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\begin{document}

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\section*{four squares and a quarter circle solution}

\hypertarget{solution_to_the_four_squares_and_a_quarter_circle_puzzle}{}\section*{{Solution to the Four Squares and a Quarter Circle Puzzle}}\label{solution_to_the_four_squares_and_a_quarter_circle_puzzle}

[[FourSquaresandaQuarterCircle.png:pic]]

\begin{quote}%
The radius of the quarter circle is $5$. What’s the total area of the four squares?


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

[[FourSquaresandaQuarterCircleLabelled.png:pic]]

Consider the [[right-angled triangle]] in the above diagram formed by joining the centre of the circle to the point on the circumference of the circle where the two smaller squares touch it. Let this have height $x$ as in the diagram, then [[Pythagoras' Theorem]] says that:

\begin{displaymath}
5^2 = x^2 + \left( \frac{5}{2} + \frac{x}{2}\right)^2 = x^2 + \frac{25}{4} + \frac{5}{2} x + \frac{x^2}{4} = \frac{25}{4} + \frac{5}{2} x + \frac{5 x^2}{4}
\end{displaymath}
Rearranging this yields

\begin{displaymath}
15 = x^2 + 2x = x(x + 2)
\end{displaymath}
This has solutions $x = 3$ and $x = -5$, but as $x$ is a length only $x = 3$ is allows.

The area that we are looking for has area:

\begin{displaymath}
2 \times \left(\frac{5}{2}\right)^2 + 2 \times \left(\frac{x}{2}\right)^2 = \frac{5^2}{2} + \frac{x^2}{2} = \frac{25}{2} + \frac{9}{2} = 17
\end{displaymath}


\end{document}