Notes
two squares and a semi-circle solution

Solution to the Two Squares and a Semi-Circle Puzzle

Solution using Chords and Pythagoras' theorem

The centre of the circle lies on the perpendicular bisectors of the chords $B C$ and $A B$. The perpendicular bisector of $B C$ passes through its midpoint $E$ and the corner $F$ while that of $A B$ passes through $D$ and $G$. Triangle $H O F$ is a right-angled isosceles triangle that is congruent to $F E B$. This establishes $O G$ has having the same length as the half side, $D B$. So $O D$ is three times the length of $D B$.

Let $x$ be the length of $A B$, then $D B$ had length $\frac{1}{2} x$ and $O G$ has length $\frac{3}{2} x$. Since the length of $O B$ is $5$, Pythagoras' theorem applied to triangle $O B D$ establishes that:

The total area of the squares is then $2 x^2 = 20$.