smc

Since the total area is $4$, the side length of square $ABCD$ is $2$. We see that since triangle $HAE$ is a right isosceles triangle with area 1, we can determine sides $HA$ and $AE$ both to be $\sqrt{2}$. Now, consider extending $FB$ and $IE$ until they intersect. Let the point of intersection be $K$. We note that $EBK$ is also a right isosceles triangle with side $2-\sqrt{2}$ and find its area to be $3-2\sqrt{2}$. Now, we notice that $FIK$ is also a right isosceles triangle (because $\angle EKB=45^{\circ }$) and find it's area to be $\frac{1}{2}$$F{I}^2$. This is also equal to $1+3-2\sqrt{2}$ or $4-2\sqrt{2}$. Since we are looking for $F{I}^2$, we want two times this.(This is a misplaced problem) That gives $\boxed{8-4\sqrt{2}}$.~TLiu