Inscribed Polygons.

Let $n$ be odd integer and $n \geq 5$ .

Using the diagram directly above we have:

$x = 2r\sin(\dfrac{\pi}{n})$ and the height $h = r\cos(\dfrac{\pi}{n}) \implies$

The area $A_{n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) r^2$

$|\overline {\rm MN}| = r + h = r(1 + \cos(\dfrac{\pi}{n}) = 2r\cos^2(\dfrac{\pi}{2n})$

Using right $\triangle{MNP}$ in the first diagram above we have:

$|\overline {\rm MN}|^2 = (2r\cos^2(\dfrac{\pi}{2n}))^2 = 21^2 + 3^2 = 450 \implies r = \dfrac{15\sqrt{2}}{2}\sec^2(\dfrac{\pi}{2n}) = \dfrac{15}{\sqrt{2}}\sec^2(\dfrac{\pi}{2n})$

Let $r_{n} = \dfrac{15}{\sqrt{2}}\sec^2(\dfrac{\pi}{2n}) \implies$

$A_{n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) r^2_{n} =$ $\boxed{\dfrac{n}{2}\sin(\dfrac{2\pi}{n})(\dfrac{225}{2})\sec^4(\dfrac{\pi}{2n})}$

Using $n = 101 \implies A_{100} =\dfrac{22725}{4}\sin(\dfrac{2\pi}{101})\sec^4(\dfrac{\pi}{202})$

$\approx \boxed{353.372164}$ .

Using $r_{n} = \dfrac{15}{\sqrt{2}}\sec^2(\dfrac{\pi}{2n}) \implies \lim_{n \rightarrow \infty} r^2_{n} = \dfrac{225}{2}$

and using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1$ we have:

$\pi\cos(\dfrac{2\pi}{n}) < \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) < \pi$ and $\pi\lim_{n \rightarrow \infty} \cos(\dfrac{2\pi}{n}) = \pi$

$\therefore$ by squeeze play theorem $\implies \lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) = \pi$

$\implies$

$A_{c} = \lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) r^2_{n} = \lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) * \lim_{n \rightarrow \infty} r^2_{n} = \dfrac{225}{2}\pi \approx \boxed{353.4291735}$ .

$\implies A_{c} - A_{101} \approx \boxed{0.05700917}$ .