2011SP-II-16

Ans: B
In $\Delta OAB$, as $OA=AB$,

$\begin{array}{rcl}
\angle AOB & = & \angle ABO \\
& = & \dfrac{1}{2}\times (180^\circ-90^\circ) \\
& = & 45^\circ
\end{array}$

In $\Delta OAD$,

$\begin{array}{rcl}
OD & = & OC \\
& = & 20\mbox{ cm}
\end{array}$

Hence, we have

$\begin{array}{rcl}
\cos \angle DOA & = & \dfrac{10}{20} \\
\angle DOA & = & 60^\circ
\end{array}$

Also,

$\begin{array}{rcl}
DA & = & \sqrt{20^2-10^2} \\
& = & 10\sqrt{3}
\end{array}$

Therefore, the area of the shaded region

$\begin{array}{cl}
= & \pi (20)^2 \times\dfrac{105^\circ}{360^\circ} – \dfrac{1}{2}(10)(10) -\dfrac{1}{2}(10)(10\sqrt{3}) \\
= & 229.916~602~5\mbox{ cm}^2 \\
\approx & 230\mbox{ cm}^2
\end{array}$