In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠POQ = 90°, area of shaded region is 114cm², find the radius of the circle. (π=3.14)
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20 will be the answer
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Let ' r ' be the radius of circle .
area of sector O-PRQ = area of triangle OPQ + are of shaded region PRQ
¢ /360 × πr^2 = 1/2 r^2 + 114
90/360 × πr^2 = 1/2 r^2 + 114
πr^2 / 4 = 1/2 r^2 + 114
πr^2 / 4 - 1/2 r^2 = 114
r^2 [ 3.14 /4 - 1/2 ] = 114
r^2 / 2 [ 3.14 / 2 - 1 ] = 114
r^2 [ 1.57 - 1 ] = 228
r^2 × 0.57 = 228
r^2 = 228 / 0.57
r^2 = 400
taking square root of both sides
r = 20 cm
area of sector O-PRQ = area of triangle OPQ + are of shaded region PRQ
¢ /360 × πr^2 = 1/2 r^2 + 114
90/360 × πr^2 = 1/2 r^2 + 114
πr^2 / 4 = 1/2 r^2 + 114
πr^2 / 4 - 1/2 r^2 = 114
r^2 [ 3.14 /4 - 1/2 ] = 114
r^2 / 2 [ 3.14 / 2 - 1 ] = 114
r^2 [ 1.57 - 1 ] = 228
r^2 × 0.57 = 228
r^2 = 228 / 0.57
r^2 = 400
taking square root of both sides
r = 20 cm
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