The lateral edge of a regular quadrangular pyramid, equal to 12 cm, forms an angle of 60 degrees

| June 23, 2021 | education

The lateral edge of a regular quadrangular pyramid, equal to 12 cm, forms an angle of 60 degrees with the base plane. Find the volume.

The side edges of the pyramid are equal, then the BMD triangle is isosceles, MB = MD, and since the MBD angle = 60, the BMD triangle is equilateral, BD = MB = 12 cm.

Since ABCD is a square, then OB = OD = BD / 2 = 12/2 = 6 cm.

In a right-angled triangle MBO, according to the Pythagorean theorem, OM ^ 2 = BM ^ 2 – OB ^ 2 = 144 – 36 = 108.

ОМ = 6 * √3 cm.

The ABD triangle is rectangular and isosceles, AB = AD, then 2 * AB ^ 2 = BD ^ 2.

AB ^ 2 = BD ^ 2/2 = 144/2.

AB = 12 / √2 = 6 * √2 cm.

Determine the area of ​​the base of the pyramid.

Sbn = AB ^ 2 = (6 * √2) ^ 2 = 72 cm2.

Let’s define the volume of the pyramid.

V = Ssc * ОМ / 3 = 72 * 6 * √3 / 3 = 144 * √3 cm3.

Answer: The volume of the pyramid is 144 * √3 cm2.



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