Exercise 12.1
Unless stated otherwise, use π = 22/7
1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
Circumference of new circle = 2π(AC + AB)
= 2π(19+9)
= 2π×28
Radius of new circle = Circumference of new circle ÷2π
Radius of new circle = 2π×28÷2π
Radius of circle = 28 cm. Ans.
2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
Area of new circle = π(AB² + AC²)
Now putting the value of AB and AC
Area of new circle = π(6² + 8²)
Area of new circle = π(36+64)
Area of new circle = π×100
Radius of new circle = √Area of new circle ÷π
Radius of new circle = √2π×100÷π
Radius of new circle = √100
Radius of new circle = 10 cm Ans.
3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Solution:
Radius of Gold scoing= 21/2 cm. = 10.5 cm
Area of Gold scoring = π(10.5)2
or, Area of Gold scoring = 22×10.5×10.5÷7
or, Area of Gold scoring = 346.5 cm2 Ans.
Radius of Red scoring = 10.5+10.5 = 21 cm.
Area of Red scoring = π(21)2
or, Area of Red scoring = 22×441÷7
or, Area of Red scoring = 1386
Radius of Blue scoring = 21 + 10.5 = 31.5 cm.
Area of Blue scoring = π(31.5)2
Area of Blue scoring = 22×31.5×31.5÷7
or, Area of Blue scoring = 3118.5 cm2 Ans.
Radius of Black scoring = 31.5 + 10.5 = 42 cm.
Area of Black scoring = π(42)2
Area of Black scoring = 22×42×42÷7
= 22×42×6 = 132×42 = 5544 cm2
Radius of white scoring = 42 + 10.5 = 52.5 cm.
Area of White scoring = π(52.5)2
or, Area of White scoring = 22×52.5×52.5÷7 = 22×52.5×7.5 = 8662.5 cm2
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
Distance covered by car in 10 minutes = speed×time
or, Distance = 60×10/60. (1 minute = 1/60 hour)
Distance = 11 km.
or, Distance = 11×1000×100 cm.
Circumference of the wheel = π×Diameter
or, Circumference = 80×22/7
No. of revolutions done by wheel = Distance/Circumference
or, No. of revolution = 11×1000×10×7÷80×22
No. of revolution = 4375 Ans.
5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units
Solution:
From question it is clear that
Circumference of the circle = Area of the circle
2πr = πr2
2 = r
r = 2 unit.
Hence correct answer or option = (A)
Exercise 12.2
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°.
From question it is clear that
Radius of the circle = r = 6 cm.
Central Angle = θ = 60°
We know that
Area of sector of a circle = πr²θ÷360°
Now putting the value of r, θ and π
Area of sector = 22×6×6×60°÷360°×7
= 22×6÷7
= 132÷7
= 18.88 cm² Ans.
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
From question it is clear that
2πr = 22 cm
r = 22÷2π
r = 22×7÷22×2
= 3.5 cm.
θ = 90°
Area of the quadrant = πr²θ÷360°
Now putting the value of π, r and θ
Area of the quadrant = 22×3.5×3.5×90°÷360°
= 11.5×3.5×3.5
= 140.875 cm² Ans.
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
From question it is clear that
Angle subtended by minute hand in 5 minutes = 360°×5÷60°
= 30°
Length of minute hand = r = 14 cm.
Area swept by minute hand in 5 minutes = πr²θ÷360
Now putting the value of r, π and θ
Area swept by minute hand in 5 minutes = 22×14×14×30°÷360°
= 359.33 cm² Ans.
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)
From question it is clear that
AB = AC = r = 10 cm.
θ = 90°
Area of sector = πr²θ÷360°
Now putting the value of r, π and θ.
Area of sector = 3.14×10×10×90°÷360°
= 314÷4
= 78.57 cm²
Area of right triangle ABC = AB×AC÷2
= 10×10÷2
= 50 cm²
Area of minor segment = Area of corresponding sector - Area of corresponding triangle = 78.57 - 50 = 28.57 cm² Ans.
Area of Circle = πr²
or, Area of Circle = 3.14×10×10
= 314 cm²
Area of major sector = Area of Circle - Area of minor sector
or, Area of major sector = 314 - 78.57 = 235.43 cm² Ans.
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:
From question it is clear that
r = 21 cm.
θ = 60°
Length of arc = Circumference of the circle×θ÷360°
= 2πr×60°÷360° = 2×22×21×60°÷360°×7 = 22 cm.
Area of the sector = πr²θ÷360° = 22×21×21×60°÷360°×7 = 231 cm².
Area of segment = Area of the sector-Area of corresponding triangle = 231 -√3×21×21÷4 = 231-190.953 = 40.047 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)
Solution:
From question it is clear that
AB = AC = r = 15 cm.
