# Solution to the [[Two Squares II]] Puzzle +-- {.image} [[TwoSquaresII.jpeg:pic]] > Two squares. What’s the shaded area? =-- ## Solution by [[Area of a Triangle]] and [[Area of a Trapezium]] +-- {.image} [[TwoSquaresIIAnnotated.jpeg:pic]] =-- With the diagram labelled as above, let the side length of the smaller square be $h$. Then the [[area of a triangle|area of triangle]] $E D A$ is $\frac{1}{2} h^2$. Since $B E$ has length $h$, $E F$ has length $12 - h$, so the [[area of a trapezium|area of trapezium]] $G F E D$ is: $$ \frac{1}{2}(12 + h)(12 - h) = \frac{1}{2}(144 - h^2) $$ The total area is therefore: $$ \frac{1}{2}(144 - h^2) + \frac{1}{2} h^2 = \frac{1}{2} \times 144 = 72 $$ ## Solution by [[Invariance Principle]] The smaller square can vary in size leading to two configurations where the shaded area can be easily found. +-- {.image} [[TwoSquaresIIInvarianceA.jpeg:pic]] =-- In this configuration, the smaller square has zero size meaning that the shaded area is half the remaining square, so has area $\frac{1}{2} \times 144 = 72$. +-- {.image} [[TwoSquaresIIInvarianceB.jpeg:pic]] =-- In this configuration both squares are the same size, meaning that the shaded area is a triangle with base $12$ and height $12$ meaning that its area is $\frac{1}{2} \times 12 \times 12 = 72$.