I did some calculations.... that seems to be the right answer! Not a proof or anything but here it is:
Take a washer w/ outer radius = 2 and inner radius = 1. The area = 2^2*pi - 1^2*pi = 3pi. Make a triangle w/ the chord and the center and cut it in half to make 2 right trianlges. Use good ol' a^2 + b^2 = c^2 and you get the chord = 2*sqrt(3).
Now take a random outer radius... lets say 4. Doing pretty much the same thing, you get an inner radius of sqrt(13). Then, 4^2*pi -sqrt(13)^2 * pi = 3pi.
Also works for outer radius of 5, but haven't tried different chord lengths.
Ok, the line is a chord of the big circle which is a tangent to the smaller circle. Then only use half of that line (up to where it touches the smaller circle)
We'll call the length of this line M for measurement.
If you take the point at which the chord touches the bigger circle and draw a line from here to the centre of both circles, you get the radius of the bigger circle. We'll call this B for big. If you now draw a line from the middle of both circles to the tangent at M then we get the radius of the small circle, or S for small. We have a right angled triangle. Thus B^2 = M^2 + S^2.
The area of the big circle is PiB^2. The area of the small circle is PiS^2. The area of the whole thing is PiB^2-PiS^2 or Pi(B^2-S^2). We know that B^2-S^2=M^2 and so the area of the washer is PiM^2.
Area of a washer
06 Jul '07 21:09
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