Is this series for Pi correct? 
The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.
Here is the series I got using Pythagorean’s theorem , is it correct?
$$\begin{align}
A&=3r^{2} + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=r\sqrt{2-\sqrt{3}}
,\\x_{n+1}&=\sqrt{2r^{2}-2r\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}}
\end{align}$$
So-for-a-unit-circle
$$\begin{align}
\pi&=3 + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{1-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=\sqrt{2-\sqrt{3}}
,\\x_{n+1}&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}
\end{align}$$

Note that, in
$x_{n+1}
=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}
$
if
$x_n = 2\sin(t)
$
then
$\begin{array}\\
x_{n+1}
&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}\\
&=\sqrt{2-2\sqrt{1-\sin^2(t)}}\\
&=\sqrt{2-2\cos(t)}\\
&=2\sqrt{\dfrac{1-\cos(t)}{2}}\\
&=2\sin(\dfrac{t}{2})\\
\end{array}
$
