# A regular triangular pyramid ABCD is given. Vertex A is removed from the straight line ВС at a distance of √3

**A regular triangular pyramid ABCD is given. Vertex A is removed from the straight line ВС at a distance of √3, and from the base plane BCD at a distance of √2. Find the volume of the pyramid ABCD.**

Since the pyramid is regular, the BCD triangle is equilateral, then the height AO passes through the point of intersection of heights, which at point O are divided in the ratio 2/1.

DO = 2 * HO.

From the right-angled triangle AOH, by the Pythagorean theorem, we determine the length of the leg OH.

OH ^ 2 = AH ^ 2 – AO ^ 2 = 3 – 2 = 1.

OH = 1 cm.Then DO = 2 * 1 = 2 cm, and DH = DO + HO = 2 + 1 = 3 cm.

We define the side of the triangle ВСD through the height DH.

DН = ВС * √3 / 2.

ВС = 2 * DH / √3 = 2 * 3 / √3 = 6 / √3 = 2 * √3 cm.

Determine the area of the base of the pyramid.

Sbn = ВС * DH / 2 = 2 * √3 * 3/2 = 3 * √3 cm2.

Let’s define the volume of the pyramid.

V = S main * AO / 3 = 3 * √3 * √2 / 3 = √6 cm3.

Answer: The volume of the pyramid is √6 cm3.