Notes
multiple semi-circles iv solution

Solution to the Multiple Semi-Circles IV Puzzle

Solution by Pythagoras' Theorem and the Area of a Circle

Let $a$, $b$, and $c$ be the radii of the circles, in ascending order. So with the points labelled as above then $O R$ has length $a$, $Q P$ and $Q R$ both have length $b$, and $O P$ has length $c$. As the three smallest semi-circles are the same size, and span across the largest semi-circle, then $c = 3 a$. Let $h$ be the length of $O Q$.

Applying Pythagoras' theorem to triangle $P Q O$ shows that

Applying it to triangle $Q O R$ shows that

Eliminating $h^2$ from these gives $c^2 - b^2 = b^2 - a^2$, so $c^2 + a^2 = 2 b^2$. The yellow region has area:

The information given in the puzzle shows that $\frac{1}{2} \pi b^2 = 10$, so the yellow region has area $2 \times 10 = 20$.