7. In $\triangle A B C$, $A C=B C, \angle A C B=90^{\circ}$, $D 、 E$ are points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
7:36.
As shown in Figure 5, rotate $\triangle C E B$ $90^{\circ}$ clockwise to get $\triangle C E^{\prime} A$. Connect $E^{\prime} D$.

It is easy to see that $A E^{\prime}=B E=4$, $\angle E^{\prime} A D=90^{\circ}$, so
$$
\begin{array}{l}
E^{\prime} D=\sqrt{A E^{\prime 2}+A D^{2}} \\
=\sqrt{4^{2}+3^{2}}=5 .
\end{array}
$$

Since $\angle D C E=45^{\circ}=\angle D C E^{\prime}$, we know $\triangle D C E \cong \triangle D C E^{\prime}$.

Therefore, $D E=E^{\prime} D=5, A B=12$.
Thus, $S_{\triangle A B C}=\frac{1}{4} A B^{2}=36$.