# Solution to the [[Four Squares and an Isosceles Triangle]] Puzzle +-- {.image} [[FourSquaresandanIsoscelesTriangle.jpeg:pic]] > Four squares and an isosceles triangle. What’s the blue area? =-- ## Solution by [[Similar Triangles]] +-- {.image} [[FourSquaresandanIsoscelesTriangleLabelled.jpeg:pic]] =-- Consider the diagram labelled as above, where point $C$ is such that angle $B \hat{C} D$ is a [[right-angle]]. Angles $C \hat{D} B$ and $G \hat{D} F$ are [[vertically opposite]], so triangles $D C B$ and $D G F$ are [[similar]]. This means that the lengths of $C D$ and $B C$ are in the ratio $1 : 2$. Angle $C \hat{A} B$ is $45^\circ$, so triangle $A C B$ is an [[isosceles]] [[right-angled triangle]] meaning that $A C$ and $C B$ have the same length. This means that the lengths of $C D$ to $C A$ are also in the ratio $1 : 2$. Since $A D$ has length $\frac{3}{2}$, this means that $C A$ has length $1$, and so also does $C B$. Using the formula for the [[area of a triangle]], triangle $A D F$ has area $\frac{1}{2} \times \frac{3}{2} \times 3$ and $A D B$ has area $\frac{1}{2} \times \frac{3}{2} \times 1$. So the total area of the blue triangle is: $$ 2\left(\frac{1}{2} \times \frac{3}{2} \times 3 + \frac{1}{2} \times \frac{3}{2} \times 1\right) = 6 $$