question_answer

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

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, : \[\therefore \]      \[PL=LR=\frac{PR}{2}\] In \[\Delta PQR,\] \[{{(PR)}^{2}}={{(PQ)}^{2}}+{{(QR)}^{2}}\] (by Pythagoras theorem) \[={{(5)}^{2}}+{{(12)}^{2}}=25+144=169={{(13)}^{2}}\] \[\Rightarrow \]\[P{{R}^{2}}={{(13)}^{2}}\]\[\Rightarrow \]\[PR=13\] Now, by theorem, if L is the mid-point of the hypotenuse PR of a right angled \[\Delta PQR,\]then \[QL=\frac{1}{2},PR=\frac{1}{2}(13)=6.5\,\,cm\]