# Solution to the Two Squares in a Triangle Puzzle +-- {.image} [[TwoSquaresinaTriangle.png:pic]] > The perimeter of each square is $\frac{1}{3}$ of the perimeter of the triangle. What fraction is shaded? =-- ## Solution by [[Area of a Triangle]] +-- {.image} [[TwoSquaresinaTriangleLabelled.png:pic]] =-- Let $x$ denote the length of the side of one of the squares. The point $O$ in the above diagram has the property that $O B$, $O D$, and $O F$ are all [[perpendicular]] to the sides of the triangle that they meet, and are all of the same length, namely $x$. Dividing the triangle into three by cutting from each vertex to $O$ shows that the area of the triangle is the same as that of three triangles each with height $x$ and whose collective bases sum to the perimeter of the original triangle, which is given as $3 \times 4 x = 12 x$. Therefore the area of the triangle is $6 x^2$, while the shaded region has area $2 x^2$. The fraction that is shaded is then $\frac{1}{3}$.