@ughaibu,
ughaibu;133973 wrote:If pi is the ratio of the circumference to the diameter, [...]
It's not clear to me how one ought to define 'diameter' for a figure other than a circle. Also, the symbol pi in this context already has a fixed meaning, so it would be better to use some other name, to avoid confusion. Rho, say.
On the other hand, a regular polygon, just like a circle, has a circumference, L, and an area, A. (The most general case, including both regular polygons and circles as special cases, would probably be a "rectifiable Jordan curve", but don't quote me on that!)
How should we define 'rho', so that it comes out as pi in the case of a circle? For a circle of radius R:
A = pi * R^2
L = 2 * pi * R
therefore:
pi = (L^2) / (4A)
so we could define, for any rectifiable Jordan curve (or whatever!), of length L, enclosing an area A:
rho = (L^2)/(4A)
For a regular polygon with n sides, we have:
L = 2n * sin(pi/n)
A = n * sin(pi/n) * cos(pi/n)
therefore:
rho = n * tan(pi/n)
For n = 3 (equilateral triangle):
rho = 3 * sqrt{3} = 5.20 approx.
For n = 4 (square):
rho = 4
For n = 6 (regular hexagon):
rho = 2 * sqrt{3} = 3.46 approx.
For any rectifiable Jordan curve whatsoever:
rho >= pi
but this is hard to prove - it is the famous "Isoperimetric Inequality". I hope I've stated it correctly. It seems so (and Wikipedia even uses the same notation, although it doesn't use the term "rectifiable Jordan curve"):
Isoperimetric inequality - Wikipedia, the free encyclopedia