Find the area of the sector inside the square ABCD.

$\alpha = 60^\circ$
 

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

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