Is the area of this pentagon $4-\sqrt 5$? 
Consider a regular pentagon with vertices (in clockwise order) $A, B, C, D, E$, let $A'$ be the point of intersection of $BD$ and $CE$, let $B'$ be the point of intersection of $CE$ and $DA$, and so on. If $\triangle AC'D'$ has area 1, what is the area of the pentagon $A'B'C'D'E'$?

I tried first to compute the are of $\triangle A'C'D'$ then by using golden triangle, I compute the are of $\triangle C'B'A'$ and $\triangle E'D'A'$, thus getting the area of the smaller pentagon $A'B'C'D'E'$ being $4-\sqrt 5$.

Let $\color{red}{C'D'=a}$ in isosceles $\triangle AC'D'$ then the angle of vertex A is given as $$\angle C'AD'=\frac{180^\circ}{5}=36^\circ\implies \angle C'AM=\frac{\angle C'AD'}{2}=18^\circ$$ Now, drop a perpendicular say $AM$ from vertex $A$ to the side $C'D'$ in $\triangle AC'D'$, we get $$\tan\angle C'AM=\frac{\frac{C'D'}{2}}{AM}$$ $$\implies \color{blue}{AM}=\frac{C'D'}{a\tan 18^\circ}=\color{blue}{\frac{a}{2\tan 18^\circ}}$$ Hence, the area of isosceles $\triangle AC'D'$ $$=\frac{1}{2}(C'D')(AM)=1\ \text{(given in the question)}$$ $$\implies \frac{1}{2}(a)\left(\frac{a}{2\tan 18^\circ}\right)=1\implies \color{blue}{a^2=4\tan 18^\circ}$$ Hence the area of regular pentagon $A'B'C'D'E'$, having each side, $C'D'=a$, $$=\frac{1}{4}(5)(a^2)\cot\left(\frac{180^\circ}{5}\right)$$ substituting the value of $a^2$, we get area of $A'B'C'D'E'$ $$=\frac{5}{4}(4\tan 18^\circ)(\cot 36^\circ)$$ $$=5\left(\sqrt{\frac{5-2\sqrt{5}}{5}}\right)\left(\sqrt{\frac{5+2\sqrt{5}}{5}}\right)=\frac{5\sqrt{25-20}}{5}$$$$=\sqrt 5$$ That is the correct answer. 

Edit Area of any regular n-gon (polygon) having each side, $a$ is given as $$\frac{1}{4}na^2\cot\left(\frac{\pi}{n}\right)$$

