Advertisement Remove all ads

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR. - Geometry

Sum

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Advertisement Remove all ads

Solution

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

\[QS = SR = \frac{1}{2}QR\]

\[{PQ}^2 + {PR}^2 = 2 {PS}^2 + 2 {QS}^2\] .......…[Apollonius theorem]

\[ \Rightarrow {11}^2 + {17}^2 = 2 \left( 13 \right)^2 + 2 {QS}^2 \]

\[ \Rightarrow 121 + 289 = 2\left( 169 \right) + 2 {QS}^2 \]

\[ \Rightarrow 410 = 338 + 2 {QS}^2 \]

\[ \Rightarrow 2 {QS}^2 = 410 - 338\]

\[ \Rightarrow 2 {QS}^2 = 72\]

\[ \Rightarrow {QS}^2 = 36\]

\[ \Rightarrow QS = 6\]

\[ \therefore QR = 2 \times QS\]

\[ = 2 \times 6\]

\[ = 12\]

Hence, QR = 12.

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Practice Set 2.2 | Q 1 | Page 43
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×