Find the area of the segment of a circle of radius $12 \,cm$ whose corresponding sector has a central angle of $60^{\circ}$ (Use $\pi=3.14$ ).

Given that, radius of a circle $(r)=12 \,cm$

and central angle of sector $OBCA$ $(\theta)=60^{\circ}$

$\therefore$ Area of sector $O B C A=\frac{\pi r^{2}}{360} \times \theta \quad$ [here, $O B C A=$ sector and $A B C A=$ segment $]$

$=\frac{314 \times 12 \times 12}{360^{\circ}} \times 60^{\circ}$

$=314 \times 2 \times 12$

$=314 \times 24=7536\, cm ^{2}$

since, $\triangle O A B$ is an isosceles triangle.

Let $\angle O A B=\angle O B A=\theta_{1}$

and$O A=O B=12 \,cm$

$\angle A O B=\theta=60^{\circ}$

$\angle O A B+\angle O B A+\angle A O B=180^{\circ} \quad\left[\because\right.$ sum of all interior angles of a triangle is $\left.180^{\circ}\right]$

$\Rightarrow \theta_{1}+\theta_{1}+60^{\circ}=180^{\circ}$

$\Rightarrow 2\theta_{1}=120^{\circ}$

$\Rightarrow {\theta}_{1}=60^{\circ}$

$\Rightarrow \theta_{1}=\theta=60^{\circ}$

So, the required $\triangle A O B$ is an equilateral triangle.

Now, area of $\triangle A O B=\frac{\sqrt{3}}{4}(\text { side })^{2} \quad\left[\because\right.$ area of an equilateral triangle $\left.=\frac{\sqrt{3}}{4}(\text { side })^{2}\right]$

$=\frac{\sqrt{3}}{4}(12)^{2}$

$=\frac{\sqrt{3}}{4} \times 12 \times 12=36 \sqrt{3}\, cm ^{2}$

Now, area of the segment of a circle

i.e.. $A B C A=$ Area of sector $O B C A-$ Area of $\triangle A O B$

$=\left(75.36-36 \sqrt{3}) \,cm ^{2}\right.$

Hence, the required area of segment of a circle is $\left(7536-36 \sqrt{3}) \,cm ^{2}\right.$.