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Miscellaneous Exercise · Q2

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

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✓ Free question

A median runs from a vertex to the midpoint of the opposite side. Using the 3D midpoint and distance formulas, the three medians have lengths 77, 34\sqrt{34} and 77.

Why this works

A median joins a vertex to the midpoint of the opposite side. So for each median we do exactly two things: find the midpoint of the opposite side, then measure the distance from the vertex to that midpoint. The vertices lie in 3D space, but the method is identical to the 2D case — every point simply carries a third coordinate.

Midpoint of (x1,y1,z1)(x_1,y_1,z_1) and (x2,y2,z2)(x_2,y_2,z_2): (x1+x22, y1+y22, z1+z22)\left( \tfrac{x_1+x_2}{2},\ \tfrac{y_1+y_2}{2},\ \tfrac{z_1+z_2}{2} \right).

Distance between them: (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.

Step-by-step solution

The vertices are A(0,0,6)A(0,0,6), B(0,4,0)B(0,4,0) and C(6,0,0)C(6,0,0).

Median from AA to side BCBC. Midpoint of BCBC:

MBC=(0+62, 4+02, 0+02)=(3, 2, 0).M_{BC} = \left( \tfrac{0+6}{2},\ \tfrac{4+0}{2},\ \tfrac{0+0}{2} \right) = (3,\,2,\,0).

AMBC=(3−0)2+(2−0)2+(0−6)2=9+4+36=49=7.AM_{BC} = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7.

Median from BB to side ACAC. Midpoint of ACAC:

MAC=(0+62, 0+02, 6+02)=(3, 0, 3).M_{AC} = \left( \tfrac{0+6}{2},\ \tfrac{0+0}{2},\ \tfrac{6+0}{2} \right) = (3,\,0,\,3).

BMAC=(3−0)2+(0−4)2+(3−0)2=9+16+9=34.BM_{AC} = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{9 + 16 + 9} = \sqrt{34}.

Median from CC to side ABAB. Midpoint of ABAB:

MAB=(0+02, 0+42, 6+02)=(0, 2, 3).M_{AB} = \left( \tfrac{0+0}{2},\ \tfrac{0+4}{2},\ \tfrac{6+0}{2} \right) = (0,\,2,\,3).

CMAB=(0−6)2+(2−0)2+(3−0)2=36+4+9=49=7.CM_{AB} = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7.

Tip

Two medians came out equal (77 each) — a hint that the triangle is isosceles.

✓Final answer

The lengths of the medians are 77, 34\sqrt{34} and 77.

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