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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    The vertices of a triangle are \[(6,0),(0,6)\] and \[(6,6)\]. The distance between its circumventer and centroid is

    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                                 \[=\sqrt{{{(4-3)}^{2}}+{{(4-3)}^{2}}}=\sqrt{2}\]


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