Here is an old geometry/measurement chestnut for you.

[polymath257]

An old problem of Hipparchus.

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The area of the half-circle ABD is the same as the area of the quarter circle. Subtract the area outside of the triangle and the quarter circle from both.
The answer is the same as the area of the triangle OAB where O is the center of the quarter circle. In other words, 1/2.

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Thanks for the Hipparchus reference.  I didn't know that and will have to look into who that was.  I don't recall where I encountered it but for whatever reason I woke up too early this morning thinking about it and had to recreate it.

I love the elegance of the solution though and how unlikely it seems when you consider the shape whose area we must find in itself.  Very satisfying I think.

There is a generalization due to al-Haytham. Take a large half-circle with diameter AB. Inscribe a triangle ABC where C is on the half-circle. Now, draw two half circles with diameters AC and BC outside of the triangle.


The question concerns the area outside the original circle and inside the two smaller circles.  

The Hipparchus figure is where the point C is on a perpendicular bisector of AB.

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The area of the two small half-circles is the same as the area of the large half circle because the triangle is a right triangle. So the area of the two 'lunes' is the same as the area of the triangle ABC.

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