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

Question 1: Find the circumference and area of a circle whose radius is: (i) \displaystyle  35 \text{ cm } (ii) \displaystyle  4.2 \text{ cm } (iii) \displaystyle  15.4 \text{ cm }

Answer:

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

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

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

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

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

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

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Question 2: Find the circumference and area of a circle whose diameter is: (i) \displaystyle  17 .5 \text{ cm } (ii) \displaystyle  5.6 \text{ cm }

Answer:

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

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

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

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

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

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

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Question 3: Taking \displaystyle  \pi = 3.14 , find the circumference and area of a circle whose radius is: (i) \displaystyle  8 \text{ cm } (ii) \displaystyle  15 \text{ cm } (iii) \displaystyle  20 \text{ cm }

Answer:

(i) \displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 8 = 50.24 \text{ cm }

\displaystyle \text{Area of circle }   = \pi r^2 = 3.14 \times 8^2 = 200.96 \text{ cm }^2

(ii) \displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 15 = 94.2 \text{ cm }

\displaystyle \text{Area of circle }   = \pi r^2 = 3.14 \times 15^2 = 706.5 \text{ cm }^2

(iii) \displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 20 = 125.6 \text{ cm }

\displaystyle \text{Area of circle }   = \pi r^2 = 3.14 \times 20^2 = 1256 \text{ cm }^2

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

Answer:

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

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

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

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

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

Answer:

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

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

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

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

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

Answer:

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

\displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 20 = 125.6 \text{ cm }

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

\displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 25 = 157 \text{ cm }

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

\displaystyle \text{Circumference }   = 2 \pi r = 2 \times 3.14 \times 7 = 43.96 \text{ cm }

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

Answer:

Area of rectangular sheet of acrylic \displaystyle  = 36 \times 25 = 900 \text{ cm }^2

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

Area of remaining sheet \displaystyle  = 900-539 = 361 \text{ cm }^2

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

Answer:

Area of rectangular ground \displaystyle  = 80 \times 35 = 2800 \text{m}^2

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

Area of remaining sheet \displaystyle  = 2800-616 = 2184 \text{m}^2

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

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

Answer:

The diameter of the circle = Side of the square

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

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

Answer:

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

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

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

Answer:

\displaystyle \text{Perimeter of square }   = 4 \times 49.5 = 198 \text{ cm }

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

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

Answer:

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

\displaystyle \text{Perimeter of the circle }   = 4 \times 27.5 = 110 \text{ cm }

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

Distance covered by the motor cycle in 1 sec \displaystyle  = \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 \displaystyle  = \frac{8.80 \times 3600}{1000} = 31.68 km/hr

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

Answer:

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

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

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

Answer:

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

Perimeter \displaystyle  = 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 \displaystyle  17 \text{ cm } and \displaystyle  11 \text{ cm } .

Answer:

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

\displaystyle  \Rightarrow Area including the path \displaystyle  = 3850 \text{m}^2

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

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

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

Answer:

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

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

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

Answer:

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

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

Area of smaller circle \displaystyle  = 3.14 \times 20^2 = 1256 \text{ cm }^2

Width of the path \displaystyle  = 30 - 20 = 10 \text{ cm }

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Question 25: Find the area of the shaded region in the adjoining figure, the radii of bigger and smaller quadrants being \displaystyle  16 \text{ cm } and \displaystyle  12 \text{ cm } respectively.

Answer:

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

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

\displaystyle  = 88 \text{ cm }^2

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Question 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 \displaystyle  28 \text{ cm } and one end is a semi-circle on AB as diameter.

Answer:

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

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

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

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

Answer:

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

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

Perimeter \displaystyle  = 5 + 15 + 3.14 \times 2 + 12 = 38.28 \text{ cm }

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

Answer:

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

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

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

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

Answer:

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

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

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

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

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

Answer:

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

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

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