A) \[2\]
B) \[\sqrt{2}\]
C) \[1\]
D) \[2\sqrt{2}\]
Correct Answer: B
Solution :
Let the vertices of a triangle be \[A(6,0),\] \[B(0,6)\] and \[C(6,6).\] Now, \[AB=\sqrt{{{6}^{2}}+{{6}^{2}}}=6\sqrt{2}\] \[BC=\sqrt{{{6}^{2}}+0}=6\] and \[CA=\sqrt{0+{{6}^{2}}}=6\] Also, \[A{{B}^{2}}=B{{C}^{2}}+C{{A}^{2}}\] Therefore, \[\Delta ABC\] is right angled at. C. So, mid point of AB is the circumcentre of \[\Delta ABC\]. \[\therefore \] Coordinate of circumcentre are \[(3,3)\]. Coordinate of centroid are, \[G\left( \frac{6+0+6}{3},\frac{0+6+6}{3} \right),\] ie, \[(4,4)\] \[\therefore \] Required distance \[=\sqrt{{{(4-3)}^{2}}+{{(4-3)}^{2}}}=\sqrt{2}\]You need to login to perform this action.
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