All that glisters may be gold

We know that the smaller pentagon has perimeter 5, so each side has length 1. We also know that each angle of the pentagon must be $\frac{3\pi}{5}$ as it is regular.

Consider the acute triangle whose base is the side of the smaller pentagon. It is an isoceles triangle with two interior angles measuring $\frac{2\pi}{5}$ (they are supplementary to the angles of the pentagon. The third angle, therefore, measures $\frac{\pi}{5}$ . We will define the length of the long sides of this triangle $x$ .

Next, consider the obtuse triangles. They have two sides $x$ and the angle between them must be $\frac{3\pi}{5}$ (vertical angles with the angles of the pentagon.) So the two other angles must measure $\frac{\pi}{5}$ each.

Considering the two triangles together, they form a larger triangle featuring two angles measuring $\frac{2\pi}{5}$ , the remaining measuring $\frac{\pi}{5}$ . The short side we know measures $x$ , and the longer sides, which must be equal, measure $s=x+1$ (Note: one of these long sides is the side of the larger pentagon!) But this triangle is similar to our smaller acute triangle, as all the angles match, so its sides are proportional:

$\frac{x}{1}=\frac{x+1}{x}$
$x^{2}=x+1$
$x^{2}-x-1=0$
$x=\frac{1\pm \sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}=\frac{1\pm \sqrt{5}}{2}$

Only one of these values is positive and so $x=\frac{1+\sqrt{5}}{2}$

We now know the length of the side of our larger pentagon is $s=x+1=\frac{1+\sqrt{5}}{2}+1$ . We also know that the area of a regular polygon is equal to $A=\frac{1}{4}ns^{2}\cot{\frac{\pi}{n}}$ . Plugging our values in, we get:

$A=\frac{1}{4}(5)(\frac{1+\sqrt{5}}{2}+1)^{2}\cot{\frac{\pi}{5}}=\boxed{11.792}$ .

By Using Cosine rule. i.e

$c^{2}= a^{2}+ b^{2}- 2ab* \cos \theta$ , where $theta$ is angle between sides 'a' and 'b'

We can easily solve all the triangles there by obtaining the answers as 11.792

The isosceles triangle above the top horizontal small pentagon side is a 36-72-72 triangle. The base is 5/5=1.
So the slant side = .5/Cos72= say X. The whole slant line from top to bottom = X+1+X.
So the bottom side of the big pentagon =2(2X+1)Cos72=2(1+Cos72) $=\color{#D61F06}{2.618034}.\\ \text{Area of an unit pentagon = 1.72048}.\\ \therefore ~\text{area of the big Pentagon =}1.72048*2.618034^2=11.7923$