Track / Stage 3 / 133 of 260 #613 of 2444

Problem 613

AMC 10/12, early questions

Geometry Difficulty 3.6 Find the answer CEMC Fermat · Canada · 2020

In the diagram, $\triangle PQR$ is right-angled at $Q$ and point $S$ is on $PR$ so that $QS$ is perpendicular to $PR$ .

If the area of $\triangle PQR$ is 30 and $PQ=5$ , the length of $QS$ is

$\frac{60}{13}$
$5$
$\frac{30}{13}$
$4$
$3$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Since $\triangle PQR$ is right-angled at $Q$ , its area equals $\frac{1}{2}\cdot PQ \cdot QR$ .
Since its area is 30 and $PQ=5$ , then $\frac{1}{2} \cdot 5 \cdot QR = 30$ and so $QR = 30 \cdot \frac{2}{5} = 12$ .
By the Pythagorean Theorem, we know that $PR^2 = PQ^2 + QR^2 = 5^2 + 12^2 = 25+144 = 169$ Since $PR>0$ , then $PR = \sqrt{169} = 13$ .
If we now consider $\triangle PQR$ as having base $PR$ and perpendicular height $QS$ , we see that its area equals $\frac{1}{2} \cdot PR \cdot QS$ .
Since its area is 30 and $PR = 13$ , then $\frac{1}{2} \cdot 13 \cdot QS = 30$ which gives $QS = 30 \cdot \frac{2}{13} = \frac{60}{13}$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.