Find the lengths of the medians of the triangle with vertices A(0,…

Question 11 of 13intermediate🔧 ApplyLong Answer4 marks

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

Correct Answer

The lengths of the medians are: $AD = 7$ units, $BE = \sqrt{34}$ units, and $CF = 7$ units.

Exercise: Miscellaneous Exercise on Chapter 11 | Q: 2 | (Page Page 8)

For More Understanding

Explanation

This question tests the understanding of medians in 3D geometry. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

Students need to first find the midpoints of all three sides using the midpoint formula, then calculate the distances from each vertex to the opposite midpoint using the distance formula in 3D.

Solution Steps

Step 1: Find the midpoint of $BC$ . Let $D$ be the midpoint of $BC$ where $B(0, 4, 0)$ and $C(6, 0, 0)$ .
$D = \left(\frac{0+6}{2},\ \frac{4+0}{2},\ \frac{0+0}{2}\right) = (3,\ 2,\ 0)$
Step 2: Find the length of median $AD$ from $A(0, 0, 6)$ to $D(3, 2, 0)$ .
$AD = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = \mathbf{7 \text{ units}}$
Step 3: Find the midpoint of $AC$ . Let $E$ be the midpoint of $AC$ where $A(0, 0, 6)$ and $C(6, 0, 0)$ .
$E = \left(\frac{0+6}{2},\ \frac{0+0}{2},\ \frac{6+0}{2}\right) = (3,\ 0,\ 3)$
Step 4: Find the length of median $BE$ from $B(0, 4, 0)$ to $E(3, 0, 3)$ .
$BE = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{9 + 16 + 9} = \sqrt{34} = \mathbf{\sqrt{34} \text{ units}}$
Step 5: Find the midpoint of $AB$ . Let $F$ be the midpoint of $AB$ where $A(0, 0, 6)$ and $B(0, 4, 0)$ .
$F = \left(\frac{0+0}{2},\ \frac{0+4}{2},\ \frac{6+0}{2}\right) = (0,\ 2,\ 3)$
Step 6: Find the length of median $CF$ from $C(6, 0, 0)$ to $F(0, 2, 3)$ .
$CF = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = \mathbf{7 \text{ units}}$

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