20. In quadrilateral $A B C D$, $\angle A B C=\angle A C D$ $=90^{\circ}, A C=20, C D=30$, diagonal $A C$ intersects $B D$ at point $E, A E=5$. Then the area of quadrilateral $A B C D$ is ( ).
(A) 330
(B) 340
(C) 350
(D) 360
(E) 370
20. D.

As shown in Figure 5, draw $B F \perp A C$ at point $F$. Let $E F=x$.
Thus, $\triangle B F E \backsim \triangle D C E$.
Given $A C=20, A E=5$, we know $C E=15, B F=2 x$.
In the right triangle $\triangle A B C$, $\angle A B C=\angle B F A=90^{\circ}$.
$$
\begin{array}{l}
\text { Then } \triangle A B F \backsim \triangle B C F \\
\Rightarrow B F^{2}=A F \cdot C F \\
\Rightarrow 4 x^{2}=(5-x)(15+x) \\
\Rightarrow x=3 .
\end{array}
$$

Therefore, $S_{\text {quadrilateral } A B C D}=S_{\triangle A B C}+S_{\triangle A C D}$
$$
=\frac{1}{2}(20 \times 6+20 \times 30)=360 \text {. }
$$