2024 AMC 10A Problem 16

"Similar" means every rectangle has the same shape: the long side divided by the short side is the same number for all of them. Call that number $r$. Finding $r$ is the key.

1. Sides from the area. Take a rectangle with short side $s$. Its long side is $rs$ and its area is $rs^2$. If the area is $S$, then $s^2 = S/r$, so
$$\text{short side} = \sqrt{S/r}, \qquad \text{long side} = \sqrt{S\,r}.$$
For example, if $r$ were $1$ (squares), both sides would be $\sqrt{S}$, as you would expect.
2. Find a spot where widths must match. The figure is drawn to scale. Rectangle $8$ lies wide (longer across than up), and it sits exactly on top of rectangles $9$ and $1$, which both stand tall. So the width of $8$ equals the width of $9$ plus the width of $1$.
3. Write that as an equation. The width of $8$ is its long side, $\sqrt{8r}$. The widths of $9$ and $1$ are their short sides, $\sqrt{9/r} = 3/\sqrt r$ and $\sqrt{1/r} = 1/\sqrt r$. So
$$\sqrt{8}\,\sqrt{r} = \frac{3}{\sqrt r} + \frac{1}{\sqrt r} = \frac{4}{\sqrt r}.$$
4. Solve for $r$. Multiply both sides by $\sqrt r$: $\sqrt8 \cdot r = 4$, so $r = \dfrac{4}{\sqrt 8} = \dfrac{4}{2\sqrt2} = \sqrt2$.
5. Double-check at another spot. Tall $16$ sits on top of tall $4$ and wide $2$. Widths: $4/\sqrt r = 2/\sqrt r + \sqrt{2r}$. Multiply by $\sqrt r$: $4 = 2 + \sqrt2\, r$, so $r = \sqrt2$ again. Good.
6. The big rectangle. Its area is the sum of all the small ones:
$$36 + 16 + 8 + 18 + 4 + 2 + 9 + 1 + 32 + 25 + 49 = 200.$$
It lies wide, so $AB$ is its long side: $AB = \sqrt{200\,r} = \sqrt{200\sqrt2}$.
7. Match the answer form. $200\sqrt2 = 100 \cdot 2\sqrt2 = 100\sqrt8$. So $AB = \sqrt{100}\cdot\sqrt{\sqrt8} = 10\sqrt[4]{8}$.

The answer is $\boxed{\textbf{(D)}\ 10\sqrt[4]{8}}$.

Similar shapes all share one ratio, so one unknown number $r$ describes every rectangle, and each side is the square root of the area times or divided by $r$. Any place where one side is split exactly by two neighbours gives an equation for $r$. Because the figure is drawn to scale, you are allowed to read off which rectangles are tall and which are wide.

A rough check helps choose among the answers. The big rectangle has area $200$ and, from the picture, is about $1.4$ times as wide as it is tall. So $AB$ is about $\sqrt{200 \cdot 1.4} = \sqrt{280}$, a bit under $17$. The choices are about $12.9$, $14.1$, $16.2$, $16.8$ and $20$. That rules out (A), (B) and (E), but (C) and (D) are too close to separate by eye, so the exact work above is still needed.