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Exercise - 12.2

Areas Related To Circle

**Question-1 :-** Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. (Take π = 22/7).

Let OACB be a sector of the circle making 60° angle at centre O of the circle. Radius of circle = 6 cm Area of sector OACB = θ/360 x πr² = 60/360 x 22/7 x 6 x 6 = 1/6 x 22/7 x 6 x 6 = 132/7 cm²

**Question-2 :-** Find the area of a quadrant of a circle whose circumference is 22 cm. (Take π = 22/7).

Let the radius of the circle be r. Circumference = 22 cm 2πr = 22 r = 22/2π r = 11/π cm Quadrant of circle will subtend 90° angle at the centre of the circle. Area of such quadrant of the circle = θ/360 x πr² = 90/360 x π x 11/π x 11/π = 1/4 x 121/π = 121/4π = (121 x 7)/(4 x 22) = 77/8 cm²

**Question-3 :-** The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (Take π = 22/7).

We know that in 1 hour = 60 minutes, the minute hand rotates 360°. In 5 minutes, minute hand will rotate = 360/60 x 5 = 30° Radius of circle = 14cm Area of sector of 30° = θ/360 x πr² = 30/360 x 22/7 x 14 x 14 = 1/12 x 22 x 2 x 14 = 154/3 cm²

**Question-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)

Let AB be the chord of the circle subtending 90° angle at centre O of the circle. Radius of circle = 10cm Area of major sector OADB = θ/360 x πr² = (360 - 90)/360 x 3.14 x 10 x 10 = 270/360 x 3.14 x 10 x 10 = 235.5 cm² Area of minor sector OACB = θ/360 x πr² = 90/360 x 3.14 x 10 x 10 = 1/4 x 3.14 x 100 = 78.5 cm² Area of ΔOAB = 1/2 x OA x OB = 1/2 x 10 x 10 = 50 cm² Area of minor segment ACB = Area of minor sector OACB − Area of ΔOAB = 78.5 − 50 = 28.5 cm²

**Question-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 (Take π = 22/7).

Radius (r) of circle = 21 cm Angle subtended by the given arc = 60° Length of arc ACB = θ/360 x 2πr = 60/360 x 2 x 22/7 x 21 = 1/6 x 2 x 22 x 3 = 22 cm Area of sector OACB = θ/360 x πr² = 60/360 x 22/7 x 21 x 21 = 1/6 x 22 x 3 x 21 = 11 x 21 = 231 cm² In ΔOAB, ∠OAB = ∠OBA (As OA = OB) ∠OAB + ∠AOB + ∠OBA = 180° 2∠OAB + 60° = 180° ∠OAB = 60° Therefore, ΔOAB is an equilateral triangle. Area of ΔOAB = √3/4 x (side)² = √3/4 x (21)² = 441√3/4 cm² Area of segment ACB = Area of sector OACB − Area of ΔOAB = (231 - 441√3/4) cm²

**Question-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)

Radius (r) of circle = 15 cm subtends angle = 60° Area of sector OPRQ = θ/360 x πr² = 60/360 x 3.14 x 15 x 15 = 1/6 x 3.14 x 15 x 15 = 117.75 cm² In ΔOPQ, ∠OPQ = ∠OQP (As OP = OQ) ∠OPQ + ∠OQP + ∠POQ = 180° 2 ∠OPQ = 120° ∠OPQ = 60° Therefore, ΔOPQ is an equilateral triangle. Area of ΔOPQ = √3/4 x (side)² = √3/4 x (15)² = 225√3/4 = 56.25 x 1.73 = 97.3125 cm² Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ = 117.75 − 97.3125 = 20.4375 cm² Area of major segment PSQ = Area of circle − Area of segment PRQ = πr² - 20.4375 = 3.14 x 15 x 15 - 20.4375 = 706.5 - 20.4375 = 686.0625 cm²

