Exercise 11.1
1Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Area of a sector of angle θ = θ360° × πr2
Area = 60360 × 227 × 6 × 6 = 16 × 227 × 36 = 1327
Area of the sector 1327 cm2 = 1867 cm2
2Find the area of a quadrant of a circle whose circumference is 22 cm.
2πr = 22 ⇒ 2 × 227 × r = 22 ⇒ r = 72 cm
Area of quadrant = 14 × πr2 = 14 × 227 × 72 × 72 = 778
Area of the quadrant 778 cm2 = 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.
The minute hand moves through 360° in 60 minutes. So, in 5 minutes it moves through 360°60 × 5 = 30°.
Area swept = 30360 × 227 × 14 × 14 = 112 × 616 = 1543
Area swept 1543 cm2 = 5113 cm2
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)
Let O be the centre and AB the chord, with ∠AOB = 90°.
Area of minor sector OAPB = 90360 × 3.14 × 10 × 10 = 78.5 cm2
Area of ΔOAB = 12 × OA × OB = 12 × 10 × 10 = 50 cm2
(i) Minor segment
Area = Area of minor sector – Area of ΔOAB = 78.5 – 50 = 28.5
Area of the minor segment 28.5 cm2
(ii) Major sector
Area = πr2 – Area of minor sector = 314 – 78.5 = 235.5
Area of the major sector 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
(i) Length of the arc
Length = θ360° × 2πr = 60360 × 2 × 227 × 21 = 22
Length of the arc 22 cm
(ii) Area of the sector
Area = 60360 × 227 × 21 × 21 = 231
Area of the sector 231 cm2
(iii) Area of the segment
In ΔOAB, OA = OB = 21 cm and ∠AOB = 60°. So, ΔOAB is equilateral.
Area of ΔOAB = √34 × 212 = 441√34 cm2
Area of segment = Area of sector – Area of ΔOAB = (231 – 441√34) cm2
Area of the segment (231 – 441√34) 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)
Area of minor sector = 60360 × 3.14 × 15 × 15 = 117.75 cm2
ΔOAB is equilateral (OA = OB and ∠AOB = 60°).
Area of ΔOAB = √34 × 152 = 1.73 × 2254 = 97.3125 cm2
Area of minor segment = 117.75 – 97.3125 = 20.4375 cm2
Area of circle = 3.14 × 225 = 706.5 cm2
Area of major segment = 706.5 – 20.4375 = 686.0625 cm2
Minor segment = 20.4375 cm2 and major segment = 686.0625 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)
Area of sector = 120360 × 3.14 × 12 × 12 = 150.72 cm2
Draw OM ⊥ AB. In ΔOAB, ∠AOM = 60°.
OM = 12 cos 60° = 6 cm ; AM = 12 sin 60° = 6√3 cm ; AB = 12√3 cm
Area of ΔOAB = 12 × AB × OM = 12 × 12√3 × 6 = 36√3 = 36 × 1.73 = 62.28 cm2
Area of segment = 150.72 – 62.28 = 88.44 cm2
Area of the segment 88.44 cm2
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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 (see Fig. 11.8). 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)
The horse grazes a quadrant (90° sector) with radius equal to the length of the rope.
(i) Rope of 5 m
Area = 14 × 3.14 × 5 × 5 = 19.625
Grazing area 19.625 m2
(ii) Rope of 10 m
New area = 14 × 3.14 × 10 × 10 = 78.5 m2
Increase = 78.5 – 19.625 = 58.875
Increase in grazing area 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 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.
(i) Total length of wire
Circumference = πd = 227 × 35 = 110 mm
Length of 5 diameters = 5 × 35 = 175 mm
Total length = 110 + 175 = 285 mm
Total length of silver wire 285 mm
(ii) Area of each sector
Each sector has angle 360°10 = 36° and radius 352 mm.
Area = 36360 × 227 × 352 × 352 = 110 × 227 × 12254 = 3854
Area of each sector 3854 mm2 = 96.25 mm2
10An 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.
8 ribs divide the circle into 8 equal sectors. Each sector has angle 360°8 = 45°.
Area = 45360 × 227 × 45 × 45 = 18 × 227 × 2025 = 2227528
Area between two consecutive ribs 2227528 cm2 ≈ 795.54 cm2
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.
Each wiper sweeps a sector of radius 25 cm and angle 115°.
Area by one wiper = 115360 × 227 × 25 × 25 = 158125252 cm2
Total area by two wipers = 2 × 158125252 = 158125126
Total area cleaned 158125126 cm2 ≈ 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)
Area = 80360 × 3.14 × 16.5 × 16.5 = 29 × 3.14 × 272.25 = 189.97
Area of the sea 189.97 km2 (approx.)
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)
The six designs are six equal segments. Each chord subtends 360°6 = 60° at the centre.
So, each triangle formed with the centre is equilateral with side 28 cm.
Area of one sector = 60360 × 227 × 28 × 28 = 410.67 cm2 (approx.)
Area of one equilateral triangle = √34 × 282 = 1.7 × 7844 = 333.2 cm2
Area of one design = 410.67 – 333.2 = 77.47 cm2 (approx.)
Area of six designs = 6 × 77.47 = 464.8 cm2
Cost = 464.8 × 0.35 = ₹ 162.68
Cost of making the designs ₹ 162.68
14Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is (A) p180 × 2πR (B) p180 × πR2 (C) p360 × 2πR (D) p720 × 2πR2
Area of sector = p360 × πR2 = p720 × 2πR2
(D) p720 × 2πR2
