# 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

Sum

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

Given: PS is the median.

⇒ QS = SR = "QR"/2

In right triangle PTR

PR2 = PT2 + TR2

= PT2 + (ST + SR)2

= PT2 + ("ST" + "QR"/2)^2

= "PT"^2 + ("ST"^2 + ("QR")/2)^2 + 2  "ST" xx ("QR"/2)

= ("PT"^2 + "ST"^2) + "QR" xx "ST" + ("QR"/2)^2

= "PS"^2 + "QR" xx "ST" + ("QR"/2)^2

In right triangle PTQ

PQ2 = PT2 + TQ2

= PT2 + (QS - ST)2

= PT2 + ("QR"/2 - "ST")^2

= PT2 + ("QR"/2)^2 - 2 xx "QR"/2 xx "ST" + "ST"^2

= (PT2 + ST2) - "QR" xx "ST" + ("QR"/2)^2

= "PS"^2 - "QR" xx "ST" + ("QR"/2)^2

Concept: Apollonius Theorem
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#### APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Practice Set 2.2 | Q 3 | Page 43