11. (6 points) As shown in the figure, $BC=3BE$, $AC=4CD$, the area of triangle $ABC$ is $\qquad$ times the area of triangle $ADE$.


【Solution】Solution: Since $B C=3 B E, A C=4 C D$, then $B C: B E=3: 1, A C: C D=4: 1$, so $S_{\triangle A B E}=\frac{1}{3} S \triangle A B C, S_{\triangle A C E}=\frac{2}{3} S_{\triangle A B C}$,
$$
S_{\triangle A D E}=\frac{3}{4} S_{\triangle A C E}=\frac{3}{4} \times \frac{2}{3} S_{\triangle A B C}=\frac{1}{2} S_{\triangle A B C},
$$

The area of triangle $A B C$ is twice the area of triangle $A D E$.
Therefore, the answer is: 2.