Find the area of the minor segment of a circle of radius 14cm, when its central angle is 60 also find the area of corresponding major segment
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Let the center of circle be O.
Let the radii of the sector be OA and OB.
Given ∠AOB = 60°.
In the ΔAOB, OA=OB, So it is isosceles.
=> ∠OAB = OBA = [180° - ∠AOB ] /2 = 60°
Hence, ΔAOB is equilateral.
Hence, AB = OA = 14 cm.
Area of ΔAOB = √3/4 * R²
Area 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
Let the radii of the sector be OA and OB.
Given ∠AOB = 60°.
In the ΔAOB, OA=OB, So it is isosceles.
=> ∠OAB = OBA = [180° - ∠AOB ] /2 = 60°
Hence, ΔAOB is equilateral.
Hence, AB = OA = 14 cm.
Area of ΔAOB = √3/4 * R²
Area 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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