Question

$OAB$ is a sector of circle with center $O$ and radius $12cm$. If $m\angle AOB={60}^{\circ }$. Find the difference between the areas of sector $AOB$ and $\Delta AOB$.

Answer 1

Given:

In, $\Delta AOB,AO=OB$ (radius of the circle)

$\Rightarrow \angle OAB=\angle OBA$ (Angles opposite to equal sides are equal)

$\Rightarrow {60}^o+\angle OAB+\angle OBA={180}^o$ (sum of angles of a triangle)

$\Rightarrow 2\angle OAB={120}^o$

$\Rightarrow \angle OAB={60}^o=\angle OBA$

$\Rightarrow AOB$ is an equilateral triangle.

So, Area $\left(\Delta AOB\right)=\frac{\sqrt{3}}{4}(side{)}^2=\frac{\sqrt{3}}{4}(12{)}^2$

$=\left(\frac{\sqrt{3}}{4}\right)144=62.28c{m}^2$

Now, Area of sector $AOB=\frac{\theta }{2\pi }\times \left(\pi r^2\right)$

$=\frac{60}{360}\times \frac{22}{7}\times 12\times 12$

$=75.42c{m}^2$

Difference $=(75.42-62.28)c{m}^2$

$=13.14c{m}^2$