Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side length $1$. The sides of the pentagon are extended to form the $10$-sided polygon shown in bold in the picture that I have attached. Find the ratio of the area of quadrilateral $A_2A_5B_2B_5$ to the area of the entire $10$-sided polygon.
Here I have attached a diagram of what I have understood so far about this problem.
Join
$A_2$ and $A_5$.
$A_3$ and $A_5$.
Note that $A_2A_3A_5$ is congruent to any of the triangle that are formed by extending the sides of the triangle. We denote the area of any such triangle by $S$.
Notice that the pentagon is partitioned into three part. One of them is the triangle $A_2A_3A_5$ with area $S$ and the other two parts, namely $A_2A_1A_5$ and $A_3A_4A_5$ are congruent (easy to check!). Let us say each of them covers an area $T$. Hence, area of the pentagon is $S+2T$.
Area of the shaded region : $2S+(S+T)=3S+T$
Area of the star: $5S+(S+2T)=6S+2T$
Clearly, the ratio is $\frac{1}{2}$