Consider a triangle PQR in which the relation | vedclass

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(C) In $	riangle PQR$,let the vertices be $P, Q, R$. Let $M$ be the midpoint of $QR$ and $N$ be the midpoint of $PR$. The medians $PM$ and $QN$ inter

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a right angle

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(C) In $	riangle PQR$,let the vertices be $P, Q, R$. Let $M$ be the midpoint of $QR$ and $N$ be the midpoint of $PR$. The medians $PM$ and $QN$ intersect at the centroid $G$.Using the property of medians,$QG = \frac{2}{3}QN$ and $GM = \frac{1}{3}PM$.In $	riangle QGM$,by the Pythagorean theorem,if $\angle QGM = 90^{\circ}$,then $QG^2 + GM^2 = QM^2$.We know $QM = \frac{1}{2}QR$,so $QM^2 = \frac{1}{4}QR^2$.Using Apollonius theorem for medians:$QN^2 = \frac{2PQ^2 + 2QR^2 - PR^2}{4}$ and $PM^2 = \frac{2PQ^2 + 2PR^2 - QR^2}{4}$.Then $QG^2 + GM^2 = \left(\frac{2}{3}QNight)^2 + \left(\frac{1}{3}PMight)^2 = \frac{4}{9}QN^2 + \frac{1}{9}PM^2$.Substituting the expressions for $QN^2$ and $PM^2$:$= \frac{4}{9} \left(\frac{2PQ^2 + 2QR^2 - PR^2}{4}ight) + \frac{1}{9} \left(\frac{2PQ^2 + 2PR^2 - QR^2}{4}ight)$$= \frac{1}{9} \left( \frac{8PQ^2 + 8QR^2 - 4PR^2 + 2PQ^2 + 2PR^2 - QR^2}{4} ight)$$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 2PR^2}{4} ight)$.Given $PR^2 = 5PQ^2 - QR^2$,substitute this:$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 2(5PQ^2 - QR^2)}{4} ight)$$= \frac{1}{9} \left( \frac{10PQ^2 + 7QR^2 - 10PQ^2 + 2QR^2}{4} ight) = \frac{1}{9} \left( \frac{9QR^2}{4} ight) = \frac{1}{4}QR^2 = QM^2$.Since $QG^2 + GM^2 = QM^2$,$	riangle QGM$ is a right-angled triangle with $\angle QGM = 90^{\circ}$.

Consider a $\triangle PQR$ in which the relation $QR^2 + PR^2 = 5PQ^2$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then,$\angle QGM$ is always

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