#### In Fig., 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.

Given: PQR is a triangle

$\angle Q = 90^{o} \; and\; QS \perp PR$

PQ = 6 cm, PS = 4 cm

In $\Delta SQP \; and \; \Delta SRQ$

$\angle S=\angle S$            (common angles and each angle is 90°)

$\angle SPQ=\angle SQR$                        (each equal to $90^{o}-\angle R$)

$\therefore \Delta SQP \sim \Delta SRQ$                   ( by AA similarity criterion)

$\Rightarrow \frac{SQ}{PS}=\frac{SR}{SQ}$

By cross multiply we get

$SQ^{2}=PS.SR \; \; \; \; \; \; \; ....(1)$

In $\Delta PSQ$ use Pythagoras theorem.

$PQ^{2}=PS^{2}+QS^{2}$

$6^{2}=4^{2}+QS^{2}$

$36-16=QS^{2}$

$20=QS^{2}$

$QS=\sqrt{20}=2\sqrt{5}cm$

Put $QS=2\sqrt{5}$ in equation (1)

$\left ( 2\sqrt{5} \right )^{2}=4 \times SR$

$\frac{20}{4}=SR$

$5cm=SR$

In $\Delta QSR$ use Pythagoras theorem

$QR^{2}=QS^{2}+SR^{2}$

$QR^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}$

$QR^{2}=20+25$

$QR=\sqrt{45}=3\sqrt{5}cm$