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A chord of a circle of radius10 cm subtends a right angle at the centre .Find the area of the corresponding 1)minor segment 2)major segment1 |
Answer» Solution :radius of a CIRCLE, AO = 10 cm ∠AOC = 90° = ½ x 10 x 10 = 50 cm² Area of sector OAECO = Ø/360° x πr² = 90°/360 x 3.14 x (10)² = 314/4 = 78.5 cm² i) Area of minor segment AECDA area of sector OAECO - area of ΔAOC = 78.5 - 50 = 28.5 cm² ii) Area of major sector OAFGCO Area of circle - Area of sector OAECO = 3.14 x (10)² - 78.5 = 314 - 78.5 = 235.5 cm² |
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