In given figure PQR is a right triangle,

Solution:

Given, ΔPQR in which ∠Q = 90°, QS ⊥ PR and PQ = 6 cm, PS = 4 cm In ΔSQP and ΔSRQ,

$\angle P S Q=\angle R S Q$ [each equal to $90^{\circ}$ ]

$\angle S P Q=\angle S Q R$ [each equal to $90^{\circ}-\angle R$ ]

$\therefore \quad \Delta S Q P \sim \Delta S R Q$

Then, $\frac{S Q}{P S}=\frac{S R}{S Q}$

$\Rightarrow$ $S Q^{2}=P S \times S R$ $\ldots$ (i)

$\begin{array}{lll}\text { In right angied } \Delta P S Q, & P Q^{2}=P S^{2}+Q S^{2} & \text { [by Pythagoras theorem] }\end{array}$

$\Rightarrow$ $(6)^{2}=(4)^{2}+Q S^{2}$

$\Rightarrow \quad 36=16+Q S^{2}$

$\Rightarrow \quad Q S^{2}=36-16=20$

$\therefore \quad Q S=\sqrt{20}=2 \sqrt{5} \mathrm{~cm}$

On putting the value of $Q S$ in Eq. (i), we get

$(2 \sqrt{5})^{2}=4 \times S R$

$\Rightarrow$ $S R=\frac{4 \times 5}{4}=5 \mathrm{~cm}$

In right angled $\triangle Q S R$, $Q R^{2}=Q S^{2}+S R^{2}$

$\Rightarrow \quad Q R^{2}=(2 \sqrt{5})^{2}+(5)^{2}$

$\Rightarrow \quad Q R^{2}=20+25$

$\therefore$$Q R=\sqrt{45}=3 \sqrt{5} \mathrm{~cm}$

Hence, $Q S=2 \sqrt{5} \mathrm{~cm}, R S=5 \mathrm{~cm}$ and $Q R=3 \sqrt{5} \mathrm{~cm}$