Exercise 11.1 Solutions
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Given: Radius (r) = 6 cm, Angle (θ) = 60°
Area of sector = (θ/360°) × πr²
= (60/360) × (22/7) × 6²
= (1/6) × (22/7) × 36
= (22/7) × 6
= 132/7
= 18.857 cm² (approx.)
Therefore, the area of the sector is 18.857 cm².
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Given: Circumference = 22 cm
We know: Circumference = 2πr
2 × (22/7) × r = 22
(44/7) × r = 22
r = 22 × (7/44)
r = 3.5 cm
A quadrant is a sector with angle 90°.
Area of quadrant = (90/360) × πr² = (1/4) × πr²
= (1/4) × (22/7) × (3.5)²
= (1/4) × (22/7) × 12.25
= (1/4) × 38.5
= 9.625 cm²
Therefore, the area of the quadrant is 9.625 cm².
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Given: Length of minute hand (r) = 14 cm
In 60 minutes, the minute hand sweeps 360°.
In 5 minutes, the angle swept = (5/60) × 360° = 30°
Area swept = (θ/360) × πr²
= (30/360) × (22/7) × 14²
= (1/12) × (22/7) × 196
= (1/12) × 22 × 28
= (1/12) × 616
= 51.33 cm² (approx.)
Therefore, the area swept by the minute hand in 5 minutes is 51.33 cm².
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 = 3.14)
Given: Radius (r) = 10 cm, Angle (θ) = 90°
(i) Area of minor segment:
Area of segment = Area of sector - Area of triangle
Area of sector = (90/360) × 3.14 × 10² = (1/4) × 3.14 × 100 = 78.5 cm²
Area of triangle = (1/2) × r² × sinθ = (1/2) × 100 × sin90° = (1/2) × 100 × 1 = 50 cm²
Area of minor segment = 78.5 - 50 = 28.5 cm²
(ii) Area of major sector:
Area of major sector = πr² - Area of minor sector
= 3.14 × 100 - 78.5
= 314 - 78.5
= 235.5 cm²
Therefore, the area of the minor segment is 28.5 cm² and the area of the major sector is 235.5 cm².
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
Given: Radius (r) = 21 cm, Angle (θ) = 60°
(i) Length of the arc:
Length of arc = (θ/360) × 2πr
= (60/360) × 2 × (22/7) × 21
= (1/6) × 2 × 22 × 3
= (1/6) × 132
= 22 cm
(ii) Area of the sector:
Area of sector = (θ/360) × πr²
= (60/360) × (22/7) × 21²
= (1/6) × (22/7) × 441
= (1/6) × 22 × 63
= (1/6) × 1386
= 231 cm²
(iii) Area of the segment:
Area of segment = Area of sector - Area of triangle
Area of triangle = (1/2) × r² × sinθ = (1/2) × 21² × sin60°
= (1/2) × 441 × (√3/2)
= (441√3)/4 cm²
Area of segment = 231 - (441√3)/4 cm²
Therefore, the length of the arc is 22 cm, area of the sector is 231 cm², and area of the segment is [231 - (441√3)/4] cm².
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 π = 3.14 and √3 = 1.73)
Given: Radius (r) = 15 cm, Angle (θ) = 60°
Area of minor sector:
= (60/360) × 3.14 × 15²
= (1/6) × 3.14 × 225
= 117.75 cm²
Area of triangle:
= (1/2) × r² × sinθ
= (1/2) × 15² × sin60°
= (1/2) × 225 × (√3/2)
= (225√3)/4
= (225 × 1.73)/4
= 389.25/4
= 97.3125 cm²
Area of minor segment:
= Area of sector - Area of triangle
= 117.75 - 97.3125
= 20.4375 cm²
Area of major segment:
= πr² - Area of minor segment
= 3.14 × 225 - 20.4375
= 706.5 - 20.4375
= 686.0625 cm²
Therefore, the area of the minor segment is 20.44 cm² and the area of the major segment is 686.06 cm².
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 π = 3.14 and √3 = 1.73)
Given: Radius (r) = 12 cm, Angle (θ) = 120°
Area of sector:
= (120/360) × 3.14 × 12²
= (1/3) × 3.14 × 144
= 150.72 cm²
Area of triangle:
= (1/2) × r² × sinθ
= (1/2) × 12² × sin120°
= (1/2) × 144 × (√3/2)
= (144√3)/4
= 36√3
= 36 × 1.73
= 62.28 cm²
Area of segment:
= Area of sector - Area of triangle
= 150.72 - 62.28
= 88.44 cm²
Therefore, the area of the segment is 88.44 cm².
