An equilateral triangle with side A rotates around the height. Find the surface area of a body of revolution.

In an equilateral triangle, the height is also the median. The body obtained by rotating the triangle around its height will be a cone.
The radius of the base of the cone will be equal to half of the side of the triangle: R = A / 2.
The generator is equal to the side of the triangle: l = A.
We calculate the area of the lateral surface of the cone:
Sside = pRl = p * A / 2 * A = pA² / 2.
Let’s calculate the area of the base (the base is a circle):
Sb = nR² = n * (A / 2) ² = nA² / 4.
We calculate the total surface area of the cone:
Spp = Sbok + Sbn = pA² / 2 + pA² / 4 = 2pA² / 4 + pA² / 4 = 3p * A² / 4