Seg Pm is a Median of ∆Pqr. If Pq = 40, Pr = 42 and Pm = 29, Find Qr. - Geometry Mathematics 2

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

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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.

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Chapter 2: Pythagoras Theorem - Problem Set 2 [Page 44]

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Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Problem Set 2 | Q 17 | Page 44

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