Class 8: Perimeter and Area of Plane Figures – Exercise 36(D) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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 $\displaystyle = 50 \times 3.5 \times 2 + \frac{1}{2} \times \{ \pi (14)^2 - \pi (10.5)^2 \} \times 2 = 619.5 \text{m}^2$