Given, area of $\triangle E A B=$ area of square $A B C D$
$E B=E D=\sqrt{130}$
Let side of square $=x$
$B M=\frac{x}{\sqrt{2}}=A M$
Area of $\triangle A E B=$ Area of $\triangle B E M$ - area of
$=\frac{1}{2} E M \times B M-\frac{1}{2} A M \times B M$
$=\frac{1}{2} B M(E M-A M)$
$=\frac{1}{2} \frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)}\right.$
$=\frac{1}{2}+\frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)=x^2}\right.$
$=\sqrt{130-\frac{x^2}{2}}=2 \sqrt{2} x+\frac{x}{\sqrt{2}}$
$130-\frac{x^2}{2}=\left(\begin{array}{c}5 x \\ \sqrt{2}\end{array}\right)^2$
$130-\frac{x^2}{2}=\begin{array}{c}25 x^2 \\ 2\end{array}$
$13 x^2=130 \Rightarrow x^2=10$
$\therefore$ Area of square $=10$
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