Question

In a circle of radius 20 cm, a chord subtends a right angle at the centre. Find the area of the major segment: (use $\pi$ = $\frac{22}{7}$)

• A. 2635.6 ${cm}^2$
• B. 1391.9 ${cm}^2$
• C. 1125.2 ${cm}^2$
• D. 1142.85 ${cm}^2$

Answer 1

The correct option is D 1142.85 ${cm}^2$

Area of minor segment (i.e. AYB) = Area of sector OAYB – Area of $△$OAB

Area of sector OAYB = $\frac{x^{\circ }}{{360}^{\circ }}$$\times$$\frac{22}{7}$$\times$ $r^2$

Since the angle subtended by the sector at the centre = ${90}^{\circ }$ and radius = 20 cm

So,
Area of sector OAYB = $\frac{{90}^{\circ }}{{360}^{\circ }}$$\times$$\frac{22}{7}$$\times$ ${20}^2$ = 314.28 ${cm}^2$

Area of $△$ OAB = $\frac{1}{2}$$\times$ OA $\times$ OB

= $\frac{1}{2}$$\times$ 20 $\times$ 20

= 200 ${cm}^2$

Area of minor segment (i.e. AYB) = 314.28 – 200 = 114.28 ${cm}^2$

Area of circle =$\frac{22}{7}$$\times$ $r^2$

=$\frac{22}{7}$$\times$ ${20}^2$

= 1257.14 ${cm}^2$

Area of major segment = [Area of circle - Area of minor Segment (AYB)]
= 1257.14 - 114.28
= 1142.85 ${cm}^2$
$\therefore$ Area of major segment = 1142.85 ${cm}^2$