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.
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