Competition problem — Geometry — Maths Olympiad Prep

Library / /113 of 168

Geometry Difficulty 2.1 Junior Find the answer

In $\triangle PQR, \angle RPQ=90^{\circ}$ and $S$ is on $PQ$ . If $SQ=14, SP=18$ , and $SR=30$ , what is the area of $\triangle QRS$ ?

Your answer

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

Solution

Since $\triangle RPS$ is right-angled at $P$ , then by the Pythagorean Theorem, $PR^{2}+PS^{2}=RS^{2}$ or $PR^{2}+18^{2}=30^{2}$ . This gives $PR^{2}=900-324=576$ , from which $PR=24$ . Since $P, S$ and $Q$ lie on a straight line and $RP$ is perpendicular to this line, then $RP$ is actually a height for $\triangle QRS$ corresponding to base $SQ$ . Thus, the area of $\triangle QRS$ is $\frac{1}{2}(24)(14)=168$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.