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 4-sided 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

1. All the angles of a square have the same value, which is $90^{\circ}$.
2. The geometrical center of a square is at the same distance from each of the four vertices.
3. The angle subtended at the center by any side is $90^{\circ}$.
4. 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 -

1. 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, $\displaystyle CE = \frac{r}{\sqrt{2}}$

2. Side of the square is $\displaystyle CD = 2 \times CE = 2 \times \frac{r}{\sqrt{2}} = r\sqrt{2}$

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

4. 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 5-sided shape (pentagon), a 6-sided 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 -

1. Is there a magic formula to obtain the value of $\mathcal{C}$ for an arbitrary sided shape?
2. 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!

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## Geometrical Constants - Part 1

Introduction What is a constant? It is an idea or an object that does not change under a set of well defined rules. What makes

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## Geometrical Constants - Part 3

In the previous part, we extended the idea of the geometrical constant $\mathcal{C}$ to a square and showed that the values of \$\mathcal{C}