smc

Since $\angle EBA=\angle FCB$ and $\angle FBC=\angle AEB$, we have $△ABE\sim △FCB$. $\frac{AB}{FC}=\frac{BE}{CB}=\frac{EA}{BF}$ $\frac{2}{FC}=\frac{\sqrt{5}}{2}=\frac{1}{BF}$ From those two equations, we find that $CF=\frac{4}{\sqrt{5}}$ and $BF=\frac{2}{\sqrt{5}}$ Now that we have $BF$ and $CF$, we can find the area of the bottom triangle $△CFB$: $\frac{1}{2}\cdot \frac{4}{\sqrt{5}}\cdot \frac{2}{\sqrt{5}}=\frac{4}{5}$ The area of left triangle $△BEA$ is $\frac{1}{2}\cdot 2\cdot 1=1$ The area of the square is $4$. Thus, the area of the remaining quadrilateral is $4-1-\frac{4}{5}=\frac{11}{5}$, and the answer is $\boxed{\frac{11}{5}}$