In the diagram, $L M$ is perpendicular to $M N$. Rectangle $W X Y Z$ has $W$ on $L M$ and $Z$ on $M N$. Also, $Y Z=1 \mathrm{~m}$, $X Y=3 \mathrm{~m}$ and $M Z=1.2 \mathrm{~m}$. What is the distance from $X$ to line $M N$, to the nearest hundredth of a metre?
(A) $2.75 \mathrm{~m}$
(B) $3.67 \mathrm{~m}$
(C) $3.15 \mathrm{~m}$
(D) $3.26 \mathrm{~m}$
(E) $3.63 \mathrm{~m}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-094.jpg?height=482&width=482&top_left_y=1930&top_left_x=1293)

Part C: Each correct answer is worth 8.
Draw a line from $X$ to $P$ on $L M$, so that $X P$ is perpendicular to $L M$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-187.jpg?height=493&width=501&top_left_y=252&top_left_x=866)

Since $\angle X P M=\angle P M N=90^{\circ}$, then $P X$ is parallel to $M N$, so the distance from $X$ to line $M N$ equals the length of $P M$.

Since $W X Y Z$ is a rectangle, then $W Z=X Y=3 \mathrm{~m}$ and $W X=Z Y=1 \mathrm{~m}$.

By the Pythagorean Theorem, $W M=\sqrt{W Z^{2}-M Z^{2}}=\sqrt{3^{2}-1.2^{2}}=\sqrt{9-1.44}=\sqrt{7.56} \mathrm{~m}$. Since $\angle X W Z=90^{\circ}$ and $P W M$ is a straight line, then $\angle P W X+\angle X W Z+\angle Z W M=180^{\circ}$, so $\angle P W X+\angle Z W M=180^{\circ}-90^{\circ}=90^{\circ}$.

But since $\triangle X P W$ is right-angled, then

$$
\angle P X W=90^{\circ}-\angle P W X=90^{\circ}-\left(90^{\circ}-\angle Z W M\right)=\angle Z W M
$$

Therefore, $\triangle X P W$ is similar to $\triangle W M Z$.

This tells us that $\frac{P W}{M Z}=\frac{X W}{W Z}$, so $P W=\frac{M Z(X W)}{W Z}=\frac{1.2(1)}{3}=0.4 \mathrm{~m}$.

Thus, $P M=P W+W M=0.4+\sqrt{7.56} \approx 3.1495 \mathrm{~m}$, which, to the nearest hundredth of a metre, equals $3.15 \mathrm{~m}$.

ANSWER: (C)