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 PN\(^{2}\) + RM\(^{2}\) = 5MN\(^{2}\).

Applying Pythagoras’ Theorem

Solution:

Given: In ∆PQR, ∠PQR = 90°.

PM = MQ and QN = NR

Therefore, PQ = 2MQ and QR = 2QN

To prove: PN\(^{2}\) + RM\(^{2}\) = 5MN\(^{2}\).

Proof:

            Statement

            Reason

1. ∆PQN, PQ\(^{2}\) + QN\(^{2}\) = PN\(^{2}\)

⟹ (2MQ)\(^{2}\) + QN\(^{2}\) = PN\(^{2}\)

⟹ 4MQ\(^{2}\) + QN\(^{2}\) = PN\(^{2}\)

1. By Pythagoras’ theorem

Given

2. ∆RQM, MQ\(^{2}\) + QR\(^{2}\) = RM\(^{2}\)

⟹ MQ\(^{2}\) + (2QN)\(^{2}\) = RM\(^{2}\)

⟹ MQ\(^{2}\) + 4QN\(^{2}\) = RM\(^{2}\)

2. By Pythagoras’ theorem

Given

3. 5MQ\(^{2}\) + 5QN\(^{2}\) = PN\(^{2}\) + RM\(^{2}\)

⟹ 5(MQ\(^{2}\) + QN\(^{2}\)) = PN\(^{2}\) + RM\(^{2}\)

3. Adding statements 1 and 2.

4. 5MN\(^{2}\) = PN\(^{2}\) + RM\(^{2}\) (Proved)

4. Applying Pythagoras’ theorem in ∆QMN.




9th Grade Math

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