Question

Consider a $\Delta PQR$ in which the relation $Q{R}^2+P{R}^2=5P{Q}^2$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then, $\angle QGM$ is always

• A. less than $45^{\circ }$
• B. obtuse
• C. a right angle
• D. acute and larger than $45^{\circ }$

Answer 1

Answer: (C)

In $\Delta PQR$
Given, $Q{R}^2+P{R}^2=5P{Q}^2$
Median $PM$ and $QN$ intersect of $G$

$QG=\frac{2}{3}QN,GM=\frac{1}{3}PM$
$\Rightarrow Q{G}^2+G{M}^2=(\frac{2}{3}QN{)}^2)+(\frac{1}{3}PM{)}^2$
$=\frac{4}{9}Q{N}^2+\frac{1}{9}P{M}^2$
$=\frac{4}{9}\left(\frac{2P{Q}^2+2Q{R}^2-P{R}^2}{4}\right)+\frac{1}{9}\left(\frac{2P{Q}^2+2P{R}^2-Q{R}^2}{4}\right)$
$=\frac{1}{9}\left[\frac{8P{Q}^2+8Q{R}^2-4P{R}^2+2P{Q}^2+2P{R}^2-Q{R}^2}{4}\right]$
$=\frac{1}{9}\left[\frac{10P{Q}^2+7Q{R}^2-2P{R}^2}{4}\right]$
$=\frac{1}{9}\left[\frac{2\left(5P{Q}^2-P{R}^2\right)+7Q{R}^2}{4}\right]$
$=\frac{1}{9}\left[\frac{2Q{R}^2+7Q{R}^2}{4}\right]$
$=\frac{1}{4}Q{R}^2=Q{M}^2$
$\therefore O{G}^2+G{M}^2=Q{M}^2$
$\therefore \angle QGM=90^{\circ }$