A nice little question:

Given a regular pentagon with side length $s$what is the relationship between the area and the side-length?

First of all a pentagon: We use triangulation to cut it into a number of triangles: With $180^\circ$ in each of the three triangles, there $3\times 180^\circ=540^\circ$ in those angles around the edges, and, as there are five of them, they are each $108^\circ$.

Next triangulate from the centre. With a plain oul pentagon we might not be sure that such a centre exists but if you start with a circle and inscribe five equidistant points along the circle, the centre of the circle serves as this centre: As everything is symmetric, each of these triangles are the same and the ‘rays’ are also the same as they are all radii. The angle at the centre is equal to $360^\circ/5=72^\circ$, and furthermore, by symmetry, the rays bisect the larger angles $108^\circ/2=54^\circ$ and so each of these triangles are $72^\circ,54^\circ,54^\circ$. Using radians, because they are nicer, $\displaystyle 54^\circ=\frac{3\pi}{10}$. Note that, where $h$ is the perpendicular height: $\displaystyle \tan\left(\frac{3\pi}{10}\right)=\frac{h}{s/2}\Rightarrow h=\frac{s}{2}\tan(3\pi/10)$.

A problem for another day is finding the exact value of $\tan\left(3\pi/10\right)$. It is $\displaystyle \frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}$ $\displaystyle \Rightarrow h=\frac{\sqrt{5}+1}{2\sqrt{10-2\sqrt{5}}}\cdot s.$

Therefore the area of one such triangle is: $\displaystyle A(\Delta)=\frac{1}{2}s\cdot \frac{\sqrt{5}+1}{2\sqrt{10-2\sqrt{5}}}\cdot s=\frac{\sqrt{5}+1}{4\sqrt{10-2\sqrt{5}}}\cdot s^2$,

Therefore the area of the pentagon is five times this: $\displaystyle A=5\cdot\frac{\sqrt{5}+1}{4\sqrt{10-2\sqrt{5}}}\cdot s^2$ $\displaystyle=\underbrace{\frac{5\sqrt{5}+5}{4\sqrt{10-2\sqrt{5}}}}_{=:\alpha}\cdot s^2$,

with $\alpha\approx 1.721$. It might be possible to simply $\alpha$ further.