Click here for First half of Exercise 16 

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:2019-01-03_21-49-08

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} ) 2019-01-03_21-58-40

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:2019-01-04_6-04-54

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.2019-01-04_6-18-20

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) 2019-01-04_6-44-17

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:2019-01-04_6-49-03

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:2019-01-04_6-53-39

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} ) 2019-01-04_7-12-42

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} ) 2019-01-04_7-20-35

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.2019-01-04_7-33-16.jpg

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.2019-01-04_7-54-20

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:2019-01-04_8-17-20

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.2019-01-04_20-16-16

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) 2019-01-04_20-30-08

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.2019-01-05_13-17-11

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.2019-01-05_13-22-29

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.2019-01-05_13-30-55

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} ) 2019-01-05_13-40-44

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) 2019-01-05_13-48-07

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:2019-01-05_13-56-34

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:2019-01-05_14-01-30

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.2019-01-05_14-06-32.jpg

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.2019-01-05_14-11-03

Answer:

AB = \sqrt{2^2 +2^2} = \sqrt{8} = 2\sqrt{2} \ m

i) Height of tunnel = 2 + \sqrt{2^2 - (\sqrt{2})^2} = 2 + \sqrt{2} = 3.414 \ m

ii) Perimeter = \frac{3}{4} (2 \pi 2) + 2\sqrt{2} = (3\pi + 2\sqrt{2} ) \ m

iii) Area of cross-section = \frac{3}{4} (\pi 2^2) + \frac{1}{2} \times 2 \times 2 = (3\pi + 2) \ m

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