Question

Consider a $\Delta PQR$ in which the relation $QR^2 + PR^2 = 5 PQ^2$ 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, $QR^2 + PR^2 = 5PQ^2$
Median $PM$ and $QN$ intersect of $G$

$QG = \frac{2}{3} QN, GM = \frac{1}{3} PM$
$\Rightarrow QG^2 + GM^2 = (\frac{2}{3} QN)^2) + (\frac{1}{3} PM)^2$
$ = \frac{4}{9} QN^2 + \frac{1}{9} PM^2$
$= \frac{4}{9}\left(\frac{2PQ^{2} + 2QR^{2}-PR^{2}}{4} \right)+\frac{1}{9}\left(\frac{2PQ^{2}+2PR^{2}-QR^{2}}{4}\right) $
$= \frac{1}{9}\left[\frac{8PQ^{2}+8QR^{2} - 4PR^{2}+2PQ^{2} + 2PR^{2} - QR^{2}}{4}\right]$
$= \frac{1}{9}\left[\frac{10PQ^{2} + 7QR^{2} - 2PR^{2}}{4}\right]$
$= \frac{1}{9}\left[\frac{2\left(5PQ^{2} - PR^{2}\right) + 7QR^{2}}{4}\right] $
$= \frac{1}{9}\left[\frac{2QR^{2}+7QR^{2}}{4}\right]$
$ = \frac{1}{4}QR^{2} = QM^{2} $
$ \therefore OG^{2} + GM^{2} = QM^{2} $
$ \therefore \angle QGM = 90^{\circ}$