In a regular triangular pyramid SABC, Q is the midpoint of the edge BC. S is the top. It is known that AB = 4

In a regular triangular pyramid SABC, Q is the midpoint of the edge BC. S is the top. It is known that AB = 4. lateral surface area = 72. Find the length of the segment SQ.

Since point Q is the midpoint of the edge BC, the segment SQ is the median of the side face SBC. Since the side face of a regular pyramid is an equilateral triangle, the median SQ is also its height.

The side faces of the regular pyramid are equal, then Sside = 3 * Ssвс.

Ssvs = 72/3 = 24 cm2.

The area of the side face SBC is equal to: Ssвс = BC * SQ / 2 = 24.

Since the triangle ABC is equilateral, then BC = AB = 4 cm.

SQ = 2 * 24/4 = 12 cm.

Answer: The length of the segment SQ is 12 cm