# Solution to the Four Triangles in a Square Puzzle +-- {.image} [[FourTrianglesinaSquare.png:pic]] > Four white equilateral triangles meet in the centre of this square. What’s the ratio of yellow to green in this design? =-- ## Solution by Areas of Special Triangles and Angle Properties +-- {.image} [[FourTrianglesinaSquareLabelled.png:pic]] =-- In the above diagram, triangle $O D B$ is [[equilateral]] so line segments $B D$ and $O B$ are the same length. Triangle $B D C$ is [[isosceles]] and [[right-angled triangle|right-angled]]. The perpendicular distance of $C$ to the side $B D$ is therefore half of the length of $B D$, so the area of this triangle is a quarter of the square of the length of $B D$. Triangle $A O B$ is also [[isosceles]]. Angle $D \hat{O} A$ is a right-angle and angle $D \hat{O} B$ is the [[interior angle]] of an [[equilateral triangle]], so is $60^\circ$. This leaves angle $B \hat{O} A$ as $30^\circ$. In the above diagram, $E$ is the point on $O B$ such that angle $O \hat{E} A$ is a right-angle, so angle $O \hat{E} A$ is $180^\circ - 90^\circ - 30^\circ = 60^\circ$ as [[angles in a triangle]] add up to $180^\circ$. Triangle $O E A$ is therefore half an [[equilateral]] triangle, meaning that $A E$ has half the length of $O A$, which is the same as $O B$ and $B D$. The area of triangle $A O B$ is therefore a quarter of the square of the length of $O B$. Triangles $B D C$ and $A O B$ therefore have the same area and so the ratio of yellow to green in the design is $1 : 1$.