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.