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Applying Pythagoras’ Theorem

Applying Pythagoras’ theorem we will prove the problem given below.

∆PQR is right-angle at Q. M and N are the midpoints of PQ and QR respectively. Prove that PN2 + RM2 = 5MN2.

Applying Pythagoras’ Theorem

Solution:

Given: In ∆PQR, ∠PQR = 90°.

PM = MQ and QN = NR

Therefore, PQ = 2MQ and QR = 2QN

To prove: PN2 + RM2 = 5MN2.

Proof:

            Statement

            Reason

1. ∆PQN, PQ2 + QN2 = PN2

⟹ (2MQ)2 + QN2 = PN2

⟹ 4MQ2 + QN2 = PN2

1. By Pythagoras’ theorem

Given

2. ∆RQM, MQ2 + QR2 = RM2

⟹ MQ2 + (2QN)2 = RM2

⟹ MQ2 + 4QN2 = RM2

2. By Pythagoras’ theorem

Given

3. 5MQ2 + 5QN2 = PN2 + RM2

⟹ 5(MQ2 + QN2) = PN2 + RM2

3. Adding statements 1 and 2.

4. 5MN2 = PN2 + RM2 (Proved)

4. Applying Pythagoras’ theorem in ∆QMN.




9th Grade Math

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