jmc

To begin with, let $DF=x$ and $FA=9-x$. $△DFA$ is a right triangle, so we can solve for $x$ by applying the Pythagorean Theorem: $x^2+9=81-18x+x^2$, so $18x=72$, or $x=4$. By applying the same argument to $△EAB$, we can see that $FA=EA=5$. Drop a perpendicular from $F$ to $EA$ and call the intersection point $P$. $PFDA$ is a rectangle, so we know that $PA=FD=4$, so $PE=5-4=1$. Furthermore, we know that $FP=DA=3$. Now, we have right triangle $△FPE$ with legs $1$ and $3$, so we can solve for $FE$ by applying the Pythagorean Theorem: $FE=\sqrt{1+9}=\boxed{\sqrt{10}}$.