In $\Delta ACO$,
$\begin{array}{rcll}
OA & = & OC & \text{(radii)} \\
\angle ACO & = & \angle CAO & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle ACO & = & \dfrac{1}{2} (180^\circ – \angle AOC) & \text{($\angle$ sum of $\Delta$)} \\
\angle ACO & = & \dfrac{1}{2} (180^\circ – 90^\circ) \\
\angle ACO & = & 45^\circ
\end{array}$
Since $\Delta OAB$ is an equilateral triangle, then $\angle AOB = 60^\circ$. In $\Delta CDO$,
$\begin{array}{rcl}
\angle COD & = & \angle AOC 0 \angle AOB \\
\angle COD & = & 90^\circ – 60^\circ \\
\angle COD & = & 30^\circ
\end{array}$
Then by applying sine law to $\Delta CDO$, we have
$\begin{array}{rcl}
\dfrac{CO}{\sin \angle CDO} & = & \dfrac{DO}{\sin \angle DCO} \\
\dfrac{CO}{\sin (180^\circ – \angle COD – \angle DCO)} & = & \dfrac{DO}{\sin \angle DCO} \\
\dfrac{12}{\sin (180^\circ – 30^\circ – 45^\circ)} & = & \dfrac{DO}{\sin 45^\circ} \\
DO & = & 8.784\ 609\ 691 \text{ cm}
\end{array}$
Hence, the area of the shaded region
$\begin{array}{cl}
= & \text{the area of sector $OBC$} – \text{the area of $\Delta OCD$} \\
= & \pi (12)^2 \times \dfrac{30^\circ}{360^\circ} – \dfrac{1}{2} \times OC \times OD \sin \angle COD \\
= & \pi \times 144 \times \dfrac{1}{12} – \dfrac{1}{2} \times 12 \times 8.784\ 609\ 691 \times \sin 30^\circ \\
= & 11.345\ 282\ 77 \text{ cm}^2 \\
\approx & 11 \text{ cm}^2
\end{array}$
