The measures of rectangle $A B C D$ are 1200 by $150 \mathrm{~m}$. In addition, $P$ is on the extension of side $B C$ and is $350 \mathrm{~m}$ away from $C$. Determine the measures of $A P, P Q, P D, C Q$ and $D P$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-033.jpg?height=383&width=442&top_left_y=1445&top_left_x=1435)
The triangle $\triangle A B P$ is a right triangle with legs $A B=1200$ and $B P=150+350=500$. By the Pythagorean Theorem, we have

$$
A P^{2}=1200^{2}+500^{2}=(144+25) \times 10^{4}=169 \times 10^{4}=\left(13 \times 10^{2}\right)^{2}
$$

so that $A P=13 \times 10^{2}=1300 \text{ m}$. Similarly, considering the right triangle $\triangle P C D$, we have

$$
D P^{2}=350^{2}+1200^{2}=\left(7^{2}+12^{2} \times 2^{2}\right)\left(5^{2} \times 10^{2}\right)=25^{2} \times 50^{2}
$$

from which $D P=1250 \text{ m}$. The triangles $\triangle P C Q$ and $\triangle P B A$ are right triangles with a common angle, so they are similar, and it follows that

$$
\frac{P Q}{P A}=\frac{P C}{P B}=\frac{C Q}{A B}
$$

Substituting the known values, we get

$$
\frac{P Q}{1300}=\frac{350}{500}=\frac{C Q}{1200}
$$

Thus,

$$
P Q=\frac{350 \times 1300}{500}=910 \text{ m} \text { and } C Q=\frac{350 \times 1200}{500}=840 \text{ m}
$$