KVPY SA Exam 2018 Question Paper

Question : 8 of 80

Marks: +1, -0

Consider a triangle PQR in which the relation

QR^2 + PR^2 = 5 PQ^2

hold. Let G be the point of intersection of medians PM and QN. Then ∠ QGM is always

Solution:

Let QR= p, PR = q, PQ =r
Given

p ^2 +q^2 = 5r^2

Now

QG ^2+GM^2=({2QN}/3)^2+({PM}/2)^2

=4{QN^2}/9+{PM}^2/9

=1/9[4 . 1/4(2r^2+2p^2-q^2)+1/4(2r^2+2q^2-p^2)

=p^2/4=QM^2

Hence Angle QGM is 90º.

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