Conjecture about areas of circular segment and polygon with equal perimeter sharing a side I was playing around with shapes and have formed a conjecture.

Length of the red circular arc $=$ total length of the $n$ green line segments
Conjecture:
$$\sup{\left(\frac{\text{Area}_1}{\text{Area}_2}\right)}=\sqrt{1-\frac{1}{n^2}}$$
Is my conjecture true, and if so, is it a known thing?
I have proven the case with $n=2$ algebraically, and (I think) I have verified the cases with $n=3$ and $n=4$, using desmos.
The case with $n=2$:
The ratio of the areas is maximized when the two green segments are the same length (this can be seen by considering the ellipse whose focal points are the ends of the black line, passing through the point where the two green segments meet).
I let the central angle of the arc be $2\theta$, and got
$$\frac{\text{Area}_1}{\text{Area}_2}=\frac{2(\sin{\theta})\sqrt{\theta^2-\sin^2{\theta}}}{2\theta-\sin{2\theta}}$$
Basic calculus shows that this function is decreasing in $0<\theta<\pi$, and the limit as $\theta\to0^+$ is indeed $\frac{\sqrt{3}}{2}$.
Applying a similar method to larger $n$ seems quite daunting.
(My background: high school math teacher.)
 A: I will give a partial answer to my own question. I believe the conjecture is true.
Lemma 1:  For any n, $\frac{(\text{Area}_1){_\text{max}}}{\text{Area}_2}\to\sup{\left(\frac{\text{Area}_1}{\text{Area}_2}\right)}$ as arc length $\to$ length of black segment.
I do not have a proof of Lemma 1 (which is why my answer is a partial answer). I will assume it is true for now.
As arc length $\to$ length of black segment, we will consider only those arc lengths that allow the green segments to be part of a regular k-gon. It is known that for a k-gon with fixed perimeter, the area is maximized when the k-gon is regular. So $\text{Area}_1$ is maximized when the green segments assume their position as part of the regular k-gon.
The diagram shows an example with $n=3$.

For convenience of notation, let $x=\frac{\pi}{k}$.
Lemma 2: $\theta\approx{x}\sqrt{n^2-1}$ for small $\theta$
Proof:

$\sin{\theta}=\frac{\sin{(nx)}}{\frac{n\sin{x}}{\theta}}$
$\frac{\sin{\theta}}{\theta}=\frac{\sin{(nx)}}{n\sin{x}}$
$\frac{\theta-\frac{\theta^3}{3!}+...}{\theta}=\frac{nx-\frac{(nx)^3}{3!}+...}{n\left(x-\frac{x^3}{3!}+...\right)}$
$1-\frac{\theta^2}{3!}+...=1-\left(\frac{n^2-1}{3!}\right)x^2+...$
$\therefore\theta\approx{x}\sqrt{n^2-1}$ for small $\theta$

Using basic trigonometry to express $\text{Area}_1$ and $\text{Area}_2$ (ignoring r because they cancel), we have
$$\frac{\text{Area}_1}{\text{Area}_2}=\frac{n(\sin{x})(\cos{x})-\frac{1}{2}\sin{(2nx)}}{\frac{1}{2}\left(\frac{n\sin{x}}{\theta}\right)^2(2\theta-\sin{(2\theta))}}$$
Using the Maclaurin series for sine, and Lemma 2, we get
$$\lim_{x\to0^+}\frac{\text{Area}_1}{\text{Area}_2}=\sqrt{1-\frac{1}{n^2}}$$
