设正方形边长为 \(a\) 并令 \(\vert AE\vert=x,~\vert AF\vert=y\)，则面积 \({\cal A}\) 可如下计算
\[
\begin{array}{l}
\left.\begin{array}{l}
\displaystyle
\frac{xy}{2}=5
\\
\displaystyle
\frac{(a-x)a}{2}=4
\\
\displaystyle
\frac{(a-y)a}{2}=3
\end{array}\right\}\Rightarrow
\left\{\begin{array}{l}
\displaystyle
x=\frac{2\sqrt{3+\sqrt{6}}}{3}\left[2\sqrt{6}-3\right]
\\
\displaystyle
y=\sqrt{3+\sqrt{6}}\left[\sqrt{6}-1\right]
\\
\displaystyle
a=2\sqrt{3+\sqrt{6}}
\end{array}\right.
\\
{\cal A}=a^2-(3+4+5)=4\sqrt{6}
\end{array}
\]