More Of Pyramids

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies$ the Volume of the pyramid $V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

The lateral surface area $P^{*} = \dfrac{nx}{2} * s,$ where $s$ is the slant height.

$V_{p} = K$ (constant) $\implies H = \dfrac{12 K}{n * cot(\dfrac{180}{n}) x^2} \implies$ $P^{*}(x) = \dfrac{1}{2 m(n) x} \sqrt{(j(n) * m(n))^2 x^6 + (24 K)^2}$ , where $m(n) = \cot(\dfrac{180}{n})$ and $j(n) = n * m(n)$

$\implies \dfrac{dP^{*}}{dx} = \dfrac{2 (j(n) * m(n))^2 x^6 - (24 K)^2}{2 m(n) \sqrt{(j(n) * m(n))^2 x^6 + (24 K)^2} * x^2} = 0$ $\implies x = (\dfrac{(24 K)^2}{2 (j(n)* m(n))^2})^{\dfrac{1}{6}}$

$\implies H = \dfrac{12 K}{(j(n)) *( \dfrac{(24 K)^2}{2 (j(n) * m(n))^2})^{\dfrac{1}{3}}}$

$\implies \tan(\lambda_{n}) = \dfrac{H}{r} = \sqrt{2} \cos(\dfrac{180}{n})$ and $\tan(\theta) = \dfrac{H}{h^{*}} = \sqrt{2} \implies \dfrac{\tan(\lambda_{n})}{\tan(\theta)} = \cos(\dfrac{180}{n}).$

Now using $n = 100 \implies \dfrac{\tan(\lambda_{100})}{\tan(\theta)} = \cos(\dfrac{9}{5}) = \boxed{0.9995}$ .

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Note: $\lim_{n \rightarrow \infty} \tan(\lambda_{n}) = \sqrt{2} * \lim_{n \rightarrow \infty} \cos(\pi/n) = \sqrt{2} = \tan(\theta)$ .

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Bonus: From above(changing degrees to radians) $\dfrac{x}{2} = r \sin(\dfrac{\pi}{n}) \implies x = 2 r \sin(\dfrac{\pi}{n})$

$\therefore V_{p}(n) = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H = \dfrac{n}{6} \sin(\dfrac{2\pi}{n}) r^2 H$ .

Let $w(n) = n * \sin(\dfrac{2\pi}{n}) \implies V_{p}(n) = \dfrac{1}{6} w(n) r^2 H$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies 2\pi * \cos(\dfrac{2\pi}{n})< n * \sin(\dfrac{2\pi}{n}) < 2\pi \implies 2\pi * \cos(\dfrac{2\pi}{n})< w(n) < 2\pi$ .

$2\pi * \lim_{n \rightarrow \infty} \cos(\dfrac{2 \pi}{n}) = 2\pi \implies \lim_{n \rightarrow \infty} w(n) = 2\pi \implies \lim_{n \rightarrow \infty} V_{p}(n) = \boxed{\dfrac{1}{3} \pi r^2 H = V_{cone}}$