Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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In the Given Figure, M is the Midpoint of Qr. ∠Prq = 90°. Prove That, Pq2 = 4pm2 – 3pr2 - Geometry

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Question

In the given figure, M is the midpoint of QR. ∠PRQ = 90°. Prove that, PQ= 4PM– 3PR2

Solution

According to Pythagoras theorem,

In ∆PRM

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

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= 4PM– 3PR2.

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