Regular polygon formula

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• June 7th 2012, 01:24 PM
Alex3914425
Regular polygon formula
Hello geniuses,

Recently, when making some calculations for people, I came across what I hope is a new formula. I have done some googling and I can't find it anywhere.
Could you take a look at this and tell me if it rings any bells? Does it even make any sense? I have applied it to a number of random shapes and they all came out exactly correct.
Please excuse me for any stupid mistakes I made. I'm not much of a math wonder ;)

The area of any regular polygon is determined by the formula

½ax*tan((180a - 360)/2a)

with a = number of sides
with x = length of a side

I figured that if anyone could help me it would be you guys. In any case, thank you for your time.
• June 7th 2012, 03:17 PM
HallsofIvy
Re: Regular polygon formula
That looks like a formula that has been found by any number of school kids. You can divide an a-gon into a isosceles triangles having base length x and vertex angle 360/a. Each of those triangles can be divided into two right triangles with "opposite side" x/2 and angle 180/a. The altitude of each isosceles triangle is the "near side" of the right triangle which has length (x/2)cot(180/a) so the area of each isoscelese triangle has area (1/2)(x)((x/2)cot(180/a)= (1/4)(x^2)cot(180/a). Although I personally would leave it that way, since cot(x)= tan(90- x), you can write that as cot(180/a)= tan(90- 180/a)= tan((90a- 180)/a)= tan((180a- 360)/2a).

Congratulations on finding that on your own!
• June 7th 2012, 03:22 PM
Plato
Re: Regular polygon formula
Quote:

Originally Posted by Alex3914425
I have done some googling and I can't find it anywhere.
Could you take a look at this and tell me if it rings any bells? Does it even make any sense? I have applied it to a number of random shapes and they all came out exactly correct.
Please excuse me for any stupid mistakes I made. I'm not much of a math wonder ;)
The area of any regular polygon is determined by the formula
½ax*tan((180a - 360)/2a)
with a = number of sides
with x = length of a si

I am not sure how hard you looked.
• June 7th 2012, 04:07 PM
Alex3914425
Re: Regular polygon formula
Quote:

That looks like a formula that has been found by any number of school kids.
Actually I was, at the time, a (high)school kid.

Quote:

You can divide an a-gon into a isosceles triangles having base length x and vertex angle 360/a. Each of those triangles can be divided into two right triangles with "opposite side" x/2 and angle 180/a.
This was indeed the basic idea which led to the formula. The only thing left needed to complete the formula is the adjacent side which is determined by tan which in turn is determined by a.
The link provided by Plato in the post above me leads to a webpage/formula which is known to me. While it does the same, it is different from mine as I didn't use any circles or radius, though I have a strong feeling this formula could easily be deviated from the one in the link, since r and R are essentially the same as the adjacent and hypotenuse from the right triangles. But even if it is, could this be seen as an original formula?

Quote:

Congratulations on finding that on your own!
I'm not sure how to take this, but for convenience I'll take it as a compliment :)
• June 7th 2012, 04:38 PM
HallsofIvy
Re: Regular polygon formula
Quote:

Originally Posted by Alex3914425
Actually I was, at the time, a (high)school kid.

This was indeed the basic idea which led to the formula. The only thing left needed to complete the formula is the adjacent side which is determined by tan which in turn is determined by a.
The link provided by Plato in the post above me leads to a webpage/formula which is known to me. While it does the same, it is different from mine as I didn't use any circles or radius, though I have a strong feeling this formula could easily be deviated from the one in the link, since r and R are essentially the same as the adjacent and hypotenuse from the right triangles. But even if it is, could this be seen as an original formula?

I'm not sure how to take this, but for convenience I'll take it as a compliment :)

Believe it or not, that was how I intended it!
• June 8th 2012, 12:57 AM
BobP
Re: Regular polygon formula
Presumably it's a typo since you say that you've used it a number of times and the results have turned out to be correct, but the formula you give can't possibly be correct. It has dimension of length rather than area. The correct formula has to contain an x squared.
• June 8th 2012, 02:08 AM
Alex3914425
Re: Regular polygon formula
Quote:

Presumably it's a typo since you say that you've used it a number of times and the results have turned out to be correct, but the formula you give can't possibly be correct. It has dimension of length rather than area. The correct formula has to contain an x squared.
I assure you there is no typo in the formula above. The way you see it here is the way it works. I don't get it. Why would the fact that x is in length make it an invalid formula? Isn't the area the product of lenght and width, after all? Why would the formula need to contain an x squared?
• June 8th 2012, 04:15 AM
BobP
Re: Regular polygon formula
Check again what you originally posted.
Try putting a=4 so that the polygon is a square. Doesn't your formula give an area of 2x ? It should be x squared.
Put a=3 so that you have an equilateral triangle and the result should be $x^{2}\sqrt{3}/4.$. Whatever regular polygon your calculating the area of, it has to contain an x squared or else the dimension is wrong.
You mention length times width, that's a length multiplied by a length, so the result will contain a length squared component.
HallsofIvy gives you the correct formula.
• June 8th 2012, 06:19 AM
Alex3914425
Re: Regular polygon formula
I was quite befoddled when I tested your example and it came out wrong, even with x filled in. I had tested my formula before on pentagons, heptagons and 9-gons and somehow they came out correct. It cannot have been mere coincidence because it went up to many decimals.
The wrong answer I get can be corrected by multiplying it by 1/2x, which, not surprisingly, leads to HallsofIvy's formula. There is one minor difference though. HallsofIvy writes "(1/4)(x^2)cot(180/a)" but I think it should be ¼a(x^2)cot(180/a). If you multiply ½ax*tan((180a - 360)/2a) by ½x you get ¼a(x^2)tan((180a - 360)/2a)=¼a(x^2)cot(180/a)
Here's an example: a pentagon with side 7.47 (a random number).
¼*5*(7.47^2)*tan(54) = 69.751125 x 1.37638192 = 96.0041873 which is correct according to online calculators. With HallsofIvy's formula the outcome would be 5 times smaller.
In any case, this formula already exists so with that my question is answered. Thanks for you help!
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