# Solution to the Three Squares Inside a Triangle Puzzle +-- {.image} [[ThreeSquaresInsideaTriangle.png:pic]] > Three squares inside a triangle. What's the sum of the two shaded areas? =-- ## Solution by [[Similar Triangles]] and [[Pythagoras' Theorem]] +-- {.image} [[ThreeSquaresInsideaTriangleLabelled.png:pic]] =-- With the points labelled as in the above diagram, triangles $D E F$ and $B C D$ are [[similar]]. The lengths of $C B$ and $C A$ are the same, and the lengths of $E F$ and $G E$ are the same. So the full length of $D A$ is the sum of the lengths of $D C$ and $C B$, and the full length of $G D$ is the same as the sum of the lengths of $G E$ and $E F$. These are sides of the white square, so $G D$ and $D A$ are the same length, and so the triangles $D E F$ and $B C D$ are actually [[congruent]]. This means that the length of $B D$ is half of that of $B F$, so is $3$. Also $E F$ and $D C$ are the same length. Writing $a$ for the length of $B C$ and $b$ for the length of $C D$, the sum of the two shaded areas is $a^2 + b^2$. Applying [[Pythagoras' theorem]] to triangle $B C D$ shows that $a^2 + b^2 = 3^2 = 9$ so the sum of the two shaded areas is $9$.