Overlapping area of congruent semicircles
- Thread starter jschwa1
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2*71.5-124=19
The overlap region is 19 units wide.
Can you use calculus?. Probably not?. We can do it without, but calc would be cool to set up an integral.
Center the leftmost circle at (71.5,0), then its equation is \(\displaystyle y=\sqrt{-x^{2}+143x}\)
The circles intersect at x=133.5
So, we double the integral because we integrate from the intersection to the edge of the circle. The other half lies in the other circle. But due to symmetry we can just double it.
\(\displaystyle 2\int_{133.5}^{143}\sqrt{-x^{2}+143x}dx=457.45 \;\ units\)
The overlap region is 19 units wide.
Can you use calculus?. Probably not?. We can do it without, but calc would be cool to set up an integral.
Center the leftmost circle at (71.5,0), then its equation is \(\displaystyle y=\sqrt{-x^{2}+143x}\)
The circles intersect at x=133.5
So, we double the integral because we integrate from the intersection to the edge of the circle. The other half lies in the other circle. But due to symmetry we can just double it.
\(\displaystyle 2\int_{133.5}^{143}\sqrt{-x^{2}+143x}dx=457.45 \;\ units\)
mmm4444bot
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Without calculus, I'm thinking that you could start with the Law of Cosines, to find the chord length.
As a matter of fact, there is a formula for the area of two circles that overlap when the circles have the same radii. If you google 'area of overlapping circles', it can be found.
What you do is find the area of the triangle inside the overlap, along with the area of the circular segments, then add them up.
What you do is find the area of the triangle inside the overlap, along with the area of the circular segments, then add them up.