The height of a regular triangular pyramid is 2√6 cm. Its side faces are inclined to the base plane at an angle of 45 degrees. Calculate the area of the side surface of the pyramid.
Consider a right-angled triangle HOD, in which the angle O is straight, and the angle H = 45, then the angle D = 45, and therefore, the triangle HOD is rectangular and isosceles, HO = DO = 2 * √6 cm. Apotheme DH = OH / Cos45 = (2 * √6) / √2 / 2 = 4 * √3 cm.
Since at the base there is a regular triangle, the height of CH is divided by the exact O into segments CO and HO, which relate as 2/1, then CH = 3 * 2 * √6 = 6 * √6 cm.
The height of a regular triangle is: CH = AB * √3 / 2.
AB = 2 * CH / √3 = 2 * 6 * √6 / √3 = 12 * √2 cm.
Determine the area of the lateral rib ABD. Savd = AB * DH / 2 = 12 * √2 * 4 * √3 / 2 = 24 * √6 cm2.
Then the lateral surface area is equal to:
Side = 3 * Savd = 3 * 24 * √6 = 72 * √6 cm2.
Answer: The lateral surface area is 24 * √6 cm2.
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