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

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

Answer:

Area of rectangle $= 60 \times 28 = 1680 \ m^2$

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

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

Hence area of the park $= 1680 + 308 = 1988 \ m^2$

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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 $36 \ m$ and $24.5 \ m$ , find, the area of the playground. $($ Take $\pi =$ $\frac{22}{7}$ $)$

Answer:

Area of rectangle $= 36 \times 24.5 = 882 \ m^2$

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

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

Therefore total area $= 882 + 471.625 = 1353.625 \ m$

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

Answer:

Let the outer radius $= r$

Circumference $= 528 \ m$

$\Rightarrow 2 \pi r = 528$

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

Therefore inner radius $= 84 - 12 = 70 \ m$

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

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

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

Answer:

Area of the rectangle $= 20 \times 15 = 300 \ cm^2$

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

Therefore Remaining area $= 300 - 38.5 = 261.5 \ m^2$

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

Answer:

Area of the square $= 10 \times 10 = 100 \ cm^2$

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

Therefore Remaining area $= 100 - 78.5 = 21.5 \ cm^2$

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

Answer:

Area of the square $= 50 \times 50 = 2500 \ m^2$

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

Therefore Remaining area $= 2500 - 1964.29 = 535.71 \ m^2$

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

Answer:

Let the radius of the park $= r$

Circumference of park $= 352 \ m$

Therefore $2 \pi r = 352$

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

Outer radius $= 56 + 7 = 63 \ m$

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

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

Answer:

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

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

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 $14 \ cm \times 7 \ cm$ . Find the area of the remaining cardboard. $($ Take $\pi =$ $\frac{22}{7}$ $)$

Answer:

Area of board $= 14 \times 7 = 98 \ cm^2$

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

Remaining area $= 98 - 77 = 21 \ cm^2$

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

Answer:

Given: $OA = 21 \ cm$ , $ABCO$ is a square

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

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

Area of square $= 21 \times 21 = 441 \ cm^2$

Therefore shaded area $= 693 - 441 = 252 \ cm^2$

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

Answer:

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

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

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

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

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

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

Therefore shaded area $= 173.25 + 308 - (481.25 - 294) = 294 \ cm^2$

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

Answer:

Total area of large semi circle

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

Area of two smaller semi circles $= 2 [$ $\frac{1}{2}$ $\times \pi (9)^2] = 81 \pi$

Let the radius of the small circle $= r$

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

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

$\Rightarrow 54r = 324$

$\Rightarrow r = 6 \ cm$

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

hence the shaded 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 $14 \ cm$ and of the smallest is $3.5 \ cm$ , find i) the length of the boundary ii) the area of the shaded region.

Answer:

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

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

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

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

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

Answer:

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

Therefore Radius $= 10 \ cm$

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

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

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

Answer:

i) Area of shaded area

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

$= 10.125 \pi - 2.25 \pi + 1.125 \pi - 5.0625 \pi = 3.9375 \pi = 12.375 \ cm^2$

ii) Therefore cost of painting shaded area $= 12.375 \times 0.25 = 3.09375 \ Rs. \approx 3 \ Rs.$

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

Answer:

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

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

Therefore shaded region $= 49 \pi - 12.25 \pi = 36.75 \pi = 115.5 \ cm^2$

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

Answer:

i) Area of quadrant $OACB$

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

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

$= 9.625 - 3.14 = 6.465 \ cm^2$

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

Answer:

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

$= 64 - 3.08 - 55.44 = 5.48 \ cm^2$

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

Answer:

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

Therefore radius of the circle $= 13 \ cm$

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

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

Answer:

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

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

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

Answer:

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

$= 280 - 154 = 126 \ cm^2$

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

$= 20 + 7 \pi + 20 + 7 \pi = 84 \ cm$

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

Answer:

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

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

Let the radius of the circle $= r$

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

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

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

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

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