A chord 10cm long is drawn in a circle whose radius is 5 root2 cm find the area of both segments.?

**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 cm^{2} - 25 cm^{2}

= 14.28 cm^{2}

Similarly, area of major segment = Area of circle - Area of minor segment

⇒ Area of major segment

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