# 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.

#### 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.

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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 1 | Page 43