  Question

A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the area of both the segments.         (Take π=3.14)

Solution Given, a chord AB of length 10 cm and radius = OA = OB = 5√2 cm. Construction: Draw OC perpendicular to AB. Now, AC = BC = 10/2 = 5 cm  [The perpendicular drawn from the centre of a circle to a chord always bisect the chord.] In triangle OAC, sin x = Perpendicular/hypotenuse = AC/ OA = 5/5√2 Similarly, ∠BOC = 45° ⇒ ∠AOB = ∠AOC + ∠BOC = 45°+ 45° = 90° We know that, area of sector Therefore, area of sector OAB Again, in triangle OAC, cos x = base/hypotenuse = OC/ OA cos 45 =  OC/5√2 Therefore, area of triangle OAB Now, area of minor segment =  Area of sector OAB - Area of triangle OAB = 39.28 cm2 - 25 cm2 = 14.28 cm2 Similarly, area of major segment = Area of circle - Area of minor segment ⇒ Area of major segment MathematicsRD SharmaStandard X

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