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\begin{document}

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\section*{square, rectangle, and two semi-circles solution}

\hypertarget{solution_to_the_square_rectangle_and_two_semicircles_puzzle}{}\section*{{Solution to the Square, Rectangle, and Two Semi-Circles Puzzle}}\label{solution_to_the_square_rectangle_and_two_semicircles_puzzle}

[[SquareRectangleandTwoSemiCircles.png:pic]]

\begin{quote}%
A square, a rectangle and two semicircles. Which shaded area is the larger?


\end{quote}
\hypertarget{solution_by_pythagoras_theorem}{}\subsection*{{Solution by [[Pythagoras' Theorem]]}}\label{solution_by_pythagoras_theorem}

[[SquareRectangleandTwoSemiCirclesLabelled.png:pic]]

In the above diagram, the points labelled $O$ and $Q$ are the centres of the upper and lower semi-circles respectively. Let $a$ be the radius of the upper semi-circle and $b$ of the lower. Then $O C$ has length $a$ and $O A$ has length $2 b - a$ so the area of rectangle $A B C O$ is $2 b a - a^2$.

Let $c$ be the length of $O F$, which is the side length of the purple square. The length of $O Q$ is $b - a$, so applying [[Pythagoras' theorem]] to triangle $Q O F$ shows that:

\begin{displaymath}
b^2 = (b - a)^2 + c^2
\end{displaymath}
which rearranges to $c^2 = 2 b a - a^2$. The square and rectangle therefore have the same area. However, the quarter circle cut out from the rectangle has smaller area than the sector cut out from the square, as can be seen in the diagram by completing the smaller circle to a full circle. Therefore the purple area is smaller than the yellow.



\end{document}