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