Sum

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ^{2 }= 4PM^{2 }– 3PR^{2}

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#### Solution

According to Pythagoras theorem,

In ∆PRM,

PR^{2} + RM^{2} = PM^{2}

⇒ RM^{2} = PM^{2} - PR^{2} ....(1)

In ∆PRQ,

PR^{2} + RQ^{2} = PQ^{2}

⇒ PQ^{2} = PR^{2} + (RM + MQ)^{2}

⇒ PQ^{2} = PR^{2} + (RM + RM)^{2}

⇒ PQ^{2} = PR^{2} + (2RM)^{2}

⇒ PQ^{2} = PR^{2} + 4RM^{2}

⇒ PQ^{2} = PR^{2} + 4(PM^{2} - PR^{2}) .....(from 1)

⇒ PQ^{2} = PR^{2} + 4PM^{2} - 4PR^{2}

⇒ PQ^{2} = 4PM^{2} - 3PR^{2}

Hence, PQ^{2 }= 4PM^{2 }– 3PR^{2}.

Concept: Right-angled Triangles and Pythagoras Property

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