Find the lengths of the medians of the triang | vedclass

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(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 $7$.

Column $I$	Column $II$
$(A)$ The centroid of the triangle formed by $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$ is	$(p)$ $(2, 2, 2)$
$(B)$ The circumcentre of the triangle formed by $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$ is	$(q)$ $(3, 1, 4)$
$(C)$ The orthocentre of the triangle formed by $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$ is	$(r)$ $(1, 1, 0)$
$(D)$ The incentre of the triangle formed by $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$ is	$(s)$ $(3, 2, 1)$

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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