, 2012

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 $\text{PQ gt;0}$.
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 $\text{TR gt;0}$.

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 $\text{PS gt;0}$.

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

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