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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?

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

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

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Here's a kind of different approach:

Let $n$ be number of sides.

$\angle DAC= \dfrac{180}{n}$

$\tan \angle DAC=\dfrac{x/2}{AD} \implies AD= \dfrac{\frac{x}{2}}{\tan \frac{180}{n}} \implies AD=\dfrac{x}{2 \tan \frac{180}{n}}$

Call $\dfrac{180}{n}=p$

Area of $\triangle ABC= \dfrac{x^2}{4 \tan p}$, you will have $n$ triangles in $n$-sided regular polygon. Area of polygon=$\dfrac{x^2 \times n}{4 \tan p}$.

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