How?Prove polygon->circle, area->max
- From: Steve Ro <junkmailjunk@xxxxxxxxxxx>
- Date: Sun, 15 Jul 2007 08:01:07 -0700
If the number of sides (N) of a regular polygon increases the area
approaches that of a circle (within which it is inscribed) - I want to
be able to show what value the area of the polygon approaches as N-
infinity for a given value of the circumference (C).
The base of the triangles (created by lines drawn from the centre to
each corner) has a length:-
Lb = C/N
The angle at the apex of each triangle is:-
theta=2pi(rad)/N
The adjacent length for each triangle (line from centre of polygon to
mid-point of a side) is:-
La = (Lb/2) / (tan (theta/2))
Area of each of the N triangles is:
An = (Lb/2) x La
i.e. (half x base) x height
Area of polygon is:-
Ap = N x An
Substituting for An and then La and Lb above and then simplifying
gives me:-
Ap = N.(X/Y)
where:
X=(C/2N)^2
Y=tan(pi/N)
How do I then show what value Ap approaches as N increases?
Inspecting the terms X and Y show that Y->0 as tan(0)=0 and X->0 and N-
infinity so I end up proving nothing - how should I go about this? IfN=1000 Ap aproaches (C^2)/4pi (from A(circle)=pi(R^2), and
C(circle)=2.pi.R)
C=2.pi.R, I thought I could differentiate Ap wrt N but I'm having to
scratch around my maths from years ago to prove this - I think I
should be able to show how
Ap = N.(X/Y) -> (C^2)/4pi as N->infinity
I may get there in the end but maybe I'm tackling it the wrong way or
I've missed something. A bit of help would be appreciated.
cheers
.
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