Question

Question 10
In given figure PQR is a right triangle, right angled at Q and QS $\perp$ PR. If PQ = 6cm and PS = 4cm, then find QS, RS and QR.

Answer 1

Given, $\Delta PQR$ in which $\angle Q={90}^{\circ }$, $QS\perp PR$ and PQ = 6cm, PS = 4cm.

In $\Delta SQP$ and $\Delta SRQ$,
$\angle PSQ=\angle RSQ$ [each equal to ${90}^{\circ }$ ]
$\angle SPQ=\angle SQR$ [each equal to ${90}^{\circ }-\angle R$]
$\therefore$ $\Delta SQP\sim \Delta SRQ$ [ by AAA similarity criterion]
Then, $\frac{SQ}{PS}=\frac{SR}{SQ}$
$\Rightarrow S{Q}^2=PS\times SR$..........(i)

In right angled $\Delta PSQ$,
$P{Q}^2=P{S}^2+Q{S}^2$ [by Pythagoras theorem]
$\Rightarrow (6{)}^2=(4{)}^2+Q{S}^2$
$\Rightarrow 36=16+Q{S}^2$
$\Rightarrow Q{S}^2=36-16=20$
$\therefore$ $QS=\sqrt{20}=2\sqrt{5}cm$
On putting the value of QS in Eq..(i), we get
$(2\sqrt{5}{)}^2=4\times SR$
$\Rightarrow SR=\frac{4\times 5}{4}=5cm$

In right angled $\Delta QSR$,
$Q{R}^2=Q{S}^2+S{R}^2$
$\Rightarrow Q{R}^2=(2\sqrt{5}{)}^2+(5{)}^2$
$\Rightarrow Q{R}^2=20+25$
$\therefore$ $QR=\sqrt{45}=3\sqrt{5}cm$
Hence, $QS=2\sqrt{5}cm,RS=5cm$ and $QR=3\sqrt{5}cm$