If the lateral edge of a regular tetrahedron is 8 cm, then it's altitude is?

The altitude is 6.53 cm

Step-by-step explanation:

A regular tetrahedron is a triangular pyramid having equilateral triangular faces, therefore we have;

Sides of the equilateral triangle = 8 cm

Given that the slant height, the edge of the tetrahedron, and half the base edge of the tetrahedron form a right triangle, we have;

The slant height, h = √(8² - (8/2)²) = √48 = 4×√3

The segment representing the altitude, H, of the tetrahedron forms a right triangle with the edge of the tetrahedron and 2/3×h

Therefore;

8² = H² + (2/3×4×√3)²

H² = 8² - (2/3×4×√3)²

H² = 64 - 64/3 = 128/3

The altitude H = √(128/3) = √6×8/3 = 6.53 cm.

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