Given is a regular triangular pyramid with a base side of 6 cm and an angle of inclination of the lateral

Given is a regular triangular pyramid with a base side of 6 cm and an angle of inclination of the lateral rib to the plane of the base is 45 degrees, and the dihedral angle at the base edge is 60 degrees. Find the volume and total surface.

In a regular triangular pyramid ABCS, consider an equilateral right-angled triangle SOK, where O is the point of projection of the apex onto the base, and K is the midpoint of the base side. Since the angle of inclination of the lateral rib to the plane of the base is 45 °, the height SO is equal to the radius of the circle inscribed in the base.

SO = r = 6 / (2√3) cm.

The volume of the pyramid is equal to one third of the product of the area of ​​the base and the height.

Let’s find the area of ​​an equilateral triangle with a side of 6 cm:

So = √3 / 4 * 6 * 6 = 9√3 cm2.

Let’s find the volume of the pyramid:

Vp = 1/3 * √3 / 4 * 6 * 6 * 6 / (2√3) = 36 cm3.

Knowing the apothem and the side of the base of the pyramid, you can find the lateral surface area.

Find the apothem by the Pythagorean theorem from the right-angled triangle SKO.

(6 / (2√3)) ^ 2 + (6 / (2√3)) ^ 2 = 2 * (6 / (2√3)) ^ 2 = 2 * 36 * / 4 * 3 = 6 cm.

Let’s find the area of ​​the lateral surface of the pyramid.

Sbp = 1/2 (6 + 6 + 6) * 6 = 18 * 3 = 54 cm2.

Let’s find the total surface area of ​​the pyramid.

Sп = 54 + 9√3 cm2.

Answer: the volume of the pyramid is 36 cubic centimeters, the total surface area is 54 + 9√3 square centimeters.