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Approximating areas of circles with polygons

Notice how if you make regular polygons with increasingly many sides, they become closer and closer to being a circle:

Since the perimeter of the unit circle is $2\pi$, we can find the perimeter of inscribed and circumscribed regular polygons around the circle to estimate the value of $\pi$: $$\begin{array}{c|c|c|c} n & \text{Inscribed} & \text{Circumscribed} & (\text{Inscribed + Circumscribed})/4 \\ \hline 6 & 6.98203 & 6.00000 & 3.23205 \\ 12 & 6.43078 & 6.21166 & 3.16061 \\ 24 & 6.31932 & 6.26526 & 3.14614 \\ 96 & 6.28543 & 6.28206 & 3.14187 \\ 768 & 6.28322 & 6.28317 & 3.14159 \\ \end{array}$$ As we can see, the value of the average of the perimeters of the inscribed and circumscribed quadrilaterals divided by $2$ approaches $\pi$ (the area of the circle) as the number of sides approaches $\infty$. This is one way to approximate $\pi$ geometrically, although not the most efficient way.

Suppose that you have a circle of radius $r.$ Inside this circle, inscribe a regular polygon with $n$ sides; we will use this polygon to approximate the area of the circle.

The polygon has apothem $a$ and side length $s.$ Draw $n$ radii connecting the center of the circle (also the center of the polygon) to each of the vertices of the polygon. This splits the polygon into $n$ triangles with base $s$ and height $a.$ Thus, the area $A_p$ of the polygon is: $$A_p = n\times \frac{1}{2}as$$ $$A_p = \frac{1}{2}aP,$$ where $P = ns$ is the perimeter of the polygon.

Now, we let $n$ grow without bound. The apothem $a$ approaches the radius $r$ of the circle, and the perimeter $P$ approaches the circle's circumference $C = 2\pi r.$ As this occurs, the regular polygon approaches the circle it is inscribed in, so it makes intuitive sense that $A_p$ approaches the area of the circle $A_c.$

We can thus intuitively derive the area of a circle (which turns out to be correct): $$A_c = \frac{1}{2}r\times 2\pi r$$ $$A_c = \pi r^2.$$