SOLUTION: what is the 2 column proof for theorem 11.8 (circumference of a circle)?

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Question 409794: what is the 2 column proof for theorem 11.8 (circumference of a circle)?
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Proving the circumference of a circle is 2%2Api%2Ar is a little tricky using a two-column proof. It's also difficult to construct a proof of the circumference without using calculus.

The best way is to consider a regular n-sided polygon and its perimeter. Suppose the distance from the center of an n-sided polygon to one of its vertices is r (where r is constant). We can construct n isosceles triangles, each one looking something like this:



where theta+=+2%2Api%2Fn. Hence, the base of the isosceles triangle is 2r%2Asin%28theta%2F2%29+=+2r%2Asin%28pi%2Fn%29 (applying the definition of sine). We have n of these segments so the perimeter of the n-gon is

2rn%2Asin%28pi%2Fn%29. Taking the limit as n goes to infinity,

C+=+lim%28n-%3Einfinity%2C+2rn%2Asin%28pi%2Fn%29%29. Let u+=+1%2Fn. This limit is equivalent to

C+=+lim%28u-%3E0%2C+2r%2Asin%28u%2Api%29%2Fu%29. Applying L'Hopital's rule this is equivalent to

C+=+lim%28u-%3E0%2C+2r%2Api%2Acos%28u%2Api%29%29+=+2%2Api%2Ar, which is the circumference of the circle.