Given that, PQ = 5 cm, QR = 12 cm and QL is a median.



\( \Large \therefore PL = \frac{PR}{2} \) ...(i)

In \( \Large \triangle PQR \)

\( \Large \left(PR\right)^{2} = \left(PQ\right)^{2}+ \left(QE\right)^{2} \)

[by Pythagoras theorem]

= \( \Large \left(5\right)^{2}+ \left(12\right)^{2} \)

= \( \Large 25 + 144 = 169 = \left(13\right)^{2} \)

=> \( \Large PR^{2} = \left(13\right)^{2} => PR=13 \)

Now, by theorem, if L is the mid-point of the hypotenuse PR of a right angled \( \Large \triangle PQR \) then

\( \Large QL = \frac{1}{2}PR = \frac{1}{2} \left(13\right) = 6.5 cm \)