There is a fun visual proof for the formula for the area of circle and it can be used to illustrate the essence of calculus. Here is a fun animation:
As you can see, you peel away the circumference of a circle and lay it down as a horizontal line. Then you peel away the next circumference of the remaining circle and lay it down horizontally on top of the first line and so on until just the center point is finally placed on top. As the thickness of these peeled away layers approached infinitesimal thinness it describes a triangle with the height which is the radius and a base which is the circumference. This allows us to derive geometric aspects of the circle with basic trigonometry.
Just for fun, I was interested in attempting to generalize this approach to derive the area of any regular polygon. So I started next with a triangle and kept the concept of it having a center point and a radius where the radius is defined as the perpendicular distance from the center to side. You can imagine peeling off perimeters to build up the triangle as with the circle.
Lets stop first to check something first. How do we know that if we unwrap these polygons they will make triangles and not these shapes instead?
With a circle it seems a bit more intuitive that each layer is a little smaller then the last in a linear fashion, but these polygons with their edges it isn’t as clear. So let’s comfort ourselves with a little test. Let’s look at a circle first, if we pick three layers at r=1, r=2 and r=3 and look at their perimeters we want and expect to see that they are related in a linear fashion to the radius:
- At r=1, the circumference (green line) is 2 Pi
- At r=2, the circumference (yellow line) is 4 Pi
- At r=3, the circumference (red line) is 6 Pi
You can see the linear relationship as expected if you plot that, or simply see that doubling r from 1 to 2 doubles the circumference from 2 to 4 and tripling r from 1 to 3 triples the circumference from 2 to 6.
Now let’s try with a polgyon and see what happens:
- At r=1, the perimeter (green line) is 8
- At r=2, the perimeter (yellow line) is 16
- At r=3, the perimeter (red line) is 24
Again we can see that the relationship between the radius and the perimeter is linear, double or triple the radius and you double or triple the perimeter.
Not an exhaustive proof, but I’m satisfied, let’s continue!
So we can do this for any polygon because we know the height of the triangle will always be the radius of the polygon we are unwrapping and the base will be the perimeter
Generalizing the solution to the area then has more to do with generalizing the solution to the perimeter for all polygons because calculated the area of the triangle is trivial. To generalize the perimeter with respect to the radius, we simply need to know how many sides the polygon has. If we know the radius we know the height of the triangle the polygon is made up of, and if we know how many side the polygon has then we know the angles of the base (interior angle = (sides-2) * 180deg / sides).
I will presume this trigonometry is well understood and so let me cut to the chase to calculate the perimeter:
Now that we have derived a general formula for the perimeter of a polygon as a function of its radius and number of sides we can now get at a generalized solution for the area:
And there you go, a generalized formula for the area of any regular polygon based on the radius and number of sides using this unwrapping approach!
.If you graph how the Perimeter and Area changes with respect to the Sides for a Radius of 1 you will see that the perimeter approaches 2 and the area approaches . So in other words, as you might expect, a polygon with infinite sides behaves like a circle.
Perimeter approaching Pi
.. but wait there’s (a lot!) more …
Quick aside, I had written this up and stuck it in a drawer a long time ago and have been more or less transcribing up to this point off of a random hard-copy print out adding these new graphics and latex formulas to illustrate along the way. Low and behold I just found the original electronic copy and I just realized that my electronic copy has a WHOLE bunch more of a deep dive beyond the hardcopy I found! So I am going to simply copy and paste the rest of this rediscovered treasure here; makes me wonder how many other things I have forgotten coming up with in the past! So if the tone, attitude and illustrations sort of shift from here, that is the difference of a couple years of maturity and writing for myself vs an audience!
Now while this approach works soundly for any regular polygon (one where the sides are all equilateral) it was unclear to me if the approach would work with irregular polygons. I first tested this on a scalene triangle. ( I need trig to solve something or other almost every day, but I have not needed the Law of Sines and Law of Cosines in long time! Lots of trig went into this, but those details are not relevant so I will skip documenting my elbow grease)
Well clearly the perimeter formula I derived for regular polygons won’t work here because the individual angles are arbitrary. The question is will the area formula work though. First we need to determine what the radius is. We can start by turning my area formula around and predicting what the answer 
Now in the case of this polygon we can find the center point by extending lines from each corner bisecting the angle as below.
Sparing all the math, one will find that it in fact the center agrees with the radius prediction! So, great, this approach works not just for regular polygons but at least some irregular ones. But what happens if the irregular polygon does not have a SINGLE central point? Lets take a look at a 2 x 4 rectangle. While you may quickly conclude that the center is simply a point in the middle of the rectangle if you return to the premise that we are peeling the layers off of the polygon you will realize that the center is actually a LINE and not a POINT. This is illustrated by seeing that the process of bisecting angles reveals two points.
The radius is clearly 1 and the length of the red center line (L) is 2. If we attempt to calculate the area using the radius we find that it doesn’t work:
At first I thought that perhaps that L may just need to be added in… but that is not the case. Let’s take a look at a parallelogram and then we can come back and see how this all can be solved.
Once again sparing all of the very tedious math to solve this out, the radius is (√2)/2, the length is 8 and the area is 10√2. As with the rectangle, the area formula fails again. 
So at this point I needed to go back to the visualization and understand how to modify the area formula such that it is compatible with both regular and irregular point‐center polygons as well as these line‐center polygons. What I needed to realize was that the stack of perimeters that I am peeling off and previously expect to form a triangle in the case of line‐center polygons actually stack up into a trapezoid and not a triangle. Visualize what you are left with for your last “peel” and it will make sense.
Rearranging the trapezoid as below you can easily derive a new formula for area
So lets go try it out!
I am confident that this will work with many types of polygons, but more testing is needed. There are certainly polygons for which I have not yet started to consider a proof for, namely those with concave corners such as this where you end up transforming from a line to two points… perhaps the line becomes a curve or a more exotic function. Who knows….
… something to work on another day… along with the 3 dimensional (>3d?) analogues of this approach to visualizing the area (volume) via skinning. Ultimately I doubt I discovered anything new in neither technique nor conclusion but the exercise of deriving, failing, visualizing and even banging out countless pages of trig on my knee riding the train crammed up next to some stranger who must of thought I was insane… I digress… as usual a good learning experience motivated by a chance encounter with an inspiring little video!
Cheers!
