The side face of the quadrangular pyramid is inclined to the base plane at an angle of 30 degrees, and the apothem is 4. Find the area of the side pyramid.
The side faces of a regular pyramid are isosceles triangles. Then the apothem MH is the height and median of the lateral face of MAD. Linear angle MHO is equal to the dihedral angle between the side face and the plane of the base.
Angle МHО = 30. Then in the right-angled triangle MOH we define the length of the leg ОН.
Cos30 = OH / MH.
OH = MH * Cos30 = 4 * √3 / 2 = 2 * √3 cm.
The segment OH is the middle line of the triangle ABD, then AD = AB = 2 * OH = 2 * 2 * √3 = 4 * √3 cm.
Determine the area of the lateral face of the MAB. Smav = AB * MH / 2 = 4 * √3 * 4/2 = 8 * √3 cm.
Then S side = 4 * Smav = 4 * 8 * √3 = 32 * √3 cm2.
Answer: The lateral surface area is 32 * √3 cm2.
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