Notes
hexagon in a quarter circle solution

Solution to the Hexagon in a Quarter Circle Puzzle

Solution by Properties of Chords, Symmetries of Regular Polygons, and Pythagoras' theorem

In the above diagram, the line segment $C D$ is a chord of the quarter circle and so its perpendicular bisector passes through the centre of the circle. It is therefore a line of symmetry of the full circle. It is also a line of symmetry of the hexagon. Reflecting in this line brings $A B$ to $F E$, and $O A$ to $O F$. Since $O A B$ is a straight line, so also must $O F E$ be. Triangle $O A F$ is therefore formed by extending two edges of a regular hexagon and so is an equilateral triangle. In particular, $O A$ has length $2$ and so $O B$ has length $4$.

The line segment $B D$ is the height of a regular hexagon which has length $\sqrt{3}$ times its side length, and so the length of $B D$ is $2 \sqrt{3}$. Applying Pythagoras' theorem to triangle $O B D$ shows that the square of the length of $O D$ is $4^2 + (2 \sqrt{3})^2 = 16 + 12 = 28$. The area of the quarter circle is therefore $7 \pi$.