9. (3 points) As shown in the figure, in trapezoid $A B C D$, if $A B=8, D C=10, S_{\triangle A M D}=10, S_{\triangle B C M}=15$, then the area of trapezoid $A B C D$ is $\qquad$ .
【Solution】Solution: $\triangle A D M, \triangle B C M, \triangle A B M$ all have the same height, so $S_{\triangle A B M}:\left(S_{\triangle A D M}+S_{\triangle B C M}\right)=8: 10=4: 5$. Given $S_{\triangle A M D}=10, S_{\triangle B C M}=15$, the area of $S_{\triangle A B M}$ is: $(10+15) \times \frac{4}{5}=20$. The area of trapezoid $A B C D$ is: $10+15+20=45$; Answer: The area of trapezoid $A B C D$ is 45. Therefore, the answer is: 45.