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

| July 23, 2021 | education

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.



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