Let s-abcd be a pyramid whose side area is twice the area of its bottom, whose height so is equal to 3, and whose sphere is the whole area of the pyramid

Let s-abcd be a pyramid whose side area is twice the area of its bottom, whose height so is equal to 3, and whose sphere is the whole area of the pyramid

Let the side length of the bottom square be a and the slant height be B
Then so & # 178; + (A / 2) &# 178; = B & # 178;
Because so = 3, B & # - A & # / 4 = 9 (*)
Moreover, the area of s side = 2S bottom, and the area of s side = 4 * (1 / 2) * AB = 2Ab, the area of s bottom = A & # 178;
Then 2Ab = 2A & # 178; that is, a = B
By substituting (*):
b²-b²/4=9
b²=12
The solution is b = a = 2 √ 3
So s bottom surface = A & # 178; = 12, s side area = 2S, bottom surface = 24
Then the total area of the pyramid = s bottom surface + s side area = 12 + 24 = 36