# Solution to the Two Rectangles Overlapping a Square Puzzle +-- {.image} [[TwoRectanglesOverlappingaSquare.png:pic]] > The yellow square has double the area of the red rectangle. How tall is the green rectangle? =-- ## Solution by [[Dissection]] and [[Area of a Rectangle]] +-- {.image} [[TwoRectanglesOverlappingaSquareLabelled.png:pic]] =-- Triangle $J H B$ is [[congruent]] to triangle $G F D$, then triangle $H F E$ is congruent to triangle $B D C$. These establish that the green rectangle has the same area as the yellow square. (This can also be established by [[shear|shears]]: first, parallel to $B D$; second, parallel to $B H$.) Triangle $B C D$ is congruent to triangle $B I J$, meaning that the length of $B C$ is the length of $B I$. Therefore, since the green rectangle has twice the area of the red, the length of $B H$ must be twice that of $B A$, namely $24$.