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

Advertisement Remove all ads
Sum

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

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

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

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
  Is there an error in this question or solution?

APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Practice Set 2.2 | Q 3 | Page 43
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×