(9) (18 points) As shown in the figure, the vertex of the parabola $y=x^{2}$ is $O, AB$ is a chord of length 2 passing through the focus $F$, and $D$ is the intersection of the perpendicular bisector of $AB$ with the $y$-axis. Find the area of quadrilateral $AOBD$.


9 Let the coordinates of points $A$ and $B$ be $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, respectively. The focus of the parabola is $F\left(0, \frac{1}{4}\right)$, and the equation of the directrix is $y=-\frac{1}{4}$. According to the definition of a parabola, we have $A F=y_{1}+\frac{1}{4}$ and $B F=y_{2}+\frac{1}{4}$. Therefore,
$$
2=A F+B F=y_{1}+y_{2}+\frac{1}{2}, \quad y_{1}+y_{2}=\frac{3}{2}.
$$
The midpoint of $A B$ is $C\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$,
$$
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{x_{2}^{2}-x_{1}^{2}}{x_{2}-x_{1}}=x_{1}+x_{2}.
$$

Thus, $k_{C D}=-\frac{1}{x_{1}+x_{2}}$, and the equation of $C D$ is
$$
y-\frac{y_{1}+y_{2}}{2}=-\frac{1}{x_{1}+x_{2}}\left(x-\frac{x_{1}+x_{2}}{2}\right).
$$

Let $x=0$, we get the coordinates of point $D$ as $D\left(0, \frac{y_{1}+y_{2}}{2}+\frac{1}{2}\right)=\left(0, \frac{5}{4}\right)$, so $O D=\frac{5}{4}$.

Since $A B$ passes through the focus $F$, the equation of $A B$ is $y-\frac{1}{4}=k_{A B} \cdot x$, i.e., $y=k x+\frac{1}{4}$. Thus, $y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+\frac{1}{2}$, so $y_{1}+y_{2}=\left(x_{1}+x_{2}\right)^{2}+\frac{1}{2}$, which implies $\left(x_{1}+x_{2}\right)^{2}=1$.
Therefore, $k_{A B}=x_{1}+x_{2}=1$ or -1, so $O D$ intersects $A B$ at a $45^{\circ}$ acute angle. Hence, the area of quadrilateral $A O B D$ is
$$
S=\frac{1}{2} A B \times O D \cdot \sin 45^{\circ}=\frac{1}{2} \cdot 2 \cdot \frac{5}{4} \cdot \frac{\sqrt{2}}{2}=\frac{5 \sqrt{2}}{8}.
$$