# Solution to the [[Four Squares and a Triangle]] Puzzle +-- {.image} [[FourSquaresandaTriangle.jpeg:pic]] > Four squares. What fraction of the total area does the yellow triangle cover? =-- ## Solution by [[Similar Triangles]] and Area of a [[Square]] and [[Triangle]] +-- {.image} [[FourSquaresandaTriangleLabelled.jpeg:pic]] =-- With the points labelled as above, since $E C A$ is a straight line, triangles $E D C$ and $C B A$ are [[similar]]. This means that the lengths of $E D$ and $D C$ are in the ratio $3 : 1$. Since the lengths of $A B$ and $C D$ add up to the length of $E D$, this means that the lengths of $E D$ to $A B$ are in the ratio $3 : 2$. Taking one unit as the length of $C D$, so then $A B$ has length $2$ and $E D$ has length $3$, the area of triangle $A E G$ is $\frac{1}{2} \times 2 \times 3 = 3$. The area of all four squares is $3 \times 4 + 9 = 21$. Therefore the yellow triangle covers $\frac{3}{21} = \frac{1}{7}$th of the total area.