The height of the cone is 2√3 cm. Find the area of the axial section of the cone, if it is a regular triangle.

The axial section of the cone is a triangle ABC, which, by condition, is equilateral, then AC = BC = AB.

Let the side length of an equilateral triangle be 2 * X cm, then AC = BC = AB = 2 * X.

The height of the OS of the cone is the height, median and bisector of the triangle ABC, then AO = BO = AB / 2 = 2 * X / 2 = X cm.

In a right-angled triangle AOB, according to the Pythagorean theorem, AC ^ 2 = AO ^ 2 + OC ^ 2.

4 * X ^ 2 = 12 + X ^ 2.

3 * X ^ 2 = 12.

X ^ 2 = 12/3 = 4.

X = OB = OA = 2 cm.

Then AO = 2 * 2 = 4 cm.

Determine the area of the axial section.

Ssection = AB * OC / 2 = 4 * 2 * √3 / 2 = 4 * √3 cm.

Answer: The axial section area is 4 * √3 cm2.