Application of pythagoras theorems (circles on sides of a right triangle)

One circle is constructed on each side of a right triangle. The center of each circle is the midpoint of the side and the side forms a diameter of the circle. The area of the triangle is 24 square units. Find the total area of the regions of the two smaller circles that lie outside the largest circle.

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What have you done so far? – GEdgar Apr 30 '12 at 23:52
You know that you're trying to use the Pythagorean theorem, which is about the length of the sides of your triangle. You're given the area, which gives you information about two of the sides of your triangle. How can you put it together to find out the length of the sides of the triangle? There is a little bit of guess-and-check involved, but it shouldn't be too hard. – chris May 1 '12 at 0:02
@chris, you can't possibly work out the length of the sides of the triangle from the information given. The only information given about the triangle is its area, and that doesn't determine its sides. – Gerry Myerson May 1 '12 at 0:04
Actually, since it's a right triangle, you can absolutely find the length of the sides. Area of a triangle is $\frac{1}{2}bh=24 \Rightarrow bh=48$. Here the base and the height are simply the $a$ and the $b$ in the Pythagorean theorem. So we're looking for two numbers (presumably integers, this really looks like a homework problem) that when multiplied give 48, and the sum of their squares is also a square number. – chris May 1 '12 at 0:07
@chris, "presumably integers" is what you didn't say in your earlier comment, and it is not justified by anything I see in the problem. – Gerry Myerson May 1 '12 at 0:12

The area is $24$ square units. The triangle is inscribed into the circle constructed over the hypotenuse. It is contained in the semicircle to one side of the hypotenuse. The inner semicircles over the legs are entirely contained within the circle over the hypotenuse. By the Pythagorean theorem, the areas of the outer semicircles over the legs add up to the area of the semicircle over the hypotenuse that contains the triangle. The parts of the semicircle over the hypotenuse that extend beyond the triangle are precisely the parts of the outer semicircles over the legs that don't extend beyond the circle over the hypotenuse. It follows that the parts of the outer semicircles over the legs that do extend beyond the circle over the hypotenuse have the same area as the triangle itself.

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I learned about this fact as the lunes of Hippocrates, but apparently those are also called the lunes of Alhazen. Both tried to square the circle. The present problem looks more tractable :) Wikipedia also recommends this book – t.b. May 1 '12 at 0:55