, 2020

We extend $RQ$ to the left until it meets $PT$ at point $U$, as shown. [[IMAGE0]] Because quadrilateral $URST$ has three right angles, then it must have four right angles and so is a rectangle. Thus, $UT = RS$ and $UR = TS = 30$. Since $UR = 30$, then $UQ = UR - QR = 30 - 18 = 12$. Now $\triangle PQU$ is right-angled at $U$. By the Pythagorean Theorem, since $PU > 0$, we have $$PU = \sqrt{PQ^2 - UQ^2} = \sqrt{13^2 - 12^2} = \sqrt{169-144} = \sqrt{25} = 5$$ Since the perimeter of $PQRST$ is 82, then $13 + 18 + RS + 30 + (UT + 5) = 82$. Since $RS = UT$, then $2\times RS = 82 - 13 - 18 - 30 - 5 = 16$ and so $RS = 8$. Finally, we can calculate the area of $PQRST$ by splitting it into $\triangle PQU$ and rectangle $URST$. The area of $\triangle PQU$ is $\frac{1}{2} \times UQ \times PU = \frac{1}{2} \times 12 \times 5 = 30$. The area of rectangle $URST$ is $RS \times TS = 8 \times 30 = 240$. Therefore, the area of pentagon $PQRST$ is $30 + 240 = 270$.