a chord of a circle of radius 10 cm subtends a right angle at the center.find the area of the corresponding minor segment and major segment.( π=3.14)? plz give solution
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Let the center be O. Let the chord be AB. Draw a diameter through O parallel to AB. Let this diameter be COD.
OA= OB = 10 cm. ΔAOB is a right angle triangle: Area = 1/2 *OA*OB= 50 m²
Sector COA + sector DOB has a total angle of 90° at O (as 180°-∠AOB).
Hence area of these wo sectors = 1/4 * area of circle = π*10²/4 = 25 π cm²
Minor segment of circle: area of semicircle - 50 cm² - 25 π cm²
= 25 π - 50 cm²
Area of major segment = π * 10² - 25 π + 50
= 75 π + 50 cm²
OA= OB = 10 cm. ΔAOB is a right angle triangle: Area = 1/2 *OA*OB= 50 m²
Sector COA + sector DOB has a total angle of 90° at O (as 180°-∠AOB).
Hence area of these wo sectors = 1/4 * area of circle = π*10²/4 = 25 π cm²
Minor segment of circle: area of semicircle - 50 cm² - 25 π cm²
= 25 π - 50 cm²
Area of major segment = π * 10² - 25 π + 50
= 75 π + 50 cm²
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