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.

Answer:

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

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

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

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