SOLUTION: what is an indirect proof of Theorem 11.9 (area of a circle)? *the area of a circle is pi times the square of the radius.

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Question 409846: what is an indirect proof of Theorem 11.9 (area of a circle)?
*the area of a circle is pi times the square of the radius.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The indirect way would be to assume that the area is not equal to pi%2Ar%5E2 and find a contradiction. However this would involve finding the actual area of a circle, so an indirect proof is unnecessary.

Instead, construct a regular n-gon with n isosceles triangles (similar to "Theorem 11.8 which I recently solved, you can find it under the solutions I posted). Instead of finding the length of the base, we find the area, and multiply by n.


Recall that the area of the triangle is r%5E2%2Asin%28theta%29%2F2 where theta+=+2%2Api%2Fn. We have n of these triangles, so the area of the n-gon is

A+=+nr%5E2%2Asin%282%2Api%2Fn%29%2F2 (this may or may not be a good formula to memorize, but it's more important to know where it came from). Take the limit as n goes to infinity:

A%28circle%29+=+lim%28n-%3Einfinity%2C+nr%5E2%2Asin%282%2Api%2Fn%29%2F2%29. Just like the last problem, if we let u+=+1%2Fn,

. Applying L'Hopital's rule,

. Since lim%28u-%3E0%2C+%282pi%2Acos%28u%2A2pi%29%29%2F2%29+=+pi%29 (by direct substitution and noting we obtain a finite and determinate number),

A%28circle%29+=+r%5E2%2Api, hence the area of a circle.