CEMC Fermat, 2016

Geometry Difficulty 3.4 AMC 10/12 Find the answer Canada

In the diagram, $\triangle PQR$ is right-angled at $P$ and $PR=12$ .
If point $S$ is on $PQ$ so that $SQ=11$ and $SR=13$ , the perimeter of $\triangle QRS$ is

Pick one

Solution

By the Pythagorean Theorem in $\triangle PRS$ , $PS = \sqrt{RS^2 - PR^2} = \sqrt{13^2 - 12^2} = \sqrt{169-144} = \sqrt{25} = 5$ since $\text{PS gt;0}$ .

Thus, $PQ = PS + SQ = 5 + 11 = 16$ .

By the Pythagorean Theorem in $\triangle PRQ$ , $RQ = \sqrt{PR^2 + PQ^2} = \sqrt{12^2 + 16^2} = \sqrt{144+256} = \sqrt{400} = 20$ since $\text{RQ gt;0}$ .

Therefore, the perimeter of $\triangle QRS$ is $RS + SQ + RQ = 13+11+20=44$ .

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