Question 1: Find the circumference and area of a circle of radius $4.2 \ cm$.

Radius $= 4.2 \ cm$

Circumference of circle $= 2 \pi r = 2 \times$ $\frac{22}{7}$ $\times 4.2 = 26.4 \ cm$

Area of circle $= \pi r^2 =$ $\frac{22}{7}$ $\times (4.2)^2 = 55.44 \ cm^2$

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Question 2: Find the circumference of a circle whose area is $301.84 \ cm^2$.

Area $= 301.84 \ cm^2$

Therefore $\pi r^2 = 301.84$

$\Rightarrow r^2 =$ $\frac{7}{22}$ $\times 301.84 = 96.04$

$\Rightarrow r = 9.8 \ cm$

Therefore circumference of the circle $= 2 \pi r = 2 \times$ $\frac{22}{7}$ $\times 9.8 = 61.6 \ cm$

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Question 3: Find the area of a circle whose circumference is $44 \ cm$.

Circumference $= 44 \ cm$

Therefore $2 \pi r = 44$

$\Rightarrow r = 2 \times$ $\frac{7}{22}$ $\times 44 = 28 \ cm$

Therefore Area $= \pi r^2 =$ $\frac{22}{7}$ $\times (28)^2 = 2764 \ cm^2$

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Question 4: The circumference of a circle exceeds the diameter by $16.8 \ cm$. Find the circumference of the circle.

Given: $Circumference - diameter = 16.8$

$\Rightarrow 2 \pi r - 2r = 16.8$

$\Rightarrow r =$ $\frac{16.8}{2\pi -2}$ $=$ $\frac{16.8 \times 7}{2 \times 22 - 2 \times 7}$ $= 3.92$

Therefore Circumference $= 2 \pi r = 2 \times$ $\frac{22}{7}$ $\times 3.92 = 24.64 \ cm$

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Question 5: A horse is tied to a pole with $28 \ m$ long string. Find the area where the horse can graze.

Area that the horse can graze $= \pi r^2 =$ $\frac{22}{7}$ $\times (28)^2 = 2464 \ m^2$

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Question 6: A steel wire when bent in the form of a square encloses an area of $121 \ cm^2$. If the same wire is bent in the form of a circle, find the area of the circle.

Area of square $= 121 \ cm^2$

Let the side of the square $= a$

$\Rightarrow a^2 = 121 \Rightarrow a = 11 \ cm$

Therefore Perimeter $= 4 \times 11 = 44 \ cm$

Let $r$ be the radius of the circle

Therefore $2 \pi r = 44$

$\Rightarrow r =$ $\frac{7 \times 44}{2 \times 22}$ $= 7 \ cm$

Therefore Area of circle $= \pi r^2 =$ $\frac{22}{7}$ $\times (7)^2 = 154 \ cm^2$

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Question 7: The diameters of the front and rear wheels of a tractor are $80 \ cm$ and $2 \ m$ respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in $1400$ revolutions.

Diameter of front wheel $= 80 \ cm$

Diameter of rear wheel $= 2 \ m$

Distance covered by front wheel $= 1400 \times 2 \times$ $\frac{22}{7}$ $\times 0.40 = 3520 \ m$

Let no of revolution that the rear wheel $= n$

Therefore $n \times 2 \times$ $\frac{22}{7}$ $\times 1 = 3520$

$\Rightarrow n =$ $\frac{7 \times 3520}{2 \times 22}$ $= 560$

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Question 8: A copper wire when bent in the form of a square encloses an area of $121 \ cm^2$ . lf the same wire is bent into the form of a circle, find the area of the circle.

Area of square $= 121 \ cm^2$

Let the side of the square $= a$

$\Rightarrow a^2 = 121 \Rightarrow a = 11 \ cm$

Therefore Perimeter $= 4 \times 11 = 44 \ cm$

Let $r$ be the radius of the circle

Therefore $2 \pi r = 44$

$\Rightarrow r =$ $\frac{7 \times 44}{2 \times 22}$ $= 7 \ cm$

Therefore Area of circle $= \pi r^2 =$ $\frac{22}{7}$ $\times (7)^2 = 154 \ cm^2$

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Question 9: The circumference of two circles are in the ratio $2: 3$. Find the ratio of their areas.

