PQR is a right triangle right anmgled at Q and $\text{QS}\perp\text{PR}.$ If PQ = 6cm and PS = 4cm, find QS, RS and QR.
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In $\triangle\text{PQR},$
$\angle\text{PQR}=90^\circ$[Given]
$\text{QS}\perp\text{PR}$
[From vertex Q to hypotenuse PR]
$QS^2 = PS \times SR ......(i)$
Now, in $\triangle\text{PSQ},$ we have
$QS^2 = PQ^2 - PS^2$
$\Rightarrow QS^2 = 6^2 - 4^2$
$\Rightarrow QS^2 = 36 - 16$
$\Rightarrow QS^2 = 20$
$\Rightarrow\text{QS}=2\sqrt{5}$
$\Rightarrow\text{QS}=\text{PS}\times\text{SR}\ .......({\text{i}})$
$\Rightarrow(2\sqrt{5})^2=4\times\text{5}$
$\Rightarrow\frac{20}{4}=\text{SR}$
$\Rightarrow\text{SR}=5\text{cm}$
Now, $\text{QS}\perp\text{PR}$
$\therefore\angle\text{QSR}=90^\circ$
$\Rightarrow\text{QR}^2=\text{QS}^2+\text{SR}^2$
$\Rightarrow\text{QR}^2=(2\sqrt{5})^2+5^2$
$\Rightarrow\text{QR}^2=20+25$
$\Rightarrow\text{QR}^2=45$
$\Rightarrow\text{QR}^2=3\sqrt{5}\text{cm}$
Hence, $\text{QS}=2\sqrt{5},$ $\text{RS}=5\text{cm}$ and $\text{QR}=3\sqrt{5}\text{cm}.$
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