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Culculate the altitude of a tetrahedron

Question:

In the tetrahedron G-ABC, GA = GB = GC = 8 and AB = BC = AC = 12. What is the altitude of the tetrahedron GO?

Solution:

The altitude GO is the line segment drawing from the vertex G and perpendicular to the plane ABC at point O.

Because GA = GB = GC, so the perpendicular point O is the center of the base triangle ABC.

Because GO is perpendicular to the plane ABC, so triangle GOB is a right triangle.

GO² = GB² - OB² = 8² - OB²

Because CA = CB = 12, so triangle CAB is an isosceles triangle.

When CE is perpendicular to AB, then CE bisects its vertex angle C and bisects the side AB in half. (This is the property of an isosceles triangle).

So AE = EB = (1/2)AB = (1/2) × 12 = 6

Same reason, the altitude BD bisects the angle B and bisects the side CA in half.

Because the triangle ABC has three equal sides, so its three interior angles are equal, that is,

angle A = angle B = angle C = 60^o

Becalse BD is an angle bisector, so angle EBO = angle FBO = (1/2)angle B = 30^o

In Right triangle EBO, EB/OB = cos30^o

So OB = EB/cos30^o = EB/sin60^o = 6 ÷ (square root of 3)/2 = 12/(square root of 3)

So GO² = 8² - OB² = 8² - [12/(square root of 3)]² = 64 - 12 × 12/3 = 64 - 48 = 16

So Go = 4

Therefore, the altitude of the tetrahedron is 4.

If you are interested in watching the video for the solution of this problem, please click the video solution of the tetrahedron problem.