The base side of a regular quadrangular pyramid is 3 cm, and each side face is inclined to the base plane at an angle

| July 2, 2021 | education

The base side of a regular quadrangular pyramid is 3 cm, and each side face is inclined to the base plane at an angle of 60 degrees, find: a) The volume of the pyramid b) the surface area and volume of the inscribed sphere.

The side faces of the pyramid are isosceles triangles. Segment ЕН – height, bisector and median of triangle DCE. The OH segment is the middle line of the ACD triangle, then OH = AD / 2 = 3/2 = 1.5 cm.

The OHE triangle is rectangular, in which the OH leg lies opposite the angle 30, then EH = 2 * OH = 2 * 1.5 = 3 cm.

Then the triangle MEH is equilateral with a side of 3 cm. OE is the height of the pyramid and an equilateral triangle, OE = MH * √3 / 2 = 3 * √3 / 2 cm.

Let’s define the volume of the pyramid. V = Sbas * OE / 3 = 9 * 3 * √3 / 6 = 4.5 * √3 cm2.

Let’s define the radius of the inscribed sphere as the radius of the circle inscribed in the equilateral triangle МHЕ.

R = MH * √3 / 6 = 3 * √3 / 6 = √3 / 2 cm.

Then Vball = 4 * π * R ^ 3/3 = 4 * π * 3 * √3 / 24 = π * √3 / 2 cm3.

Sball = 4 * π * R ^ 2 = 4 * π * 3/4 ​​= π * 3 cm2.

Answer: The volume of the pyramid is 4.5 * √3 cm2, the area of ​​the ball is π * 3 cm2, the volume of the ball is π * √3 / 2 cm3.



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