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\begin{document}

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\section*{striped dodecagon solution}

\hypertarget{solution_to_the_striped_dodecagon_puzzle}{}\section*{{Solution to the Striped Dodecagon Puzzle}}\label{solution_to_the_striped_dodecagon_puzzle}

[[StripedDodecagon.png:pic]]

\begin{quote}%
This dodecagon is regular. What's the total shaded area?


\end{quote}
\hypertarget{solution_by_triangle_area}{}\subsection*{{Solution by [[Triangle Area]]}}\label{solution_by_triangle_area}

[[StripedDodecagonLabelled.png:pic]]

With the points labelled as above, consider the upper shaded region. This can be decomposed into two triangles, $E A B$ and $E B D$. Triangle $E A B$ is [[congruent]] to triangle $E F B$, so the shaded region has the same area as the [[quadrilateral]] $D E F B$. This region can be decomposed into triangles $D E F$ and $D F B$. As the line segments $D F$ and $B H$ are [[parallel]], sliding the apex of triangle $D F B$ from $B$ along to $O$ does not change its area - the ``height'' of the apex above $D F$ remains constant. So triangles $D F B$ and $D F O$ have the same area. Thus the shaded region has the same area as the [[kite]] $D E F O$. This in turn is one sixth of the area of the full dodecagon, so when combined with the lower shaded region then the shaded region is one third of the full dodecagon.

To find the area of the dodecagon, consider the triangles $A O K$ and $A O L$. The first of these, $A O K$, is [[equilateral]] and so $A K$ has length $6$. Then since the line segment $A K$ cuts $O L$ perpendicularly, triangle $A O L$ has base $O L$ of length $6$ and height half of $A K$ of length $3$, so has area $9$. This is one twelfth of the area of the whole polygon, which therefore has area $108$.

The shaded region then has area $36$.



\end{document}