A nice little question:

*Given a regular pentagon with side length *, *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 in each of the three triangles, there in those angles around the edges, and, as there are five of them, they are each .

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 , and furthermore, by symmetry, the rays bisect the larger angles and so each of these triangles are .

Using radians, because they are nicer, . Note that, where is the perpendicular height:

.

A problem for another day is finding the exact value of . It is

Therefore the area of one such triangle is:

,

Therefore the area of the pentagon is five times this:

,

with . It might be possible to simply further.

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