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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

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Question

M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM2 + RN2 = 5 MN2
(ii) 4 PM2 = 4 PQ2 + QR2
(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2

Sum
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Solution


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 = NQ+ RQ

= NQ+ ( QM + RM )2

= NQ + QM + RM + 2QM .RM

= MN + RM + 2QM . RM   .......(ii)

Adding (i) and (ii) we get, 

PM + RN = MN + PN + 2PN.NQ + MN2  + RM + 2QM. RM

PM + RN = 2MN + PN + RM + 2PN.NQ + 2QM.RM 

PM + RN = 2MN + NQ + QM + 2(QN) + 2(QM)

PM + RN = 2MN + MN + 2MN

PM2 + RN2  = 5MN                

Hence Proved.

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

PM2 = PQ2 + MQ2

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

4PM2 = 4PQ2 + 4.`(1/2 "QR")^2` ...[ MQ = `1/2` QR ]

4PM2 = 4PQ2 + 4PQ + 4 . `1/4` QR2

4PM2 = 4PQ2  + QR

Hence Proved.

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

RN2 = NQ2 + RQ2

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

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

4RN2 = 4QR2 + 4 .`1/4` PQ2

4RN2 = PQ2 + 4QR2

Hence Proved.

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

RN+ NQ+ RQ

= NQ+ ( QM + RM )2

= NQ + QM + RM + 2QM .RM

= MN + RM + 2QM . RM .......(ii)

Adding (i) and (ii) we get,

PM + RN = MN + PN + 2PN . NQ + MN2  + RM + 2QM. RM

PM + RN = 2MN + PN + RM + 2PN . NQ + 2QM . RM 

PM + RN = 2MN + NQ + QM + 2(QN) + 2(QM)

PM + RN = 2MN + MN + 2MN

PM + RN = 5MN

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

4( PM2 + RN2 ) = 4.5. `[ ( 1/2 "PQ" )^2 + ( 1/2 "RQ" )^2 ]   ....[ ∵ "NQ" = 1/2 "PQ" , "MQ" = 1/2 "QR" ]`

4 ( PM2 + RN2 ) = 5PR2

Hence Proved.

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Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 164]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 4 | Page 164

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