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

Asked by Pragya Singh | 1 year ago |  144

##### Solution :-

For the given triangle ABC. Let AD, BE and CF are the medians We know that, median divides the line segment into two equal parts, so D is the midpoint of BC, therefore,

Coordinates of point D = $$\dfrac{0+6}{2}, \dfrac{4+0}{2}, \dfrac{0+0}{2}$$

Coordinates of point D = (3,2,0)

AD =$$\sqrt{0– 3)^2 + (0 – 2)^2 + (6– 0)^2}$$

AD = $$\sqrt{9+4+36}$$

AD = $$\sqrt{49}$$

Similarly, E is the midpoint of AC,

Coordinates of point E = $$\dfrac{0+6}{2}, \dfrac{0+0}{2}, \dfrac{0+6}{2}$$

Coordinates of point E = (3,0,3)

AC = $$\sqrt{3– 0)^2 + (0 – 4)^2 + (3– 0)^2}$$

AC = $$\sqrt{9+16+9}$$

AC = $$\sqrt{34}$$

Similarly, F is the midpoint of AB,

Coordinates of point F =$$\dfrac{0+0}{2}, \dfrac{0+4}{2}, \dfrac{6+0}{2}$$

Coordinates of point F = (0,2,3)

CF =$$\sqrt{6– 0)^2 + (0 – 2)^2 + (0– 3)^2}$$

CF = $$\sqrt{36+4+9}$$

CF = $$\sqrt{49}$$

CF = 7

Therefore, the lengths of the medians of the triangle ABC we obtain are,
7, $$\sqrt{34}$$, 7

Answered by Abhisek | 1 year ago

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