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) $15.4 \ cm$

(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) $5.6 \ cm$

(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) $20 \ cm$

(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) $132 \ cm$

(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) $24.64 \ m^2$

(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) $153.86 \ cm^2$

(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.

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.

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 $28 \ cm$.

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?

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?

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?

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

(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 $176 \ m$ ?

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.

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.

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.

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 $11 \ cm$.

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.

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.

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.

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.

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.

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 $\pi = 3.14$).

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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Question 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.

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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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 $28 \ cm$ and one end is a semi-circle on AB as diameter.

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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Question 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 $\pi = 3.14$).

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:

(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:

(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:

(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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