Triangle $ABC$ is an isosceles right triangle with $AC=BC=1$ and $\angle C=90^\circ$, and $D$ is the midpoint of $\overline{AB}$. The figure consists of the sector with center $C$ and radius $CA$ (bounded by $\overline{CA}$, $\overline{CB}$ and the arc $AB$) together with the semicircle with center $D$ and radius $DA$ on the same side of $\overline{AB}$ as $C$. What is the area of this figure?

1. Since $AB$ is the hypotenuse of right triangle $ABC$ and $D$ is its midpoint, $C$ lies on the circle centered at $D$ with radius $DA$ (Theorem 29), so the $D$-sector (a semicircle, since $A,D,B$ are collinear) passes exactly through $C$.
2. $[ABC]=\frac12\cdot AC\cdot BC=\frac12$, and $AB=\sqrt2$, so $DA=DB=\frac{\sqrt2}2$.
3. The $C$-sector is a quarter circle (since $\angle ACB=90^\circ$) of radius $CA=1$: area $=\frac14\pi(1)^2=\frac{\pi}4$. The $D$-sector is a semicircle of radius $\frac{\sqrt2}2$: area $=\frac12\pi\left(\frac{\sqrt2}2\right)^2=\frac{\pi}4$.
4. The two sectors overlap exactly in $\triangle ABC$: the $C$-sector's only points on the $C$-side of line $AB$ are $\triangle ABC$ itself (the rest of the $C$-sector is the segment beyond $AB$, on the opposite side from $C$), while the $D$-sector lies entirely on the $C$-side of $AB$.
5. So the total (union) area is $\frac{\pi}4+\frac{\pi}4-\frac12=\boxed{\dfrac{\pi-1}2}$.