Notes
chord splitting a square in half solution

Solution to the Chord Splitting a Square in Half Puzzle

Solution by Symmetry and lengths in an isosceles right-angled triangle

In the above diagram, point $O$ is the centre of the semi-circle, so $O F$ and $O E$ are radii. Point $M$ is the midpoint of the chord $E F$, so the line $O M$ is perpendicular to $E F$. Point $G$ is where the line extending $O M$ meets $C D$.

As the line segment $E F$ splits the square in half, point $M$ is also the middle of the square. So rotating the square $90^\circ$ anticlockwise about $M$ rotates $O G$ exactly onto $E F$. This shows that $O E G F$ is itself a square, and so angle $E \hat{O} F$ is a right-angle. Hence triangle $E O F$ is an isosceles right-angled triangle and so $O E$ is $\frac{6}{\sqrt{2}} = 3\sqrt{2}$ in length.

The area of the semi-circle is then $\frac{1}{2} \pi (3 \sqrt{2})^2 = 9 \pi$.

Solution by Invariance principle

The chord $E F$ can be drawn at any angle. Two possibilities make the radius of the semi-circle straightforward to deduce.

In this case, the point $E$ coincides with $B$ and so to split the square in half the chord must be a diagonal of the square and the sides $O B$ and $O D$ are radii of the semi-circle.

In this case, the chord is drawn horizontally. The isosceles right-angled triangle $O B E$ has hypotenuse a radius and other sides of length $3$, establishing the radius as $3 \sqrt{2}$.