# Solution to the Three Squares in a Semi-Circle Puzzle +-- {.image} [[ThreeSquaresinaSemiCircle.png:pic]] > The two smallest squares each have area $6$. How long is the red line? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[ThreeSquaresinaSemiCircleLabelled.png:pic]] =-- Let $a$ be the side length of the smallest squares, so that $a^2 = 6$, and let $b$ be the side length of the larger one. As the [[perpendicular bisector]] of a [[chord]] passes through the centre of the circle, the continuation of $D E$ passes through the centre of the circle at $O$. Triangle $B D C$ is [[right-angled triangle|right-angled]] with $B D$ of length $a$ and $D C$ of length $b - 2 a$. The desired length, $B C$, is the hypotenuse of this triangle so by [[Pythagoras' theorem]] the square of its length is given by: $$ a^2 + (b - 2 a)^2 = 5a^2 - 4 a b + b^2 $$ Let $r$ be the radius of the circle. Triangle $B D O$ is right-angled with side lengths $a$, $b + a$, and $r$, so by [[Pythagoras' theorem]]: $$ r^2 = a^2 + (b + a)^2 = 2 a^2 + 2 a b + b^2 $$ Triangle $O E C$ is also right-angled and its side lengths are $b$, $b - a$, and $r$ so: $$ r^2 = (b - a)^2 + b^2 = a^2 - 2 a b + 2 b^2 $$ Subtracting the second equation from the first gives: $$ 0 = a^2 + 4 a b - b^2 $$ Adding this to the equation for the square of the length of $C B$ gives that that quantity is $6 a^2$ which is $36$. Therefore, the length of $C B$ is $6$.