Notes
two circles and two semi-circles solution

Solution to the Two Circles and Two Semi-Circles Puzzle

Solution by Intersecting Chords Theorem or Pythagoras' Theorem

As in the above diagram, let $a$, $b$, and $c$ be the radii of the three sizes of circle. The relationship between them is given by the intersecting chords theorem. To apply this, the largest semi-circle needs to be completed to a full circle. The cross chord splits into two segments of length $b$, while the extension of the radius of the largest circle splits into $c - 2 a$ and $c + 2 a$. The intersecting chords theorem then shows that:

This can also be established using Pythagoras' theorem applied to a right-angled triangle with sides $2a$, $b$, and $c$.

The yellow and white regions combined form the middle sized semi-circle and two of the smallest circles, so this has area:

The black and white regions combined form the largest size semi-circle and so has area $\frac{1}{2} \pi c^2$.

Since $c^2 = b^2 + 4 a^2$, these two regions have the same area. Therefore the yellow and black regions have the same area, which is $16$.