【Analysis】As shown in the figure, since $\mathrm{AC} / / \mathrm{DE}$, we have $S_{\triangle A C F}=S_{\triangle A C D}=S_{\triangle A C E}$
$$
\begin{array}{l}
\text { Clearly } \quad S_{\triangle A C F}=S_{\triangle A C B} \\
\therefore S_{\triangle A C B}=S_{\triangle A C D}=S_{\triangle A C E} \\
\therefore E A=A B, \quad B C=C D \\
\therefore S_{\triangle B D E}=2 S_{\triangle A B D}=4 S_{\triangle A B C}=4 \times \frac{1}{2} \times 5 \times 3=30 \\
\therefore S_{A C D E}=S_{\triangle B D E}-S_{\triangle A B C}=30-\frac{1}{2} \times 5 \times 3=22.5
\end{array}
$$
15. As shown in the figure, a rectangular piece of paper is folded along the line AC, with vertex B landing at point F. For the second fold, fold along line DE through point F, such that the crease DE is parallel to AC. If AB = 5, BC = 3, then the area of trapezoid ACDE is $\qquad$.