Given $\angle AQP=130^\circ$, $\angle PQR = 180^\circ - 130^\circ = 50^\circ$ (linear pair). Given $\angle BRP=140^\circ$, $\angle PRQ = 180^\circ - 140^\circ = 40^\circ$ (linear pair). In $\triangle PQR$, the sum of angles is $180^\circ$, so $\angle QPR + \angle PQR + \angle PRQ = 180^\circ$. Substituting the values, $\angle QPR + 50^\circ + 40^\circ = 180^\circ$, which gives $\angle QPR + 90^\circ = 180^\circ$. Therefore, $\angle QPR = 90^\circ$. Since $\triangle PQR$ is a right-angled triangle at P, we can use the Pythagorean theorem: $PQ^2 + PR^2 = QR^2$. Substituting the given lengths, $16^2 + 12^2 = QR^2$. This simplifies to $256 + 144 = QR^2$, so $400 = QR^2$. Taking the square root, $QR = \sqrt{400} = 20$ cm.