9. As shown in Figure 1, in isosceles $\triangle ABC$, $AB=AC$, all triangles in the figure are similar, the area of the smallest triangle is $1$, and the area of $\triangle ABC$ is 40. Then the area of trapezoid $DBCE$ is ( ).
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24
9. E.

Notice that the ratio of the area of the smallest triangle to $\triangle A B C$ is $\frac{1}{40}$, and all triangles are similar. Thus, the similarity ratio is $\frac{1}{\sqrt{40}}$.
From the diagram, $\frac{D E}{B C}=\frac{4}{\sqrt{40}} \Rightarrow \frac{S_{\triangle A D E}}{S_{\triangle A B C}}=\frac{16}{40}$.
Given $S_{\triangle A B C}=40$, therefore
$S_{\text {figure } D B C E}=40-16=24$.