Question

$OAB$ is a sector of the circle with centre $O$ and radius $12cms$. If $m\angle AOB={60}^o$, find the difference between the areas of sectors $AOB$ and $△OAB$.

Answer 1

Since $OA=OB$, so $\Delta OAB$ is an equilateral triangle, therefore, let

$\angle A=\angle B=x$

Since the sum of the angles of a triangle is ${180}^{\circ }$.

$x+x+{60}^{\circ }={180}^{\circ }$

$2x={180}^{\circ }-{60}^{\circ }$

$2x={120}^{\circ }$

$x={60}^{\circ }$

The area of sector $OAB$ is,

$A=\pi r^2\frac{\theta }{360}$

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

$=75.42c{m}^2$

The area of $\Delta OAB$ is,

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

$=\frac{\sqrt{3}}{4}\times 12\times 12$

$=62.28c{m}^2$

The difference in the area is,

$A=75.42-62.28$

$A=13.14c{m}^2$