In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QS}$ is an altitude to the hypotenuse $PR$. If $PQ = 6$ and $PS = 4$,find $QS$,$QR$,and $RS$.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QS \perp PR$.
By the property of geometric mean in a right-angled triangle,$PQ^2 = PS \cdot PR$.
Given $PQ = 6$ and $PS = 4$,we have $6^2 = 4 \cdot PR$,which implies $36 = 4 \cdot PR$,so $PR = 9$.
Since $PR = PS + RS$,we have $9 = 4 + RS$,so $RS = 5$.
Using the altitude theorem,$QS^2 = PS \cdot RS = 4 \cdot 5 = 20$,so $QS = \sqrt{20} = 2\sqrt{5}$.
In $\Delta QSR$,by the Pythagorean theorem,$QR^2 = QS^2 + RS^2 = 20 + 5^2 = 20 + 25 = 45$.
Thus,$QR = \sqrt{45} = 3\sqrt{5}$.