# 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$$^{2}$$.

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’ theoremGiven 2. ∆RQM, MQ$$^{2}$$ + QR$$^{2}$$ = RM$$^{2}$$⟹ MQ$$^{2}$$ + (2QN)$$^{2}$$ = RM$$^{2}$$⟹ MQ$$^{2}$$ + 4QN$$^{2}$$ = RM$$^{2}$$ 2. By Pythagoras’ theoremGiven 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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