The base side of a regular quadrangular pyramid is 16, and its side edge is 10. Find the area of the entire

The base side of a regular quadrangular pyramid is 16, and its side edge is 10. Find the area of the entire surface of the pyramid.

SABCD is a regular quadrangular pyramid, then at its base lies the square ABCD (AB = BC = CD = AD = 16), all side edges are equal (SA = SB = SC = SD = 10), which means that all side faces are the same isosceles triangles with a side length of 10 and a base of 16.
The total surface area of ​​the pyramid is:
S = Sb. + S side.
1. Find the area of ​​the base. The base area will be the area of ​​square ABCD. The area of ​​the square is:
S = a²,
where a is the length of the side of the square.
Sosn. = AB² = 16² = 256.
2. The lateral surface area of ​​a regular pyramid is equal to:
S side. = Psn. * H / 2,
where Psn. – base perimeter, h – apothem.
2.1. The perimeter of the square is:
P = 4a;
Rosn. = 4AB = 4 * 16 = 64.
2.2. Consider △ ASD: SA = SD = 10, AD = 16.
Let’s draw the height SH. Since △ ASD is isosceles, then SH is both the height and the median: AH = DH = AD / 2 = 16/2 = 8.
In △ AHS, by the Pythagorean theorem, we find the length SH:
SH = √ (SA² – AH²) = √ (10² – 8²) = √ (100 – 64) = √36 = 6.
3. Find the total surface area of ​​the SABCD pyramid:
S = 256 + 64 * 6/2 = 256 + 64 * 3 = 256 + 192 = 448.
Answer: S = 448.