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 $\sqrt{\pi}$?



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:

