2Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution & Reference:
Circumference = 2πr = 22 cm
⇒ r = 22/(2π) = 11/π cm
Area of quadrant = (1/4) × πr² = (1/4) × π × (11/π)² = (1/4) × 121/π
Using π = 22/7, Area = (1/4) × 121 × 7/22 = (121 × 7)/(4 × 22) = 847/88 = 9.625 cm² Answer: 9.625 cm²
3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution & Reference:
In 60 minutes, minute hand completes 360°
In 5 minutes, angle swept = (5/60) × 360° = 30°
Radius (r) = 14 cm
Area swept = (30/360) × πr² = (1/12) × π × 14² = (1/12) × π × 196 = 49π/3 cm²
Using π = 22/7, Area = 49/3 × 22/7 = (7 × 22)/3 = 154/3 ≈ 51.33 cm² Answer: 51.33 cm² (approx.)
4A 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)
Solution & Reference:
Given: Radius (r) = 10 cm, Angle (θ) = 90°
(i) Area of minor segment = Area of sector - Area of triangle
Area of sector = (90/360) × πr² = (1/4) × 3.14 × 100 = 78.5 cm²
Area of ΔOAB = (1/2) × OA × OB = (1/2) × 10 × 10 = 50 cm² (since ∠AOB = 90°)
Area of minor segment = 78.5 - 50 = 28.5 cm²
(ii) Area of major sector = πr² - Area of minor sector = 3.14 × 100 - 78.5 = 314 - 78.5 = 235.5 cm² Answer: (i) 28.5 cm² (ii) 235.5 cm²
5In 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
(ii) Area of sector = (θ/360) × πr² = (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = (1/6) × 22 × 63 = 231 cm²
(iii) Area of segment = Area of sector - Area of ΔOAB
Since θ = 60° and OA = OB, ΔOAB is equilateral with side 21 cm
Area of equilateral triangle = (√3/4) × side² = (1.73/4) × 441 = (1.73 × 110.25) = 190.7325 cm²
Area of segment = 231 - 190.7325 = 40.2675 cm² ≈ 40.27 cm² Answer: (i) 22 cm (ii) 231 cm² (iii) 40.27 cm²
6A 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)
Solution & Reference:
Given: Radius (r) = 15 cm, Angle (θ) = 60°
Area of sector = (60/360) × πr² = (1/6) × 3.14 × 225 = 117.75 cm²
Area of ΔOAB = (√3/4) × r² = (1.73/4) × 225 = 97.3125 cm²
Area of minor segment = 117.75 - 97.3125 = 20.4375 cm²
Area of circle = πr² = 3.14 × 225 = 706.5 cm²
Area of major segment = 706.5 - 20.4375 = 686.0625 cm² Answer: Minor segment: 20.44 cm², Major segment: 686.06 cm²
7A 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)
Solution & Reference:
Given: Radius (r) = 12 cm, Angle (θ) = 120°
Area of sector = (120/360) × πr² = (1/3) × 3.14 × 144 = 150.72 cm²
To find area of ΔOAB:
Draw OM ⊥ AB. Then ∠AOM = 60°
OM = r cos 60° = 12 × 0.5 = 6 cm
AM = r sin 60° = 12 × (√3/2) = 6√3 cm
AB = 2 × AM = 12√3 cm
Area of ΔOAB = (1/2) × AB × OM = (1/2) × 12√3 × 6 = 36√3 = 36 × 1.73 = 62.28 cm²
Area of segment = 150.72 - 62.28 = 88.44 cm² Answer: 88.44 cm²
8A 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)
Solution & Reference:
(i) The horse can graze in a quadrant of radius 5 m
Area = (1/4) × πr² = (1/4) × 3.14 × 25 = 19.625 m²
(ii) With 10 m rope, area = (1/4) × 3.14 × 100 = 78.5 m²
Increase in area = 78.5 - 19.625 = 58.875 m² Answer: (i) 19.625 m² (ii) 58.875 m²
9A 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. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
(ii) Each sector has angle = 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 Answer: (i) 285 mm (ii) 96.25 mm²
10An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution & Reference:
Radius (r) = 45 cm
Angle between ribs = 360°/8 = 45°
Area between ribs = (45/360) × πr² = (1/8) × (22/7) × 2025 = (1/8) × 6364.2857 ≈ 795.5357 cm² Answer: 795.54 cm² (approx.)
11A 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 & Reference:
Radius (r) = 25 cm, Angle (θ) = 115°
Area cleaned by one wiper = (115/360) × πr² = (115/360) × (22/7) × 625
= (115 × 22 × 625) / (360 × 7) = (1,581,250) / (2,520) ≈ 627.48 cm²
Total area cleaned by two wipers = 2 × 627.48 = 1254.96 cm² Answer: 1254.96 cm²
12To 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)
13A 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)
Solution & Reference:
Radius (r) = 28 cm
Each design is a segment with angle = 360°/6 = 60°
Area of one segment = Area of sector - Area of equilateral triangle
Area of sector = (60/360) × πr² = (1/6) × (22/7) × 784 = (1/6) × 2464 = 410.67 cm²
Area of equilateral triangle = (√3/4) × r² = (1.7/4) × 784 = 333.2 cm²
Area of one segment = 410.67 - 333.2 = 77.47 cm²
Total area of 6 designs = 6 × 77.47 = 464.82 cm²
Cost = 464.82 × 0.35 = ₹162.687 ≈ ₹162.69 Answer: ₹162.69
14Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
Solution & Reference:
From the formula: Area of sector = (θ/360) × πr²
Here, θ = p and r = R
So, Area = (p/360) × πR²
Looking at options:
(A) (p/180) × πR²
(B) (p/180) × πR
(C) (p/360) × πR²
(D) (2p/720) × πR² = (p/360) × πR²
Both (C) and (D) are mathematically equivalent, but (C) is the standard form. Answer: (C) (p/360) × πR²
SChapter Summary
Reference: Page 160
Length of an arc of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × 2πr
Area of a sector of a circle with radius r and angle with degree measure θ is (θ/360) × πr²
Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle