Class 9: Perimeter and Area of a Circle – Exercise 16.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the circumference and area of a circle of $\displaystyle \text{Radius } 4.2 \text{ cm }$ .

Answer:

$\displaystyle \text{Radius } = 4.2 \text{ cm }$

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

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

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

Answer:

$\displaystyle \text{Area } = 301.84 \text{ cm}^2$

$\displaystyle \text{Therefore } \pi r^2 = 301.84$

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

$\displaystyle \Rightarrow r = 9.8 \text{ cm }$

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

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

Answer:

$\displaystyle \text{Circumference } = 44 \text{ cm }$

$\displaystyle \text{Therefore } 2 \pi r = 44$

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

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

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

Answer:

$\displaystyle \text{Given: Circumference - diameter} = 16.8$

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

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

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

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

Answer:

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

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

Answer:

$\displaystyle \text{Area of square } = 121 \text{ cm}^2$

$\displaystyle \text{Let the side of the square } = a$

$\displaystyle \Rightarrow a^2 = 121 \Rightarrow a = 11 \text{ cm }$

$\displaystyle \text{Therefore Perimeter } = 4 \times 11 = 44 \text{ cm }$

Let $\displaystyle r$ be the radius of the circle

$\displaystyle \text{Therefore } 2 \pi r = 44$

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

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

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

Answer:

Diameter of front wheel $\displaystyle = 80 \text{ cm }$

Diameter of rear wheel $\displaystyle = 2 \text{ m }$

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

$\displaystyle \text{Let no of revolution that the rear wheel }= n$

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

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

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

Answer:

$\displaystyle \text{Area of square } = 121 \text{ cm}^2$

$\displaystyle \text{Let the side of the square } = a$

$\displaystyle \Rightarrow a^2 = 121 \Rightarrow a = 11 \text{ cm }$

$\displaystyle \text{Therefore Perimeter } = 4 \times 11 = 44 \text{ cm }$

Let $\displaystyle r$ be the radius of the circle

$\displaystyle \text{Therefore } 2 \pi r = 44$

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

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

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

Answer:

Let the radius of first circle $\displaystyle = r_1$

Let the radius of second circle $\displaystyle = r_2$

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

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

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

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

Answer:

When a square is inscribed in the circle

Therefore diameter $\displaystyle = 10\sqrt{2}$

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

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

When the circle is inscribed in the square

Diameter $\displaystyle = 10 \text{ cm }$

$\displaystyle \text{Therefore } \text{Radius } = 5 \text{ cm }$

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

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

Answer:

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

$\displaystyle \text{Therefore } r_1 + r_2 = 140$ … … … … … i)

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

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

$\displaystyle \text{Therefore } 2\pi (r_1-r_2) = 88$

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

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

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

$\displaystyle \Rightarrow 2r_1 = 154$

$\displaystyle \Rightarrow r_1 = 77 \text{ cm }$

$\displaystyle \text{Therefore } r_2 = 140 - r_1 = 140 - 77 = 63 \text{ cm }$

Diameter of circles are $\displaystyle 154 \text{ cm }$ and $\displaystyle 126 \text{ cm }$

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

Answer:

$\displaystyle \text{Area of square } = 64 \text{ cm}^2$

$\displaystyle \Rightarrow$ Side of square $\displaystyle = 8 \text{ cm }$

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

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

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

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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 $\displaystyle Rs. 2640$ at the rate of $\displaystyle Rs. 12$ per meter. Then, the field is to be thoroughly ploughed at the cost of $\displaystyle Rs. 0.50 per \text{ m }$ . What is the amount required to plough the field? $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

Cost of fencing $\displaystyle = 2640 Rs.$

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

$\displaystyle \text{Therefore } 2 \pi r = 220$

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

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

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

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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.

Answer:

$\displaystyle \text{Let the side of the square } = a$

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

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

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

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

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

Answer:

Let the radius of the circle $\displaystyle = r$

Area of the park $\displaystyle = 100 \times 120 = 12000 \text{ m}^2$

Are of circle $\displaystyle = \pi r^2$

$\displaystyle \text{Therefore } 12000 = \pi r^2 = 8700$

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

$\displaystyle \Rightarrow \pi r^2 = 3300$

$\displaystyle \Rightarrow r^2 = 1050$

$\displaystyle \text{Therefore } r = 32.40 \text{ m }$

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

Answer:

Let the radius of the circle $\displaystyle = r$

$\displaystyle \text{Therefore } \pi r^2 = \pi (8)^2 + \pi (6)^2$

$\displaystyle \Rightarrow r^2 = 64 + 36$

$\displaystyle \Rightarrow r = 10 \text{ cm }$

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

Answer:

Let the radius of the circle $\displaystyle = r$

$\displaystyle \text{Therefore } 2\pi r = 2\pi (19) + 2\pi (9)$

$\displaystyle \Rightarrow r = 19 + 9 = 28 \text{ cm }$

$\displaystyle \text{Therefore } \text{Area } = \pi (28)^2 = 2464 \text{ cm}^2$

$\displaystyle \text{Circumference } = 2 \pi (28) = 176 \text{ cm }$

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

Answer:

Let the radius of the wheel $\displaystyle = r$

$\displaystyle \text{Therefore } 2 \pi r \times 450 = 1000$

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

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

Answer:

Let the radius of the inner circle $\displaystyle = r$

$\displaystyle \text{Therefore } \pi (21)^2 - \pi (r)^2 = 770$

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

$\displaystyle \Rightarrow \pi r^2 = 616 \text{ cm}^2$

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

$\displaystyle \Rightarrow r = 14 \text{ cm }$

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

Answer:

No. of revolutions per second $\displaystyle = 5$

Radius of the wheel $\displaystyle = 42 \text{ cm }$

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

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

$\displaystyle \text{Hence distance covered in one hour } = 13.20 \times 3600 = 47520 \text{ m or } 47.52 k\text{ m }$

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

Answer:

Number of disks by length $\displaystyle = 22$

No of disks by breadth $\displaystyle = 4$

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

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

Answer:

Let the side of the equilateral triangle $\displaystyle = a$

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

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

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

$\displaystyle \Rightarrow a = 22 \text{ cm }$

$\displaystyle \text{Therefore Perimeter } = 66 \text{ cm }$

Let the radius of the circle $\displaystyle = r$

$\displaystyle \text{Therefore } 2\pi r = 66$

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

$\displaystyle \text{Area of circle } = \pi (10.5)^2 = 346.5 \text{ cm}^2$

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Question 23: A plot is in the form of a rectangle $\displaystyle ABCD$ having semi-circle-on $\displaystyle BC$ as shown in the adjoining figure. $\displaystyle AB =60 \text{ m }$ and $\displaystyle BC = 28 \text{ m }$ , find the area of the plot.

Answer:

Area of rectangle $\displaystyle = 60 \times 28 = 1680 \text{ m}^2$

Diameter of semi circle $\displaystyle = \frac{28}{2} = 14 \text{ m }$

Therefore Area of semi circle $\displaystyle = \frac{1}{2} \times \pi (14)^2 = \frac{1}{2} \times \frac{22}{7} \times (14)^2 = 308 \text{ m}^2$

Hence area of the park $\displaystyle = 1680 + 308 = 1988 \text{ m}^2$

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Question 24: A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are $\displaystyle 36 \text{ m }$ and $\displaystyle 24.5 \text{ m }$ , find, the area of the playground. $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

Area of rectangle $\displaystyle = 36 \times 24.5 = 882 \text{ m}^2$

$\displaystyle \text{Radius of semicircle } = \frac{24.5}{2} = 12.25 \text{ m }$

$\displaystyle \text{Therefore Area of 2 semi circles } = 2 [ \frac{1}{2} \pi (12.25)^2 ] = 471.625 \text{ m}^2$

$\displaystyle \text{Therefore total Area } = 882 + 471.625 = 1353.625 \text{ m }$

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Question 25: The outer circumference of a circular race-track is $\displaystyle 528 \text{ m }$ . The track is everywhere $\displaystyle 14 \text{ m }$ wide. Calculate the cost of leveling the track at the rate of $\displaystyle 50$ paise per square meter. $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

Let the outer $\displaystyle \text{Radius } = r$

$\displaystyle \text{Circumference } = 528 \text{ m }$

$\displaystyle \Rightarrow 2 \pi r = 528$

$\displaystyle \Rightarrow r = \frac{528 \times 7}{2 \times 22} = 84 \text{ m }$

Therefore inner $\displaystyle \text{Radius } = 84 - 12 = 70 \text{ m }$

Therefore area of track $\displaystyle = \pi (84)^2 - \pi (70)^2 = 6776 \text{ m}^2$

Cost of leveling $\displaystyle = 6776 \times 0.5 = 3388 Rs.$

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Question 26: A rectangular piece is $\displaystyle 20 \text{ m }$ long and $\displaystyle 15 \text{ m }$ wide. From its four corners, quadrants of radii $\displaystyle 3.5 \text{ m }$ have been cut. Find the area of the remaining part.

Answer:

$\displaystyle \text{Area of the rectangle } = 20 \times 15 = 300 \text{ cm}^2$

$\displaystyle \text{Area of quadrants } = 4 [ \frac{1}{4} \times \pi \times 3.5^2 ] = 38.5 \text{ m}^2$

$\displaystyle \text{Therefore Remaining Area } = 300 - 38.5 = 261.5 \text{ m}^2$

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Question 27: Four equal circles, each of $\displaystyle \text{Radius } 5 \text{ cm }$ , touch each other as shown in the adjoining figures. Find the area included between them. $\displaystyle ($ Take $\displaystyle \pi = 3.14)$

Answer:

$\displaystyle \text{Area of the square } = 10 \times 10 = 100 \text{ cm}^2$

$\displaystyle \text{Area of quadrants } = 4 [ \frac{1}{4} \times \pi \times 5^2 ] = 78.5 \text{ cm}^2$

$\displaystyle \text{Therefore Remaining Area } = 100 - 78.5 = 21.5 \text{ cm}^2$

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Question 28: Four cows are tethered at four corners of a square plot of side $\displaystyle 50 \text{ m }$ , so that they just cannot reach one another. what area will be left ungrazed?

Answer:

Area of the square $\displaystyle = 50 \times 50 = 2500 \text{ m}^2$

$\displaystyle \text{Area of quadrants } = 4 [ \frac{1}{4} \times \pi \times 25^2 ] = 1964.29 \text{ m}^2$

Therefore Remaining $\displaystyle \text{Area } = 2500 - 1964.29 = 535.71 \text{ m}^2$

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Question 29: A road which is $\displaystyle 7 \text{ m }$ wide surrounds a circular park whose circumference is $\displaystyle 352 \text{ m }$ . Find the area of the road.

Answer:

Let the radius of the park $\displaystyle = r$

Circumference of park $\displaystyle = 352 \text{ m }$

$\displaystyle \text{Therefore } 2 \pi r = 352$

$\displaystyle \Rightarrow r = \frac{352 \times 7}{2 \times 22} =56 \text{ m }$

Outer $\displaystyle \text{Radius } = 56 + 7 = 63 \text{ m }$

$\displaystyle \text{Therefore area of the road } = \pi (63)^2 - \pi (56)^2 = \frac{22}{7} \times (63^2 - 56^2) = 2618 \text{ m}^2$

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Question 30: Four equal circles, each of radius a, touch each other. show that the area between the is $\displaystyle \frac{6}{7} a^2$ . $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

Area of rectangle = (2a) \times (2a) = 4a^2

$\displaystyle \text{Area of quadrants } = 4 [ \frac{1}{4} \times \pi \times a^2 ] = \frac{22}{7} a^2$

$\displaystyle \text{Therefore Remaining Area } = 4a^2 - \frac{22}{7} a^2 = \frac{6}{7} a^2$

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Question 31: Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions $\displaystyle 14 cm \times 7 \text{ cm }$ . Find the area of the remaining cardboard. $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

Area of board $\displaystyle = 14 \times 7 = 98 \text{ cm}^2$

Area of two cut outs $\displaystyle = 2 [ \pi 3.5^2 ] = 77 \text{ cm}^2$

Remaining $\displaystyle \text{Area } = 98 - 77 = 21 \text{ cm}^2$

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Question 32: In the adjoining figure, a square $\displaystyle OABC$ is inscribed in a quadrant $\displaystyle ODBE$ of a circle. If $\displaystyle OA=21 \text{ cm }$ , find the area of the shaded region.

Answer:

$\displaystyle \text{Given: } OA = 21 \text{ cm }$ , $\displaystyle ABCO$ is a square

$\displaystyle \text{Therefore } \text{Radius } = 21\sqrt{2} \text{ cm }$

$\displaystyle \text{Hence area of quadrant } = \frac{1}{4} [ \pi \times (21\sqrt{2})^2 ] = \frac{1}{4} \times \frac{22}{7} \times 2 \times 21 \times 21 = 693 \text{ cm}^2$

$\displaystyle \text{Area of square } = 21 \times 21 = 441 \text{ cm}^2$

Therefore shaded $\displaystyle \text{Area } = 693 - 441 = 252 \text{ cm}^2$

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Question 33: In the adjoining figure, $\displaystyle ABC$ is a right angled triangle in which $\displaystyle \angle A = 90^o, AB = 21 \text{ cm }$ and $\displaystyle AC = 28 \text{ cm }$ . Semi-circles are described on $\displaystyle AB, BC$ and $\displaystyle AC$ as diameters. Find the area of the shaded region.

Answer:

$\displaystyle BC = \sqrt{AB^2 + AC^2}$

$\displaystyle = \sqrt{21^2 + 28^2} = \sqrt{1225} = 35 \text{ cm }$

$\displaystyle \text{Area of small semi circle with diameter } AB = \frac{1}{2} [ \pi \Big( \frac{21}{2} \Big)^2 ] = 173.25 \text{ cm}^2$

$\displaystyle \text{Area of large semi circle with diameter } AC = \frac{1}{2} [ \pi \Big( \frac{28}{2} \Big)^2 ] = 308 \text{ cm}^2$

$\displaystyle \text{Area of large semi circle with diameter } BC = \frac{1}{2} [ \pi \Big( \frac{35}{2} \Big)^2 ] = 481.25 \text{ cm}^2$

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} \times 21 \times 28 = 294 \text{ cm}^2$

$\displaystyle \text{Therefore shaded Area } = 173.25 + 308 - (481.25 - 294) = 294 \text{ cm}^2$

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Question 34: In the adjoining figure, $\displaystyle AB = 36 \text{ cm }$ and $\displaystyle O$ is mid-point of $\displaystyle AB$ . Semi-circles are drawn on $\displaystyle AB , AO$ and $\displaystyle OB$ as diameters. A circle with center $\displaystyle C$ touches all the three circles. Find the area of the shaded region.

Answer:

Total area of large semi circle

$\displaystyle = \frac{1}{2} (\pi (18)^2) = 162 \pi$

Area of two smaller semi circles

$\displaystyle = 2 [ \frac{1}{2} \times \pi (9)^2] = 81 \pi$

Let the radius of the small circle $\displaystyle = r$

Therefore, $\displaystyle (9+r)^2 = 9^2 + (18-r)^2$

$\displaystyle \Rightarrow 81 + r^2 + 18r = 81 +324 + r^2 -36r$

$\displaystyle \Rightarrow 54r = 324$

$\displaystyle \Rightarrow r = 6 \text{ cm }$

Therefore area of small circle $\displaystyle = \pi r^2 = \pi (6)^2 = 36 \pi$

hence the shaded $\displaystyle \text{Area } = 162 \pi -36 \pi -81 \pi = 45 \pi$

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Question 35: In the adjoining figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is $\displaystyle 14 \text{ cm }$ and of the smallest is $\displaystyle 3.5 \text{ cm }$ , find i) the length of the boundary ii) the area of the shaded region.

Answer:

$\displaystyle \text{Length of boundary } = \frac{1}{2} [2 \pi (7) ] + \frac{1}{2} [2 \pi (1.75) ] + \frac{1}{2} [2 \pi (3.5) ] + \frac{1}{2} [2 \pi (1.75) ]$

$\displaystyle = 7\pi + 1.75 \pi + 3.5 \pi +1.75 \pi = 14 \pi = 14 \times \frac{22}{7} = 44 \text{ cm }$

$\displaystyle \text{Area } \text{Therefore shaded } = \frac{1}{2} [ \pi (7)^2 ] - 2 \times \frac{1}{2} [ \pi (1.75)^2 ] + \frac{1}{2} [ \pi (3.5)^2 ]$

$\displaystyle = 24.5 \pi - 3.0625 \pi + 6.125 \pi = 27.5625 \pi = 27.5625 \times \frac{22}{7} = 86.625 \text{ cm}^2$

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Question 36: In the adjoining figure, $\displaystyle O$ is the center of a circular arc and $\displaystyle AOB$ is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. $\displaystyle ($ Take $\displaystyle \pi = 3.14)$

Answer:

$\displaystyle AB = \sqrt{12^2 + 16^2} = \sqrt{400} = 20 \text{ cm }$

$\displaystyle \text{Therefore } \text{Radius } = 10 \text{ cm }$

$\displaystyle \text{Perimeter } = \frac{1}{2} [ 2 \pi (10) ] + 12+16 = 10 \pi + 28 = 59.42 \text{ cm }$

$\displaystyle \text{Shaded region } = \frac{1}{2} [ \pi (10)^2 ] - \frac{1}{2} \times 12 \times 16 = 50\pi -96 = 61.1428 \text{ cm}^2$

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Question 37: In the adjoining figure, there are three semicircles, $\displaystyle A, B$ and $\displaystyle C$ having diameter $\displaystyle 3 \text{ cm }$ each, and another semicircle $\displaystyle E$ having a circle $\displaystyle D$ with diameter $\displaystyle 4.5 \text{ cm }$ are shown. Calculate: (i) the area of the shaded region (ii) the cost of painting the shaded region at the rate of $\displaystyle 25 paise per \text{ cm}^2$ , to the nearest rupee.

Answer:

i) Area of shaded area

$\displaystyle = [ \frac{1}{2} \pi (4.5)^2 ] - 2 \times [ \frac{1}{2} \pi (1.5)^2 ] + [ \frac{1}{2} \pi (1.5)^2 ] - [\pi (2.25)^2]$

$\displaystyle = 10.125 \pi - 2.25 \pi + 1.125 \pi - 5.0625 \pi = 3.9375 \pi = 12.375 \text{ cm}^2$

ii) Therefore cost of painting shaded $\displaystyle \text{Area } = 12.375 \times 0.25 = 3.09375 Rs. \approx 3 Rs.$

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Question 38: In the adjoining figure $\displaystyle AB$ and $\displaystyle CD$ are two diameters of a circle perpendicular to each other and $\displaystyle OD$ is the diameter of the smaller circle. lf $\displaystyle OA =7 \text{ cm }$ , find the area of the shaded region.

Answer:

Area of larger circle $\displaystyle = \pi (7)^2 = 49 \pi$

Area of smaller circle $\displaystyle = \pi (3.5)^2 = 12.25 \pi$

Therefore shaded region $\displaystyle = 49 \pi - 12.25 \pi = 36.75 \pi = 115.5 \text{ cm}^2$

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Question 39: In the adjoining figure, $\displaystyle OACB$ is a quadrant of a circle with center $\displaystyle O$ and $\displaystyle \text{Radius } 3.5 \text{ cm }$ . If $\displaystyle OD$ is $\displaystyle 2 \text{ cm }$ , find the area of the i) quadrant $\displaystyle OACB$ an ii) shaded region.

Answer:

i) Area of quadrant $\displaystyle OACB$

$\displaystyle = \frac{1}{4} [ \pi (3.5)^2 ] = 9.625 \text{ cm}^2$

ii) $\displaystyle \text{Area of shaded region } = 9.625 - \frac{1}{4} [ \pi (2)^2 ]$

$\displaystyle = 9.625 - 3.14 = 6.465 \text{ cm}^2$

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Question 40: For each of the two opposite corners of a square of side $\displaystyle 8 \text{ cm }$ , a quadrant of a circle of $\displaystyle \text{Radius } 1.4 \text{ cm }$ is cut. Another circle of $\displaystyle \text{Radius } 4.2 \text{ cm }$ is also cut from the center as shown in the figure. Find the area of the remaining shaded portion of the square. $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

$\displaystyle \text{Shaded Area } = 8 \times 8 - 2 [ \frac{1}{4} \pi (1.4)^2 ] - \pi (4.2)^2$

$\displaystyle = 64 - 3.08 - 55.44 = 5.48 \text{ cm}^2$

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Question 41: Find the area of the shaded region in the adjoining figure, if $\displaystyle AC = 24 cm, BC = 10 \text{ cm }$ , and $\displaystyle O$ is center of the circle. $\displaystyle ($ Take $\displaystyle \pi = 3.14)$

Answer:

$\displaystyle AB = \sqrt{10^2 + 24^2} = 26 \text{ cm }$

Therefore radius of the circle $\displaystyle = 13 \text{ cm }$

Hence the shaded $\displaystyle \text{Area } = \pi (13)^2 - \frac{1}{4} \pi (13)^2 - \frac{1}{2} \times 10 \times 24 = 145.33 \text{ cm}^2$

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Question 42: In the adjoining figure, $\displaystyle OABC$ is a square of side $\displaystyle 7 \text{ cm }$ . If $\displaystyle OAPC$ is a quadrant of a circle with center $\displaystyle O$ , then find the area of the shaded region. $\displaystyle ($ Take $\displaystyle \pi = \frac{22}{7} )$

Answer:

$\displaystyle \text{Shaded Area } = 7 \times 7 - \frac{1}{4} \pi (7)^2$

$\displaystyle = 49 - \frac{1}{4} \times \frac{22}{7} \times 49 = 10.5 \text{ cm}^2$

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Question 43: In the adjoining figure, $\displaystyle ABCD$ is a rectangle, having $\displaystyle AB = 20 \text{ cm }$ and $\displaystyle BC = 14 \text{ cm }$ . Two sectors of $\displaystyle 180^o$ have been cut off. Calculate: i) the area of the shaded region ii) the length of the boundary of the shaded region.

Answer:

$\displaystyle \text{Area of the shaded region } = 20 \times 14 - 2 \times [ \frac{1}{2} \pi (7)^2 ]$

$\displaystyle = 280 - 154 = 126 \text{ cm}^2$

$\displaystyle \text{Perimeter } = 20 + \frac{1}{2} [ 2 \pi (7) ] + 20 + \frac{1}{2} [2 \pi (7) ]$

$\displaystyle = 20 + 7 \pi + 20 + 7 \pi = 84 \text{ cm }$

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Question 44: A circle is inscribed in an equilateral triangle $\displaystyle ABC$ is side $\displaystyle 12 \text{ cm }$ , touching its Sides as shown in the adjoining figure. Find the radius of the inscribed circle and the area of the shaded part.

Answer:

$\displaystyle AO = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}$

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} \times 12 \times 6\sqrt{3} = 36 \sqrt{3}$

Let the radius of the circle $\displaystyle = r$

$\displaystyle \text{Therefore } 36\sqrt{3} = \frac{1}{2} \times 1.2 \times r \times 3$

$\displaystyle \Rightarrow r = 2\sqrt{3}$

$\displaystyle \text{Therefore } \text{Area of circle } = \pi (2\sqrt{3})^2 = 12 \pi$

$\displaystyle \text{Therefore shaded Area } = 36 \sqrt{3} - 12 \times \frac{22}{7} = 24.639 \text{ cm}^2$

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Question 45: In the adjoining figure, shows the cross-section of railway tunnel. The $\displaystyle \text{Radius } OA$ of the circular part is $\displaystyle 2 \text{ m }$ . If $\displaystyle \angle AOB = 90^o$ , calculate: (i) the height of the tunnel (ii) the perimeter of the cross-section (iii) the area of the cross-section.