As a student you were most likely forced to remember the area and circumferece of a circle. Suppose we wanted to estimate this using some elementary means and suppose further that we're trying to show this to someone who has never seen calculus and has no idea what an integral is. We'll just use some plane geometry and some light (very light!) trigonometry to do this. So, the question becomes, **can we estimate the area and circumference of a circle using just geometry (and some light trigonometry)?**

Indeed, we can! First, let's note that we can approximate a circle using regular polygons of an increasing number of sides. Remember, regular polygons are shapes which have sides of the same length. See the cool gif below for a visual "proof" of this fact:

**Note, for all of what follows, we'll assume the circle is a unit circle; that is, a circle with radius 1**.

Neat. So if we take a regular polygon with a large number of sides (say, 100) then its circumference and area are probably fairly close to that of a circle. Easy-peasy. The hard part is finding the circumference and area of these things.

For this, we'll use some facts that we're going to assume the reader can work out for themselves.

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First, *because we are inscribed in a unit circle* the *lengths of the sides* of the regular $$n$$-sided polygon are given by
\[l = 2\sin\left(\frac{\pi}{n}\right)\]
You can show this using trig and noting that the distance from the edge of the polygon to the center of the polygon is 1 in this case. (If you wanted to make this larger by a factor of $$\alpha$$ then you'd also multiply the length of the side by $$\alpha$$).

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Next, the reader should be familliar with the nortion of the Apothem. The apothem is the line from the center of one of the sides of the polygon to the center of the polygon, and the way that we can find the apothem for an $$n$$-sided polygon is given by \[a = \cos\left(\frac{\pi}{n}\right)\] in our case. (Again, multiply this by $$\alpha$$ if you scale the polygon up.) This can be shown with some trig as well.

Once we have the length of the side and the apothem, you might realize that we can draw lines from each edge to the edge across from it, as well as draw every apothem, and our polygon is now a collection of $$2*n$$ right triangles. We'll need this to find the area.

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**We're ready now.**

First, let's look at the **perimeter** of our polygon. The perimeter will just be the sum of however many sides there are on the polygon times the side length:
\[P = ln = 2n\sin\left(\frac{\pi}{n}\right)\]
If we knew calculus, we could take a limit and notice that we get exactly $$2\pi$$ which is, in the case of the unit circle, the correct answer. Because we don't know calculus, we'll just plug this into our calculator with $$n = 100$$ and note that we get 6.2821518..., which is around 0.001 away from the actual solution. If we plug in 1000, we get to within 0.00001 of the actual solution. That's not bad!

The **area** is also fairly easy to get once we know we have all of those triangles. Recall that the area of a right triangle with legs of length $$a, b$$ is given by $$\frac{1}{2}ab$$. We have $$2*n$$ right triangles whose legs are length $$\sin\left(\frac{\pi}{n}\right)$$ (half of a side) and $$\cos\left(\frac{\pi}{n}\right)$$ (the apothem). The area of one triangle is then
\[A_{tri} = \frac{1}{2}\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}\right)\]
That means that the area of the polygon is $$2n$$ times this, or
\[A_{poly} = \frac{1}{2}n\left(2\sin\left(\frac{\pi}{n}\right)*\cos\left(\frac{\pi}{n}\right)\right)\]
This complicated looking thing simplifies if we remember some trig identities, though! Recall that $$\sin(2a) = 2\sin(a)\cos(a)$$, which is exactly what's on the right side of our solution above! Simplifying, we get
\[A_{poly} = \frac{n}{2}\sin\left(\frac{2\pi}{n}\right)\]
If we were doing calculus, we'd note here that the limit of this calculation is exactly $$\pi$$. Since we're not in the business of doing too much calculus in this post, we'll just put in a big number like 1,000. We get 3.141571982. Not a terrible approximation!