Ans: A
Let $OA=r\text{ cm}$. Then we have
Let $OA=r\text{ cm}$. Then we have
$\begin{array}{rcl}
2r + 2\pi r \times \dfrac{90^\circ}{360^\circ} & = & 12 \\
2r + \dfrac{\pi}{2} r & = & 12 \\
r ( 2 + \dfrac{\pi}{2}) & = & 12 \\
r & = & 3.360\ 594\ 921 \\
r & \approx & 3.36
\end{array}$
Therefore, $OA\approx 3.36\text{ cm}$.
