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Pastimes : INTERMEDIATE ALGEBRA
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To: Nazbuster who wrote (82)9/20/1999 12:12:00 PM
From: Lazarus_Long  Read Replies (1) of 96
 
I thought this thread had died.

Let R = radius of outer circle
Let r = radius of inner circle
Then area of ring A = pi*(R^2 - r^2) [1]

Let the length of the chord be L and let a = L/2
At the point of tangency, the chord is perpendicular to the smaller circles radius (drawn from the center of the smaller circle to the point of tangency).

The hypotenuse of this right triangle is R, so R^2 = r^2 + a^2.
So R^2 - r^2 = a^2. Substituting into [1] above,
A = pi*a^2, so

A = (pi/4)*L^2.

Since L = 1 meter, A = pi/4.
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