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

In a regular triangular pyramid SABC, point-P is the midpoint of the edge AB, S is the vertex. It is known BC = 5, and SP = 6. Find the area of the side surface of the pyramid.

In a regular triangular pyramid, an equilateral triangle lies at the base. Then AB = BC = SD = 5 cm.

The side faces of the pyramid are equilateral triangles. Point P is the middle of AB, then SP is the median of the triangle SAB, and since it is isosceles, then the height of the triangle.

Then Ssab = AB * SB / 2 = 5 * 6/2 = 15 cm2.

Since all the side faces of the pyramid are equal triangles, then Sside = 3 * Ssab = 3 * 15 = 45 cm2.

Answer: The lateral surface area is 45 cm2.