Question

A regular polygon can be divided into a finite number of triangles the sum of whose area is the area of the polygon. If we do the same with a circle of radius `r`, treating it as a polygon with `oo` sides, does that mean `oo xx 0 = 2pir^2` ?

Answer 1

No

Explanation:

First off, you got the area of the circle right, and comparing it to the area of the regular polygon, but the area of the polygon is not `oo xx 0`.

A regular polygon can be split up into isoceles triangles, where the number of isosceles triangles equals the number of sides of the polygon.

`"Area of a triangle" = 1/2ab sinC`

`a=b=1` (for a circle with a radius of 1 unit)

`C=360/n or (2pi)/n`

`"Total area of triangles"=(nsin(360/n))/2 or (nsin((2pi)/n))/2`

`"Total area of triangles"~~"Area of circle"`

`(nsin(360/n))/2~~pi` (using degrees)
`(nsin((2pi)/n))/2~~pi` (using radians)

(For a circle with a radius of 1 unit, `pir^2=pi`)

We know that for small values of `theta`, `sintheta~~theta`

So:
`sin(360/n)~~(2pi)/n`
`sin((2pi)/n)~~(2pi)/n`

`lim_(n->oo)(nsin(360/n))/2=pi`
`lim_(n->oo)(nsin((2pi)/n))/2=pi`

A visual proof that uses the angle instead of a number of whole triangles can be found (using degrees) here, or with radians here

Answer 2

No, but...

Explanation:

Using transfinite and infinitesimal arithmetic...

If you divide the circle into `omega` segments with each arc of length `(2pir)/omega`, then the area of each segment is:

`1/2 xx "base" xx "height" = 1/2 * (2pir)/omega * r + O(1/omega^2) = (pir^2)/omega+O(1/omega^2)`

So the total area of the circle is:

`omega * ((pir^2)/omega+O(1/omega^2)) = pir^2+O(1/omega)`

the standard part of which is `pir^2`

Of course a more modest approximation is given with a finite number of segments...