9. (18 points) As shown in Figure 1, given the parabola $y=x^{2}$ with vertex $O$, $AB$ 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 $AOBD$.

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9. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, with the focus $F\left(0, \frac{1}{4}\right)$, and the directrix equation as $y=-\frac{1}{4}$.
By the definition of a parabola, we have
$$
\begin{array}{l}
2=A F+B F=y_{1}+y_{2}+\frac{1}{2} \\
\Rightarrow y_{1}+y_{2}=\frac{3}{2} .
\end{array}
$$

Let the midpoint of $A B$ be $C\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Then
$$
\begin{array}{l}
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{x_{2}^{2}-x_{1}^{2}}{x_{2}-x_{1}}=x_{1}+x_{2} \\
\Rightarrow k_{C D}=-\frac{1}{x_{1}+x_{2}} .
\end{array}
$$

Then the line $l_{C D}: y-\frac{y_{1}+y_{2}}{2}=-\frac{1}{x_{1}+x_{2}}\left(x-\frac{x_{1}+x_{2}}{2}\right)$.
Let $x=0$, we get $O D=\frac{y_{1}+y_{2}}{2}+\frac{1}{2}=\frac{5}{4}$.
Since the chord $A B$ passes through the focus $F$, then
$$
\begin{array}{l}
l_{A B}: y=k x+\frac{1}{4} \\
\Rightarrow y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+\frac{1}{2} \\
\Rightarrow\left(x_{1}+x_{2}\right)^{2}=1 \Rightarrow k_{A B}= \pm 1 .
\end{array}
$$

Therefore, $O D$ intersects $A B$ at a $45^{\circ}$ acute angle.
Thus, $S_{\text {quadrilateral } A O B D}=\frac{1}{2} A B \cdot O D \sin 45^{\circ}$
$$
=\frac{1}{2} \times 2 \times \frac{5}{4} \times \frac{\sqrt{2}}{2}=\frac{5 \sqrt{2}}{8} .
$$