Circle problem from last night

[Eric Roode <eric@verne.myxa.com>]

WARNING: spoiler ensues.



It's much simpler than that, Geoff. :-)

If you assume that whoever told you that you had enough information to 
solve the problem was correct, then you know that the area of the
ring-shaped floor is independent of the radii of the inner and outer
circles. 

One can imagine a huge pair of circles, almost exactly the same size,
such that the chord is 70 feet, or one can imagine the inner circle
being tiny, with the same 70-foot chord.

If you extrapolate this latter case to the extreme, where the inner
circle is a mere point, then the 70-foot chord becomes the diameter of 
the outer circle.  r^2 * pi == 35^2 * pi. 


 > The final equations
 > 
 > A = (PI * R^2) - (PI * r^2)
 > r^2 + n^2 = R^2
 > 
 > A = (PI * (r^2 + n^2) ) - (PI * r^2)
 > A = (PI * r^2) + (PI * n^2) - (PI * r^2)
 > A = ((PI * r^2) - (PI * r^2)) + (PI * n^2)
 > A = (PI * n^2)
 > 
 > And the actual area for a chord of 70 feet
 > 
 > A = PI * 35^2
 > A = PI * 1225
 > A = 3848.4510006474967171167381445174

-- 
----------------------------------------------------------------------
Eric J. Roode                                            eric@myxa.com
Senior Software Engineer, Myxa Corporation
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