Ans: $10\pi\text{ cm}$
The length of $\overparen{AB}$
$\begin{array}{cl}
= & \pi \times (12)^2 \times \dfrac{150^\circ}{360^\circ} \\
= & 10 \pi \text{ cm}
\end{array}$
The length of $\overparen{AB}$
$\begin{array}{cl}
= & \pi \times (12)^2 \times \dfrac{150^\circ}{360^\circ} \\
= & 10 \pi \text{ cm}
\end{array}$
