# Solution to the Three Squares and a Semi-Circle Puzzle +-- {.image} [[ThreeSquaresandaSemiCircle.png:pic]] > The smallest of these three squares has area $7$. What's the total shaded area? =-- ## Solution by [[Properties of Chords]] and [[Pythagoras' Theorem]] +-- {.image} [[ThreeSquaresandaSemiCircleLabelled.png:pic]] =-- In the diagram above, the point labelled $O$ is the centre of the semi-circle. Let the squares have side lengths $a$, $b$, and $c$ in increasing order, so that $a^2 = 7$. The area of the shaded region is $c^2 - b^2 - a^2 = c^2 - b^2 - 7$. As one of the sides of the middle square forms a [[chord]] across the circle, $F E$ is half of that side and is [[perpendicular]] to $O E$. Therefore triangle $O F E$ is [[right-angled triangle|right-angled]] and has side lengths $a$, $\frac{b}{2}$, and $\frac{c}{2}$. Applying [[Pythagoras' theorem]] shows that: $$ \begin{aligned} \left(\frac{c}{2}\right)^2 &= \left(\frac{b}{2}\right)^2 + a^2 \\ c^2 &= b^2 + 4 a^2 \\ \end{aligned} $$ Then the shaded region has area $c^2 - b^2 - 7 = 21$.