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CEMC Cayley, 2014

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

Pick one

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 $\text{PR gt;0}$ .

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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