so hmm notice the picture below
recall, a square has all angles to 90°, thus, the diagonals bisecting any of the vertices, will make a 45° on each
anyhow, [tex]\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad
\begin{cases}
\theta=\textit{central angle in degrees}\\
r=radius\\
----------\\
\theta=90\\
r=3
\end{cases}[/tex]
