Four, (Full score 25 points) As shown in Figure 1, EFGH is an inscribed quadrilateral of square $ABCD$, $\angle BEG$ and $\angle CFH$ are both acute angles. Given that $EG=3, FH=4$, and the area of quadrilateral $EFGH$ is 5. Find the area of square $ABCD$.

保留源文本的换行和格式，直接输出翻译结果如下：

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Four, (Full score 25 points) As shown in Figure 1, EFGH is an inscribed quadrilateral of square $ABCD$, $\angle BEG$ and $\angle CFH$ are both acute angles. Given that $EG=3, FH=4$, and the area of quadrilateral $EFGH$ is 5. Find the area of square $ABCD$.
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In square $ABCD$, perpendicular lines are drawn from $E, F, G, H$ to the opposite sides, forming rectangle $PQRT$. Let the side length of $ABCD$ be $a$, $PQ=b$, and $QR=c$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
b=\sqrt{3^{2}-a^{2}}, \\
c=\sqrt{4^{2}-a^{2}} .
\end{array}
$$

From $S_{\triangle AEH}=S_{\triangle TEH}, S_{\triangle YEF}=S_{\triangle PEF}, S_{\triangle CFG}=$ $S_{\triangle QFG}, S_{\triangle DGH}=S_{\triangle RGY}$, we get
$$
a^{2}+b=2.5.
$$
$$
a^{2}+\sqrt{3^{2}-a^{2}} \cdot \sqrt{4^{2}-a^{2}}=10 \text {, }
$$

Thus, $5 a^{2}=44, a^{2}=\frac{44}{5}$.
Therefore, $S_{\text {square } ABCD}=\frac{44}{5}$.