# Solution to the Three Congruent Rectangles II Puzzle +-- {.image} [[ThreeCongruentRectanglesII.png:pic]] > The rectangles are congruent, each with area $9$. What’s the area of the triangle? =-- ## Solution by [[Similar Triangles]] and [[Area of a Triangle]] +-- {.image} [[ThreeCongruentRectanglesIILabelled.png:pic]] =-- Consider the diagram labelled as above. Since $A B$ and $B K$ are the same length, and $A K G$ is a straight line, by [[similar triangles]] then $H K$ and $H G$ are also the same length as each other. This means that $H G D K$ is a square, and the area of triangle $G D K$ is half of that of $H G D K$. Similarly, $G F E D$ is a rectangle, and the area of triangle $G D F$ is half of that of it. Therefore, the area of [[trapezium]] $F G K D$ is half that of rectangle $F H K E$. Triangle $D K A$ can be viewed as having base $D K$, in which case its [[area of a triangle|perpendicular height]] is the length of $B K$. This means that it has the same area as triangle $D K B$, which is half of the area of $D K B C$. In total, then, triangle $F G A$ has the same area as one of the congruent rectangles, namely $9$. ## Solution by [[Similar Triangles]] and [[Area of a Triangle]] With the diagram labelled as above, let $a$ and $b$ be the lengths of the sides of the rectangles, with $a \lt b$. As the line $A F$ passes through $D$, triangles $A C D$ and $D E F$ are [[similar]]. The horizontal and vertical lengths of these triangles are, respectively, $a + b$ and $b$ for triangle $A C D$ and $b - a$ and $a$ for $D E F$. Therefore: $$ \begin{aligned} a + b : b &= b - a : a \\ \frac{a + b}{b} &= \frac{b - a}{a} \\ (a + b) a &= (b - a)b \\ a^2 + a b &= b^2 - a b \\ 2 a b &= b^2 - a^2 \end{aligned} $$ The rectangles have area $9$, so $a b = 9$. This means that $b^2 - a^2 = 18$. Triangles $A J K$ and $A I G$ are likewise similar, so since $A J$ and $J K$ have the same length, $A I$ and $I G$ are also the same length, which is $a + b$. Therefore, $F G$ has length $2b - (a + b) = b - a$. The area of the shaded triangle $A F G$ is therefore: $$ \begin{aligned} \frac{1}{2} (b - a) \times (b + a) &= \frac{1}{2} (b^2 - a^2) \\ &= \frac{1}{2} \times 18 \\ &= 9 \end{aligned} $$