Is this true? The area of a circle can be approximated by a polygon

[John Clement Husain]

TL;DR

Does the limit as n approaches infinity of the area of an n-sided polygon equal to the area of a circle?

Hello everyone!
I have been looking for a general equation for any regular polygon and I have arrived at this equation:

$$\frac{nx^{2}}{4}tan(90-\frac{180}{n})$$

Where x is the side length and n the number of sides.

So I thought to myself "if the number of sides is increased as to almost look like a circle, does it result in the area of a circle?"

Is this:

$$\lim_{n\to\infty} \frac{nx^{2}}{4}tan(90-\frac{180}{n}) = \frac{C^{2}}{4\pi}$$

true?

 

[willem2]

$$tan(90-\frac{180}{n}) $$
should become smaller with increasing n, to get a finite limit.

 

[PeroK]

Summary: Does the limit as n approaches infinity of the area of an n-sided polygon equal to the area of a circle?

Hello everyone!
I have been looking for a general equation for any regular polygon and I have arrived at this equation:

$$\frac{nx^{2}}{4}tan(90-\frac{180}{n})$$

Where x is the side length and n the number of sides.

So I thought to myself "if the number of sides is increased as to almost look like a circle, does it result in the area of a circle?"

Is this:

$$\lim_{x\to\infty} \frac{nx^{2}}{4}tan(90-\frac{180}{n}) = \frac{C^{2}}{4\pi}$$

true?

Why are you taking the limit as $x \rightarrow \infty$?

If you keep $x$ fixed, then the area is infinite as $n \rightarrow \infty$. If you want your polygon to tend to a finite shape, then you need $x$ and $n$ to be related.

Note that using $l$ for the length of a side might have been more conventional.

It might be simpler to look at the angle, $\theta = 2\pi / n$ at the centre of the polygon and have a fixed distance to the vertices, $r$, say.

Then you let $n \rightarrow \infty$ and see what happens to the limit of the area of the polygons. Note that the length of the sides of the polygon will tend to $0$ in this case.

 

[mfb]

I extended the title to more accurately describe the topic.

 

[PeroK]

PS what's perhaps more interesting is to show that the length of the perimeter of the polygons, $nl$, tends to $2\pi r$.

 

[John Clement Husain]

Why are you taking the limit as $x \rightarrow \infty$?

Oh! right, that was a typo.[/QUOTE]

 

[PeroK]

Oh! right, that was a typo.

You still need to calculate $x$ in terms of $n$.

 

[mfig]

It's easier to see if you find the area in terms of one variable relating to a circle. Take for instance the area of a regular polygon in terms of the circumradius r and number of sides n.

$A = \frac{r^2 n sin(\frac{2\pi}{n})}{2}$

The limit of this formula as the number of sides $n\rightarrow\infty$ is the familiar formula for the area of a circle.