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:

pentagon

We use triangulation to cut it into a number of triangles:

pentagon1

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:

pentagon2

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.

pentagon3

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.

 

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