θ = 60°
Triangle inscribed with centre will be equilateral triangle.
Area of sector = πr²θ÷360°
= 3.14×15×15×60°÷360°
= 1.57×15×5
= 117.75 cm²
Area of corresponding triangle = √3r²÷4
= √3×15×15÷4
= 1.73×225÷4
= 97.3125
Area of minor segment = 117.75 - 97.3125 = 20.4375 cm²
Area of the Circle = πr²
3.14×15×15 = 706.5 cm²
Area of major segment = Area of the Circle - Area of minor 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)
Solution:
From question it is clear that
r = AB = AC = 12 cm.
θ = 120°
Consider in right triangle ADB.
AD = AB×cos60°
AD = 12×1÷2
or, AD = 6.
Similarly,
BD = AB×sin60°
or, BD = 12×√3÷2
or, BD = 6×√3
BC = 2×BD = 12√3.
Area of sector ABC = πr²θ÷360°
Now putting the value
or, Area of sector ABC = 3.14×12²×120°÷360°
or, Area of sector ABC = 150.72 cm².
Area of triangle ABC = AD×BC÷2
or, Area of triangle ABC = 6×12×√3÷2
or, Area of triangle ABC = 62.28 cm².
Area of the corresponding segment of the Circle = Area of sector ABC - Area of triangle ABC
or, Area of the corresponding segment of the Circle = 150.72 - 62.28 = 88.44 cm² Ans.
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. 12.11). Find
(i) the area of that part of the field in which the horse can graze.
Fig. 12.11
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Solution:
From question it is clear that
(i)
r = 5 m.
θ = 90°
Grazing area of the field = πr²θ÷360°
or, Grazing area of the field = 3.14×5×5×90°÷360°
or, Grazing area of the field = 19.625 m².
(ii)
When
r = 10 m.
Grazing area of the field = πr²θ÷360°
or, Grazing area of the field = 3.14×10×10×90°÷360°
or, Grazing area of the field = 78.5 cm²
The increase in the grazing area = 78.5 - 19.625 = 58.875 m².
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.
Fig. 12.12
Solution:
(i)
Total length of the silver wire = π×diameter + 5×diameter
or, total length of the wire = (22×35÷7) + 5×35
or, total length of the wire = 110 + 175
or, total length of the wire = 285 mm.
(ii)
The area of each sector of the brooch = π(d/2)²(360°/10)÷360°
or, The area of the sector of the brooch = 22×35×35×360°÷7×2×2×10×360°
or, The area of the sector of the brooch = 96.25 mm².
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.
Fig. 12.13
Solution:
From question it is clear that
r = 45 cm.
θ = 360°÷8 = 45°
Area between two consecutive ribs = πr²θ÷360°
or, Area between two consecutive ribs = 3.14×45²×45°÷360°
or, Area between two consecutive ribs = 3.14×45×45÷8
or, Area between two consecutive ribs = 794.8125 cm² Ans.
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.
Solution:
From question it is clear that
r = 25 cm.
θ = 115°
Total area cleaned at each sweep of the blades = 2×πr²θ÷360°
or, Total area cleaned at each sweep of the blades = 2×3.14×25×25×115°÷360°
or, Total area cleaned at each sweep of the blades = 1,253.82 cm² Ans.
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)
Solution:
From question it is clear that
r = 16.5 km.
θ = 80°
The area of the sea over which the ships are warned = πr²θ÷360°
or, The area of the sea over which the ships are warned = 3.14×16.5²×80°÷360°
or, The area of the sea over which the ships are warned = 3.14×16.5×16.5×2÷9
or, The area of the sea over which the ships are warned = 189.97 km² Ans.
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)
Fig. 12.14
From question it is clear that
r = 28 cm.
θ = 360°÷6 = 60°
Area of the sector = πr²θ÷360°
or, Area of the sector = 22×28²×60°÷360°×7
or, Area of the sector = 22×4×28÷6
or, Area of the sector = 410.67 cm²
Area of the corresponding equilateral triangle = √3×28²÷4 (side = r = 28 cm)
or, Area of the corresponding equilateral triangle = √3×7×28
or, Area of the corresponding equilateral triangle = 1.7×7×28
or, Area of the corresponding equilateral triangle = 332.2 cm²
Area of design = Area of the sector - Area of the corresponding equilateral triangle
or, Area of design = Area of the sector - Area of the corresponding equilateral triangle
or, Area of design = 410.67 - 332.2
or, Area of design = 78.47 cm².
Total area of design = 6×78.47
Cost of design = Rs. 0.35×6×78.47
or, Cost of design = Rs. 164.79 Ans.
14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
Solution:
The correct answer is (D)
Which is just equal to πr²θ÷360°
