Question

The adjacent figure shows two arcs $P$ and $Q.$ Arc $P$ is part of the circle with centre $O$ and radius $OA=10cm$. Arc $Q$ is a part of the circle with centre $M$ and radius $AM=5cm,$ where $M$ is the mid-point of $AB.$ Area enclosed by the arc $P$ and chord, $AB$ is

• A. $25\sqrt{3}-\frac{50\pi }{3}c{m}^2$
• B. $\frac{50\pi }{3}+25\sqrt{3}c{m}^2$
• C. $\frac{50\pi }{3}-25\sqrt{3}c{m}^2$
• D. None of these

Answer 1

The correct option is C

$\frac{50\pi }{3}-25\sqrt{3}c{m}^2$

In $\Delta OMA$, we have
$\sin \theta =\frac{AM}{OA}=\frac{5}{10}=\frac{1}{2}\Rightarrow \theta ={30}^{\circ }\therefore \angle AOB=2\times \theta ={60}^{\circ }$

$\Rightarrow \cos {30}^{\circ }=\frac{OM}{OA}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OM}{10}$
$\therefore OM=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}cm$

Area of $△AOB=\frac{1}{2}\times AB\times OM=25\sqrt{3}$

Area enclosed between arc $P$ and chord $AB$ is Area of sector $OAPB-$ Area of $△AOB$
$=\pi \times {10}^2\times \frac{2\theta }{{360}^{\circ }}-25\sqrt{3}$

$=\pi \times 100\times \frac{60}{360}-25\sqrt{3}$

$=\frac{50\pi }{3}-25\sqrt{3}c{m}^2$