Ans: (b) $4\pi\text{ cm}^2$
- In $\Delta OPS$ and $\Delta ORS$,
$\begin{array}{rcll}
OP & = & OR & \text{(given)} \\
OS & = & OS & \text{(common side)} \\
PS & = & RS & \text{($S$ is the mid-point of $PR$)}
\end{array}$$\therefore \Delta OPS \cong \Delta ORS$ (S.S.S.).
- Since $O$ is the centre of the circle, we have
$\begin{array}{rcll}
\angle POQ & = & 2 \times \angle PRQ & \text{($\angle$ at centre twice $\angle$ at $\unicode{x2299}^{ce}$)} \\
& = & 2 \times 10^\circ & \\
& = & 20^\circ
\end{array}$Note that $\angle POQ = \angle ROQ = 20^\circ$ (corr. $\angle$s, $\cong \Delta$s),
therefore $\angle POR = 40^\circ$.
Hence, the area of the sector $OPQR$
$\begin{array}{cl}
= & \pi \times 6 \times 6 \times \dfrac{40^\circ}{360^\circ} \\
= & 4 \pi \text{ cm}^2
\end{array}$
