Areas Related to Circles Ex 12.2 Solutions
Question - 1 : - Find the area of a sector of a circle withradius 6 cm if angle of the sector is 60°.
Answer - 1 : -
It is given that the angle of the sector is60°
We know that the area of sector = (θ/360°)×πr2
∴ Area of the sector with angle 60° = (60°/360°)×πr2 cm2
= (36/6)π cm2
= 6×22/7 cm2 = 132/7 cm2
Question - 2 : - Find the area of aquadrant of a circle whose circumference is 22 cm.
Answer - 2 : -
Circumference of the circle, C = 22 cm (given)
It should be noted that a quadrant of a circleis a sector which is making an angle of 90°.
Let the radius of the circle = r
As C = 2πr = 22,
R = 22/2π cm = 7/2 cm
∴ Area of the quadrant = (θ/360°) × πr2
Here, θ = 90°
So, A = (90°/360°) × π r2 cm2
= (49/16) π cm2
= 77/8 cm2 = 9.6 cm2
Question - 3 : - The length of theminute hand of a clock is 14 cm. Find the area swept by the minute hand in 5minutes.
Answer - 3 : -
Length of minute hand = radius of the clock(circle)
∴ Radius (r) of the circle = 14 cm (given)
Angle swept by minute hand in 60 minutes =360°
So, the angle swept by the minute hand in 5minutes = 360° × 5/60 = 30°
We know,
Area of a sector = (θ/360°) × πr2
Now, area of the sector making an angle of 30°= (30°/360°) × πr2 cm2
= (1/12) × π142
= (49/3)×(22/7) cm2
= 154/3 cm2
Question - 4 : - A chord of a circleof radius 10 cm subtends a right angle at the centre. Find the area of thecorresponding:
(i) minor segment
(ii) major sector.(Use π = 3.14)
Answer - 4 : - 
Question - 5 : - In a circle ofradius 21 cm, an arc subtends an angle of 60° at the centre.
Answer - 5 : -
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
Given,
Radius = 21 cm
θ = 60°
(i) Length of an arc = θ/360°×Circumference(2πr)
∴ Length of an arc AB = (60°/360°)×2×(22/7)×21
= (1/6)×2×(22/7)×21
Or Arc AB Length = 22cm
(ii) It is given that the angle subtend by the arc = 60°
So, area of the sector making an angle of 60°= (60°/360°)×π r2 cm2
= 441/6×22/7 cm2
Or, the area of the sector formed by the arcAPB is 231 cm2
(iii) Area of segment APB = Area of sector OAPB – Area of ΔOAB
Since the two arms of the triangle are theradii of the circle and thus are equal, and one angle is 60°, ΔOAB is anequilateral triangle. So, its area will be √3/4×a2 sq. Units.
Area of segment APB = 231-(√3/4)×(OA)2
= 231-(√3/4)×212
Or, Area of segment APB = [231-(441×√3)/4] cm2
Question - 6 : - A chord of a circleof radius 15 cm subtends an angle of 60° at the centre. Find the areas of thecorresponding minor and major segments of the circle. (Use π = 3.14 and √3 =1.73)
Answer - 6 : - 
Question - 7 : - A chord of a circleof radius 12 cm subtends an angle of 120° at the centre. Find the area of thecorresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Answer - 7 : - 
Question - 8 : - A horse is tied to apeg at one corner of a square shaped grass field of side 15 m by means of a 5 mlong rope (see Fig. 12.11).
Answer - 8 : -
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 π = 3.14)
Solution:
As the horse is tied at one end of a squarefield, it will graze only a quarter (i.e. sector with θ = 90°) of the fieldwith radius 5 m.
Here, the length of rope will be the radius ofthe circle i.e. r = 5 m
It is also known that the side of square field= 15 m
(i) Area of circle = πr2 = 22/7 × 52 =78.5 m2
Now, the area of the part of the field wherethe horse can graze = ¼ (the area of the circle) = 78.5/4 = 19.625 m2
(ii) If the rope is increased to 10 m,
Area of circle will be = πr2 =22/7×102 =314 m2
Now, the area of the part of the field wherethe horse can graze = ¼ (the area of the circle)
= 314/4 = 78.5 m2
∴ Increase in the grazing area = 78.5 m2 – 19.625m2 = 58.875 m2
Question - 9 : - A brooch is madewith silver wire in the form of a circle with diameter 35 mm. The wire is alsoused in making 5 diameters which divide the circle into 10 equal sectors asshown in Fig. 12.12. Find:
(i) the total lengthof the silver wire required.
(ii) the area ofeach sector of the brooch.
Answer - 9 : -
Find:
(i) the total lengthof the silver wire required.
(ii) the area ofeach sector of the brooch.
Diameter (D) = 35 mm
Total number of diameters to be considered= 5
Now, the total length of 5 diameters thatwould be required = 35×5 = 175
Circumference of the circle = 2πr
Or, C = πD = 22/7×35 = 110
Area of the circle = πr2
Or, A = (22/7)×(35/2)2 =1925/2 mm2
(i) Total length of silver wire required = Circumference ofthe circle + Length of 5 diameter
= 110+175 = 185 mm
(ii) Total Number of sectors in the brooch = 10
So, the area of each sector = total area ofthe circle/number of sectors
∴ Area of each sector = (1925/2)×1/10 = 385/4 mm2
Question - 10 : - An umbrella has 8 ribs which are equallyspaced (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.
Answer - 10 : - 
Solution:
The radius (r) of the umbrella when flat = 45cm
So, the area of the circle (A) = πr2 =(22/7)×(45)2 =6364.29 cm2
Total number of ribs (n) = 8
∴ The area between the two consecutive ribs of the umbrella = A/n
6364.29/8 cm2
Or, The area between the two consecutive ribsof the umbrella = 795.5 cm2