1. As shown in Figure 1, divide a line segment of length 1 into two segments of lengths $x$ and $y$. Then, bend the segment of length $x$ into a semicircle $A C B$, and fold the segment of length $y$ into three sides $(B D, D E, E A)$ of a rectangle $A B D E$, forming a closed "curved polygon" $A C B D E A$. The maximum value of the area of this curved polygon is $\qquad$
$-1 \cdot \frac{1}{2(\pi+4)}$.
Let the radius of the circle be $r$, and the width of the rectangle be $h$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x=\pi r, \\
1-x=2 r+2 h
\end{array}\right. \\
\Rightarrow r=\frac{x}{\pi}, h=\frac{1}{2}\left(1-x-\frac{2 x}{\pi}\right) .
\end{array}
$$

Therefore, the area of the curved shape is
$$
\begin{aligned}
S & =\frac{1}{2} \pi \cdot \frac{x^{2}}{\pi^{2}}+\frac{1}{2} \cdot \frac{2 x}{\pi}\left(1-x-\frac{2 x}{\pi}\right) \\
& =\frac{x}{\pi}-\frac{(4+\pi) x^{2}}{2 \pi^{2}} \\
& =\frac{4+\pi}{2 \pi^{2}}\left(\left(\frac{\pi}{4+\pi}\right)^{2}-\left(x-\frac{\pi}{4+\pi}\right)^{2}\right) .
\end{aligned}
$$

Thus, when $x=\frac{\pi}{\pi+4}$, $S_{\text {max }}=\frac{1}{2(\pi+4)}$.