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

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

Since point P is the midpoint of side AB, segment SP is the median of triangle SAB.

The SAB triangle is isosceles, since the pyramid is regular, then SP is the height of the SAB triangle.

The sides of the ABC triangle are equal, since the base is an equilateral triangle.

Then Ssa = AB * SP / 2 = 5 * 6/2 = 15 cm2.

The side faces of the pyramid are equal triangles, then S side = 3 * Ss = 3 * 15 = 45 cm2.

Answer: The area of the side surface of the pyramid is 45 cm2.