Notes
subdivided hexagon iv solution

Solution to the Subdivided Hexagon IV Puzzle

Solution by Area of a Triangle, Properties of a Regular Hexagon, Angles in Parallel Lines, Vertically Opposite Angles, and Similar Triangles

With the diagram labelled as above, first consider triangle $A D H$. From the properties of a regular hexagon, this has area one third of the outer hexagon. Triangle $A G H$ has area one twelfth of the outer hexagon, so triangle $A D G$ has area one quarter of the outer hexagon.

Since the lengths of $E C$ and $C B$ are in the ratio $2 : 1$, so also the areas of triangles $A E C$ and $A C B$ are in the same ratio.

Triangles $A B C$ and $D E C$ are similar because their angles correspond, using angles in parallel lines and vertically opposite angles, and the length scale factor is $2$, so triangle $D E C$ has area four times that of $A B C$.

Triangles $F G E$ and $F A B$ are also similar also with length scale factor $2$, so area scale factor $4$, while triangles $A G E$ and $F G E$ are congruent, so have the same area. This means that triangle $A E B$ has area twice that of triangle $A G E$.

Summarising, triangle $A C E$ has area twice that of triangle $A B C$, so triangle $A E B$ has area three times it. Then triangle $A G E$ has area half of that, so area $\frac{3}{2}$ times that of triangle $A B C$. Triangle $D C E$ has area four times that of triangle $A B C$, so the total area of triangle $A G D$ is $2 + \frac{3}{2} + 4 = \frac{15}{2}$ times that of triangle $A B C$.

Since this is one quarter of the area of the hexagon, triangle $A B C$ has area one thirtieth of that of the hexagon. As a fraction of the full hexagon, the shaded region therefore has area: