Find the lengths of the medians of the triang | vedclass

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Question

(A) Let $AD, BE,$ and $CF$ be the medians of the given triangle $ABC$ with vertices $A(0,0,6)$,$B(0,4,0)$,and $C(6,0,0)$.Since $AD$ is the median,$D$

Vedclass · Accepted Answer

$7, \sqrt{34}, 7$

Vedclass · Answer

(A) Let $AD, BE,$ and $CF$ be the medians of the given triangle $ABC$ with vertices $A(0,0,6)$,$B(0,4,0)$,and $C(6,0,0)$.Since $AD$ is the median,$D$ is the mid-point of $BC$.Coordinates of point $D = \left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right) = (3,2,0)$.$AD = \sqrt{(0-3)^{2} + (0-2)^{2} + (6-0)^{2}} = \sqrt{9+4+36} = \sqrt{49} = 7$.Since $BE$ is the median,$E$ is the mid-point of $AC$.Coordinates of point $E = \left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right) = (3,0,3)$.$BE = \sqrt{(3-0)^{2} + (0-4)^{2} + (3-0)^{2}} = \sqrt{9+16+9} = \sqrt{34}$.Since $CF$ is the median,$F$ is the mid-point of $AB$.Coordinates of point $F = \left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right) = (0,2,3)$.$CF = \sqrt{(6-0)^{2} + (0-2)^{2} + (0-3)^{2}} = \sqrt{36+4+9} = \sqrt{49} = 7$.Thus,the lengths of the medians of $\Delta ABC$ are $7, \sqrt{34},$ and $7$.

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

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