CEMC Fermat, 2013

We calculate the area of quadrilateral $STRQ$ by subtracting the area of $\triangle PTS$ from the area of $\triangle PQR$.

Let $PT=x$.

Then $PR=PT+TR=x+271$.

Since $PQ=PR=x+271$ and $SQ=221$, then $PS = PQ - SQ = (x+271)-221=x+50$.

By the Pythagorean Theorem in $\triangle PTS$, we have $$\begin{aligned} PT^2 + TS^2 & = PS^2 \\ x^2 + 120^2 & = (x+50)^2\\ x^2 + 14400 & = x^2 + 100x + 2500 \\ 11900 & = 100x \\ x & = 119\end{aligned}$$ Therefore, $\triangle PTS$ has $PT = x = 119$, $TS=120$, and $PS = x+50=169$.

Since $\triangle PTS$ is right-angled at $T$, then its area is $\frac{1}{2}(PT)(TS) = \frac{1}{2}(119)(120) = 7140$.

Furthermore, in $\triangle PQR$, we have $PR = PQ = x+271 = 390$.

Now, $\triangle PQR$ is isosceles, so when we draw a median $PX$ from $P$ to the midpoint $X$ of $QR$, it is perpendicular to $QR$.

Since $X$ is the midpoint of $QR$ and $QR = 300$, then $QX = \frac{1}{2}QR=150$.

We can use the Pythagorean Theorem in $\triangle PXQ$ to conclude that $$PX = \sqrt{PQ^2-QX^2} = \sqrt{390^2-150^2} = \sqrt{21600} = 360$$ since $\text{PX gt;0}$.

Since $PX$ is a height in $\triangle PQR$, then the area of $\triangle PQR$ is $\frac{1}{2}(QR)(PX)=\frac{1}{2}(300)(360) = 54\,000$.

Finally, the area of $STRQ$ is the difference in the areas of these two triangles, or $54\,000-7140$, which equals $46\,860$.