In the right-angled $\triangle PQR$ with $\angle PQR=90^\circ$, we are given $PQ=12$ cm and $QR=16$ cm. First, find the length of the hypotenuse PR using the Pythagorean theorem: $PQ^2 + QR^2 = PR^2$. So, $12^2 + 16^2 = PR^2$, which means $144 + 256 = PR^2$. Thus, $PR^2 = 400$, and $PR = \sqrt{400} = 20$ cm. The area of $\triangle PQR$ can be calculated as $\frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 12 \times 16 = 96$ cm$^2$. The area can also be expressed as $\frac{1}{2} \times PR \times QM$, where QM is the altitude to the hypotenuse. So, $96 = \frac{1}{2} \times 20 \times QM$. This simplifies to $96 = 10 \times QM$. Therefore, $QM = \frac{96}{10} = 9.6$ cm.