As @person has shown, the circular segment portion of each region is negligible in the infinite product, so we can regard each region as the remaining triangle.

Call the lengths of the line segments $m_1, m_2, m_3, ..., m_n$.

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n\left(\prod\limits_{k=1}^n m_k \right)^2$

It has been shown by @Bob Dobbs in a related question, that if $p=\dfrac{\sqrt{a^2-1}}{a}$, which means $\color{blue}{a=\sqrt{\dfrac{1}{1-p^2}}}$, then

$$\color{blue}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n m_k=1}$$

In this question we set the mean area of the regions as $\dfrac{1}{1-p^2}$, which means $\color{red}{a=\sqrt{\dfrac{n}{\pi}}\sqrt{\dfrac{1}{1-p^2}}}$, so

$$\color{red}{\lim\limits_{n\to\infty}\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}=1}$$

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}\left(\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}\right)^2$

$=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}$

$=\lim\limits_{n\to\infty}\dfrac{\left(\sin{\dfrac{2\pi}{n}}\right)^n}{\left(\dfrac{2\pi}{n}\right)^n}$

By L'Hopital's rule, the last expression equals $1$.

As @person has shown, the circular segment portion of each region is negligible in the infinite product, so we can regard each region as the remaining triangle.

Call the lengths of the line segments $m_1, m_2, m_3, ..., m_n$.

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n\left(\prod\limits_{k=1}^n m_k \right)^2$

It has been shown in a related question, that if $p=\dfrac{\sqrt{a^2-1}}{a}$, which means $\color{blue}{a=\sqrt{\dfrac{1}{1-p^2}}}$, then

$$\color{blue}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n m_k=1}$$

In this question, the mean area of the regions is $\dfrac{1}{1-p^2}$, which means $\color{red}{a=\sqrt{\dfrac{n}{\pi}}\sqrt{\dfrac{1}{1-p^2}}}$, so

$$\color{red}{\lim\limits_{n\to\infty}\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}=1}$$

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}\left(\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}\right)^2$

$=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}$

$=\lim\limits_{n\to\infty}\dfrac{\left(\sin{\dfrac{2\pi}{n}}\right)^n}{\left(\dfrac{2\pi}{n}\right)^n}$

By L'Hopital's rule, the last expression equals $1$.

As @person has shown, the circular segment portion of each region is negligible in the infinite product, so we can regard each region as the remaining triangle.

Call the lengths of the line segments $m_1, m_2, m_3, ..., m_n$.

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n\left(\prod\limits_{k=1}^n m_k \right)^2$

It has been shown by @Bob Dobbs in a related question, that if $p=\dfrac{\sqrt{a^2-1}}{a}$, which means $\color{blue}{a=\sqrt{\dfrac{1}{1-p^2}}}$, then

$$\color{blue}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n m_k=1}$$

In this question we set the mean area of the regions as $\dfrac{1}{1-p^2}$, which means $\color{red}{a=\sqrt{\dfrac{n}{\pi}}\sqrt{\dfrac{1}{1-p^2}}}$, so

$$\color{red}{\lim\limits_{n\to\infty}\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}=1}$$

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}\left(\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}\right)^2=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}=\lim\limits_{n\to\infty}\dfrac{\left(\sin{\dfrac{2\pi}{n}}\right)^n}{\left(\dfrac{2\pi}{n}\right)^n}$

By L'Hopital's rule, the last expression equals $1$.

As @person has shown, the circular segment portion of each region is negligible in the infinite product, so we can regard each region as the remaining triangle.

Call the lengths of the line segments $m_1, m_2, m_3, ..., m_n$.

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n\left(\prod\limits_{k=1}^n m_k \right)^2$

It has been shown by @Bob Dobbs in a related question, that if $p=\dfrac{\sqrt{a^2-1}}{a}$, which means $\color{blue}{a=\sqrt{\dfrac{1}{1-p^2}}}$, then

$$\color{blue}{\lim\limits_{n\to\infty}\prod\limits_{k=1}^n m_k=1}$$

In this question we set the mean area of the regions as $\dfrac{1}{1-p^2}$, which means $\color{red}{a=\sqrt{\dfrac{n}{\pi}}\sqrt{\dfrac{1}{1-p^2}}}$, so

$$\color{red}{\lim\limits_{n\to\infty}\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}=1}$$

$L=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}\left(\dfrac{\prod\limits_{k=1}^n m_k}{\left(\sqrt{\dfrac{n}{\pi}}\right)^n}\right)^2$

$=\lim\limits_{n\to\infty}\left(\dfrac{1}{2}\sin{\dfrac{2\pi}{n}}\right)^n \left(\sqrt{\dfrac{n}{\pi}}\right)^{2n}$

$=\lim\limits_{n\to\infty}\dfrac{\left(\sin{\dfrac{2\pi}{n}}\right)^n}{\left(\dfrac{2\pi}{n}\right)^n}$

By L'Hopital's rule, the last expression equals $1$.