In a regular quadrangular pyramid, the side of the base is 12, the apothem of the pyramid is 10.

In a regular quadrangular pyramid, the side of the base is 12, the apothem of the pyramid is 10. Find the volume of the inscribed sphere.

A ball inscribed in a regular pyramid touches it in the center O, the intersection of the diagonals of the base and apothems of the pyramid.

The section of the pyramid according to the apothem KSH is an isosceles triangle with lateral sides equal to the apothems, SK = SH = 10 cm and the base KH = AD = 12 cm.

The section of a ball is a circle inscribed in an isosceles triangle.

Let us define the area of ​​a triangle KSH by Heron’s theorem. The half-perimeter of the triangle is: p = (10 + 10 + 12) / 2 = 16 cm.

Then Sksh = √16 * (16 – 10) * (16 – 10) * (16 – 12) = √2304 = 48 cm2.

Then R = S / p = 48/16 = 3 cm.

Let’s define the volume of the ball.

V = 4 * n * R3 / 3 = 4 * n * 27/3 = 36 * n cm3.

Answer: The volume of the ball is 36 * n cm3.