Archimedes and the Area of a Circle
Date: 09/17/97 at 17:49:19 From: Bob Calabrese Subject: Pre-Calculus The question is, how do you find the area of a circle without pi? I've tried to do some derivations, but nothing seems to work.
Date: 09/23/97 at 12:34:54 From: Doctor Rob Subject: Re: Pre-Calculus The way Archimedes did it was to approximate the area of the circle by the area of an inscribed regular polygon. This method only requires that you be able to take square roots. Let the circle have radius R. Then the square inscribed in it will have side s = Sqrt*R. Its area will be A = 2*R^2. The octagon inscribed in it will have side s = Sqrt[2 - Sqrt]*R. Its area is A = 2*Sqrt*R^2. Here is a table showing the sides and areas of the (2^n)-gon for n = 1, 2, 3, 4, 5, 6: n s[n] = side of (2^n)-gon A[n] = area of (2^n)-gon = ================================ ============================= 1 2*R 0 2 Sqrt*R 2*R^2 = 1*s*R 3 Sqrt[2-Sqrt]*R 2*Sqrt*R^2 = 2*s*R 4 Sqrt[2-Sqrt[2+Sqrt]]*R 4*Sqrt[2-Sqrt]*R^2 = 4*s*R 5 Sqrt[2-Sqrt[2+Sqrt[2+Sqrt]]]*R 8*Sqrt[2-Sqrt[2+Sqrt]]*R^2 = 8*s*R 6 Sqrt[2-Sqrt[2+Sqrt[2+Sqrt[2+Sqrt]]]*R ... = 16*s*R This pattern continues, with n-3 plus signs in s[n] for a (2^n)-gon, and A[n] = 2^(n-2)*s[n-1]*R. For a 1024-gon, n = 10, you already get the approximation A/R^2 = 3.14157294 ~=~ pi, which is good to five significant figures. For n = 20, you get 11 significant figures. For n = 30, you get 17 significant figures. How accurate does your area need to be? -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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