18. As shown in the figure, $A B C D$ is a square with a side length of $\mathbf{6}$, $A D G H$ is a trapezoid, points $E$ and $F$ are the midpoints of $A D$ and $G H$ respectively, $H F=6, E F=4, E F \perp G H$. Connect $H E$ and extend it to intersect $C D$ at point $I$, draw $I J \perp H A$, then $I J=$ $\qquad$ -
【Analysis】As shown in the figure, connect $\mathrm{AI}$, extend $\mathrm{CD}$ to intersect $\mathrm{GH}$ at $\mathrm{K}$
It is easy to know that $K D=4, G K=3$, so $G D=5$, similarly $A H=5$
Also, since $H K=9, D E=3, K D=4, I D: I K=D E: H K$, we know $I D=2$. Let $I J=a, J A=b$, by the Pythagorean theorem
we have $\left\{\begin{array}{l}a^{2}+b^{2}=A J^{2} \\ a^{2}+(b+5)^{2}=H I^{2}\end{array}\right.$
Also by the Pythagorean theorem $\left\{\begin{array}{l}A J^{2}=I D^{2}+D A^{2}=40 \\ H I^{2}=I K^{2}+K H^{2}=117\end{array}\right.$

So $\left\{\begin{array}{l}a^{2}+b^{2}=40 \\ a^{2}+(b+5)^{2}=117\end{array} \Rightarrow 10 b+25=77 \Rightarrow b=5.2\right.$
$$
\Rightarrow a^{2}+27.04=40 \Rightarrow a^{2}=12.96 \Rightarrow a=3.6
$$