# Solution to the [[Three Squares]] Puzzle +-- {.image} [[ThreeSquares.png:pic]] > The areas of the three squares are given. How long is the red line? =-- ## Solution by Properties of [[Right-Angled Triangles]] and [[Congruent]] Triangles +-- {.image} [[ThreeSquaresLabelled.png:pic]] =-- With the points labelled as above, the side lengths of the squares are $4\sqrt{2}$, $3\sqrt{2}$, and $5. Point $B$ is such that triangle $F B A$ is [[right-angled triangle|right-angled]], and so also point $E$ is such that $F E D$ is also right-angled. Angles $D \hat{F} E$ and $A \hat{F} B$ are equal, as both add to angle $B \hat{F} D$ to make a right-angle. Therefore, triangles $F B A$ and $F E D$ are [[congruent]]. Let line segment $F E$ have length $a$ and $E D$ have length $b$. Then these are also the lengths of line segments $F B$ and $A B$, respectively. Translating line segment $F E$ along $F B$ shows that the lengths of $A B$ and $F E$ add up to $4 \sqrt{2}$, similarly the difference of the length of $F B$ and of $E D$ is $3 \sqrt{2}$. That is, $a + b = 4 \sqrt{2}$ and $a - b = 3 \sqrt{2}$. Hence $a = \frac{7\sqrt{2}}{2}$. Triangle $F E C$ is right-angled and [[isosceles]] with shorter side length $a = \frac{7\sqrt{2}}{2}$. Its diagonal is therefore $\sqrt{2} a = \frac{7 \sqrt{2}}{2} \times \sqrt{2} = 7$. Hence line segment $F C$ has length $7$.