Notes
three circles and two semi-circles in a circle solution

Solution to the Three Circles and Two Semi-Circles in a Circle Puzzle

Solution by Pythagoras' Theorem

As in the above diagram, let $b$ be the radius of the smallest circle, $a$ of the middle, and $r$ of the outer. Then by considering the horizontal radius, $r = a + 2 b$. Applying Pythagoras' Theorem to the blue triangle yields $r^2 = a^2 + (a + b)^2$. Putting these together gives:

So either $a = - b$ or $a = 3 b$. As $a$ and $b$ are lengths, neither can be negative and so it must be the case that $a = 3b$. Then $r = 5 b$.

The area of the large circle is $\pi r^2$ and the shaded area is $\pi a^2 + 3 \pi b^2$. So the area of the large circle is $25 \pi b^2$ and of the shaded area is $12 \pi b^2$. The fraction that is shaded is therefore $\frac{12}{25}$.