The lateral edge of a regular quadrangular pyramid of length 2 is inclined to the base

The lateral edge of a regular quadrangular pyramid of length 2 is inclined to the base plane at an angle of 30 degrees. Find the volume of the pyramid.

The volume of the pyramid is found by the formula:
V = Sh / 3,
where S is the area of ​​the base, h is the height of the pyramid.
By condition, the angle between the side edge and the base is 30 degrees. Since the pyramid is correct, the angles SAC, SBD, SCA and SDB are 30 degrees.
From the top of the pyramid S draw the height SO to the base of the pyramid. Since the pyramid is correct, there is a square at its base. SO falls in the center of the intersection of the square’s diagonals (then O divides AC and BD in half).
1. Consider a triangle SOC: SOC = 90 degrees, SCO = SCA = 30 degrees, SC = 2 – hypotenuse. Opposite the angle of 30 degrees, there is a leg, which is 2 times less than the hypotenuse, then:
SO = SC / 2 = 2/2 = 1.
So the height of the pyramid is 1.
2. By the Pythagorean theorem:
OS = √ (SC ^ 2 – SO ^ 2) = √ (2 ^ 2 – 1 ^ 2) = √ (4 – 1) = √3.
3. Since O divides the AC in half, then:
AO = OS = AC / 2;
AC = 2OC = 2√3.
4. Consider a triangle ADC: AD = DC = x – legs, AC = 2√3 – hypotenuse.
By the Pythagorean theorem:
AD ^ 2 + DC ^ 2 = AC ^ 2;
x ^ 2 + x ^ 2 = (2√3) ^ 2;
2x ^ 2 = 12;
x ^ 2 = 12/2;
x ^ 2 = 6;
x = √6.
Then AD = DC = x = √6.
5. The base area is equal to:
S = a ^ 2 = AD ^ 2 = (√6) ^ 2 = 6.
6. Find the volume:
V = Sh / 3 = 6 * 1/3 = 2.
Answer: V = 2.