The height of a regular triangular pyramid is h, and the dihedral angle at the base side is 45 °

| May 27, 2021 | education

The height of a regular triangular pyramid is h, and the dihedral angle at the base side is 45 °. Find the surface area of the pyramid.

Since the pyramid is regular, the ABC triangle is equilateral.

The triangle DOH is rectangular and isosceles, since the angle DHO = 45, then OH = DO = h cm.

In an equilateral triangle, the intersection of heights is the center of the inscribed circle. Then OH = R = BC / 2 * √3 = h.

BC = 2 * h * √3 cm.

Determine the area of ​​the base.

Sb = ВС ^ 2 * √3 / 4 = (2 * h * √3) ^ 2 * √3 / 4 = h ^ 2 * 3 * √3 cm2.

From the right-angled triangle DOH, we determine the length of the hypotenuse DH.

DH ^ 2 = DO ^ 2 + HO ^ 2 = h ^ 2 + h ^ 2 = 2 * h ^ 2.

DН = h * √2 cm.

Determine the area of ​​the side face of the pyramid.

Sdvs = ВС * DH / 2 = 2 * h * √3 * h * √2 / 2 = h ^ 2 * √6 cm2.

Side = 3 * Sdvs = 3 * h ^ 2 * √6 cm2.

Then Spol = S main + S side = h ^ 2 * 3 * √3 + 3 * h ^ 2 * √6 = h ^ 2 * 3 * √3 * (1 + √2) cm2.

Answer: The total surface area is h ^ 2 * 3 * √3 * (1 + √2) cm2.



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