In PQM , PM ^ 2 = PQ ^ 2 + QM ^ 2 —(1) QM= 1 2 QR ⟹ QM ^ 2 = 1 4 QR ^ 2 Substituting for QM ^ 2 = 1 4 QR ^ 2 in (1), PM ^ 2 ...

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Pythagoras Theorem

4 PM2 = 4 PQ2...

Question

$4 P M^{2} = 4 P Q^{2} + Q R^{2}$

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Solution

In $△ P Q M$ ,
${P M}^{2} = {P Q}^{2} + {Q M}^{2}$ ---(1)

$Q M = \frac{1}{2} Q R ⟹ {Q M}^{2} = \frac{1}{4} {Q R}^{2}$
Substituting for ${Q M}^{2} = \frac{1}{4} {Q R}^{2}$ in (1),

${P M}^{2} = {P Q}^{2} + \frac{1}{4} {Q R}^{2} 4 {P M}^{2} = {4 P Q}^{2} + {Q R}^{2}$
Hence proved.

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