Notes
three congruent rectangles solution

Solution to the Three Congruent Rectangles Puzzle

Solution by Dissection and Congruent Triangles

In the above diagram, $G$, $I$, $B$, and $D$ are where the the yellow rectangle meets the edges of the figure. Line segments $L B$ and $G K$ are horizontal, while $I L$ and $D K$ are vertical.

Triangles $G M I$ and $G H I$ are congruent, as are triangle $B L I$ and $B A I$, triangles $B N D$ and $B C D$, and triangles $G K D$ and $G F D$. Therefore the sum of the areas of triangles $G H I$, $B L I$, $B N D$, and $G K D$ is the same as that of triangles $G M I$, $B A I$, $B C D$, and $G F D$.

The latter comprises the region of the figure that is outside the yellow rectangle, but since the yellow rectangle is congruent to the blue and red, its area is half of the full diagram. Therefore triangles $G H I$, $B L I$, $B N D$, and $G K D$ cover the same area as the yellow rectangle.

But those triangles cover the yellow rectangle with an overlap, namely the rectangle $L M K N$. Therefore, that rectangle must have zero area and so points $B$ and $G$ are at the same height.

The distance between $B$ and $G$ is therefore the same as the bottom side of the diagram, which is $F$. This is also the diagonal of the yellow rectangle, so is also the length of line segment $I D$.

Solution by Similar Triangles and Pythagoras' Theorem

With the points labelled as above, let line segment $H G$ have length $a$, $A C$ have length $b$, and $H I$ have length $c$. Then $I A$ has length $4 - c$.

Triangles $G H I$ and $I A B$ are similar since they are both right-angled and angles $G \hat{I} H$ and $A \hat{I} B$ add up to $90^\circ$. Therefore:

which rearranges to $c(4 - c) = a b$.

Since the rectangles are congruent, $G I$ has length $2$ and $I B$ has length $a + b$. Therefore, applying Pythagoras' theorem to triangles $G H I$ and $I A B$ produces:

Putting those together:

Therefore $c(4 - c) = 3$ which rearranges to $c^2 - 4 c + 3 = 0$ and this has solutions $c = 1$ and $c = 3$. Note that $4 - c$ is the other solution, so without loss of generality take $c = 1$.

Then $a^2 = 4 - c^2 = 3$, so $a = \sqrt{3}$ and likewise $b = \sqrt{3}$. This means that $a + b = 2 \sqrt{3}$, so the rectangles have side lengths $2$ and $2 \sqrt{3}$. Their diagonal is then $\sqrt{2^2 + (2 \sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.