1. As shown in the figure, in $\odot O$, $\overparen{A D B} = 90^{\circ}$, chord $A B = a$, and a circular arc with center $B$ and radius $B A$ intersects $\odot O$ at another point $C$. Then the area $S$ of the crescent shape (the shaded part in the figure) formed by the two circular arcs is $S=$ . $\qquad$
ii. $1 \cdot \frac{1}{2} a^{2}$.
It is known that $AC$ is the diameter of $\odot O$, connect $BC$ (figure not shown). Then $\triangle ABC$ is an isosceles right triangle:
$\therefore OA=\frac{1}{2} AC=\frac{1}{2} \sqrt{2} a$.
$\therefore$ The area of sector $ABC$ $S_{1}=\frac{\pi}{4} \cdot AB^{2}=\frac{\pi}{4} a^{2}$,
The upper half area of $\odot O$ $S_{2}=\frac{1}{2} \pi \cdot OA^{2}=\frac{\pi}{4} a^{2}$.
$\mathrm{F}$ is, $S_{2}=S_{1}$.
Subtract the common part (the area of the non-shaded segment $AC$) from both sides, we get $S=S_{\triangle ABC}=\frac{1}{2} a^{2}$.