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\(^{2}\).
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. |
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