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

In a regular triangular pyramid SABC Q is the midpoint of the edge AB, S is the vertex. It is known that BC = 7, and the area of the side surface of the pyramid is 42. Find SQ.

Since the pyramid is regular, all its lateral faces are equal, then Sasb = Sside / 3 = 42/3 = 14 cm2.

At the base of the pyramid is an equilateral triangle, then AO = BC = AC = 7 cm.

Triangle ASB is isosceles, SA = SB, then the segment SQ is the height and median of the triangle.

Sаsb = AB * SQ / 2.

SQ = 2 * Sasb / AB = 2 * 14/7 = 4 cm.

Answer: The height of the segment SQ is 4 cm.