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ही 5 Angle of o Sectos A 2R ave. s qu-‘u;",;!f कह रक्त है asea and सिव्िडट 2?= |
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Answer» Let the center of circle be O.Let the radii of the sector be OA and OB.Given∠AOB = 60°. In theΔAOB, OA=OB, So it is isosceles. => ∠OAB = OBA = [180° -∠AOB ] /2 = 60°Hence,ΔAOB is equilateral. Hence, AB = OA = 14 cm. Area ofΔAOB =√3/4 * R²Area of sector OAB =π R² * (60°/360°) =π R² / 6 Area of minor segment AB : πR²/ 6 -√3/4 R² = 22/7 * 14² /6 - √3 /4 * 14² cm² = 17.75 cm² |
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