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three congruent rectangles ii solution

Solution to the Three Congruent Rectangles II Puzzle

Solution by Similar Triangles and Area of a Triangle

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 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:

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: