ABC is an equilateral triangle .
So, 
We know that, Area of an equilateral triangle 
A/Q, 
Here, 
and 
The area of the shaded region = Area of an equilateral triangle ABC – Area of 3 sector
11. On a square handkerchief , nine circular designs each of radius 7 cm are made (see Fig. 12.29) . Find the area of the remaining portion of the handkerchief .
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