A regular pentagon with area $2\sqrt{5}+2$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center, creating a smaller pentagon. What is the area of the new pentagon?

1. As in the main problem, folding the vertices into the center produces a regular pentagon whose area is $\dfrac{3-\sqrt5}2$ times the area of the original pentagon.
2. For the area $2\sqrt5+2$, the new area is $(2\sqrt5+2)\cdot\dfrac{3-\sqrt5}2=(\sqrt5+1)(3-\sqrt5)=\boxed{2\sqrt5-2}$.