Find the lengths of the medians of the triang | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $C(6,0,0)$.

Similar Questions

The incenter and centroid of the triangle,whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$,are respectively given by

Let $ABCD$ be a parallelogram and $E$ be the mid-point of $AB$. If $P$ is the point of intersection of $DE$ and $AC$,then $\frac{DP}{PE} + \frac{AP}{PC} = $

The points $A(-1, 2, 3)$,$B(2, -3, 1)$,and $C(3, 1, -2)$:

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

The points $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$ form a

Vedclass Test Series

Exam Paper Generator

Online Exam Module