3. In trapezoid $A B C D$; $A D / / B C, A C 、 B D$ intersect at point $O, \triangle A O D$ and $\triangle A O B$ have areas of 9 and 12, respectively. Then the area of trapezoid $A B C D$ is $\qquad$
3. 49.

As shown in Figure 10, since
$$
\begin{array}{l}
\frac{S_{\triangle A O D}}{S_{\triangle A U B}}=\frac{O D}{O B}=\frac{O A}{O C} \\
=\frac{S_{\triangle A O B}}{S_{\triangle B C C}},
\end{array}
$$

thus $S_{\triangle A B B}^{2}=S_{\triangle M D} \cdot S_{\triangle B O C}$,
which means $S_{\triangle B O C}=\frac{12^{2}}{9}=16$.
Furthermore, since $S_{\triangle O 1 D}=S_{\triangle A O B}=12$, therefore,