# Unused Cycles

## May 24, 2008

### Interesting way to derive the area formula for a circle

Filed under: Mathematics — Kevin @ 2:19 pm

I was bored today and looking through some old notebooks of mine and came across an interesting derivation of the circle area formula using some geometry and the concept of a limit. This idea dates back to my freshman or sophomore year.

Consider an N-gon that is broken up into N identical isosceles triangles with their unique angle touching the center. Let the side that is shared by each triangle have length r. This description might be difficult to understand, so I drew the case for N=5, a pentagon.

Clearly if we let N approach infinity, we get smoother and smoother polygons until we finally get a circle, as shown in the picture below.

So, if we’d like to find the area of the N-gon, we can find the area of each triangle we’ve created and multiply by N. Using that and letting N approach infinity should give us the area of a circle.

Now the inner angle on each of these pentagons is $2\pi /N$ radians, which leaves each of the outer angles to be

$\displaystyle\psi=\frac{\pi-2\pi/N}{2}=\pi/2-\pi/N$,

in radians. As a sanity check, as N goes to infinity here, the angle goes to $\pi/2$, which is expected.

Recall that the area for a triangle is $a=\frac 1 2 b h$, where b is the base width and h is the height of the triangle. We can find the base by the law of sines. That is:

$\begin{array}{rcl}\displaystyle\frac{\sin\theta}{b}&\displaystyle=&\displaystyle\frac{\sin\psi}{r}\vspace{0.3 cm}\\\displaystyle\frac{\sin(2\pi/N)}{b}&\displaystyle=&\displaystyle\frac{\sin(\pi/2-\pi/N)}{r}\end{array}$.

This gives us b, but we still don’t have h. Luckily, we can form a right triangle out of half of each isosceles triangle. This new triangle has hypotenuse r and legs h and b/2. Thus

$\displaystyle \sin\psi=\frac h r$.

Putting things together, we get that

$\displaystyle \frac{\sin\theta}{b}=\frac h{r^2}$,

or after rearrangement,

$\begin{array}{rcl}\displaystyle A&=&\displaystyle Na\vspace{0.3 cm}\\&\displaystyle=&\displaystyle N\frac{bh}{2}\vspace{0.3 cm}\\&=&N\left(\frac{r^2\sin\theta}{2}\right)\vspace{0.3 cm}\\&=&N\left(\frac{r^2\sin(2\pi/N)}{2}\right)\end{array}$

If we now take the limit as N goes to infinity, we get an indeterminate form. We can, however, apply L’Hopital’s rule after some rearrangement:

$\begin{array}{rcl}\displaystyle\lim_{N\rightarrow\infty}\frac{r^2\sin(2\pi/N)}{2N^{-1}}&=&\displaystyle\frac{r^2}{2}\lim_{N\rightarrow\infty}\frac{\cos(2\pi/N)(-2\pi N^{-2})}{-N^{-2}}\vspace{0.3 cm}\\&=&\displaystyle\frac{r^2}{2}\lim_{N\rightarrow\infty}2\pi\cos(2\pi/N)\vspace{0.3 cm}\\&\displaystyle=&\displaystyle\pi r^2\end{array}$

Taking the limit gives the familiar formula.

Part of my purpose for this posting is to test out how Blogspot handles mathematical formulas. As far as I can tell, there’s no easy way to do so. I had to use an image renderer here to render the formulas, and I’m not too impressed with it. I’ll be experimenting with MathML here in the next few days to see how it renders.

Update: Fixed all of the math to go with WordPress’s built in $\LaTeX$ capabilities. Looks pretty nice.

1. […] of a Circle So I was thinking about my previous area problem and the thought occurred to me that the same method could also be used to derive the circumference […]

Pingback by Circumference of a Circle « Unused Cycles — May 26, 2008 @ 2:06 am

2. Won’t the following work just as well?

∫_0^2π▒〖 ∫_0^r▒〖rdrdθ = 〗〗 ∫_0^2π▒r^2/2 dθ= r^2/2 θ,Evaluated between 0 and 2π
= (2πr^2)/2= πr^2

I copied and pasted the above from Microsoft Office 2007 equation editor, but obviously did not work very well!
It is the double integral of rdrdθ.
Limits on first integral are: 0 and 2π
Limits on second integral are: 0 and r

Comment by Tom — January 31, 2009 @ 7:19 pm

• That does indeed work. In fact, you can use triple integrals to find volumes of spheres and hyperspheres. This is just a fun way of going about doing the derivation and is by no means the most practical way of doing it 🙂

Comment by Kevin — February 3, 2009 @ 9:39 am

3. wow!!interesting!!

Comment by divs — August 25, 2009 @ 12:02 pm