Question

4. M and N are the mid-points of the sides QR
and PQ respectively of a A PQR, right-angled
at Q. Prove that :
(i) PM? + RN2 = 5 MN2​

Answer 1

Answer:

M Andn Are the Mid-points of the Sides Qr and Pq Respectively of A Pqr, Right-angled at Q. Prove That: Pm2 + Rn2 = 5 Mn2 4 Pm2 = 4 Pq2 + Qr2 4 Rn2 = Pq2 + 4 Qr2 4 (Pm2 + Rn2) = 5 Pr2 - Mathematics.

Answer 2

Step-by-step explanation:

We draw, PM, MN, NR

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 the 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 the 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.

(ii) We consider the ΔPQM, and applying the Pythagoras theorem we get,

PM2 = PQ2 + MQ2

4PM2 = 4PQ2 + 4MQ2             ...[ Multiply both sides by 4]

4PM2 = 4PQ2 + 4.( 1/2 QR )2 ...[ MQ =

12 QR ]

4PM2 = 4PQ2 + 4PQ + 4 .

12 QR2

4PM2 = 4PQ2  + QR2

Hence Proved.

(iii) We consider the ΔRQN, and applying the Pythagoras theorem we get,

RN2 = NQ2 + RQ2

4RN2 = 4NQ2 + 4QR2  ...[ Multiplying both sides by 4]

4RN2 = 4QR2 + 4 .(1/2 PQ)2 ...[ NQ = 12 PQ ]

4RN2 = 4QR2 + 4 .

14 PQ2

4RN2 = PQ2 + 4QR2

Hence Proved.

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

PM2 = PQ2 + MQ2

= ( PN + NQ )2 + MQ2

= PN2 + NQ2 + 2PN.NQ + MQ2  

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

Now, we consider the ΔRNQ, and applying the 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

4( PM2 + RN2 ) = 4.5. (NQ2 + MQ2)

4( PM2 + RN2 ) = 4.5.

[(12PQ)2+(12RQ)2]  

....[∵NQ=12PQ, MQ=12QR]

4 ( PM2 + RN2 ) = 5PR2

Hence Proved.