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$A=\pi r^2$

## Prerequisites

• For any regular polygon, $A=\frac{lns}{2}$ , where $A$ is the area of the polygon, $l$ is the length of the apothem, $n$ is the number of sides, and $s$ is the length of each side.
• Given a circle inscribed in a regular polygon, the radius of that circle is equal to the apothem of the polygon.
• $\pi$ is the ratio of a circle's circumference to its diameter.
• Constant multiple rule of limits

## Proof A circle of radius $r$ inscribed in a pentagon and a dodecagon (left and right, respectively)

Construct a circle of radius $r$ . Construct an n-sided polygon such that the circle is inscribed in the polygon. Then the apothem of the polygon is equal to $r$ . Let $A_c$ represent the area of the circle and $A_p$ represent the area of the polygon. Let $P$ represent the perimeter of the polygon and $C$ represent the circumference of the circle. Then:

$A_p=\frac{rns}{2}\approx A_c$

$P=ns\approx C$

Further, let $n$ increase without bound. Then:

$\lim_{n\to\infty}\frac{rns}{2}=A_c\ ,\ \lim_{n\to\infty}ns=C$

Since $\pi$ is the ratio of a circle's circumference to its diameter:

$\pi=\frac{C}{2r}=\frac{\lim\limits_{n\to\infty}ns}{2r}=\lim_{n\to\infty}\frac{ns}{2r}$

$A_c=\lim_{n\to\infty}\frac{rns}{2}=\lim_{n\to\infty}\frac{r^2ns}{2r}=r^2\lim_{n\to\infty}\frac{ns}{2r}=\pi r^2$

QED

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