In the previous part, we obtained a mathematical constant $\mathcal{C}$ for an equilateral triangle, and its value was approximately $5.196$. In the second part, we will continue the calculations in the same manner for a 4sided polygon, which is a square.
The Sqaure
A square is a closed geometrical shape having four sides having equal lengths, making an angle of $90^{\circ}$ with adjacent sides. Let us write down some basic facts that we obtained by careful observation
 All the angles of a square have the same value, which is $90^{\circ}$.
 The geometrical center of a square is at the same distance from each of the four vertices.
 The angle subtended at the center by any side is $90^{\circ}$.
 The line segment passing through the center cuts the opposite side in half, making an angle of $90^{\circ}$ on the opposite side.
Let us have a look at the diagram 
Following the steps of the previous part, we can obtain the following quantities 

Since $\theta = 90^{\circ}$ in this case, we can use the basic trigonometric formula: $\sin{\theta} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$ to get $$CE = r\sin{\left(\theta/2\right)} = r\sin{\left(45^{\circ}\right)}.$$
From the trigonometric tables, we get $\displaystyle \sin{45^{\circ}} = \frac{1}{\sqrt{2}}$. Therefore, $\displaystyle CE = \frac{r}{\sqrt{2}}$ 
Side of the square is $\displaystyle CD = 2 \times CE = 2 \times \frac{r}{\sqrt{2}} = r\sqrt{2}$

Perimeter of a square is $\displaystyle 4\times CD = 4\sqrt{2} \times r$.

Constant $\displaystyle \mathcal{C} = \frac{\text{Perimeter}}{OC} = \frac{4\sqrt{2}r}{r} = 4\sqrt{2} = 5.656 \cdots$
So, its clear that the value of $\mathcal{C} = 5.656 \cdots$ for the square is slightly greater than that of the triangle which was $5.196 \cdots.$
The reader is encouraged to repeat this exercise for a 5sided shape (pentagon), a 6sided shape (hexagon) and so on, to obtain the values of $\mathcal{C}$ in each case. For the sake of brevity, here we only show the values without calculations 
No of Sides  Shape  Value of $\mathcal{C}$ 

3  Triangle  $5.196 \cdots$ 
4  Square  $5.656 \cdots$ 
5  Pentagon  $5.877 \cdots$ 
6  Hexagon  $6$ 
7  Heptagon  $6.074 \cdots$ 
$\cdots$  $\cdots$  $\cdots$ 
99  Enneacontagon  $6.282130 \cdots$ 
100  Hectogon  $6.282151 \cdots$ 
Looking at the table above, it is clear that as the number of sides of the shapes increase, the value of $\mathcal{C}$ also increases, but the increase is not linear! It seems like its getting stable around 6.282.
Now, we ask the following two intriguing questions 
 Is there a magic formula to obtain the value of $\mathcal{C}$ for an arbitrary sided shape?
 What is so special about this number $6.282 \cdots$ ?
We shall find out the answers to these questions in the third (final) part of the series. Stay tuned!