# Solution to the [[Three Squares IV]] Puzzle +-- {.image} [[ThreeSquaresIV.png:pic]] > Three squares. The green square has area 16. What’s the area of the blue square? =-- ## Solution by [[Transformation]] and [[Congruent Triangles]] +-- {.image} [[ThreeSquaresIVLabelled.png:pic]] =-- With the points labelled as above, triangles $D E F$ and $D B C$ are related by a rotation of $90^\circ$ about point $D$. They are therefore [[congruent]], so line segments $D E$ and $D B$ have the same length. This means that $D$ lies on the diagonal $G H$, and this further means that $D$ is the same distance from $F$ as it is from $A$. Therefore the side length of the blue square is $\sqrt{2}$ times the side length of the green square, and so the area of the blue square is twice the area of the green square, meaning that it has area $2 \times 16 = 32$. ## Solution by [[Invariance Principle]] The point $C$ in the above diagram must lie on the base of the black square, and with that constraint there are some configurations that make the solution simple to see. +-- {.image} [[ThreeSquaresIVInvarianceA.png:pic]] =-- In this configuration, the blue and black squares coincide, while the diagonal of the green square is clearly one side of both. +-- {.image} [[ThreeSquaresIVInvarianceB.png:pic]] =-- In this configuration, the blue and black squares are alongisde and [[congruent]]. The diagonal of the green square is one side of the black square, so also one side of the blue.