  Question

Find the difference of the areas of the two segments of a circle formed by a chord of length 5cm subtending an angle of 90degree at the centre

Solution Let r be the radius of the circle and AB be the chord, which subtend angle of 90 at centre O. AB = 5 cm in the right angled triangle OAB, using Pythagoras theorem: OA2+OB2=AB2 r2+r2=52 2r2=25 r=5√2cm Therefore area of the circle = πr2=227×252=39.2857 cm2 The area of the minor segment (shaded area) = area of the sector OAB - area of the triangle OAB area of the sector OAB = πr2×90360=39.2857×14=9.8214 cm2 Area of triangle OAB = 12×OA×OB=12r2=12×252=254=6.25 cm2 The area of minor segment = 9.8214 - 6.25 = 3.5714 sq cm ……(1) The area of the major segment = area of the circle - area of the minor segment = 39.2857 -  3.5714 = 35.7143 sq cm     ….(2) Difference of the areas of two segments = 35.7143 - 3.5714 = 32.1429 cm2  All India Test Series

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