1. As shown in the figure, divide a line segment of length 1 into segments $x$ 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 $A B D E$ to form a closed "curved polygon" $A C B D E A$. The maximum value of the area of this curved polygon is . $\qquad$
The perimeter of a semicircle $\pi r=x \Rightarrow r=\frac{x}{\pi}$, the length of the rectangle $2 r$, and the width $\frac{y-2 r}{2}$. Then the area of the curved shape $S=\frac{1}{2} \pi r^{2}+r(y-2 r)=\frac{x^{2}}{2 \pi}+\frac{x(1-x)}{\pi}-\frac{2 x^{2}}{\pi^{2}}=\frac{2 \pi x-(\pi+4) x^{2}}{2 \pi^{2}}$, so when $x=\frac{\pi}{\pi+4}$, $S_{\text {max }}=\frac{1}{2(\pi+4)}$.