What is the official proof (if there is any) for the area of a circle of radius 'r' ?
I remember in my school days they simply told that area of a circle of radius 'r' is $\pi*r^{2}$.
The teacher also told, $\pi$ = $\frac{circumference}{diameter}$
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1Do you know calculus? – Calvin Lin Apr 12 '13 at 02:36
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See: http://www.artofproblemsolving.com/LaTeX/Examples/AreaOfACircle.pdf – Amzoti Apr 12 '13 at 02:38
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do you have some other definition for $\pi$? Or is this asking why it should be proportional to $r^2$ in the first place? – Robert Mastragostino Apr 12 '13 at 02:41
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They never prove that the area of a circle is $\pir^{2}$ in any text book (for e.g., the text book of CLASS X). They say it is $\pir^{2}$. – HOLYBIBLETHE Apr 12 '13 at 03:25
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@ Calvin Lin : circle is introduced in class 8 if i'm right. calculus takes the $\pi$ from there. I'm talking of basics. If basics are not convincing how do we proceed further. – HOLYBIBLETHE Apr 12 '13 at 03:30
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Related question: http://math.stackexchange.com/q/44631/5676 – Peter Taylor Nov 01 '13 at 16:56
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I refer you to this article of the AMS.
The idea is this: approximate the circle within and without by polygons, so that the area of the circle is sandwiched. Then see what happens when you take polygons with more and more sides. The results are the same, and give the area of the circle.
This also gives the following pleasant heuristic. Regular polygons can be made of identical triangles from their center. For a 'nearly infinite'-sided polygon inscribed in a circle of radius $r$, each of the triangles will have altitude essentially $r$. The circumference of the circle is $2 \pi r$, so putting all these triangles together will give a triangle with are $\frac{1}{2} \cdot (r) \cdot (2 \pi r)$ from $\frac{1}{2} \cdot \text{height} \cdot \text{base}$, heuristically.
davidlowryduda
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