SOLUTIONS
Unless stated otherwise, use π= $22/7$1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. Answer: Area of sector = $(θ/360)$ ×π $r^2$ = $60/360$ × $22/7 × 6 × 6$ $cm^2$ =$132/7$ $cm^2$2. Find the area of a quadrant of a circle whose circumference is 22 cm. Answer : Angle of quadrant = $90^0$ = $(90/360)$ ×π $r^2$ = $1/4$ ×π $r^2$ Given: circumference of circle = 22 cm $ 2πr=22 $ $ 2× 22/7 ×r=22 $ $ r=7/2$ cm Therefore ,Angle of quadrant = $1/4$ ×π $r^2$ = ${1/4} ×22/7× 7/2 × 7/2 $ = $77/8$ $cm^2$3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. Answer : Degree swept by minute hand in 60 minutes = $360^0$ So degree swept by minute hand in 1 minute = $360/60 $ =$6^0$ So, degree swept by minute hand in 5 minute = $6×5$ = $30^0$ So, area swept by minute hand in 5 minutes= $(30/360)$ ×π $r^2$ = $(30/360) ×22/7 ×14×14 $ $cm^2$ = $154/3 $ $cm^2$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 segment (iii) minor sector (iv) major sector. (Use π= 3.14) Answer Area of minor sector (AOBP) = $(θ/360)$ ×π $r^2$ = $(90/360)×22/7 × 10×10$ = 78.57 $cm^2$ Area of Major sector (AOBQ) = π $r^2$ - area of minor sector = $3.14 ×10×10 - 78.57 $ = $314 - 78.57 $ = 235.43 $cm^2$ Area of Minor segment(ABP) = Area of minor sector (AOBP) -Area of ∆ AOB Area of ∆ABC = $1/2 × base × height$ Height of ∆ABC = $sin 45^0 = h/r $ = $1/√2 = h/10 $ = $10/√2 = h $ = $h= 5√2 $ Base of ∆ABC = $√(r^2+r^2)$ = $√(2r^2)$ = $r√2$ = $10√2$ Area of ∆ABC = $1/2 ×10√2 ×5√2 $ = 50 $cm^2$ Area of Minor segment(ABP)= 78.57 - 50 = 28.58 $cm^2$ Area of Major segment(AQB) = Area of circle -Area of minor segment (ABP) = $3.14 ×10×10- 28.58$ = $314- 28.58$ = $285.42$ $cm^2$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. Solution: (i) Let $l$ be the length of arc $ l/{2πr} =θ/360$ $l= 60/360 × 2×22/7×21 $ $l= 22 $ $cm$ (ii) Area of minor sector form by the arc= $A/{πr^2} = θ/360 $ $A = 60/360 × 22/7×21×21$ $A = 1/6 × 22×3×21$ $A = 11×21$ $A = 231$ $cm^2$ Area of major sector = $πr^2 - area of minor sector$ Area of major sector = $22/7×21×21 - 231$ = $1155$ $cm^2$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) Ans: Area of Minor segment(ABP) = Area of minor sector (AOBP) -Area of ∆ AOB Area of minor sector (AOBP) = $(θ/360)$ ×π $r^2$ = $(60/360) ×22/7 × 15 × 15$ =$(1/6) ×22/7 × 15 × 15$ =$11/7 × 5 × 15$ =$117.86$ $cm^2$ Area of ∆ AOB = $1/2 × base × height $ Let height of the triangle be $h$. Therefore $sin 60^0 = h/r$ => $h= √3/2 × r$ => $h= {15√3}/2$ Base of triangle => $cos 60^0 = b/r$ => $b = r ×{cos60^0}$ => $b = 15 × 1/2 $ Base of triangle => $2 × 15 × 1/2$ => 15 $cm$ Area of ∆ AOB = $1/2 × 15 × {15√3}/2 $ = 97.31 $cm^2$ Area of Minor segment(ABP) = $117.86 - 97.31 $ $cm^2$ = $20.55 $ $cm^2$ Area of Major segment(AQB) = Area of circle -Area of minor segment (ABP) = $π r^2 - 5.36 $ = $22/7 ×15×15 - 20.77 $ = 686.37 $cm^2$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) Ans: Area of Minor segment(ABP) = Area of minor sector (AOBP) -Area of ∆ AOB Area of minor sector (AOBP) = $(θ/360)$ ×π $r^2$ =$(120/360) ×22/7 × 12 ×12$ =$(1/3) ×22/7 × 12 ×12$ =$22/7 × 4 ×12$ = 150.72 $cm^2$ Area of ∆ = $1/2 × base × height $ Height => $h=r×sin 30^0$ => $12 × 1/2$ => $6$ $cm$ and base = $2r× cos30^0$ => $2× 12 ×√3/2 $ => $ 20.76 $ $cm$ Area of ∆ = $1/2 × 20.76 × 6$ $cm^2$ = $62.28$ $cm^2$ Area of Minor segment(ABP) = Area of minor sector (AOBP) -Area of ∆ AOB = $(150.72 - 62.28)$ $cm^2$ = $88.44$ $cm^2$ Area of Major segment(AQB) = Area of circle -Area of minor segment (ABP) = $3.14× 12×12 - 88.44 $ $cm^2$ = 363.72 $cm^2$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) Ans: (i) area of field in which the horse can graze => area of sector = $(90/360)$ ×π $r^2$ =$(90/360) ×3.14×5×5$ =$(1/4) ×3.14×5×5$ =$19.625$ $m^2$ (ii) Increase in grazing area if the rope were 10 m long. =(area of sector with rope 10 $m$) - (area of sector with rope 5 $m$ long) area of sector with rope 10 $m$ = $(90/360) ×3.14 ×10×10 $ = $78.5$ $m^2$ Increase in grazing area if the rope were 10 m long = $78.5 - 19.625$ $m^2$ = $58.875$ $m^2$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. 12.12. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. 10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13).Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Ans: Angle which the consecutive ribs make = $360/8$ = $ 45^0$ Area between consecutive ribs = $(45/360)$ ×3.14 × 45×45 $ =$(1/8)$ ×22/7 × 45×45 $ =$44550/28$ $cm^2$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. Ans: Area of swept by each wiper = $(θ/360)$ ×π $r^2$ = $(115/360) ×22/7 ×25×25 $ = $(23/72) ×22/7 ×25×25 $ = $(23/36) ×11/7 ×25×25 $ = $627.50 $ $cm^2$ Total area swept = $2 ×627.50$ $cm^2$ = $1255$ $cm^2$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) 13. A round table cover has six equal designs as shown in Fig. 12.14. 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) 14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is (A) $p/180$ × 2 $πR$ (B) $p/180$ × $πR^2$ (C) $p/360$ × 2 $πR$ (D) $p/720$ × 2 $πR^2$