My Math Forum Why is area of circle exactly divisible by circumference?

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 November 22nd, 2013, 01:09 PM #1 Senior Member   Joined: Jan 2013 Posts: 209 Thanks: 3 Why is area of circle exactly divisible by circumference? I feel like I've been abused by education when they taught me x^2 + y^2 = radius^2, but they don't know why. Let's move on to more advanced subjects, and if you like you can understand the most basic things in college. What business do they have teaching calculus before finishing circles? More generally, why do the volumes and surfaces of n dimensional circles/spheres/hyperspheres alternate and are defined as integer and pi and factorial ratios of previous volumes/surfaces? http://en.wikipedia.org/wiki/Unit_sphere If pi is not an integer, then why is it the only non-integer factor of the volumes/surfaces which are exactly divisible by each other (in the equations on that Unit_sphere page)? If pi is not an integer, then why is the Cooley Tukey Fast Fourier Transform both periodic and recursive in a binary search? Why are there so many right triangles you can fit onto a circle, but so few are widely known? I had to discover http://en.wikipedia.org/wiki/Thales%27_theorem on my own, that the diameter of a circle is a hypotenuse of right triangle with every point on the circle. Looks like lorentz to me, but thats another subject. Why is the world ignoring this very basic question as if somebody knows the answer? I don't think anyone knows more about what they're teaching in elementary school.
 November 22nd, 2013, 02:30 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 For a circle, consider a very small sector... its area (as it approximates a triangle), is given approximately by c*r/2, where c is the length of the lesser arc corresponding to the sector. For the entire circle, which is made up of such sectors, that implies area of circle = c*r/2, where c is the circumference of the circle. For a sphere, consider a very slender "cone" with spherical base... its volume (as it approximates a cone), is given approximately by s*r/3, where s is the area of the spherical base. For the entire sphere, ... , one gets volume of sphere = s*r/3, where s is the surface area of the sphere. In general, $V_{\small n}\,=\,A_{\small n}r/n.$ The topic title isn't quite getting it right.
 November 22nd, 2013, 02:41 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Why is area of circle exactly divisible by circumference x² + y² = r² can be explained with the Pythagorean theorem (of which there are many proofs).
 November 22nd, 2013, 05:24 PM #4 Senior Member   Joined: Jan 2013 Posts: 209 Thanks: 3 Re: Why is area of circle exactly divisible by circumference Thanks, but I don't think that gets to the core of it yet, since you said approximately, but it's exactly divisible (both have a factor of pi)... I've made some progress... Why the Cooley Tukey Fast Fourier Transform is able to compute in n*log_base_2(n) cycles instead of n^2 as the first kind of FFT discovered... The hypotenuse of every right triangle is both a radius and diameter as can be seen recursively in a half size circle whose radius hangs off of the midpoint of the bigger radius and always ends at zero height and mirrors the lower half of the big radius because the midpoint of a line is the midpoint in each dimension, so every right triangle can be divided into 4 right triangles of the same shape, and 2 of these back to back form the double angle of the smaller circle. You need both: http://en.wikipedia.org/wiki/Thales_theorem http://en.wikipedia.org/wiki/Pythagorean_theorem
 November 22nd, 2013, 10:50 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 I used "approximately", but the error can be as small as you like, which is why the formula V = Ar/n is exact. For r = 1, that means A/V is an integer. For a unit circle, circumference/area = 2?/? = 2. For a unit sphere, one gets surface area/volume = 3, etc. The hypotenuse of a right-angled triangle can, of course, be whatever you like, depending on what diagram you choose to use.

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