The height of a regular quadrangular pyramid is √6, and the lateral rib is inclined at an angle of 60

| September 1, 2021 | education

The height of a regular quadrangular pyramid is √6, and the lateral rib is inclined at an angle of 60 degrees to the base plane, find the lateral rib and the area of the lateral pyramid

Consider a right-angled triangle MOA in which the leg MO = √6 cm, and the MAO angle = 60.

Then the hypotenuse is SinMAO = MO / MA.

Sin60 = √6 / MA.

MA = √6 / (√3 / 2) = 2 * √6 / √3 = 2 * √18 / 3 = 2 * √2 cm.

Angle AMO = 180 – 90 – 60 = 30. Then the leg AO lies opposite angle 30, and is equal to half the length of AM.

AO = MA / 2 = 2 * √2 / 2 = √2 cm.

The diagonal of the base is: AC = AO * 2 = 2 * √2 cm.

Then the side of the base will be equal to: AB = BC = CD = AD = AC / √2 = 2 * √2 / √2 = 2 cm.

Let’s define the apothem of the pyramid.

MH ^ 2 = MO ^ 2 + OH ^ 2 = (√6) ^ 2 + 1 ^ 2 = 7.

MH = √7 cm.

Let us determine the lateral surface area through the semiperimeter and apothem.

p = (AB + BC + CD + AD) / 2 = 8/2 = 4 cm.

Sside = p * MH = 4 * √7 cm2.

Answer: The area of ​​the lateral surface of the pyramid is 4 * √7 cm2.



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