Notes
two squares in a semi-circle solution

Solution to the Two Squares in a Semi-Circle Puzzle

Solution by Pythagoras' Theorem, Properties of Chords, and Isosceles Right-Angled Triangles

In the above diagram, the point labelled $O$ is the centre of the circle and $D$ is the midpoint of the chord $A B$. Point $F$ is directly below $D$, and $E$ is where the horizontal from $A$ meets the vertical from $D$. Let $a$ be the length of a side of the blue square and $b$ of the yellow square.

The horizontal distance from $A$ to $B$ is $a + b$, so the length of $A E$ is $\frac{a + b}{2}$. The height of $D$ above the diameter is the average of the heights of $A$ and $B$, so also is $\frac{a + b}{2}$. As the perpendicular bisector of a chord passes through the centre of the circle, angle $A \hat{D} O$ is a right-angle. This means that angles $A \hat{D} E$ and $F \hat{D} O$ add up to $90^\circ$ so triangles $A E D$ and $D F O$ are similar. Since $A E$ and $D F$ have the same length, they are actually congruent. This means that $A D$ and $D O$ are also the same length. Triangle $A D O$ is therefore isosceles, as is triangle $B D O$, and so triangle $A O B$ is a right-angled isosceles triangle. The length of $O B$ is the radius of the semi-circle, which is $8$, so the square of the length of $A B$ is $8^2 + 8^2 = 128$.

Triangle $A C B$ is also right-angled with $A C$ of length $\sqrt{2} a$ and $C B$ of length $\sqrt{2}b$. Applying Pythagoras' theorem to this triangle gives:

which gives the total area of the squares as $64$.

Solution by Invariance Principle

In this special case, the two squares are drawn to be the same size. The diagonal of one of the squares is a radius of the circle, so is of length $8$. The side length is therefore $\frac{8}{\sqrt{2}}$ and so each has area $32$. The total area is then $64$.