**Question-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)

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST. SV = VT Radius of circle (OS) = 12cm In ΔOVS, OV/OS = cos 60° OV/12 = 1/2 OV = 6 cm SV/SO = sin 60° SV/12 = √3/2 SV = 6√3 cm ST = 2SV = 2 x 6√3 = 12√3 cm Area of ΔOST = 1/2 x ST x OV = 1/2 x 12√3 x 6 = 36√3 = 36 x 1.73 = 62.28 cm² Area of sector OSUT = θ/360 x πr² = 120/360 x 3.14 x 12 x 12 = 1/3 x 3.14 x 144 = 150.72 cm² Area of segment SUT = Area of sector OSUT − Area of ΔOST = 150.72 − 62.28 = 88.44 cm²

**Question-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 (see Fig.). 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)

It can be observed that the horse can graze a sector of 90° in a circle of 5 m radius. (i) Area that can be grazed by horse = Area of sector OACB = θ/360 x πr² = 90/360 x 3.14 x 5 x 5 = 1/4 x 3.14 x 25 = 19.625 m² (ii) Area that can be grazed by the horse when length of rope is 10 m long = θ/360 x πr² = 90/360 x 3.14 x 10 x 10 = 1/4 x 3.14 x 100 = 78.5 m² Increase in grazing area = (78.5 − 19.625) m² = 58.875 m²

**Question-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. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch. (Take π = 22/7).

(i) Total length of wire required will be the length of 5 diameters and the circumference of the brooch. Radius of circle = 35/2 mm Circumference of brooch = 2πr = 2 x 22/7 x 35/2 = 110 mm Length of wire required = 110 + 5 × 35 = 110 + 175 = 285 mm (ii) It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle. Therefore, area of each sector = θ/360 x πr² = 36/360 x 22/7 x 35/2 x 35/2 = 1/10 x 11 x 5 x 35/2 = 385/4 mm²

**Question-10 :-** An umbrella has 8 ribs which are equally spaced (see Fig.). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. (Take π = 22/7)

There are 8 ribs in an umbrella. Radius of flat circle = 45cm The area between two consecutive ribs is subtending (360°/8 = 45°) at the centre of the assumed flat circle. Area between two consecutive ribs of circle = θ/360 x πr² = 45/360 x 22/7 x 45 x 45 = 11/28 x 2025 = 22275/28 cm²

**Question-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. (Take π = 22/7)

It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius. Area of such sector = θ/360 x πr² = 115/360 x 22/7 x 25 x 25 = 23/72 x 22/7 x 625 = 158125/252 cm² Area swept by 2 blades = 2 x 158125/252 = 158125/126 cm²

**Question-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)

It can be observed from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius. Area of sector OACB = θ/360 x πr² = 80/360 x 3.14 x 16.5 x 16.5 = 2/9 x 3.14 x 16.5 x 16.5 = 189.97 cm²

**Question-13 :-** A round table cover has six equal designs as shown in Fig. 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)

It can be observed that these designs are segments of the circle. Consider segment APB. Chord AB is a side of the hexagon. Radius of cover = 28cm Each chord will substitute 360°/6 = 60° at the centre of the circle. In ΔOAB, ∠OAB = ∠OBA (As OA = OB) ∠AOB = 60° ∠OAB + ∠OBA + ∠AOB = 180° 2∠OAB = 180° − 60° = 120° ∠OAB = 60° Therefore, ΔOAB is an equilateral triangle. Area of ΔOAB = √3/4 x (side)² = √3/4 x (28)² = 196 x 1.7 = 333.2 cm² Area of sector OAPB = θ/360 x πr² = 60/360 x 22/7 x 28 x 28 = 1/6 x 22 x 4 x 28 = 1232/3 cm² Area of segment APB = Area of sector OAPB − Area of ΔOAB = (1232/3 - 333.2)cm² Therefore, area of designs = 6 x (1232/3 - 333.2) = 2464 - 1999.2 = 464.8 cm² Cost of making 1 cm² designs = ₹ 0.35 Cost of making 464.76 cm² designs = 464.8 x 0.35 = ₹ 162.68

**Question-14 :-** Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is

(i) P/180 x 2πR (ii) P/180 x πR² (iii) P/360 x 2πR (iv) P/720 x 2πR²

We know that area of sector of angle θ = θ/360 x πR² Area of sector of angle P = P/360 x πR² = P/720 x 2πR² Hence, (D) is the correct answer.

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