a chord of 10cm long is drawn in a circle of radius 5✓2cm. find the area of the both of segments
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In △OAB,
OA² + OB² = 50 + 50 = 100
AB² = 10² =100
i.e., OA² + OB² = AB² which satisfies the Pythagoras theorem. Therefore, θ = 90°
Area of minor segment = area of sector - Area of △AOB
= θ/360° x πr² -1/2 x OA x OB
= 90°/360° x 22/7 x 5root2 × 5root2 - 1/2 x 5root2 x5root2
= 39.29 - 25 = 14.29 cm²
Area of major segment = area of circle - area of minor segment
= πr² - 14.29 = 22/7 x 5root2 x 5root2 − 14.29
= 142.85cm²
See the figure also.
Hope it helps.
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