# The height of a regular triangular pyramid is 2√6 cm. Its side faces are inclined to the base

**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.