Question

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

Answer 1

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.