Question

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

Answer 1

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

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

$\therefore$ Area of sector OBCA
$=\frac{\pi r^2}{360}\times \theta$ [here, OBCA = sector and ABCA = segment]

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

$=3.14\times 2\times 12$

$=3.14\times 24=75.36c{m}^2$

Since, $\Delta OAB$ is an isosceles triangle.

Let $\angle OAB=\angle OBA={\theta }_1$

and OA = OB = 12cm

$\angle AOB=\theta ={60}^{\circ }$

$\therefore \angle OAB+\angle OBA+\angle AOB={180}^{\circ }$
[$\because$ Sum of all interior angles of a triangle is ${180}^{\circ }$]

$\Rightarrow {\theta }_1+{\theta }_1+{60}^{\circ }={180}^{\circ }$

$\Rightarrow 2{\theta }_1={120}^{\circ }$

$\Rightarrow {\theta }_1={60}^{\circ }$

$\therefore {\theta }_1=\theta ={60}^{\circ }$

So, the required $\Delta AOB$ is an equilateral triangle.

Now, area of $\Delta AOB=\frac{\sqrt{3}}{4}(side{)}^2$
$[\because areaofanequilateraltriangle=\frac{\sqrt{3}}{4}(side{)}^2]$

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

$=\frac{\sqrt{3}}{4}\times 12\times 12=36\sqrt{3}c{m}^2$

Now, area of the segment of a circle i.e

ABCA = Area of sector OBCA - Area of $\Delta AOB$

$=(75.36-36\sqrt{3})c{m}^2$

Hence, the required of segment of a circle is:
$(75.36-36\sqrt{3})c{m}^2$.