18. As shown in the figure, $\mathrm{ABCD}$ is a square with a side length of 20, $\mathrm{E}$ is a point on $\mathrm{AD}$, and $\mathrm{AE}=15$. Connect $\mathrm{BE}$, draw $\mathrm{AF} \perp \mathrm{BE}$, $\mathrm{FG} \perp \mathrm{BC}$, then the length of $\mathrm{FG}$ is $\qquad$ -
【Analysis】As shown in the figure, extend $\mathrm{GF}$ to intersect $\mathrm{AD}$ at $\mathrm{H}$
By the Pythagorean theorem, we know $B E=25$
$$
S_{\triangle A B E}=\frac{1}{2} \times A B \times A E=\frac{1}{2} \times B E \times A F \Rightarrow A F=12
$$

By the Pythagorean theorem, we know $E F=9$
$$
S_{\triangle A E F}=\frac{1}{2} \times A F \times E F=\frac{1}{2} \times A E \times H F \Rightarrow H F=7.2
$$

Therefore, $F G=20-7.2=12.8$