Area of a circle without calculus

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  • #26
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1) draw a circle of radious 1
2) inscribe 6 equilateral triangles.
3) count the sides close to the circumference and you get 6 and divide this be 2 for the first approximation. it equals 3
4) divide a 60 degree segment in half and draw the 2 new chords. compute the length of the chords.
5) add the 12 chords and divide by 2 to get 3.105 for the 2nd approximation
6) Divide the 1/12 chords in half and compute the new chord length.
7 compute the total chord length of the 24 chords and divide in half for the 3rd approximation.
etc
 
  • #27
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From the above approach I obtained the following

6 chords pi = 3
12 chords pi = 3.105828541
24 chords pi = 3.132628611
48 chords pi = 3.139350197
 
  • #28
Svein
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divide a ... segment in half and draw the 2 new chords - compute the length of the chords.
And how do you do that without calculus?
 
  • #29
Orodruin
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Also:
  1. That method computes the area, but the approach is suggestively similar to a Riemann sum. It is just an area integral of a piecewise constant function.
  2. You obtain the area, but it does not show that the area is ##\pi##. If you have a numerical value for ##\pi## obtained from elsewhere, you can just see it approaching. You need to show that the limit of the series is equal to your given value (which probably is also a different series).
 
  • #31
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It is a right triangle.
As you described things, "it" isn't a right trangle. When you divide one of the original 60° sectors in two, the two new chords are presumably points on the circle. Each of the new triangles is isosceles, with a 30° angle at the top (radiating from the circle's center) and two base angles of 75°. These aren't right triangles. The two triangles whose hypotenuses are the new chords, and whose bases are half the length of the old chord (before splitting the equilateral triangle into two). Those two small triangles are right triangles, each with an acute angle of 15°.
 
  • #32
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Also:
  1. That method computes the area, but the approach is suggestively similar to a Riemann sum. It is just an area integral of a piecewise constant function.
  2. You obtain the area, but it does not show that the area is ##\pi##. If you have a numerical value for ##\pi## obtained from elsewhere, you can just see it approaching. You need to show that the limit of the series is equal to your given value (which probably is also a different series).
pi is simply the length of one half of a unit circle.

At the process goes on 1/2 the sum of the lengths of the chords approaches the length of half the circumference.
As you described things, "it" isn't a right triangle. When you divide one of the original 60° sectors in two, the two new chords are presumably points on the circle. Each of the new triangles is isosceles, with a 30° angle at the top (radiating from the circle's center) and two base angles of 75°. These aren't right triangles. The two triangles whose hypotenuses are the new chords, and whose bases are half the length of the old chord (before splitting the equilateral triangle into two). Those two small triangles are right triangles, each with an acute angle of 15°.
Yes everything you say is correct. My point is that you can use these right triangles to compute the length of the two new chords in terms of the old chord. This results in a approximation of pi and it can be done over and over to any degree of accuracy desired.
 
  • #33
Orodruin
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pi is simply the length of one half of a unit circle.

At the process goes on 1/2 the sum of the lengths of the chords approaches the length of half the circumference.

Yes. And how do you show that your sum actually approaches this number? As I said, you need to show that your series converge to the same number.
 
  • #34
Baluncore
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Is finding the area of a circle not called “squaring the circle” ?
Was calculus not devised to make it possible to “square the circle” ?
What has changed that now makes calculus no longer necessary for the task it was created to solve ?
 
  • #35
PeterDonis
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Is finding the area of a circle not called “squaring the circle” ?

No. "Squaring the circle" refers to one of the classic problems in straightedge-and-compass Euclidean geometry, the problem of constructing, using straightedge and compass alone, a square with area equal to that of a given circle. This problem is not solvable by straightedge and compass alone (briefly, because ##\pi## is transcendental, and transcendental numbers cannot be constructed with straightedge and compass alone).

Was calculus not devised to make it possible to “square the circle” ?

No.
 
  • #36
Baluncore
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classic problems in straightedge-and-compass
You are ignoring the requirement that it be done "using only a finite number of steps" with straightedge-and-compass.
Taking an infinite number of steps in one step is what calculus made possible.
 
  • #37
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You are ignoring the requirement that it be done "using only a finite number of steps" with straightedge-and-compass.
Taking an infinite number of steps in one step is what calculus made possible.
I don't think @PeterDonis's intent was to provide a complete description of "squaring the circle." It was to dispute your claim that finding the area of a circle was "squaring the circle." I agree with Peter -- it isn't.
 
  • #38
PeterDonis
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Taking an infinite number of steps in one step is what calculus made possible.

This is perfectly true, and has nothing to do with what I was saying. As @Mark44 has said, I was disputing your claim that the term "squaring the circle" means "finding the area of a circle". It doesn't; the term "squaring the circle" has a much more specific meaning, which I described, and which has nothing whatever to do with using calculus to derive the formula for the area of a circle in terms of its radius.
 

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