My Math Forum The Area of the region of the Equilateral Triangle that lies Outside the Circle

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 March 23rd, 2019, 08:44 AM #1 Newbie   Joined: Mar 2019 From: Coquitlam Posts: 1 Thanks: 0 The Area of the region of the Equilateral Triangle that lies Outside the Circle I'm considering a circle of diameter D superimposed over the equilateral triangle of side length D, with their centres coinciding. What is the total area of the three regions of the triangle that lie outside the circle?
March 23rd, 2019, 03:55 PM   #2
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Joined: Jul 2011
From: Texas

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Hopefully, I interpreted your problem statement correctly.

I looked at this problem to see if there was a way to calculate the desired area by means of elementary geometry. If there is, someone will no doubt point out a simpler method.

Having failed any elementary method, I turned to using the calculus of polar graphs to determine it.

I can tell you the exact solution is not nice as far as simplification is concerned.

If you are interested, I determined the total area of the three small "triangular" looking regions to be ...

$A = \dfrac{\pi+2(\sqrt{3}-\sqrt{2})-6\arctan\left(\frac{\sqrt{2}}{2}\right)}{8} \cdot D^2$

... attached is a sketch of the diagram used.
Attached Images
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 Tags area, circle, equilateral, lies, region, triangle

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