4.9. Each diagonal of the convex pentagon $A B C D E$ cuts off a triangle of unit area from it. Calculate the area of the pentagon $A B C D E$.
4.9. Since $S_{A B E}=S_{A B C}$, then $E C \| A B$. The other diagonals are also parallel to the corresponding sides. Let $P$ be the intersection point of $B D$ and $E C$. If $S_{B P C}=x$, then $S_{A B C D E}=S_{A B E}+S_{E P B}+S_{E D C}+S_{B P C}=3+x$ (since $S_{E P B}=S_{A B E}=1$, as $A B P E$ is a parallelogram). Since $S_{B P C}: S_{D P C}=B P: D P=S_{E P B}: S_{E P D}$, then $x:(1-x)=1: x$, which means $x=(\sqrt{5}-1) / 2$ and $S_{A B C D E}=(\sqrt{5}+5) / 2$.