Note: Take \pi = \frac{22}{7} until and unless stated otherwise.

Question 1: Find the circumference and area of a circle whose radius is:

(i) 35  \ cm (ii) 4.2  \ cm (iii) 15.4 \ cm

Answer:

(i) Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 35 = 220 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 35^2 = 3850 \ cm^2

(ii) Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 4.2 = 26.4 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 35^2 = 55.44 \ cm^2

(iii) Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 15.4 = 96.8 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 15.4^2 = 745.36 \ cm^2

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Question 2: Find the circumference and area of a circle whose diameter is:

(i) 17 .5  \ cm (ii) 5.6  \ cm

Answer:

(i) Radius = \frac{d}{2} = \frac{17.5}{2} = 8.75 \ cm

Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 8.75 = 55 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 8.75^2 = 240.625 \ cm^2

(ii) Radius = \frac{d}{2} = \frac{5.6}{2} = 2.8 \ cm

Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 2.8 = 17.6 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 2.8^2 = 24.64 \ cm^2

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Question 3: Taking \pi = 3.14 , find the circumference and area of a circle whose radius is:

(i) 8  \ cm (ii) 15  \ cm (iii) 20 \ cm

Answer:

(i) Circumference = 2 \pi r = 2 \times 3.14  \times 8 = 50.24 \ cm

Area of circle = \pi r^2 = 3.14  \times 8^2 = 200.96 \ cm^2

(ii) Circumference = 2 \pi r = 2 \times 3.14  \times 15 = 94.2 \ cm

Area of circle = \pi r^2 = 3.14  \times 15^2 = 706.5 \ cm^2

(iii) Circumference = 2 \pi r = 2 \times 3.14  \times 20 = 125.6 \ cm

Area of circle = \pi r^2 = 3.14  \times 20^2 = 1256 \ cm^2

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Question 4: Find the area of a circle whose circumference is: (i) 55  \ cm (ii) 132  \ cm

Answer:

(i) Radius = \frac{Circumference}{2 \pi} = \frac{55}{2 \times \frac{22}{7}} = 8.75 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 8.75^2 = 240.625 \ cm^2

(ii) Radius = \frac{Circumference}{2 \pi} = \frac{132}{2 \times \frac{22}{7}} = 21 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times 21^2 = 1386 \ cm^2

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Question 5: Find the radius and circumference of a circle whose area is: (i) 154  \ cm^2 (ii) 24.64  \ m^2

Answer:

(i) Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{154}{\frac{22}{7}}} = 7 \ cm

Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 7 = 44 \ cm

(ii) Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{24.64}{\frac{22}{7}}} = 2.8 \ cm

Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 2.8 = 17.6 \ cm

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Question 6: Taking \pi = 3.14 , find the radius and circumference of a circle whose area is: (i) 1256  \ cm (ii) 1962.5  \ cm^2 (iii) 153.86  \ cm^2

Answer:

(i) Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{1256}{3.14}} = 20 \ cm

Circumference = 2 \pi r = 2 \times 3.14 \times 20 = 125.6 \ cm

(ii) Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{1962.5}{3.14}} = 25 \ cm

Circumference = 2 \pi r = 2 \times 3.14 \times 25 = 157 \ cm

(iii) Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{153.86}{3.14}} = 7 \ cm

Circumference = 2 \pi r = 2 \times 3.14 \times 7 = 43.96 \ cm

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Question 7: A rectangular sheet of acrylic is 36  \ cm by 25  \ cm . From it, 56 circular buttons, each of diameter 3.5  \ cm have been cut out. Find the area of the remaining sheet.

Answer:

Area of rectangular sheet of acrylic = 36 \times 25 = 900 \ cm^2

Area of 56 buttons = 56 \times \frac{22}{7} \times 1.75^2 = 539 \ cm^2

Area of remaining sheet = 900-539 = 361 \ cm^2

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Question 8: A rectangular ground is 80  \ m long and 35  \ m broad. In the middle of the ground, there is a circular tank of radius 14  \ m . Find the cost of turfing the remaining portion at the rate of Rs. 21.50 sq. meter.36d9

Answer:

Area of rectangular ground = 80 \times 35 = 2800 \ m^2

Area of circular tank = \frac{22}{7} \times 14^2 = 616 \ m^2

Area of remaining sheet = 2800-616 = 2184 \ m^2

Cost of turfing = 2184 \times 21.50 = 46956 Rs.

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Question 9: Find the area of a circle inscribed in a square of side 28  \ cm .

Answer:36d8

The diameter of the circle = Side of the square

Area of circle = \frac{22}{7} \times 14^2 = 616 \ cm^2

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Question 10: A wire is in the form of a circle of radius 28  \ cm . It is straightened and bent into the form of a square. What is the length of the side of the square?

