Question

M and N are the midpoints of the sides QR and PQ respectively of ∆ , right angled

at Q . Prove that PM2 + RN2 = 5MN2​

Answer 1

Answer:

Construct a line to joining point N to M

InΔPQM

PQ

2

+QM

2

=PM

2

....(1)

InΔNQR

NQ

2

+QR

2

=NR

2

...(2)

andInΔNQM

NQ

2

+QM

2

=NM

2

....(3)

Addingeq(1)and(2)

PM

2

+NR

2

=(PQ

2

+QR

2

)+(QM

2

+QN

2

)

PM

2

+NR

2

=(2QN)

2

+(2QM)

2

+NM

2

$[∵MandNismidpointPQandQRrespectively]$

PM

2

+NR

2

=4(QN

2

+4QM

2

)+NM

2

Answer 2

We draw, PM, MN, NR

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since M and N are the mid-points of the sides QR and PQ respectively, therefore, PN = NQ, QM = RM

(i) First, we consider the ΔPQM, and applying Pythagoras theorem we get,

PM2= PQ2 + MQ2

= ( PN + NQ )2 + MQ2

= PN + NQ2 + 2PN . NQ + MQ2

= MN2+ PN2 + 2PN.NQ ...[From, ΔMNQ, MN2 = NQ2 + MQ2] ......(i)

Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,

RN2 = NQ2 + RQ2

= NQ2 + ( QM + RM )2

= NQ2 + QM2 + RM2 + 2QM .RM

= MN2 + RM2 + 2QM . RM .......(ii)

Adding (i) and (ii) we get,

PM2 + RN2 = MN2 + PN2 + 2PN.NQ + MN2 + RM2 + 2QM. RM

PM2 + RN2 = 2MN2 + PN2 + RM2 + 2PN.NQ + 2QM.RM

PM2 + RN2 = 2MN2 + NQ2 + QM2 + 2(QN2 ) + 2(QM2 )

PM2 + RN2 = 2MN2 + MN2 + 2MN2

PM 2 + RN2 = 5MN2 Hence Proved.