Question

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and $\sqrt3$ = 1.73)

Answer 1

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT

In ΔOVS,

$\frac{OV}{OS} = cos 60^{\degree}$

$\frac{ OV}{ 12} = \frac{1}2$

$OV = 6 \,cm$

$\frac{SV}{ SO} = sin 60^{\degree} = \frac{\sqrt3} 2$

$\frac{SV}{ 12} = \frac{\sqrt3}{2}$

$SV = 6 \sqrt3\,cm$

$ST = 2SV = 2 \times 6 \sqrt3 = 12 \sqrt3 \,cm$

Area of ΔOST = $\frac{1}2 \times ST \times OV$

= $\frac{1}2 \times 12 \sqrt3 \times 6$

= $36\sqrt3 = 36 \times 1.73 = 62.28 \,cm^2$

Area of sector OSUT = $\frac{120^{\degree} }{ 360^{\degree}} \times \pi (12)^2$

= $\frac{1}3 \times 3.14 \times 144 = 150.72 \,cm^2$

Area of segment SUT = Area of sector OSUT - Area of ΔOST
= 150.72 - 62.28
= 88.44 cm$^2$