Given that, PR = 12 cm, QR = 6 cm and PL = 8cm
Now, in right angled ΔPLR, using Pythagoras theorem,
(Hypotenuse)2=(Perpendicular)2+(Base)2
⇒PR2=PL2+LR2
⇒LR2=PR2−PL2=(12)2−(8)2
⇒LR2=144−64=80
⇒LR=√80=4√5cm
∵LR=LQ+QR⇒LQ=LR−QR=(4√5−6)cm
Now, area of ΔPLR,
A1=12×LR×PL
=12×(4√5)×8
=16√5cm2
Again, area of ΔPLQ,
A2=12×LQ×PL
=12×(4√5−6)×8
=(16√5−24)cm2
Now, Area of ΔPLR= Area of ΔPLQ + Area of ΔPQR
⇒16√5=(16√5−24)+Area of ΔPQR
⇒ Area of ΔPQR=24cm2
⇒12×PR×QM=24
⇒12×12×QM=24
∴QM=4cm