Hello, Quick!

I was trying to find an equation to find the area of any regular polygon

and I came up with this: $\displaystyle A \:=\:\frac{s^2n}{4}\tan\left(90 -\frac{180}{n}\right)$

$\displaystyle n$ = number of sides

$\displaystyle s$ = length of each side

and $\displaystyle 90 - \frac{180}{s}$ is degrees

I think this equation works but I'm not sure . . . does it?

. . . Yes! I think I know how you reasoned this out . . . *Nice work!* Code:

A
*
/:\
/ : \
/ θ:θ \
/ : \
/ :h \
/ : \
/ : \
* - - - * - - - *
B D s/2 C

We have $\displaystyle n$ isosceles triangle as shown above.

$\displaystyle \angle BAC = \frac{360^o}{n}\quad\Rightarrow\quad \theta = \frac{180^o}{n}$

The height $\displaystyle h$ of the triangle is: $\displaystyle h \;= \;\frac{s}{2}\cdot\tan C \;= \;\frac{s}{2}\cdot\tan(90^o - \theta)$

The area of the triangle is: $\displaystyle A_{\Delta}\;=\;\frac{1}{2}bh\;=\;\frac{1}{2}(s)[\frac{s^2}{2}\tan(90^o - \theta)] \;=\;\frac{s^2}{4}\tan(90 - \theta)$

And $\displaystyle n$ triangles have an area of: $\displaystyle A\;=\;n \times \frac{s^2}{4}\tan(90^o - \theta) \;= \;\frac{s^2n}{4}\tan\left(90^o - \frac{180^o}{n}\right)

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