Let the radius of first circle $= r_1$

Let the radius of second circle $= r_2$

Therefore $\frac{2\pi r_1}{2 \pi r_2}$ $=$ $\frac{2}{3}$

$\Rightarrow$ $\frac{r_1}{r_2}$ $=$ $\frac{2}{3}$

Therefore Ratios of their areas $=$ $\frac{\pi {r_1}^2}{\pi {r_2}^2}$ $= \Big($ $\frac{r_1}{r_2}$ $\Big)^2 =$ $\frac{4}{3}$

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Question 10: The side of a square is $10 \ cm$. Find the area of circumscribed and inscribed circles.

When square is inscribed in the circle

Therefore diameter $= 10\sqrt{2}$

Therefore radius of circle $= 5\sqrt{2}$

Therefore are of circle $= \pi (5\sqrt{2})^2 = 50\pi = 157.143 \ cm^2$

When the circle is inscribed in the square

Diameter $= 10 \ cm$

Therefore radius $= 5 \ cm$

Therefore Area of circle $= \pi (5)^2 = 25 \pi = 78.57 \ cm^2$

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Question 11: The sum of the radii of two circles is $140 \ cm$ and the difference of their circumferences is $88 \ cm$. Find the diameters of the circles.

Let the two radii be $r_1$ and $r_2$

Therefore $r_1 + r_2 = 140$ … … … … … i)

$2\pi r_1 - 2\pi r_2 = 88$ … … … … … ii)

From i) and ii) $r_2 = 140 - r_1$

Therefore $2\pi (r_1-r_2) = 88$

$\Rightarrow 2 \pi (r_1 - (140 - r_1)) = 88$

$2\pi (2r_1 - 140) = 88$

$2r_1 = \Big($ $\frac{88}{2} \times \frac{22}{7}$ $\Big) + 140$

$\Rightarrow 2r_1 = 154$

$\Rightarrow r_1 = 77 \ cm$

Therefore $r_2 = 140 - r_1 = 140 - 77 = 63 \ cm$

Diameter of circles are $154 \ cm$ and $126 \ cm$

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Question 12: Find the area of the circle in which a square of area $64 \ cm^2$ is inscribed. $($ Take $\pi = 3.14)$

Area of square $= 64 \ cm^2$

$\Rightarrow$ Side of square$= 8 \ cm$

$\Rightarrow$ Diameter of circle $= 8\sqrt{2} \ cm$

$\Rightarrow$ Radius of the circle $= 4\sqrt{2} \ cm$

Hence Area of circle $= \pi (4\sqrt{2})^2 = 100.48 \ cm^2$

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Question 13: A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be $Rs. \ 2640$ at the rate of $Rs. \ 12$ per meter. Then, the field is to be thoroughly ploughed at the cost of $Rs. \ 0.50 \ per \ m$. What is the amount required to plough the field? $($ Take $\pi =$ $\frac{22}{7}$ $)$

Cost of fencing $= 2640 \ Rs.$

Circumference of the field $=$ $\frac{2640}{12}$ $= 220 \ m$

Therefore $2 \pi r = 220$

$\Rightarrow r =$ $\frac{220 \times 7}{2 \times 22}$ $= 35 \ m$

Area of the field $= \pi r^2 = \pi (35)^2 = 3850 \ m^2$

Therefore cost of ploughing $= 3850 \times 0.5 = 1925 \ Rs.$

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Question 14: If square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Let the side of the square $= a$

Therefore diameter of circle $= a\sqrt{2}$

Therefore radius of the circle $=$ $\frac{a}{\sqrt{2}}$

Therefore ratios of their areas $=$ $\frac{\pi (\frac{a}{\sqrt{2}})^2}{a^2}$ $=$ $\frac{\pi}{2}$

Hence the ratios of their area is $\pi : 2$

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Question 15: A park is in the form of a rectangle $120 \ m \times 100 \ m$. At the center of the park there is a circular lawn. The area of park excluding lawn is $8700 \ m^2$. Find the radius of the circular lawn. $($ Take $\pi =$ $\frac{22}{7}$ $)$

