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Geometry Difficulty 3.5 AMC 10/12 Find the answer Canada

In $\triangle PQR$ , $\angle RPQ = 90^\circ$ and $S$ is on $PQ$ .

If $SQ=14$ , $SP=18$ , and $SR=30$ , then the area of $\triangle QRS$ is

Pick one

Solution

Solution 1

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 $PR^2 = 30^2-18^2 = 900 - 324 = 576$ , from which $PR=24$ , since $PR>0$ .

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 $\frac{1}{2}(24)(14)=168$ .

Solution 2

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 $PR^2 = 30^2-18^2 = 900 - 324 = 576$ , from which $PR=24$ , since $PR>0$ .

The area of $\triangle QRS$ equals the area of $\triangle RPQ$ minus the area of $\triangle RPS$ .

Since $\triangle RPQ$ is right-angled at $P$ , its area is $\frac{1}{2}(PR)(PQ)=\frac{1}{2}(24)(18+14)=12(32)=384$ .

Since $\triangle RPS$ is right-angled at $P$ , its area is $\frac{1}{2}(PR)(PS)=\frac{1}{2}(24)(18)=12(18)=216$ .

Therefore, the area of $\triangle QRS$ is $384-216=168$ .

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