9. As shown in Figure 1, in the convex

quadrilateral $A B C D$,
$$
\begin{array}{l}
A B=3, B C=4, C D \\
=13, A D=12,
\end{array}
$$
$\angle A B C=90^{\circ}$. Then the area $S$ of quadrilateral $A B C D$ is ( ).
(A) 30
(B) 36
(C) 40
(D) 48
(E) 58.5
9. B.

As shown in Figure 1, connect $A C$. In the right triangle $\triangle A B C$,
$$
\angle A B C=90^{\circ}, A B=3, B C=4 \text {. }
$$

Therefore, $A C=\sqrt{A B^{2}+A C^{2}}=5$.
In $\triangle A C D$, $C D=13, A D=12$, and
$$
A C^{2}+A D^{2}=C D^{2} \text {, }
$$

Thus, $A C \perp A D$.
Then, $S=S_{\triangle A B C}+S_{\triangle A C D}=\frac{3 \times 4}{2}+\frac{5 \times 12}{2}=36$.