2. In trapezoid $A B C D$, it is known that $A B / / C D$, $\angle A B C=90^{\circ}, B D \perp A D, B C=5, B D=13$. Then the area of trapezoid $A B C D$ is $\qquad$ .
2. $\frac{1565}{24}$.

As shown in Figure 2, from the given conditions we know
$$
C D=\sqrt{13^{2}-5^{2}}=12, \angle B D C=\angle A B D \text {. }
$$

Therefore, Rt $\triangle B C D \backsim$ Rt $\triangle A D B$
$$
\Rightarrow \frac{A B}{13}=\frac{13}{12} \Rightarrow A B=\frac{169}{12} \text {. }
$$

Thus, $S_{\text {trapezoid } 1 B C D}=\frac{1}{2}\left(\frac{169}{12}+12\right) \times 5=\frac{1565}{24}$.