Since the height of any given triangle is s/2tan(180/n) the area of any triangle there will be;

A = (s/2) [s/2tan(180/n)] = s²/(4tan[180/n])

So the total area is number of sides times the area of any of the triangles which gives the formula you have above. But there was no calculus needed.

]]>Edit: I found it!

*If a triangle has no right angle,And the area needs to be seen,Multiply half by two of the sides,And the sin of the angle between.*

There's a little treat for anyone who likes browsing through old topics.

]]>Even Tandog has a trick formula for polygon areas. There's also an extract from Shakespeare's long forgotten play Henry X11 part 5 and finally... the secret identites of the superheros are revealed.]]>

Do you know of any links.

I'll check around first...

...I found it at one web page, but no explanation...

This looks even more general!!]]>

I just worked it out with my

silly new calculus of a hexagon.(this thread)

This result is for any regular

polygon (all sides equal and all angles equal).

It works for a square of length 5.

And for a pentagon of side length 5, it says 43.01, but I don't know if it is right.