question_answer

The vertices of triangle ABC be \[\mathbf{A}\left( \mathbf{0},\mathbf{0},\mathbf{6} \right),\mathbf{B}\left( \mathbf{0},\mathbf{4},\mathbf{0} \right)\]and \[\left( \mathbf{6},\mathbf{0},\mathbf{9} \right)\] then length of medians of triangles be:

A) \[6,6,6\]                                   

B) \[6,\sqrt{34},6\]     

C) \[7,\sqrt{34},7\]             

D) None of these

Correct Answer: C

Solution :

[c] \[D\equiv \left( 3,2,0 \right)\]     \[E\equiv \left( 3,0,3 \right)\]             \[AD=\sqrt{{{(3)}^{2}}+{{(2)}^{2}}+{{(6)}^{2}}}=\sqrt{49}=7\] \[F=(0,2,3)BE=\sqrt{{{3}^{2}}+{{4}^{2}}+{{3}^{2}}}=\sqrt{34}\]             \[CF=\sqrt{{{6}^{2}}+{{2}^{2}}+{{3}^{2}}}=7\] Thus. length medians be \[(7,\sqrt{34},7).\] Hence, option [c] is correct