Notes
triangle in four rectangles solution

Triangle in Four Rectangles

Solution by Area of a Rectangle and Area of a Triangle

With the points labelled as above, rectangles $B C D J$ and $J D E I$ have the same width, so since they have the same area then they must have the same height. Then rectangle $A B I H$ has twice the height of these rectangles, so its width is half. That is, the length of $A B$ is half that of $B C$. Finally, the width of $H E F G$ is three times that of $A B I H$, so the height of $H E F G$ is one third of that of $A B I H$.

Let $a$ be the length of $A B$, and $b$ the length of $C D$. Then $B C$ has length $2 a$ and $A H$ has length $2 b$. The length of $H G$ is one third of that of $A H$, so is $\frac{2}{3} b$. As the area of each rectangle is $12$ then $2 a b = 12$ so $a b = 6$.

Triangle $A B G$ has base $a$ and height $\frac{8}{3} b$ so has area $\frac{4}{3} a b = 8$. Triangle $B C D$ has base $2 a$ and height $b$ so has area $a b = 6$. Triangle $G F D$ has base $3 a$ and height $\frac{5}{3} b$ so has area $\frac{5}{2} a b = 15$. The total area of these triangles is $6 + 8 + 15 = 29$. The total area of the rectangle is $3 a \times \frac{8}{3} b = 8 a b = 48$. This leaves $19$ for the central triangle.