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Let us say that there is a circle in a square. The circle has a radius $r$ while the square has a side length of $2r$.

Ok now we know that $\pi=4\frac{A_c}{A_s}$, where $A_c$ and $A_s$ are the area of the square and the circle, respectively.

Just for fun, I thought I would see if there was a relation between the perimeter and the value of $\pi$. Turns out that $\pi=4\frac{P_c}{P_s}$ where P is perimeter or circumference.

Now we know that $\pi$ is constant so we could set these equal expressions equal to each other $4\frac{A_c}{A_s}=4\frac{P_c}{P_s}$. Dividing out the $4$, we get that:

$\frac{A_c}{A_s}=\frac{P_c}{P_s}$.

But this doesn't make any sense? How can the perimeters and the areas equal?

Oh and yes, I did come across this after learning a bit about monte carlo methods.

Viradeus
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  • See https://math.stackexchange.com/questions/2030161/how-to-relate-areas-of-circle-square-rectangle-and-triangle-if-they-have-same/2031209#2031209, please. – Michael Hoppe Oct 30 '20 at 12:26

2 Answers2

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One way to think about it is to consider a small piece of perimeter (circumference), and join its two endpoints to the common centre of the square and the circle.

Notice how the perpendicular distance from the centre to that piece of perimeter is always $r$, for both the square and the circle.

For a piece of the square perimeter of length $s$, the area of the triangle it forms with the centre is

$$a = \frac 12 r\cdot s$$

For a piece of the circumference of arc length $s$, the area of the sector it forms with the centre is

$$a = \pi r^2 \cdot \frac{s}{2\pi r} = \frac 12 r\cdot s$$

So the area each small piece of perimeter (circumference) contributes to the total area is the same for both cases.

This is not saying that "the perimeters and the areas equal", but the ratio of perimeters and the ratio of the areas are equal $\left(\frac4\pi\right)$.

peterwhy
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The perimeters and areas are not "equal".

You have two ratios, which are dimensionless numbers, and are equal. Nothing special.