General Polygons - Problem Solving

We shall first study regular polygons. A regular polygon is a polygon in which all sides have the same length and all angles are equal in measure. A rhombus is not a regular polygon, though all sides are equal. This is because all angles are not equal. Let $S$ be the side length of an $n$-sided polygon ($n$-gon), $R$ the circumradius, and $r$ the inradius.

An $n$-gon is made up of $n$ isosceles triangles with base $S,$ sides $R,$ and angle between the two sides measuring $\frac{2\pi} n.$

For relations between various lengths and angles, the right-angled triangle, which is half of the isosceles triangle, is helpful. So we get the relation $S=2\times R\sin\frac \pi n ~\text{ or }~ R=\dfrac S{2\times \sin\frac \pi n }$ and $S=2\times r\tan\frac \pi n ~\text{ or }~ r=\dfrac {S\times \cot\frac \pi n } 2.$ The base angles are $\frac{(n-2)\times \pi}{2n},$ while the angle between adjacent sides is $\frac{(n-2)\times\pi}{n}.$

The sketch below shows how the isosceles triangles are placed with a common vertex, coinciding sides, and the angles. Apart from the solution of the Isosceles triangle, the bases of $n$-gon solution, the lengths of diagonals $D_2, D_3, D_4, \ldots$ are important.

For an $n$-gon, there will be $n$ vertices $A_0, A_1, \ldots, A_{n-2}, A_{n-1}.$ Then $(n-1)$ lines can be drawn from $A_0$ to othere vertices. Out of these, $A_0A_1$ and $A_0A_{n-1}$ are adjacent sides. So there are only $(n-3)$ diagonals. If $n$ is even, there will be one big diagonal, and all the $(n-2)$ remaining ones will be in $\frac{n-4} 2$ pairs, as shown in the sketch. If $n$ is odd, all $(n-3)$ diagonals will be in $\frac{n-3} 2$ pairs. What is true for vertex $A_0$ is true for each of the $n$ vertices.

Next, we have to find out the lengths of diagonals.

The diagonal $D_n$ between $A_0A_k$ also forms an isosceles triangle with sides $R$ and the angle $\alpha=\frac{2\pi} n\times k.$ Now we can solve these as we had solved for sides.