Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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In the Given Figure Seg Ps is the Median of ∆Pqr and Pt ⊥ Qr. Prove That, P R 2 = P S 2 + Q R × S T + ( Q R 2 ) 2 P Q 2 = P S 2 − Q R × S T + ( Q R 2 ) 2 - Geometry

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Question

In the given figure seg PS is the median of ∆PQR and PT ⊥ QR. Prove that,

\[{PR}^2 = {PS}^2 + QR \times ST + \left( \frac{QR}{2} \right)^2\]

\[{PQ}^2 = {PS}^2 - QR \times ST + \left( \frac{QR}{2} \right)^2\]

Solution

According to Pythagoras theorem, in ∆PTQ

\[{PQ}^2 = {PT}^2 + {QT}^2 . . . \left( 1 \right)\]

In ∆PTS

\[{PS}^2 = {PT}^2 + {TS}^2 . . . \left( 2 \right)\]

In ∆PTR

\[{PR}^2 = {PT}^2 + {RT}^2 . . . \left( 3 \right)\]

In ∆PQR, point S is the midpoint of side QR.

\[QS = SR = \frac{1}{2}QR . . . \left( 4 \right)\]

\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow {PR}^2 = 2 {PS}^2 + 2 {QS}^2 - {PQ}^2 \]

\[ \Rightarrow {PR}^2 = {PS}^2 + {PS}^2 + {QS}^2 + {QS}^2 - {PQ}^2 \]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( {PT}^2 + {TS}^2 \right) + {QS}^2 + \left( \frac{QR}{2} \right)^2 - \left( {PT}^2 + {QT}^2 \right) \left( \text{From} \left( 1 \right), \left( 2 \right) \text{and} \left( 4 \right) \right)\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {PT}^2 + {TS}^2 + {QS}^2 - {PT}^2 - {QT}^2 \]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 + \left( {TS}^2 - {QT}^2 \right)\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 + \left( TS + QT \right)\left( TS - QT \right)\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 + \left( QS \right)\left( TS - QT \right)\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 + QS \times TS - QS \times QT\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + QS \times TS + {QS}^2 - QS \times QT\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + QS \times TS + QS\left( QS - QT \right)\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + QS \times TS + QS \times TS\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + 2QS \times TS\]

\[ \Rightarrow {PR}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + QR \times TS\]

Hence ,

\[{PR}^2 = {PS}^2 + QR \times ST + \left( \frac{QR}{2} \right)^2\]

Now,

\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2 \left( \text{by Apollonius theorem} \right)\]

\[ \Rightarrow {PQ}^2 = 2 {PS}^2 + 2 {QS}^2 - {PR}^2 \]

\[ \Rightarrow {PQ}^2 = {PS}^2 + {PS}^2 + {QS}^2 + {QS}^2 - {PR}^2 \]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( {PT}^2 + {TS}^2 \right) + {QS}^2 + \left( \frac{QR}{2} \right)^2 - \left( {PT}^2 + {RT}^2 \right) \left( \text{From} \left( 2 \right), \left( 3 \right) \text{and} \left( 4 \right) \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {PT}^2 + {TS}^2 + {QS}^2 - {PT}^2 - {RT}^2 \]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 - \left( {RT}^2 - {TS}^2 \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 - \left( TS + RT \right)\left( RT - TS \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 - \left( TS + RT \right)\left( RS \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 - \left( TS + RT \right)\left( QS \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 + {QS}^2 - QS \times TS - QS \times RT\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - QS \times TS + {QS}^2 - QS \times RT\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - QS \times TS - QS\left( RT - QS \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - QS \times TS - QS\left( RT - SR \right)\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - QS \times TS - QS \times TS\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - 2QS \times TS\]

\[ \Rightarrow {PQ}^2 = {PS}^2 + \left( \frac{QR}{2} \right)^2 - QR \times TS\]

\[{PQ}^2 = {PS}^2 - QR \times ST + \left( \frac{QR}{2} \right)^2\]
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 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Practice Set 2.2 | Q: 3 | Page no. 43
Solution In the Given Figure Seg Ps is the Median of ∆Pqr and Pt ⊥ Qr. Prove That, P R 2 = P S 2 + Q R × S T + ( Q R 2 ) 2 P Q 2 = P S 2 − Q R × S T + ( Q R 2 ) 2 Concept: Apollonius Theorem.
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