Find the volume of a regular tetrahedron with edge a

The tetrahedron has side faces and a base of equilateral triangles.

In triangle ABC, all interior angles are 60.

Then Sop = AB * AC * Sin60 / 2 = a * a * √3 / 4 = a ^ 2 * √3 / 4 cm2.

Also Sosn = AC * BH / 2.

BH = 2 * Sosn / AC = 2 * (a ^ 2 * √3 / 4) / a = a * √3 / 2 cm.

The height of the BH is also the median, then the point O divides it in a ratio of 2/1.

ОВ = 2 * ВН / 3 = 2 * (a * √3 / 2) / 3 = a * √3 / 3 cm.

In a right-angled triangle BOD, according to the Pythagorean theorem, OD ^ 2 = BD ^ 2 – OB ^ 2 = a ^ 2 – a ^ 2/3 = a ^ 2 * 2/3.

OD = a * √2 / √3 = a * √6 / 3.

Then V = Sax * OD / 3 = (a ^ 2 * √3 / 4) * (a * √6 / 3) / 3 = a ^ 3 * √2 / 12 cm3.

Answer: The volume of the tetrahedron is a3 * √2 / 12 cm3.