# Solution to the Square Overlapping a Quarter Circle Puzzle +-- {.image} [[SquareOverlappingaQuarterCircle.png:pic]] > The three coloured sections here have the same area. What's the total area of the square? =-- ## Solution by [[Area of a Triangle]] and [[Pythagoras' Theorem]] +-- {.image} [[SquareOverlappingaQuarterCircleLabelled.png:pic]] =-- With the points labelled as above, let $x$ be the length of $O A$ so the area of the square is $x^2$. As the coloured sections have the same area, the area of triangle $O A B$ is a third of that of the square, so from the [[area of a triangle]], $A B$ has length $\frac{2}{3} x$. Since $O B$ is a radius of the circle, it has length $13$ and so applying [[Pythagoras' theorem]] to $O A B$ gives: $$ 13^2 = x^2 + \frac{4}{9} x^2 = \frac{13}{9} x^2 $$ This shows that the area of the square is $x^2 = 9 \times 13 = 117$.