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Areas Related To Circles, Problems based on, Perimeter and area of a circle, , 1
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Problems based on, Perimeter and area of a circle, Q1) A wheel rotates 25000 times to cover a distance of 90 km. Find its radius., Solution :, Rotation of the wheel = 25000 times, Distance covered, = 90 km = 90x1000 m = 90x1000x100 cm, Distance covered in one rotation = 9000000/25000 = 360 cm, Circumference of the wheel = Distance covered in one rotation., Let ‘r’ be the radius of the wheel., 2πr = 360 cm, 2 x 22 x r = 360cm, 7, r = 360x7 cm, 2x22, = 180 x 7, 22, = 90 x 7, 11, = 630, 11, , ?, , r = 57.27 cm Hence, the radius of the wheel is 57.27cm, , 2
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Problems based on, Perimeter and area of a circle, Q2) The diameter of a cart wheel is 21 cm. How many revolutions will it make in, moving 1.32 km?, Solution: Let the radius of the cart wheel be ’r’., Thus, r = Diameter = 21 cm, 2, 2, 21 cm, Circumference of the cart wheel = 2πr = 2x22x21 = 462 = 66 cm, 7 2, 7, Converting 1.32 Km into cm, we get,, 1.32 Km = 1.32x1000 = 1320 m = 1320 x 100 = 132000 cm, Number of revolutions = Total distance covered, Circumference (Distance covered by, 1 round of the cart wheel), = 132000 cm, 66 cm, , = 2000, Hence, the cart wheel will make 2000 revolutions in moving 1.32 km., 3
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Problems based on, Perimeter and area of a circle, Q3) A wheel of a bicycle makes 6 revolutions per second. If the diameter of the, wheel is 80 cm, find its speed., Solution: Let the radius of the wheel be denoted as ‘r’., Thus, r = Diameter = 80 = 40 cm, 2, 2, Circumference of the wheel = 2 πr = 2x22x40 = 251.42cm, 7, Distance covered in 1 revolution = circumference = 251.42cm, Distance covered in 6 revolutions, = 6 x Distance covered in 1 revolution = 6 x 251.42 = 1508.52 cm, = 1508.52 m = 15.0852 m, 100, Speed = Distance = 15.0852 m, Time, 1 second, Hence, the speed of the wheel is 15.08m/second., , 4
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Problems based on, Perimeter and area of a circle, Q4) Find the radius of the circle whose perimeter and area are numerically, equal., Solution: Let ‘r’ be the radius of the circle., Then, its area = πr2 and, its perimeter = 2πr, It is given that the area of the circle is numerically equal to its perimeter., Thus,, πr2 = 2πr, => πr2 - 2πr = 0, => πr(r-2) = 0, => Either πr = 0, or, r-2 = 0, r = 0 (rejected) or, r=2, Hence, the radius of the circle is, , 2 units, 5
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Problems based on, Perimeter and area of a circle, , 6
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Problems based on, Perimeter and area of a circle, , 7
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Problems based on, Perimeter and area of a circle, , 8
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Problems based on, Perimeter and area of a circle, , 9
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Problems based on, Perimeter and area of a circle, , 10
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Problems based on, Perimeter and area of a circle, , 11
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Problems based on, Perimeter and area of a circle, , 12
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Problems based on, Perimeter and area of a circle, , 20, , 20, , 13
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Problems based on, Perimeter and area of a circle, , 14
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Thank you, , 15
