The height of a regular triangular pyramid is 3, and the lateral edge SB is inclined to the base

| February 10, 2021 | education

The height of a regular triangular pyramid is 3, and the lateral edge SB is inclined to the base plane at an angle of 45 degrees. Find the edge length of the base.

Since the pyramid is correct, its height passes through the intersection point of the heights of the equilateral triangle ABC. Then the triangle BОС is rectangular, and since, by condition, the angle СBО = 450, then the triangle BОС is isosceles, BО = СO = 3 cm.

In the ABC triangle, the heights, at the intersection point, are divided in the ratio of 2/1, starting from the top, then BO / НO = 2 / 1. BO = НO / 2 = 3/2 cm.

ВН = ВO + HO = 3 + 3/2 = 9/2 = 4.5 cm

BH is the height of an equilateral triangle, then BH = AC * √3 / 2.

AC = 2 * ВН / √3 = (2 * 9/2) / √3 = 9 / √3 = 3 * √3 cm.

Answer: The length of the edge of the base is 3 * √3 cm.



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