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Is value of $\pi = 4$?

I know that you can take area out of a square without changing it's perimeter. Now, here's this problem:

Draw a circle with dia = 1;

Draw a square around with perimeter = 4, side = 1;

Remove the corners, still perimeter is same.

repeat this thing infinite times, you have to see the image below.

https://i.stack.imgur.com/qUYei.jpg

I know there is something missing, but what is it? I am not a mathematician, not even the worst one. Please be kind and answer in English as much as possible. Thanks

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Anubhav
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    Hello Anubhav, it turns out your question has already been asked here before. You can find a link to the earlier question above. You didn't do anything wrong; it's just that we like to avoid duplicate questions. – Zev Chonoles Mar 23 '12 at 07:36
  • How kind of our mod! –  Mar 23 '12 at 07:38
  • yes. you want to down vote? @KannappanSampath – Anubhav Mar 23 '12 at 07:40
  • I am not sure I understand what you mean. Let me clarify: I don't down vote exact duplicates! I wanted to compliment Zev for being kind enough to clarify, which I thought was appropriate and hence the clarification. –  Mar 23 '12 at 07:45
  • @KannappanSampath Okay. But I still don't get it. Those answers are still in "Mathematics" and I have a young brat(12 year old) to satisfy who wouldn't take any more mathematical terms. – Anubhav Mar 23 '12 at 07:54
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    @AnubhavSaini The corners don't become smoothed out, so you could imagine that it is like pressing a crinkled paper to the table. The area of the smoothed out paper would be more than the area of the table covered by the crinkled paper. (And yes, this comparison has some weaknesses, as there is no overlap in your example, a stretched/compressed rubber band is more accurate, but it might get you the idea.) – Phira Mar 23 '12 at 09:48
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    And I really don't see why this question has to be always asked with the ugly picture. – Phira Mar 23 '12 at 10:32
  • @AnubhavSaini: Take a look at this other question which contains more accessible arguments about why $\pi\neq 4$. – Zev Chonoles Mar 23 '12 at 15:16
  • An easier solution to this question was posted a few days ago under the other question: http://math.stackexchange.com/questions/12906/is-value-of-pi-4/1933209#1933209 . Pictures need to be added. It's very simple idea. When we zoom in to a tiny arc of circle cut out by removed tiny corner, the circle's tiny arcs become indistinguishable from line segments (lines are circles with "infinite" radius). The correct approxi has 2 sides that mark off the arc be close to the arc (almost 180 deg triangle), but the incorrect approx above always has these tiny triangles as right triangles. – Jose_X Sep 25 '16 at 21:55
  • ... for the 90 deg (incorrrect) approx, the 2 leg lengths added do not approx the 3rd leg (hypotenuse, arc) unless one of legs is very small. For correct approx, the 2 legs always sum to just less than the 3rd leg (arc). Eg of incorrect approx that almost works: At the very top of the circle, when we zoom, we get a right "triangle" (remember, 3rd leg is the arc that looks like a line) where say 1 leg is .01 and the other is .000001. The sum here approximates the arc "side". a^2+b^2=c^2, we have a=c approx when b is near 0. – Jose_X Sep 25 '16 at 22:01
  • .. OK, but at the 45deg angle, let's say after a bunch of steps, we'd find .001 for each of 2 legs. the arc (3rd leg) is not .002, their sum. Not even close. It is .0014... No matter how much we zoom, we get these large percent differences around much of the circle. – Jose_X Sep 25 '16 at 22:06
  • ..[sorry, this is so long]. For the correct case however, no matter where we pick around the circle. the 2 legs sum is basically the length of 3rd arc leg. The overall relative error goes to zero as the arcs get straighter and straighter. This is basically the approx of circumference by using inscribed regular polygon. Here each little triangle created gets flatter and flatter. The angle opposite the arc approaches 180 deg (unlike in the incorrect case where it is always 90 deg). – Jose_X Sep 25 '16 at 22:10
  • inscribed regular polygon -> circumscribed regular polygon – Jose_X Sep 25 '16 at 22:17
  • Picture response here: http://math.stackexchange.com/a/1991404/10963 – Jose_X Oct 30 '16 at 12:59

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