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Geometry Difficulty 1.3 Junior Find the answer Canada

In the diagram, $\triangle PQR$ has $\angle RPQ = 90^\circ$ , $PQ=10$ , and $QR=26$ .The area of $\triangle PQR$ is

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Solution

Since $\triangle PQR$ is right-angled at $P$ , we can use the Pythagorean Theorem. We obtain $PQ^2 + PR^2 = QR^2$ or $10^2 + PR^2 = 26^2$ . This gives $PR^2 = 26^2 - 10^2 = 676 - 100 = 576$ and so $PR = \sqrt{576} = 24$ , since $PR>0$ . Since $\triangle PQR$ is right-angled at $P$ , its area equals $\frac{1}{2}(PR)(PQ) = \frac{1}{2}(24)(10) = 120$ .

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