In a regular quadrangular pyramid, the height is 4√3, and the dihedral angle at the base is 60 degrees. find the total surface area of the pyramid.
The dihedral angle at the base is equal to the linear angle МHО = 60.
In a right-angled triangle MOH, tg60 = MO / OH.
OH = MO / tg60 = 4 * √3 / √3 = 4 cm.
Angle OMH = (90 – 60) = 30, then the length of the OH leg is equal to half the length of the hypotenuse MH, then MH = 2 * OH = 2 * 4 = 8 cm.
The OH segment is the middle line of the triangle ABC, then AB = 2 * OH = 2 * 4 = 8 cm.
Determine the area of the triangle MBC.
Smvs = MH * BC / 2 = 8 * 8/2 = 32 cm2.
Since the side faces of the pyramid are equal in size, then Sside = 4 * Smvs = 4 * 32 = 128 cm2.
Sbn = AB ^ 2 = 82 = 64 cm2.
Then Sпов = Sсн + Sbok = 64 + 128 = 192 cm2.
Answer: The surface area of the pyramid is 19 cm2.
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