#### Question

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

^{}

#### Solution

According to Pythagoras theorem,

In ∆PRM

\[{PR}^2 + {RM}^2 = {PM}^2 \]

\[ \Rightarrow {RM}^2 = {PM}^2 - {PR}^2 . . . \left( 1 \right)\]

\[ \Rightarrow {RM}^2 = {PM}^2 - {PR}^2 . . . \left( 1 \right)\]

In ∆PRQ

\[{PR}^2 + {RQ}^2 = {PQ}^2 \]

\[ \Rightarrow {PQ}^2 = {PR}^2 + \left( RM + MQ \right)^2 \]

\[ \Rightarrow {PQ}^2 = {PR}^2 + \left( RM + RM \right)^2 \]

\[ \Rightarrow {PQ}^2 = {PR}^2 + \left( 2RM \right)^2 \]

\[ \Rightarrow {PQ}^2 = {PR}^2 + 4 {RM}^2 \]

\[ \Rightarrow {PQ}^2 = {PR}^2 + 4\left( {PM}^2 - {PR}^2 \right) \left( \text{from} \left( 1 \right) \right)\]

\[ \Rightarrow {PQ}^2 = {PR}^2 + 4 {PM}^2 - 4 {PR}^2 \]

\[ \Rightarrow {PQ}^2 = 4 {PM}^2 - 3 {PR}^2\]

Hence, PQ^{2 }= 4PM^{2 }– 3PR^{2}.

Is there an error in this question or solution?

Solution In the Given Figure, M is the Midpoint of Qr. ∠Prq = 90°. Prove That, Pq2 = 4pm2 – 3pr2 Concept: Pythagoras Theorem.