All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length $AB$? $\textbf{(A) }4+4\sqrt5\qquad$ $\textbf{(B) }10\sqrt2\qquad$ $\textbf{(C) }5+5\sqrt5\qquad$ $\textbf{(D) }10\sqrt[4]{8}\qquad$ $\textbf{(E) }20$
Solution
Every small rectangle is similar to the enclosing one, so all of them have the same ratio $k=\dfrac{\text{long side}}{\text{short side}}$. A rectangle of area $a$ therefore has short side $\sqrt{a/k}$ and long side $\sqrt{ak}$.
Compare the heights of the left and the right column of the figure. On the left, the rectangle $36$ stands on its short side (height $\sqrt{36k}=6\sqrt k$) and the rectangle $32$ lies flat (height $\sqrt{32/k}=\dfrac{4\sqrt2}{\sqrt k}$). On the right, the rectangle $18$ lies flat (height $\dfrac{3\sqrt2}{\sqrt k}$) and the rectangle $49$ stands up (height $7\sqrt k$). Both columns have the height of the enclosing rectangle, so $$6\sqrt k+\frac{4\sqrt2}{\sqrt k}=\frac{3\sqrt2}{\sqrt k}+7\sqrt k\ \Longrightarrow\ \frac{\sqrt2}{\sqrt k}=\sqrt k\ \Longrightarrow\ k=\sqrt2 .$$
The areas of the eleven rectangles add up to $36+16+8+18+4+2+9+1+32+25+49=200$. If $AB=w$ is the long side of the enclosing rectangle, its height is $w/k$, so $w\cdot\dfrac wk=200$.
Then $w^2=200k=200\sqrt2$, so $w^4=80000=10^4\cdot8$ and $AB=w=10\sqrt[4]{8}$.
The rectangle $ABCD$ with $AB>BC$ is cut into six smaller rectangles, each of which is similar to $ABCD$. The figure is drawn to scale, and each number is the area of the rectangle it labels. The areas are $32,\,16,\,8,\,4,\,2,\,2$. What is the length $AB$?
$\textbf{(A) }4+4\sqrt5\qquad$ $\textbf{(B) }10\sqrt2\qquad$ $\textbf{(C) }5+5\sqrt5\qquad$ $\textbf{(D) }10\sqrt[4]{8}\qquad$ $\textbf{(E) }20$ 
