In a regular triangular pyramid SABC, the R is the midpoint of the edge AB. S is the vertex.

In a regular triangular pyramid SABC, the R is the midpoint of the edge AB. S is the vertex. It is known that BC = 7, SR = 6 Find the lateral surface area

The side faces of a regular triangular pyramid are isosceles triangles. Since point R is the midpoint of the side of the base AB, it is the median of triangle SAB, and since it is isosceles, then it is also its height.

Then the area of the triangle SAB will be equal to:

Ssa = AB * SR / 2 = 7 * 6/2 = 21 cm2.

The side faces of the pyramid are equal triangles, therefore their areas are equal, then Side = 3 * Ssa = 3 * 21 = 63 cm2.

Answer: The area of the side surface of the pyramid is 63 cm2.