The lateral edges of a regular quadrangular pyramid are 5 and are inclined to the base plane

The lateral edges of a regular quadrangular pyramid are 5 and are inclined to the base plane at an angle whose sine is 0.6. Find the volume of the pyramid.

In the right-angled triangle of the ВOМ, we determine the lengths of the legs OB and OM.

SinB = OM / BM.

OM = BM * SinB = 5 * 0.6 = 3 cm.

Then, OB ^ 2 = BM ^ 2 – OM ^ 2 = 25 – 9 = 16.

OB = 4 cm.

Since there is a square at the base of the pyramid, its diagonals are equal and at the point of intersection they are divided in half, then the ВOС triangle is rectangular and equilateral.

СВ ^ 2 = OB ^ 2 + OS ^ 2 = 16 + 16 = 32.

СВ = √32 = 4 * √2 cm.

The area of the base of the pyramid is: Sbn = CB^2 = 32 cm2. Then V = Sosn * ОМ / 3 = 32 * 3/3 = 32 cm3.

Answer: The volume of the pyramid is 32 cm3.