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Question
Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).
Solution
The midpoint of the line segment joining the points B(0, 4, 0) and C(6, 0, 0) is D `((0 + 6)/2, (4 + 0)/2, (0 + 0)/ 2)` or (3, 2, 0).
The coordinates of point A are (0, 0, 6).
Length of median AD of triangle ABC
= `sqrt((3 - 0)^2 + (2 - 0)^2 + (0 - 6)^2)`
= `sqrt(9 + 4 + 36)`
= `sqrt49`
= 7
The coordinates of C and A are (6, 0, 0) and (0, 0, 6)
The mid-point of AC is E `((0 + 6)/2, (0 + 0)/2, (0 + 6)/2)` or E (3, 0, 3)
And coordinates of B are (0, 4, 0).
Length of median BE of triangle ABC
= `sqrt((3 - 0)^2 + (0 - 4)^2 + (3 - 0)^2)`
= `sqrt(9 + 16 + 9)`
= `sqrt34`
The coordinates of points A and B are (0, 0, 6), (0, 4, 0) respectively.
∴ The midpoint of AB is F`((0 + 0)/2, (0 + 4)/2, (6 + 0)/2)` or F (0, 2, 3).
Length of median CF of triangle ABC
= `sqrt((6 - 0)^2 + (0 - 2)^2 + (0 + 3)^2)`
= `sqrt(36 + 4 + 9)`
= `sqrt49`
= 7
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