Notes
semi-circles in a rectangle solution

Solution to the Semi-Circles in a Rectangle Puzzle

Solution by Angle Between a Radius and Tangent, Similar Triangles, and Pythagoras' Theorem

With the points labelled as above, $E$ and $H$ are the centres of their respective semi-circles.

Since $D B$ is tangent to the semi-circle centred at $E$, angle $D \hat{G} E$ is the angle between a radius and tangent and so is $90^\circ$. Triangle $D G E$ is therefore a right-angled triangle. Triangle $D A E$ is also right-angled. The line segment $D E$ is the hypotenuse of both, and the sides $E G$ and $E A$ are radii of the same circle so have the same length. Triangles $D G E$ and $D A E$ are therefore congruent which means that the line segments $D A$ and $D G$ have equal length. Since $J$ and $G$ are evenly spaced along $D B$, $D B$ has length $\frac{3}{2} \times 5 = \frac{15}{2}$. Applying Pythagoras' theorem to triangle $A D B$ then shows that the length of $A B$ is:

Triangle $B G E$ is also right-angled. It shares angle $G \hat{B} E$ with triangle $A B D$ and so these triangles are similar with $B G$ corresponding to $B A$. Therefore the lengths of $E B$ and $E G$ are in the ratio $3 : 2$, meaning that the points $E$ and $F$ divide $A B$ in the ratio $2 : 2 : 1$ and so the radius of the semi-circle is $\frac{2}{5}$ of the length of $A B$, namely $\sqrt{5}$.

The area of the two semi-circles is therefore $5 \pi$.