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

AI Thread Summary
The discussion centers on approximating the area of a circle using a regular polygon's area formula as the number of sides increases. The initial equation presented is questioned for its approach to limits, particularly regarding the relationship between side length and the number of sides. A more conventional method involves expressing the area in terms of the circumradius and the number of sides, leading to a clearer limit that converges to the area of a circle as the number of sides approaches infinity. Ultimately, the area of a regular polygon can indeed approximate the area of a circle when analyzed correctly. The conversation emphasizes the importance of relating the side length to the number of sides for accurate calculations.
John Clement Husain
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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?
 
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$$tan(90-\frac{180}{n}) $$
should become smaller with increasing n, to get a finite limit.
 
John Clement Husain said:
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.
 
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I extended the title to more accurately describe the topic.
 
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PS what's perhaps more interesting is to show that the length of the perimeter of the polygons, ##nl##, tends to ##2\pi r##.
 
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PeroK said:
Why are you taking the limit as ##x \rightarrow \infty##?

Oh! right, that was a typo.[/QUOTE]
 
John Clement Husain said:
Oh! right, that was a typo.

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