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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.
- GO
^{2}= GB^{2}- OB^{2}= 8^{2}- OB^{2} - 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 GO
^{2}= 8^{2}- OB^{2}= 8^{2}- [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.