# Solution to the [[Three Squares and an Equilateral Triangle]] Puzzle +-- {.image} [[ThreeSquaresandanEquilateralTriangle.png:pic]] > Three squares and an equilateral triangle. What's the area of the pink square? =-- _There is a hidden assumption here that the lower edge is a straight line._ ## Solution by [[Angles in a Triangle]], [[Isosceles Triangles]], and [[Lengths in a Square]] +-- {.image} [[ThreeSquaresandanEquilateralTriangleLabelled.png:pic]] =-- With the points labelled as above, angle $A \hat{C} D = 45^\circ$ and angle $A \hat{C} B = 60^\circ$, so angle $D \hat{C} B = 15^\circ$. Then since angle $D \hat{C} E = 90^\circ$, angle $B \hat{C} E = 75^\circ$. Angle $A \hat{B} C = 60^\circ$ and angle $A \hat{B} E = 90^\circ$, so angle $C \hat{B} E = 30^\circ$. Then angle $C \hat{E} B = 180^\circ - 30^\circ - 75^\circ = 75^\circ$. Thus triangle $B C E$ is [[isosceles]]. This means that $B E$ and $B C$ are the same length, then also $B E$ and $A C$ are the same length. So the side length of the right-hand square is the same length as the diagonal of the pink square. The area of the pink square is therefore half the area of the right-hand square, thus is $4$.