Find the volume of a regular quadrangular pyramid, if the side of the base is m = 5 cm

Find the volume of a regular quadrangular pyramid, if the side of the base is m = 5 cm, the flat angle at the apex is α = 60 °.

The base of a regular quadrangular pyramid is a square, the side edges are equal to each other, which means that the side faces are isosceles triangles. The planar angle at the top of the pyramid is the angle between the side edges that belong to the same side face. Since a flat coal at the apex is 60 °, the side faces are equilateral triangles, therefore, the side edge is equal to the side of the base:

l = m = 5 cm – lateral rib.

Area of ​​a square with a side equal to 5 cm:

Sb = 5 ^ 2 = 25 cm2.

The diagonal of this square is:

d ^ 2 = m ^ 2 + m ^ 2 = 25 + 25 = 50;

d = 5√2 cm.

The diagonals of the square at the intersection point are halved.

From a right-angled triangle, consisting of the height of the pyramid, half of the diagonal of the base and the side edge, we can find the height:

h ^ 2 = l ^ 2 – (d / 2) ^ 2 = 52 – (25 * 2/4) = 25 – 25/2 = 25/2;

h = 5 / √2 cm.

The volume of the pyramid is equal to one third of the product of the base area by the height:

V = Sbn * h / 3 = 25 * (5 / √2) / 3 = 125 / 3√2 ≈ 29.46 cm3.