Notes
six squares in a hexagon solution

Solution to the Six Squares in a Hexagon Puzzle

Solution by Properties of a Regular Hexagon, Isosceles Triangles, and Equilateral Triangles

Consider the diagram labelled as above.

Using the decomposition of a hexagon into triangles, the rectangle in the inner hexagon can be seen to be $\frac{2}{3}$ of that inner hexagon.

The trapezium $A B E F$ is divided into the half square $A G F$ and the two right-angled triangles $A H G$ and $G E F$. The interior angle of a regular hexagon is $120^\circ$, so angle $G \hat{A} B$ is $15^\circ$. Then angle $B \hat{G} A$ is $75^\circ$ and so angle $E \hat{G} F$ is also $15^\circ$. Therefore triangles $A H G$ and $G E F$ are congruent.

To find their area, consider the isosceles triangle $G D F$. This has apex angle $30^\circ$. Placing two such triangles next to each other, as in the diagram below, we can see that the perpendicular height of $F$ above $G D$ is half the length of $G F$. The area of triangle $G D F$ is therefore half that of $A G F$.

This means that the combined area of triangles $A B G$ and $G E F$ is $\frac{1}{3}$ of the area of trapezium $A B E F$. Equivalently, the area of triangle $A G F$ is $\frac{2}{3}$ of that trapezium.

So the six squares comprise $\frac{2}{3}$ of the outer ring in the hexagon, meaning that the six squares and the rectangle together make up $\frac{2}{3}$ of the area of the entire hexagon.