Given a regular triangular pyramid of the side of the base of which is 34 cm and the area of the lateral surface

| August 5, 2021 | education

Given a regular triangular pyramid of the side of the base of which is 34 cm and the area of the lateral surface is twice the area of the base. find the height of the pyramid.

In an equilateral triangle ABC, draw the height AH and determine its length.

AH = a * √3 / 2, where a is the length of the side of the triangle. AH = 34 * √3 / 2 = 17 * √3 cm.

Determine the area of ​​the base of the pyramid. Sbn = ВС * АН / 2 = 34 * 17 * √3 / 2 = 289 * √3 cm2.

Then, by condition, S side = 2 * Sb = 2 * 289 * √3 = 578 * √3 cm2.

Then the area of ​​the side face is equal to: Svsd = Sside / 3 = 578 * √3 / 3 cm2.

Svsd = BC * DH / 2, then DH = 2 * Svsd / BC = 2 * (578 * √3 / 3) / 34 = 34 * √3 / 3 cm.

Since the triangle is regular, OH = AH / 3 = 17 * √3 / 3.

Then from a right-angled triangle DOH, DO ^ 2 = DH ^ 2 – OH ^ 2 = (34 * √3 / 3) ^ 2 – (17 * √3 / 3) ^ 2 = 1156/3 – 289/3 = 867 / 3 = 289.

DO = 17 cm.

Answer: The height of the pyramid is 17 cm.



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