Area of a Regular Polygon I was trying to find the area of a regular polygon in terms of n, the side length and s, the number of sides.
Because there are $s$ sides number of isosceles triangles in a regular polygon, I decided to work out the area of an isosceles triangle in terms of $A$, the unique angle and $a$, the unique side:
$$Area=\frac{1}{2}ab\sin{C}$$
$$b=\frac{\sin{C}\times a}{\sin{A}}$$
(Sine rule, $C=B$)
$$b=\frac{\sin{\frac{180-A}{2}}\times a}{\sin{A}}$$
$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$
$$C=\frac{180-A}{2}=90-\frac{A}{2}$$
$$Area=\frac{{(\sin({90-\frac{A}{2}})\times{a})}^{2}}{\sin{A}}$$
And that was where I got to in finding the area of an isosceles triangle. Then I tried to find the area of the whole regular polygon:
$$Area=s\times\frac{{(\sin({90-\frac{A}{2})}\times{n})}^{2}}{\sin{A}}$$
Where $s$ is the number of sides and $n$ replaces $a$
$$A=\frac{360}{s}$$
$$Area=s\times\frac{{(\sin({90-\frac{\frac{360}{s}}{2}})\times{n})}^{2}}{\sin{\frac{360}{s}}}=s\times\frac{\sin^{2}({90-\frac{180}{s}})\times{n}^{2}}{\sin{\frac{360}{s}}}$$
$$Area=\frac{s{n}^{2}\sin^{2}({90-\frac{180}{s}})}{\sin{\frac{360}{s}}}$$
However, when I tested this formula it was wrong. Can someone tell me where I've gone wrong?


$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$

It seems that you dropped $\frac{1}{2}$ here. :-)
