# Solution to the [[Six Squares in a Hexagon]] Puzzle +-- {.image} [[SixSquaresinaHexagon.png:pic]] > Six identical squares and a smaller rectangle are fitted into this regular hexagon. What fraction of the hexagon do they cover? =-- ## Solution by Properties of a [[Regular Hexagon]], [[Isosceles Triangles]], and [[Equilateral Triangles]] +-- {.image} [[SixSquaresinaHexagonLabelled.png:pic]] =-- 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 $A G F$. +-- {.image} [[TwoThirtyIsoscelesTriangles.png:pic]] =-- 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.