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Pastimes : INTERMEDIATE ALGEBRA -- Ignore unavailable to you. Want to Upgrade?

To: Nazbuster who wrote (82)	9/19/1999 10:22:00 PM
From: Little Poo	Read Replies (1) \| Respond to of 96

the radius, x, of the inner circle is perpendicular to the tangent chord. so we get a right triangle with legs 1/2 (half the chord)and x, and hypotenuse r (the radius of the outer circle. so, x^2+(1/2)^2=r^2.....and x=sqrt(r^2-1/4). (this requires believing that the inner circle and outer intersect at the midpoint of the curve. whaddya think? back to the bills....

To: Nazbuster who wrote (82)	9/20/1999 12:12:00 PM
From: Lazarus_Long	Read Replies (1) \| Respond to 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 = pia^2, so A = (pi/4)*L^2. Since L = 1 meter, A = pi/4.