# At the base of the triangular pyramid mavs lies a regular triangle abc with a side equal to √2, ma = √2.

**At the base of the triangular pyramid mavs lies a regular triangle abc with a side equal to √2, ma = √2. the side faces of the pyramid have equal areas. find the volume of the pyramid.**

Since a regular triangle lies at the base of the pyramid, AB = BC = AC = √2 cm.

The area of a regular triangle is determined by the formula:

Sop = a ^ 2 * √3 / 4, where a is the side length of a regular triangle.

Sbasn = AB ^ 2 * √3 / 4 = (√2) ^ 2 * √3 / 4 = √3 / 2 cm2.

Let’s omit the height in the triangle ABC.

Since there is a regular triangle at the base of the pyramid, the height BH is also the median of the triangle, which means AH = CH = √2 / 2.

By the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2 = (√2) ^ 2 – (√2 / 2) ^ 2 = 2 – 1/2 = 3/2. AB = √3 / 2.

According to the properties of a regular triangle, the height at the point of their intersection is divided by a ratio of 2/1 starting from the top. Then BО = (√3 / 2) * 2/3 = 2 * √3 / 3 * √2 = 2 / √6.

Consider a right-angled triangle MOB. Let us define, according to the Pythagorean theorem, the leg MO, which is the height of the pyramid.

MO ^ 2 = MB ^ 2 – BO ^ 2 = (√2) ^ 2 – (2 / √6) ^ 2 = 2 – 2/3 = 4/3.

MO = 2 / √3.

Then the volume of the pyramid will be equal to:

V = Sbase * MO / 3 = (√3 / 2 * 2 / √3) / 3 = 1/3 cm3.

Answer: The volume of the pyramid is 1/3 cm3.