A quadrilateral has vertices $P(2a,b)$, $Q(a,3b)$, $R(-2a,-b)$, and $S(-a,-3b)$, where $a$ and $b$ are integers with $a>b>0$. The area of $PQRS$ is $50$. What is $a+b$?

1. By the Shoelace theorem, $[PQRS]=2|x_Py_Q-y_Px_Q|$. With $P=(2a,b)$ and $Q=(a,3b)$, this gives
   $$[PQRS]=2|2a\cdot3b-b\cdot a|=10ab.$$
2. Setting $10ab=50$ gives $ab=5$. Since $a>b>0$ are integers, $(a,b)=(5,1)$.
3. Therefore, $a+b=6$.