Find the area of the sector inside the square ABCD.

$\alpha = 60^\circ$
 

$\begin{align} \theta & = 180^\circ - 2\alpha \\ \\ & = 180^\circ - 2(60^\circ) \\ \\ & = 60^\circ \end{align}$
 

$\begin{align} A & = \dfrac{\pi r^2 \theta_{deg}}{360^\circ} \\ \\ & = \dfrac{\pi(10^2)(60^\circ)}{360^\circ} \\ \\ & = 52.36 ~ \text{cm}^2 \end{align}$