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. 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)
The horse is tied at one corner of a square field, so it can graze in a quadrant of a circle.
(i) With rope length 5 m:
Area = (1/4) × πr² = (1/4) × 3.14 × 5² = (1/4) × 3.14 × 25 = 19.625 m²
(ii) With rope length 10 m:
Area = (1/4) × πr² = (1/4) × 3.14 × 10² = (1/4) × 3.14 × 100 = 78.5 m²
Increase in grazing area:
= 78.5 - 19.625 = 58.875 m²
Therefore, the area in which the horse can graze with a 5 m rope is 19.625 m², and the increase in grazing area when the rope is 10 m long is 58.875 m².
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. 11.9. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
Given: Diameter = 35 mm, so radius (r) = 35/2 = 17.5 mm
(i) Total length of silver wire:
Circumference of circle = πd = (22/7) × 35 = 110 mm
Length of 5 diameters = 5 × 35 = 175 mm
Total length = 110 + 175 = 285 mm
(ii) Area of each sector:
The circle is divided into 10 equal sectors, so angle of each sector = 360°/10 = 36°
Area of each sector = (36/360) × πr² = (1/10) × (22/7) × (17.5)²
= (1/10) × (22/7) × 306.25
= (1/10) × 962.5
= 96.25 mm²
Therefore, the total length of silver wire required is 285 mm, and the area of each sector is 96.25 mm².
10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Given: Radius (r) = 45 cm, Number of ribs = 8
The ribs divide the circle into 8 equal sectors.
Angle of each sector = 360°/8 = 45°
Area between two consecutive ribs = Area of one sector
= (45/360) × πr²
= (1/8) × (22/7) × 45²
= (1/8) × (22/7) × 2025
= (1/8) × 6364.29
= 795.54 cm² (approx.)
Therefore, the area between two consecutive ribs is 795.54 cm².
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.
Given: Length of wiper blade (r) = 25 cm, Angle (θ) = 115°
Each wiper cleans a sector of a circle.
Area cleaned by one wiper:
= (115/360) × π × 25²
= (115/360) × (22/7) × 625
= (115 × 22 × 625) / (360 × 7)
= 1581250 / 2520
= 627.48 cm² (approx.)
Total area cleaned by both wipers:
= 2 × 627.48 = 1254.96 cm²
Therefore, the total area cleaned at each sweep of the blades is 1254.96 cm².
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 π = 3.14)
Given: Radius (r) = 16.5 km, Angle (θ) = 80°
Area of sector = (80/360) × πr²
= (2/9) × 3.14 × (16.5)²
= (2/9) × 3.14 × 272.25
= (2/9) × 854.865
= 189.97 km² (approx.)
Therefore, the area of the sea over which ships are warned is 189.97 km².
13. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use √3 = 1.7)
Given: Radius (r) = 28 cm, Number of designs = 6
The designs divide the circle into 6 equal sectors, each with angle = 360°/6 = 60°
Area of one sector:
= (60/360) × πr² = (1/6) × (22/7) × 28²
= (1/6) × (22/7) × 784
= (1/6) × 2464
= 410.67 cm²
Area of equilateral triangle with side 28 cm:
= (√3/4) × side² = (1.7/4) × 28²
= 0.425 × 784
= 333.2 cm²
Area of one design (segment):
= Area of sector - Area of triangle
= 410.67 - 333.2
= 77.47 cm²
Total area of 6 designs:
= 6 × 77.47 = 464.82 cm²
Cost of making designs:
= 464.82 × 0.35 = ₹162.69
Therefore, the cost of making the designs is ₹162.69.
14. Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
The formula for area of a sector is:
Area = (p/360) × πR²
Comparing with the options:
(A) p/180 × 2πR → This is for arc length, not area
(B) p/180 × πR² → Incorrect denominator
(C) p/360 × 2πR → This is for arc length
(D) p/720 × 2πR² → Incorrect formula
None of the options exactly match the correct formula. The closest is option (B) but with wrong denominator.
The correct formula should be: (p/360) × πR²