Let the radius of the circle $= r$

Area of the park $= 100 \times 120 = 12000 \ m^2$

Are of circle $= \pi r^2$

Therefore $12000 = \pi r^2 = 8700$

$\Rightarrow \pi r^2 = 12000 - 8700$

$\Rightarrow \pi r^2 = 3300$

$\Rightarrow r^2 = 1050$

Therefore $r = 32.40 \ m$

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Question 16: The radii of two circles are $8 \ cm$ and $6 \ cm$ respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

Let the radius of the circle $= r$

Therefore $\pi r^2 = \pi (8)^2 + \pi (6)^2$

$\Rightarrow r^2 = 64 + 36$

$\Rightarrow r = 10 \ cm$

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Question 17: The radii of two circles are $19 \ cm$ and $9 \ cm$ respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.

Let the radius of the circle $= r$

Therefore $2\pi r = 2\pi (19) + 2\pi (9)$

$\Rightarrow r = 19 + 9 = 28 \ cm$

Therefore Area $= \pi (28)^2 = 2464 \ cm^2$

Circumference $= 2 \pi (28) = 176 \ cm$

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Question 18: A car travels 1-kilometer distance in which each wheel makes $450$ complete revolutions. Find the radius of its wheels.

Let the radius of the wheel $= r$

Therefore $2 \pi r \times 450 = 1000$

$\Rightarrow r =$ $\frac{1000 \times 7}{2 \times 450 \times 22}$ $= 0.3535 \ m \ or \ 35.35 \ cm$

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Question 19: The area enclosed between the concentric circles is $770 \ cm^2$. If the radius of the outer circle is $21 \ cm$, find the radius of the inner circle.

Let the radius of the inner circle $= r$

Therefore $\pi (21)^2 - \pi (r)^2 = 770$

$\Rightarrow \pi r^2 =$ $\frac{22}{7}$ $\times 21^2 - 770$

$\Rightarrow \pi r^2 = 616 \ cm^2$

Therefore $r^2 = \frac{7}{22} \times 616 = 196$

$\Rightarrow r = 14 \ cm$

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Question 20: The wheel of a car is making $5$ revolutions per second. If the diameter of the wheel is $84 \ cm$, find its speed in $km/hr$. Give your answer, correct to nearest km.

No. of revolutions per second $= 5$

Radius of the wheel $= 42 \ cm$

Therefore circumference of the wheel $= 2 \pi r = 2 \times$ $\frac{22}{7}$ $\times 42 = 264 \ cm$

Hence the distance covered in 1 second $= 5 \times 264 = 1320 \ cm \ or \ 13.20 \ m$

Hence distance covered in one hour $= 13.20 \times 3600 = 47520 \ m \ or \ 47.52 \ km$

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Question 21: A sheet is $11 \ cm$ long and $2 \ cm$ wide. Circular pieces of $0.5 \ cm$ in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.

Number of disks by length $= 22$

No of disks by breadth $= 4$

Therefore total number of disks that can be cut $= 4 \times 22 = 88$

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Question 22: A copper wire when bent in the form of an equilateral triangle has area $121\sqrt{3} \ cm^2$. If the same wire is bent into the form of a circle, find the area enclosed.

Let the side of the equilateral triangle $= a$

Therefore $h =$ $\sqrt{a^2 - (\frac{a}{2})^2}$ $=$ $\frac{\sqrt{3}}{2}$ $a$

Therefore $121 \sqrt{3} =$ $\frac{1}{2}$ $\times a \times$ $\frac{\sqrt{3}}{2}$ $a$

$\Rightarrow a^2 = 2^2 \times 11^2$

$\Rightarrow a = 22 \ cm$

Therefore perimeter $= 66 \ cm$

Let the radius of the circle $= r$

Therefore $2\pi r = 66$

$\Rightarrow r =$ $\frac{7 \times 66}{2 \time 22}$ $= 10.5 \ cm$

Area of circle $= \pi (10.5)^2 = 346.5 \ cm^2$

Click here for Second half of Exercise 16