What is the area of a polygon with n equal length sides?

Question

What is the area and circumference of a polygon with n equal sides? (triangle, square, pentagon all the way to a circle)

It doesn't matter if it's based on the radius (let's call it r) or the length n.

EDIT: I ment regular polygon.

Answer 1

Assuming the polygon is regular ...

HINT: Cut it into $n$ isosceles triangles. Then compute the area of each of those triangles, and add them up.

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SOLUTION:

Now you know the area formula for a triangle, in particular an isosceles triangle. If the top angle is $\theta = \frac{2\pi}{n}$ and the side length is $r$, then the height is $r\cos{(\theta/2)}$ and the base is $2r\sin{(\theta/2)}$, so the area of the isosceles triangle is $r^2\cos{(\theta/2)}\sin{(\theta/2)}$.

This makes the area of the polygon $nr^2\cos{(\theta/2)}\sin{(\theta/2)}$.

For the perimeter, add the bases up: $2nr\sin{(\theta/2)}$.

Answer 2

One way to get these values is by calculating the length $c$ of the sides of your regular n-gon. Asumme the radius of the circumcircle is $r$. Then each side of the $n$-gon is the side of a triangle with third point the middle of the circumcircle and length $r, r$ and $c$. The angle bounded by the to sides of length $r$ is $2\pi/n$, so by the law of cosines you have $$ c^2 = r^2+r^2 - 2r^2 \cos(\frac{2\pi}{n}) = 2r^2(1-\cos(\frac{2\pi}{n}))$$ From this you can calculate $c$ and with $nc$ you know the circumference. The area is just $n$ times the area of the triangle, for which there are many well known formulas.