Ans: C
Let $\theta$ be the angle at centre of the sector. Then we have
Let $\theta$ be the angle at centre of the sector. Then we have
$\begin{array}{rcl}
\pi (12)^2 \times \dfrac{\theta}{360^\circ} & = & 48\pi \\
\theta & = & 120^\circ
\end{array}$
Therefore, the perimeter of the sector
$\begin{array}{cl}
= & 2\pi (12) \times \dfrac{120^\circ}{360^\circ} + 2(12) \\
= & 49.132~741~23 \\
\approx & 49.1\text{ cm}
\end{array}$
