# Solution to the [[Three Triangles in a Square]] Puzzle +-- {.image} [[ThreeTrianglesinaSquare.png:pic]] > Inside this square are three triangles of equal area. How long is the red line? =-- ## Solution by [[Area of a Triangle]] and [[Pythagoras' Theorem]] +-- {.image} [[ThreeTrianglesinaSquareAnnotated.png:pic]] =-- In the diagram above, $E F G$ is a straight line parallel to the side of the square. Triangles $C F B$ and $D C H$ have the same area and the same length base since $C B$ and $D C$ are sides of the square, so by the formula for the [[area of a triangle]] their heights must be the same. Hence $F G$ and $D H$ have the same length. Then since $D H$ and $H A$ make up one side of the square, which is the same length as $E G$, $E F$ must have the same length as $H A$, namely $4$. This means that the area of triangle $A H F$ is $\frac{1}{2} \times 4 \times 4 = 8$, and so this is the area of all the triangles. Let $x$ be the side length of the square. Then $D H$ has length $x - 4$, so since the orange triangle has area $8$, we have: $ 8 = \frac{1}{2}x (x - 4) $ so $x^2 - 4x = 16$. Let $y$ be the length of $C H$. Applying [[Pythagoras' theorem]] to triangle $D H C$ shows that: $$ y^2 = x^2 + (x - 4)^2 = 2 x^2 - 8 x + 16 = 2(x^2 - 4x) + 16 = 2 \times 16 + 16 = 48 $$ Hence $y = \sqrt{48} = 4\sqrt{3}$.