Answer:

Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 28 = 176 \ cm

Side of the square = \frac{176}{4} = 44 \ cm

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Question 11: A wire is in the form of a square of side 49.5  \ cm . It is straightened and bent into a circle. What is the radius of the circle so formed?

Answer:

Perimeter of square = 4 \times 49.5 = 198 \ cm

Radius = \frac{198}{2 \pi} = \frac{198}{2 \times \frac{22}{7}} =  31.5 \ cm

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Question 12: A wire when bent in the form of a square, encloses an area of 756.25  \ cm^2 . If the same wire is bent to form a circle, what will be the radius of the circle so formed?

Answer:

Side of the square = \sqrt{756.25} = 27.5 \ cm

Perimeter of the circle = 4 \times 27.5 = 110 \ cm

Radius of the circle = \frac{110}{2 \pi} = \frac{110}{2 \times \frac{22}{7}} = 17.5 \ cm

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Question 13: The radius of a wheel is 63  \ cm . Find the distance travelled by it m: (i) one revolution (ii) 200 revolutions

Answer:

(i) Distance traveled in one revolution = 2 \pi r = 2 \times \frac{22}{7} \times 63 = 396 \ cm = 3.96 \ m

(ii) Distance traveled in 200 revolution = 200 \times 3.96 = 792 \ m

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Question 14: How many revolutions would a cycle wheel of diameter 40  \  cm make to cover a distance of 176  \  m ?

Answer:

Circumference of the wheel = 2 \pi r = 2 \times \frac{22}{7} \times 20 = 125.71 \ cm = 1.2571 \ m

Number of revolution = \frac{176}{1.2571} = 140

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Question 15: The wheel of motor-cycle, 70  \ cm in diameter. is making 40 revolutions in every 10 seconds. Find the speed of the motor-cycle in km per hour.

Answer:

Distance covered by the motor cycle in 1 sec = \frac{40 \times 2 \pi r}{10} = \frac{40 \times 2 \times \frac{22}{7} \times 35}{10} =  880 \ cm = 8.80 \ m

Speed of motor cycle = \frac{8.80 \times 3600}{1000} = 31.68 km/hr

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Question 16: The wheel of a car rotates 1000 times in travelling a distance of 1.65  \ km . Find the diameter of the wheel.

Answer:

Circumference of the wheel = \frac{1650}{1000} = 1.65 \ m

Diameter of the wheel = 2 \times \frac{1.65}{2 \pi} = \frac{1.65}{2 \times  \frac{22}{7}} = 0.525 \ m = 52.5 \  cm

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Question 17: The shape of a park is a rectangle bounded by semi-circles at the ends, each of radius 17 .5 \ m , as shown in the adjoining figure. Find the area and the perimeter of the park.36d7

Answer:

Area = \frac{22}{7} \times  17.5^2 + 35 \times 40 = 2362.5 \ m^2

Perimeter = 2 \times \frac{22}{7} \times 17.5 + 40 + 40 = 190 \ m

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Question 18: Find the area of the space enclosed by two concentric circles of radii 17  \ cm and 11  \ cm .

Answer:

Enclosed space = \frac{22}{7} \times 17^2 - \frac{22}{7} \times 11^2 = 528 \ cm^2

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Question 19: The diameter of a circular park is 52  \ m . On its outside, there is a path 4 \ m   wide, running around it. Find the cost of turfing the path at Rs. 22.50 per square meter.

Answer:

Enclosed space = \frac{22}{7} \times 30^2 - \frac{22}{7} \times 26^2 = 704 \ m^2

Cost of turfing = 704 \times 22.50 = 15840 \ Rs.

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Question 20: A circular field of radius 34  \ m has a circular path of uniform width of 5  \ m along and inside its boundary. Find the cost of paving the path at Rs. 36 per sq. meter.

Answer:

Enclosed space = \frac{22}{7} \times 34^2 - \frac{22}{7} \times 29^2 = 990 \ m^2

Cost of turfing = 990 \times 36 = 35640 \ Rs.

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Question 21: The area of a circular pond is 346.5  \ m^2 . A path of width 3.5  \ m runs all around it, on the outside. Calculate the area of the path.

Answer:

Radius = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{346.5}{\frac{22}{7}}} = 10.5 \ m

Area of the path = \frac{22}{7} \times 14^2 - \frac{22}{7} \times 10.5^2 = 269.5 \ m^2

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Question 22: The area of a circular grass lawn is 2464  \ m^2 . A path of uniform width was laid all around it, on the outside. The area of the path is 1386  \ m^2 . Find (i) the radius of the grass lawn; (ii) the width of the path.

