The Pillar of Chios

'The Pillar of Chios' printed from https://nrich.maths.org/

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The whole shape is made up of a rectangle, two semicircles on $AB$ and on $DC$ together making one circle, and the two semicircles on $AD$ and $BC$ making another circle.

Excellent solutions were sent in by a pupil from Dr Challoner's Grammar School, Amersham and Nisha Doshi ,Y9, The Mount School, York. Here is one of their solutions: Take: $AB=2x, AD=2y$.

\begin{eqnarray} \mbox{Total area of shape} &=& \pi x^2 + \pi y^2 + (2x \times\ 2y)\\ &=& \pi x^2 + \pi y^2 + 4xy. \end{eqnarray}

By Pythagoras Theorem

\begin{eqnarray} AC^2 &=& AD^2 + DC^2\\ &=& (2x)^2 +\ (2y)^2\\ &=& 4(x^2 + y^2)\\ \mbox{Area of big circle} &=& \pi(\mbox{AC}/2)^2\\ &=& \pi(x^2 + y^2)\\ \mbox{Area of crescents} &=& \mbox{Area of shape - Area of big circle}\\ &=& \pi x^2 + \pi y^2 + 4xy - \pi (x^2 + y^2)\\ &=& 4xy\\ \mbox{Area of rectangle} &=& 2x \times\ 2y\\ &=& 4xy \end{eqnarray}

so the sum of the areas of the four crescents is equal in area to the rectangle $ABCD$.