2012PP-II-15

Ans: C
Let $r\mbox{ cm}$ be the radius of the sector $OABC$.

$\begin{array}{rcl}
\dfrac{1}{2} r^2 & = & 12 \\
r & = & \sqrt{24} \\
& = & 2\sqrt{6}
\end{array}$

Therefore the area of the segment $ABC$

$\begin{array}{cl}
= & \dfrac{1}{4} \pi (\sqrt{24})^2 -12 \\
= & 6\pi -12 \\
= & 6(\pi – 2) \mbox{ cm}^2
\end{array}$