Question

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of ${60}^{\circ }$.

Answer 1

Area of minor segment = Area of sector- Area of $\Delta OAB$

In $\Delta OAB$

Refer image,

$\theta ={60}^0$

$OA=OB=r=12cm$

$\angle B=\angle A=x[\angle s$ opp. to equal sides are equal

$\Rightarrow \angle A+\angle B+\angle O={180}^0$

$\Rightarrow x+x+{60}^0={180}^0$

$\Rightarrow 2x={180}^0-{60}^0$

$\Rightarrow x=\frac{{120}^0}{2}={60}^0$

$\therefore \Delta OAB$ is equilateral $\Delta$ with each side (a)

$=12cm$

Area of the equilateral $\Delta =\frac{\sqrt{3}}{4}{a}^2$

Area of minor segment = Area of the sector - Area of $\Delta OAB$

$=\frac{\pi r^2\theta }{{360}^0}-\frac{\sqrt{3}}{4}{a}^2$

$=\frac{3.14\times 12\times 12\times {60}^0}{{360}^0}-\frac{\sqrt{3}}{4}\times 12\times 12$

$=6.28\times 12-36\sqrt{3}$

$\therefore$ Area of minor segment $=(75.36-36\sqrt{3})c{m}^2$