Answer:

Radius of circular grass lawn = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{2464}{\frac{22}{7}}} = 28 \ m

Therefore 1386 = Area \ including \ the \ path - 2464

\Rightarrow Area including the path = 3850 \ m^2

Radius of circular grass lawn + Path  = \sqrt{\frac{Area}{\pi}} = \sqrt{\frac{3850}{\frac{22}{7}}} = 35 \ m

Therefore the width of the path = 35 - 28 = 7 \ m

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Question 23: The outer edge of a circular running track is 264  \ m long. It is widened by 10.5  \ m all around on the outside. Find the area of the widened part.

Answer:

Radius of circular running track  = \frac{Circumference}{2 \pi} = \frac{264}{2 \times \frac{22}{7}} = 42 \ cm

Area of the path = \frac{22}{7} \times (42+10.5)^2 - \frac{22}{7} \times 42^2 = 3118.5 \ m^2

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Question 24: The area of a ring is 1570  \ cm^2 and the radius of the outer circle is 30  \ cm . Find: (i)  the area of the smaller circle (ii) the width of the ring (Take \pi = 3.14 ).

Answer:

Given: \pi (30)^3 - \pi r^2 = 1570

\Rightarrow r = \sqrt{30^2 - \frac{1570}{3.14}} = 20 \ cm

Area of smaller circle = 3.14 \times 20^2 = 1256 \ cm^2

Width of the path = 30 - 20 = 10 \ cm

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36d6Question 25: Find the area of the shaded region in the adjoining figure, the radii of bigger and smaller quadrants being 16  \ cm and 12  \ cm respectively.

Answer:

Area of the shaded region = \frac{1}{4} \times (\pi (16)^2 - \pi (12)^2)

= \frac{1}{4} \times ( \frac{22}{7} \times  (16)^2 - \frac{22}{7} \times (12)^2)

= 88 \ cm^2

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36d5Question 26: The perimeter and the area of the shaded region in the adjoining figure, it being given that OAB is an equilateral triangle of side 28  \ cm and one end is a semi-circle on AB as diameter.

Answer:

Perimeter of the figure = \pi r + 28 + 28 = \frac{22}{7} \times 14 + 56 = 100 \ cm

Area of the shaded region = \frac{1}{2} \times \frac{22}{7} \times (14)^2 + \frac{1}{2} \times 28 \times 24.248 = 647.472 \ cm^2

Note: The height of the triangle = \sqrt{28^2 - 14^2} = 24.248 \ cm

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36d4Question 27: Find the perimeter and the area of the adjoining figure, it being given that AB \parallel CD, AB = 12  \ cm, CD = 15  \ cm, BD = 4 \ cm and one end of the figure is a semi-circle with BD as diameter (Take \pi = 3.14 ).

Answer:

Area of the figure = 12 \times 4 + \frac{1}{2} \times 3.14 \times (2)^2 + \frac{1}{2} \times 3 \times 4 = 60.28 \ cm^2

BE = \sqrt{3^2 + 4^2} = 5 \ cm

Perimeter = 5 + 15 + 3.14 \times 2 + 12 = 38.28 \ cm

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Question 28: Find the area of the shaded region in each of the following figures:36d3

Answer:

(i) Area of the shaded region = 21^2 - \pi ( \frac{21}{2} )^2 = 94.5 \ m^2

(ii) Area of the shaded region = 21^2 - \frac{1}{4} \pi ( \frac{10.5}{2} )^2 \times 4 = 354.375 \ m^2

(iii) Area of the shaded region = 20 \times 7 - \pi ( \frac{7}{2} )^2 = 101.5 \ m^2 

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Question 29: Find the area of the shaded region in each of the following figures:36d2

Answer:

(i) Area of the shaded region = 3.5 \times 12 \times 2 + \{\pi \times 7^2 - \pi \times 3.5^2 \} \times \frac{1}{2} = 141.75 \ m^2

(ii) Area of the shaded region = 50 \times 3.5 \times 2 + \frac{1}{2} \times \{ \pi (14)^2 - \pi (10.5)^2 \} \times 2 = 619.5 \ m^2

(iii) Area of the shaded region = \frac{1}{2} \{ \pi (10.5)^2 - \pi (5.25)^2 \} \times 2 = 259.875 \ m^2

(iv) Area of the shaded region = \frac{1}{2} \times \pi (10.5)^2 - \frac{1}{2} \times \pi (3.5)^2 \times 2 = 134.75 \ m^2

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Question 30: Find the area of the shaded region in each of the following figures:36d1

Answer:

(i) a + \frac{a}{2} + \frac{a}{2} = 42 \Rightarrow a = 21 \ cm

Area of the shaded region = 21 \times 21 + 2 \times \pi (10.5)^2 = 1134 \ cm^2

(ii) Area of the shaded region = \frac{1}{2} \times (21)^2 + 2 \times \frac{1}{2} \pi \times (10.5)^2 = 1039.5 \ cm^2

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