The base of the pyramid is a square with a side of 12 s. The height of the pyramid is 8 cm. Find the surface area of the pyramid.

The pyramid SABCD, at the base is the square ABCD, then AB = BC = CD = AD = 12 cm, SO = 8 cm. The base of the height SO is located at the intersection of the diagonals of the square, in the center of the inscribed and circumscribed circles. Since the diagonals of the square are halved by the intersection point, then AO = OS.
1. Find the length of the diagonal AC by the Pythagorean theorem from the triangle ADC:
AC = √ (AD ^ 2 + CD ^ 2);
AC = √ (12 ^ 2 + 12 ^ 2) = √ (144 + 144) = √2 * 144 = 12√2 (cm).
Then:
AO = OC = AC / 2 = 12√2 / 2 = 6√2 (cm).
2. By the Pythagorean theorem from the triangle AOS, we find the length AS:
AS = √ (AO ^ 2 + SO ^ 2);
AS = √ ((6√2) ^ 2 + 8 ^ 2) = √ (72 + 64) = √136 (cm).
3. Using Heron’s formula, we find the area of ​​the side face (ASD):
S = (p – b) * √p (p – a),
where p is the semi-perimeter, a is the base, b is the side.
Semi-perimeter:
p = (12 + √136 + √136) / 2 = (12 + 2√136) / 2 = 6 + √136.
S = (6 + √136 – √136) * √ (6 + √136) (6 + √136 – 12) = 6 * √ (6 + √136) (√136 – 6) = 6 * √ (136 – 36) = 6 * √100 = 6 * 10 = 60 (cm ^ 2).
4. Find the area ABCD:
S = a ^ 2,
where a is the side length of the square.
S = 12 ^ 2 = 144.
5. The total surface area of ​​the SABCD pyramid is:
Sп = Sside + Sbn;
Sп = 4 * 60 + 144 = 240 + 144 = 348 (cm ^ 2).
Answer: Sп = 348 cm ^ 2.