# In a regular quadrangular pyramid, the height is 4√3, and the dihedral angle

**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.