1. As shown in Figure 3, in the rectangular paper piece $A B C D$, $A B=4$, $A D=9$, point $E$ is on side $A B$, and $B E=1$. The paper is folded so that point $D$ coincides with point $E$, resulting in the crease $GH$. Then the area of the overlapping part $E F G H$ of the paper is
$=1 . \frac{295}{24}$. 
Let $D H=x$. Then $E H=x, A H=9-x$. 
In the right triangle $\triangle A E H$, we have 
$$
3^{2}+(9-x)^{2}=x^{2} \text {. }
$$

Solving for $x$ gives $x=5$, so $D H=5$. Thus, $A H=4$. 
Then, $\triangle A E H \sim \triangle B F E \sim \triangle C^{\prime} F G$ 
$$
\begin{array}{l}
\Rightarrow A E: A H: E H=B F: B E: F E=C^{\prime} F: C^{\prime} G: F G \\
\Rightarrow B F=\frac{3}{4}, F E=\frac{5}{4}, B G=\frac{16}{3}, F G=\frac{55}{12} .
\end{array}
$$

Therefore, $S_{\text {quadrilateral } E F G H}=S_{\text {quadrilateral } A B G H}-S_{\triangle A E H}-S_{\triangle B E F}$
$$
\begin{array}{l}
=2(A H+B G)-\frac{3}{2} A H-\frac{1}{2} B F \\
=\frac{1}{2} A H+2 B G-\frac{1}{2} B F=\frac{295}{24} .
\end{array}
$$