Basel Problem - Area of $\frac 16$ of Circle with Radius $\sqrt{\pi}$. There are several proofs to the solution of the well-known Basel Problem, i.e.
$$\sum_{n=1}^\infty \frac 1{n^2}=\frac {\pi^2}6$$

Is is possible to create a geometrical interpretation of this identity in the form of the area of $\frac 16$ of a circle with radius $\sqrt{\pi}$?

 A: (For the best expierience- please use a compass)
I. An approximation for $\sqrt{\pi}$ :

II. How to split a circle into 6 equal part:


EDIT: Epilog and history: 
I have start w/ some famous square roots (without a compass):
 
On this way we can win by hand roots like $\sqrt{3}$; $\sqrt{10}$ since $=\sqrt{ 3^2+1^2}$ 

Note:  $\sqrt{3}<\sqrt{\pi}$ and $\pi<\sqrt{10}$

or the golden ratio- we need add to $1$ with the compass the $\sqrt{5}$ (or vice versa) and split in the middle:

A: 
TOPIC:  Area of $\frac{1}{6}$ of Circle with Radius $\sqrt{\pi}$

Last time I show how to get an easy approximation for radius via Pythagorean theorem:
$$ (\sqrt{\pi})^2\geq1.7^2+0.5^2
$$
and have give a suggestion to do different- let me show for $\pi\leq\frac{22}{7}$ (by using intercept theorem for $\frac{\sqrt{7}}{7}$) and fully geometrical solving this time:

A: 
Topic: Radius $\sqrt{\pi}$

Today (and this gonna be my last post- unless I get requests) with the golden ratio:

$\phi=\frac{1+\sqrt{5}}{2}$ ; Property : $\phi^2=\phi+1$

I found this pythagorean relation:
$$
(\sqrt{\pi})^2\leq\phi^2+(\frac{\phi}{4})^2+0.6^2
$$

Accuracy:
$$$$ \begin{align} \frac{17}{16}\phi^2+0.36\approx3.141661\\    \pi\approx3.141593 \\  { For-comparison:}
\frac{22}{7}\approx3.14\color{red}{2857} \\  \end{align}
