Six identical rectangles with height $h$ and width $w$ are arranged as shown. Line segment $P Q$ intersects the vertical side of one rectangle at $X$ and the horizontal side of another rectangle at $Z$. If right-angled $\triangle X Y Z$ has $Y Z=2 X Y$, then $\frac{h}{w}$ equals
(A) $\frac{2}{3}$
(D) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) $\frac{3}{8}$
(E) $\frac{3}{4}$
Solution 1

We label the bottom left corner as $R$ and label various side lengths as $h$ and $w$ :

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-150.jpg?height=325&width=629&top_left_y=2285&top_left_x=802)

Since the diagram is made up of rectangles, then $X Y$ is parallel to $P R$, which tells us that $\angle Y X Z=\angle R P Q$. Also, $Y Z$ is parallel to $R Q$, which tells us that $\angle X Z Y=\angle P Q R$.

Therefore, $\triangle P R Q$ is similar to $\triangle X Y Z$.

Thus, $\frac{R Q}{P R}=\frac{Y Z}{X Y}$.

But $Y Z=2 X Y, R Q=3 w$ and $P R=4 h$.

This tells us that $\frac{3 w}{4 h}=\frac{2 X Y}{X Y}$ or $\frac{3}{4} \cdot \frac{w}{h}=2$ or $\frac{h}{w}=\frac{3}{8}$.

Solution 2

As in Solution 1, we label the bottom left corner as $R$ and label various side lengths as $h$ and $w$. Since $Y Z=2 X Y$, then the slope of line segment $X Z$ is $\frac{X Y}{Y Z}=\frac{X Y}{2 X Y}=\frac{1}{2}$.

Since $P R=4 h$ and $R Q=3 w$, then the slope of line segment $P Q$ is $\frac{P R}{R Q}=\frac{4 h}{3 w}$.

Since line segment $X Z$ is a portion of line segment $P Q$, then the slopes of these two line segments are equal, so $\frac{4 h}{3 w}=\frac{1}{2}$ and so $\frac{h}{w}=\frac{1}{2} \cdot \frac{3}{4}=\frac{3}{8}$.

ANsWER: (C)