# The angle between the height of a regular quadrangular pyramid and its side edge is 60 °

**The angle between the height of a regular quadrangular pyramid and its side edge is 60 ° find the total surface area of the pyramid if its height is 10 cm.**

Consider a right-angled triangle AOK, in which the angle O is straight, A – 60, and the angle K = 180 – 90 – 60 = 30. Then the leg AO will be equal to AO = KO / tg60 = 10 / √3.

Since there is a square at the base of the pyramid, AB = AC / √2 = (2 * AO) / √2 = 20 / √6.

Sbn = AB ^ 2 = (20 / √6) 2 = 200/3 = cm2.

Apothem KM will be equal to KM = √ (KO ^ 2 + OM ^ 2) = √ (100 + 100/6) = 10 * √ (7/6).

Let us find the area of the lateral surface, which is equal to half the product of the apothem and the perimeter of the base.

Side = (4 * AB * KM) / 2 = (4 * 20 / √6 * 10 * √ (7/6) / 2 = 200 * √7 / 6 = = 200 * √7 / 3.

Then the total area of the pyramid is: S = 200/3 + 200 * √7 / 3 = 200 * (1/3 + √7 / 3) = 200 * (1 + √7) / 3.

Answer: S = 200 * (1 + √7) / 3.