Question

The radius of the circle is 7 cm and the angle subtended by the arc is 60°. If the area of $\Delta$OAB is $21c{m}^2$, then find the area of segment ABP.

Answer 1

Given:
$\text{Area of}\Delta OAB=21c{m}^2$

$\text{Area of sector OAPB}=\frac{60}{360}\times \pi r^2=\frac{60}{360}\times \frac{22}{7}\times 7^2=25.7{cm}^2$
So,
Area of segment APB
= Area of sector OAPB - Area of triangle OAB
$=25.7-21=4.7{cm}^2$

Therefore, area of segment APB$=4.7{cm}^2$