# Solution to the [[Subdivided Hexagon IV]] Puzzle +-- {.image} [[SubdividedHexagonIV.jpeg:pic]] > What fraction of this regular hexagon is shaded? =-- ## Solution by Area of a [[Triangle]], Properties of a [[Regular Hexagon]], [[Angles in Parallel Lines]], [[Vertically Opposite Angles]], and [[Similar Triangles]] +-- {.image} [[SubdividedHexagonIVLabelled.jpeg:pic]] =-- 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: $$ \frac{1}{30} + \frac{4}{30} + \frac{3}{2} \times \frac{1}{30} + \frac{1}{12} = \frac{2 + 8 + 3 + 5}{60} = \frac{18}{60} = \frac{3}{10} $$