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\begin{document}

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\section*{five squares ii solution}

\hypertarget{solution_to_the_five_squares_ii_puzzle}{}\section*{{Solution to the [[Five Squares II]] Puzzle}}\label{solution_to_the_five_squares_ii_puzzle}

[[FiveSquaresII.jpeg:pic]]

\begin{quote}%
What fraction of the total area is shaded?


\end{quote}
\hypertarget{solution_by_shearing}{}\subsection*{{Solution by [[Shearing]]}}\label{solution_by_shearing}

[[FiveSquaresIISheared.jpeg:pic]]

The required [[shears]] are as shown in the above two diagrams:

\begin{enumerate}%
\item Shear the left-hand blue triangle horizontally from the right to the left of the base of the second square, keeping its horizontal top in place.
\item Shear the right-hand yellow triangle horizontally from the right to the left of the top of the fifth square, keeping its horizontal base in place.
\item Split the resulting yellow triangle along the left-hand edge of the fifth square, and merge the left-hand part with the orange triangle.
\item Shear the resulting orange-and-yellow triangle along the top of the fourth square.

\end{enumerate}
Each square is then half white and half coloured, meaning that the original shaded area is half of the total area.

\hypertarget{solution_by_area_of_a_triangle}{}\subsection*{{Solution by [[Area of a Triangle]]}}\label{solution_by_area_of_a_triangle}

Taking the ``base'' of the blue triangle as its top edge, its ``base'' is the side of the first square and its ``height'' is the height of that same square, so its area is half that of the first square.

The purple triangle has area one half of the second square.

Therefore, the blue and purple areas combine to give half of the first two squares.

For the other three squares, consider the unshaded, white regions.

The first white region in the middle square has area one half of the middle square.

In the fourth square, the upper white triangle has ``base'' the top edge of that square and its ``height'' is the height of that same square, so its area is half the fourth square.

The last white triangle has the same relationship to the fifth square.

Hence the white regions in the third, fourth, and fifth squares cover half of those squares.

Overall, therefore, half of the total area is shaded.



\end{document}