In the figure,$PQR$ is a right triangle right-angled at $Q$ and $QS \perp PR$. If $PQ = 6 \, cm$ and $PS = 4 \, cm$,find $QS$,$RS$,and $QR$.

(N/A) Given,$\Delta PQR$ in which $\angle Q = 90^{\circ}$,$QS \perp PR$,$PQ = 6 \, cm$,and $PS = 4 \, cm$.
In right-angled $\Delta PSQ$,by Pythagoras theorem:
$PQ^2 = PS^2 + QS^2$
$6^2 = 4^2 + QS^2$
$36 = 16 + QS^2$
$QS^2 = 20$
$QS = \sqrt{20} = 2\sqrt{5} \, cm$.
Since $\Delta PSQ \sim \Delta QSR$,we have:
$\frac{PS}{QS} = \frac{QS}{RS}$
$QS^2 = PS \times RS$
$20 = 4 \times RS$
$RS = 5 \, cm$.
In right-angled $\Delta QSR$,by Pythagoras theorem:
$QR^2 = QS^2 + RS^2$
$QR^2 = 20 + 5^2$
$QR^2 = 20 + 25 = 45$
$QR = \sqrt{45} = 3\sqrt{5} \, cm$.
Thus,$QS = 2\sqrt{5} \, cm$,$RS = 5 \, cm$,and $QR = 3\sqrt{5} \, cm$.