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 `angleQGM` is always

Options

A. less than `45^(@)`
B. obtuse
C. a right angle
D. acute and larger than `45^(@)`

Answer

C

Explanation

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)+2p^(2)-q^(2))]`
`(p^(2))/4=QM^(2)`
Hence Angle QGM is `90^(@)`.