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Problem 492

AMC 10/12, early questions

Geometry Difficulty 3.0 Prove it CEMC Hypatia · Canada · 2012

Quadrilateral $PQRS$ is constructed with $QR=51$ , as shown. The diagonals of $PQRS$ intersect at $90^{\circ}$ at point $T$ , such that $PT=32$ and $QT=24$ .

Calculate the length of $PQ$ .
Calculate the area of $\triangle PQR$ .
If $QS:PR=12:11$ , determine the perimeter of quadrilateral $PQRS$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

In $\triangle PTQ$ , $\angle PTQ=90^{\circ}$ .

Using the Pythagorean Theorem, $PQ^2=32^2+24^2$ or $PQ^2=1024+576=1600$ and so $PQ=\sqrt{1600}=40$ , since $PQ>0$ .
In $\triangle QTR$ , $\angle QTR=90^{\circ}$ .

Using the Pythagorean Theorem, $51^2=TR^2+24^2$ or $TR^2=2601-576=2025$ and so $TR=\sqrt{2025}=45$ , since $TR>0$ .

Since $PT=32$ and $TR=45$ , then $PR=PT+TR=32+45=77$ .

In $\triangle PQR$ , $QT$ is perpendicular to base $PR$ and so $\triangle PQR$ has area $\frac{1}{2}\times(PR)\times(QT)=\frac{1}{2}\times77\times24=924$ .
From part (b), the length of $PR$ is 77.

Since $QS:PR=12:11$ , then $\dfrac{QS}{77}=\dfrac{12}{11}$ or $QS=77\times\dfrac{12}{11}=84$ .

So then $TS=QS-QT=84-24=60$ .

Using the Pythagorean Theorem in $\triangle PTS$ , $PS=\sqrt{32^2+60^2}$ or $PS=\sqrt{4624}=68$ , since $PS>0$ .

Using the Pythagorean Theorem in $\triangle RTS$ , $RS=\sqrt{45^2+60^2}$ or $RS=\sqrt{5625}=75$ , since $RS>0$ .

Thus, quadrilateral $PQRS$ has perimeter $40+51+75+68$ or 234.

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