Question

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

Answer 1

Solve for area of the given segment

Given,

1. Radius of the circle $r=12cm$
2. Central angle formed by sector $\theta ={60}^{°}$

From the figure we can say that,

Area of segment $ABCA=$Area of sector $OABC-$ Area of $△OAB$

Step 1: Find area of $△OAB$

According to the figure, $\angle OAB+\angle OBA=180°-60°=120°$

$\therefore \angle OAB=\angle OBA=\frac{120°}{2}=60°[\because OA=OB]$

Hence, $△OAB$ forms equilateral triangle of side $s=12cm$

$\therefore$Area of $△OAB=\frac{\sqrt{3}}{4}{\left(s\right)}^2$

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

$\Rightarrow$Area of $△OAB=62.35c{m}^2$

Step 2: Find area of sector $OABC$

According to the figure, radius of the circle $r=12cm$ and

central angle formed by sector $\theta ={60}^{°}$

We know that, area of sector$=\frac{\theta }{360}\pi r^2$

$\therefore$Area of sector $OABC=\frac{60}{360}\times 3.14\times {\left(12\right)}^2$

$=\frac{1}{6}\times 3.14\times 144$

$\Rightarrow$Area of sector $OABC=75.36c{m}^2$

Step 3: Find area of segment $ABCA$

Area of segment $ABCA=$Area of sector $OABC-$ Area of $△OAB$

$=75.36-62.35$

$\Rightarrow$Area of the segment $ABCA=13.01c{m}^2$

Hence, area of segment $ABCA$ is $13.01c{m}^2$