question_answer

In the given figure below, \[\angle PQR=90{}^\circ \]and QL is a median, \[PQ=5\,\,cm\] and \[QR=12\,\,cm.\]Then, QL is equal to                                                                                                                                     [CDS 2013]

A) \[5\,\,cm\]                     

B) \[5.5\,\,cm\]                    

C) \[6\,\,cm\]                     

D) \[6.5\,\,cm\]

Correct Answer: D

Solution :

Given that, \[PQ=5\,\,cm,\]

\[QR=12\,\,cm\]

and QL is a median.

\[\therefore \]      \[PL=LR=\frac{PR}{2}\]                                    (i)

In \[\Delta PQR,\]

\[(P{{R}^{2}})={{(PQ)}^{2}}+{{(QR)}^{2}}\]

[by Pythagoras theorem]

\[={{(5)}^{2}}+{{(12)}^{2}}\]

\[=25+144=169={{(13)}^{2}}\]

\[\Rightarrow \]\[P{{R}^{2}}={{(13)}^{2}}\]\[\Rightarrow \]\[PR=13\]

Now, by theorem, it L is the mid-point of the hypotenuse PR of a right angled \[\Delta PQR.\]

Then,    \[QL=\frac{1}{2}PR=\frac{1}{2}(13)=6.5\,\,cm\]