Hushar Mulga
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Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 17 | Page 44
Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR

solution

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

\[QM = MR = \frac{1}{2}QR\]

\[{PQ}^2 + {PR}^2 = 2 {PM}^2 + 2 {QM}^2 \left( \text{by Apollonius theorem} \right)\]
\[ \Rightarrow \left( 40 \right)^2 + \left( 42 \right)^2 = 2 \left( 29 \right)^2 + 2 {QM}^2 \]
\[ \Rightarrow 1600 + 1764 = 1682 + 2 {QM}^2 \]
\[ \Rightarrow 3364 - 1682 = 2 {QM}^2 \]
\[ \Rightarrow 1682 = 2 {QM}^2 \]
\[ \Rightarrow {QM}^2 = 841\]
\[ \Rightarrow QM = 29\]
\[ \Rightarrow QR = 2 \times 29\]
\[ \Rightarrow QR = 58\]

Hence, QR = 58.

Explanation:- 

In ∆PQR, point M is the midpoint of side QR. QM = MR = (1/2)QR

By Apollonius theorem, PQ^2 + PR^2 = 2PM^2 + 2QM^2

=> (40)^2 + (42)^2 = 2(29)^2 + 2QM^2 => 1600 + 1764 = 1682 + 2QM^2 => 1682 = 2QM^2 => QM^2 = 841 => QM = 29 => QR = 2×29 => QR = 58

Hence, QR = 58.

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 17 | Page 46