Question

A chord 10 cm long is drawn in a circle whose radius is $5\sqrt{2}cm$. Find the area of both the segments. (Take $\pi =3.14$)

Answer 1

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