Given a regular triangular pyramid SABC, at the base lies a regular triangle ABC. AP is the median of the base

| February 26, 2021 | education

Given a regular triangular pyramid SABC, at the base lies a regular triangle ABC. AP is the median of the base, O is the point of intersection of the medians, SO is the height of the pyramid. AP = 6, SO = √5. Find the area of the triangle SBC.

Since the pyramid is correct, AB = BC = AC = SA = SB = SC.

Let’s use the formula for the median length in a regular triangle:

AR = CB * √3 / 2.

СВ = AP / (√3 / 2) = 6 / (√3 / 2) = 12 / √3 = 4 * √3.

Consider a right-angled triangle SOP, in which the leg SO = √5, and the leg OP is equal to the third part of the AP by the property of the medians, which are divided at the point of intersection in the ratio 2/1, starting from the vertex.

OP = AP / 2 = 6/3 = 2 cm.

Then SP ^ 2 = SO ^ 2 + OP ^ 2 = (√5) ^ 2 + 2 ^ 2 = 5 + 4 = 9.

SP = 3 cm.

Determine the area of the triangle CSB

Scsb = CB * SP / 2 = 4 * √3 * 3/2 = 6 * √3 cm2.

Answer: Scsb = 6 * √3 cm2.



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