# Solution to the Two Squares Overlapping a Third Square Puzzle +-- {.image} [[TwoSquaresOverlappingaThirdSquare.png:pic]] > Two of these three squares have area $4$. What's the area of the blue triangle? =-- ## Solution by [[Lengths in Equilateral Triangles]] and [[Lengths in Squares]] +-- {.image} [[TwoSquaresOverlappingaThirdSquareLabelled.png:pic]] =-- In the above diagram, the point labelled $C$ is directly across from $B$, so that angle $A \hat{C} B$ is a [[right-angle]]. As the red squares have area $4$, they have side length $2$ and diagonal $2\sqrt{2}$. The length of line segment $B C$ is half of a diagonal of the red squares, namely $\sqrt{2}$, so triangle $A B C$ is a [[right-angled triangle]] with one side half of the hypotenuse. This means that it is half of an [[equilateral triangle]] and so $A C$ has length $\sqrt{3}$ times the length of $B C$, so has length $\sqrt{6}$. Since $C D$ also has length $\sqrt{2}$, the full length of the side of the outer square is $\sqrt{6} + \sqrt{2}$. The length of $E F$ is then $\sqrt{6} + \sqrt{2} - 2\sqrt{2} = \sqrt{6} - \sqrt{2}$. The area of triangle $E G F$ is therefore: $$ \frac{1}{2} \times (\sqrt{6} + \sqrt{2}) (\sqrt{6} - \sqrt{2}) = \frac{1}{2} (6 - 2) = 2 $$