What's the Area?

Let $m(\angle{ACB}) = \dfrac{(n - 2)180}{n}$ be one of the interior angles of the larger $n-gon$ .

$\sin(\dfrac{(n - 2)180}{n}) = \sin(180 - \dfrac{360}{n}) = \sin(\dfrac{360}{n}) = \sin(\dfrac{2\pi}{n})$ .

Using the law of sines $\implies \dfrac{c}{\sin(\dfrac{2\pi}{n})} = \dfrac{a}{\sin(\alpha)} \implies$ $\sin(\alpha) = \dfrac{a\sin(\dfrac{2\pi}{n})}{c} \implies h^{*} = \dfrac{ab\sin(\dfrac{2\pi}{n})}{c} \implies$ $\sum_{j = 1}^{n} A_{j} = n\dfrac{1}{2}(c)(\dfrac{ab\sin(\dfrac{2\pi}{n})}{c}) = \dfrac{n}{2} ab\sin(\dfrac{2\pi}{n})$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \pi\cos(\dfrac{\pi}{n}) < \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) < \pi \implies \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} A_{j} = \boxed{\pi ab}$ $\implies \sum_{j = 1}^{n} A_{j}$ converges to the area of an ellipse.

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} A_{j} = \pi ab = \pi(a^2 + 2ab - 12b^2) \implies a^2 + ab - 12b^2 = 0 \implies (a - 3b)(a + 4b) = 0$ since $a,b > 0 \implies \dfrac{a}{b} = \boxed{3}$ .