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2.   Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

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Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now,

Let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

 \left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=(0,2,3)

Vertices of E =

\left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=(3,2,0)

    Vertices of the F =

\left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=(3,0,3)

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

CD=\sqrt{}CD=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7

AE=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt{9+4+36}=\sqrt{49}=7

BF=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}

Hence lengths of the median are 7,7 \:and\:\sqrt{34}.

Posted by

Pankaj Sanodiya

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