Exercise 11.1 Solutions
Areas Related to Circles
(Unless stated otherwise, take π = 22/7)
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 a sector = (θ/360°) × πr²
= (60/360) × (22/7) × 6² = (1/6) × (22/7) × 36
= (22/7) × 6 = 132/7 cm².
Answer: The area of the sector is 132/7 cm².
2. Find the area of a quadrant of a a circle whose circumference is 22 cm.
Step 1: Find the radius (r).
Circumference = 2πr = 22 => 2 × (22/7) × r = 22 => r = 7/2 cm.
Step 2: Find the area of the quadrant.
A quadrant is 1/4 of a circle, so the angle θ = 90°.
Area = (1/4) × πr² = (1/4) × (22/7) × (7/2)²
= (1/4) × (22/7) × (49/4) = 77/8 cm².
Answer: The area of the quadrant is 77/8 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.
The minute hand acts as the radius (r) = 14 cm.
Step 1: Find the angle swept in 5 minutes.
In 60 minutes, the minute hand sweeps 360°.
In 1 minute, it sweeps 360/60 = 6°.
In 5 minutes, it sweeps 6° × 5 = 30°.
Step 2: Find the area of the sector.
Area = (θ/360°) × πr² = (30/360) × (22/7) × 14²
= (1/12) × (22/7) × 196 = 154/3 cm².
Answer: The area swept is 154/3 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: r = 10 cm, θ = 90°.
Step 1: Find the area of the minor sector.
Areasector = (90/360) × 3.14 × 10² = (1/4) × 314 = 78.5 cm².
Step 2: Find the area of the triangle.
Areatriangle = 1/2 × base × height = 1/2 × r × r = 1/2 × 10 × 10 = 50 cm².
(i) Area of minor segment = Areasector – Areatriangle = 78.5 – 50 = 28.5 cm².
(ii) Area of major sector = Area of circle – Area of minor sector
= (πr²) – 78.5 = (3.14 × 10²) – 78.5 = 314 – 78.5 = 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: r = 21 cm, θ = 60°.
(i) Length of the arc = (θ/360°) × 2πr
= (60/360) × 2 × (22/7) × 21 = (1/6) × 44 × 3 = 22 cm.
(ii) Area of the sector = (θ/360°) × πr²
= (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = 231 cm².
(iii) Area of the segment
Since θ = 60°, the triangle formed is equilateral. Area = (√3/4) × side²
Areatriangle = (√3/4) × 21² = 441√3/4 cm².
Areasegment = Areasector – Areatriangle = (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: r = 15 cm, θ = 60°.
Areasector = (60/360) × 3.14 × 15² = (1/6) × 3.14 × 225 = 117.75 cm².
Areatriangle (equilateral) = (√3/4) × 15² = (1.73/4) × 225 = 97.3125 cm².
Area of minor segment = 117.75 – 97.3125 = 20.4375 cm².
Areacircle = πr² = 3.14 × 15² = 706.5 cm².
Area of major segment = Areacircle – Areaminor segment = 706.5 – 20.4375 = 686.0625 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: r = 12 cm, θ = 120°.
Areasector = (120/360) × 3.14 × 12² = (1/3) × 3.14 × 144 = 150.72 cm².
Areatriangle = 1/2 × r² × sin(θ) = 1/2 × 12² × sin(120°) = 72 × sin(60°) = 72 × (√3/2) = 36√3.
= 36 × 1.73 = 62.28 cm².
Area of segment = Areasector – Areatriangle = 150.72 – 62.28 = 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 can graze in a quadrant of a circle (θ = 90°).
(i) Grazing area with 5 m rope:
r = 5 m. Area = (1/4) × πr² = (1/4) × 3.14 × 5² = 19.625 m².
(ii) Increase in grazing area:
New radius R = 10 m. New Area = (1/4) × πR² = (1/4) × 3.14 × 10² = 78.5 m².
Increase in Area = New Area – Old Area = 78.5 – 19.625 = 58.875 m².
9. A brooch is made with silver wire in the form of a circle… Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
>Diameter = 35 mm, so radius r = 17.5 mm.
(i) Total length of wire = (Circumference) + (5 × Diameter)
= (2πr) + (5 × 35) = (2 × 22/7 × 17.5) + 175 = 110 + 175 = 285 mm.
(ii) Area of each sector
The circle is divided into 10 equal sectors. Angle of each sector θ = 360°/10 = 36°.
Area = (36/360) × πr² = (1/10) × (22/7) × (17.5)² = 96.25 mm².
10. An umbrella has 8 ribs which are equally spaced… Find the area between the two consecutive ribs of the umbrella.
>The umbrella forms a circle with radius r = 45 cm. It has 8 ribs, creating 8 equal sectors.
Angle between two consecutive ribs (θ) = 360° / 8 = 45°.
Area between ribs = Area of one sector = (θ/360°) × πr²
= (45/360) × (22/7) × 45² = (1/8) × (22/7) × 2025 = 22275/28 cm².
Answer: The area is 22275/28 cm² (or approx. 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.
Area cleaned by one wiper is the area of a sector.
r = 25 cm, θ = 115°.
Area of one sector = (115/360) × (22/7) × 25² = 158125/252 cm².
Total area for two wipers = 2 × (158125/252) = 158125/126 cm².
Answer: The total area is 158125/126 cm² (or approx. 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)
This is the area of a sector with r = 16.5 km and θ = 80°.
Area = (80/360) × 3.14 × (16.5)² = (2/9) × 3.14 × 272.25 = 189.97 km².
Answer: The area of the sea 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)
>Each design is a segment. There are 6 equal designs, so the angle of each sector is 360°/6 = 60°.
Step 1: Find the area of one segment.
r = 28 cm, θ = 60°.
Areasector = (60/360) × (22/7) × 28² = (1/6) × 2464 = 1232/3 cm².
Areatriangle (equilateral) = (√3/4) × 28² = 1.7/4 × 784 = 333.2 cm².
Areasegment = (1232/3) – 333.2 ≈ 410.67 – 333.2 = 77.47 cm².
Step 2: Find the total area of 6 designs and the cost.
Total Area = 6 × 77.47 = 464.82 cm².
Cost = Total Area × Rate = 464.82 × 0.35 = ₹162.68.
Answer: The cost of making the designs is ₹162.68.
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 standard formula for the area of a sector is (θ/360) × πR².
Here, θ = p. So the formula is (p/360) × πR².
Let’s check the options.
(A) and (C) are formulas for arc length.
(B) has 180 in the denominator.
(D) is (p/720) × 2πR². If we simplify this by cancelling the 2, we get (p/360) × πR².
Answer: (D) (p/720) × 2πR².
