# The lateral edge of a regular quadrangular pyramid forms an angle of 45 degrees with its height.

**The lateral edge of a regular quadrangular pyramid forms an angle of 45 degrees with its height. Calculate the volume of the pyramid if the length of the lateral edge is 2 cm.**

Consider a right-angled triangle MOB in which one of the acute angles is 45, then this triangle is also isosceles. OB = OM, then MB ^ 2 = OB ^ 2 + OM ^ 2 = 2 * OB ^ 2.

2 * OB ^ 2 = 4.

ОМ ^ 2 = 4/2 = 2.

ОМ = ОВ = √2 cm.

Diagonal ВD is divided in half at point O, then ВD = 2 * ОВ = 2 * √2 cm.

In a right-angled triangle ABD AD = AB as the sides of a square, then 2 * AB ^ 2 = BD ^ 2 = 8.

AB ^ 2 = 8/2 = 4.

AB = 2 cm.

Determine the area of the base of the pyramid. Sbn = AB ^ 2 = 2 ^ 2 = 4 cm2.

Let’s define the volume of the pyramid. V = Sbasn * ОМ / 3 = 4 * √2 / 3 cm3.

Answer: The volume of the pyramid is 4 * √2 / 3 cm3.