In a regular quadrangular pyramid, the height is 2, the side edge is 5. Find its volume.

At the base of a regular quadrangular pyramid lies a square, the height of the pyramid from its top drops to the center of the base, which is the intersection point of the square’s diagonals. Consider a right-angled triangle formed by legs – the height of the pyramid and half the diagonal of the base, and the hypotenuse – by the side edge. The sum of the squares of the legs is equal to the square of the hypotenuse, hence the half of the diagonal of the base of the pyramid can be found as the square root of the difference between the squares of the side edge and the height:
d / 2 = √ (5 ^ 2-2 ^ 2) = √ (25-4) = √21;
d = 2√21.
The area of ​​the base of the pyramid is equal to half the square of its diagonal: Sbase = d ^ 2/2 = 4 * 21/2 = 2 * 21 = 42.
The volume of a regular quadrangular pyramid is equal to a third of the product of the base area by the height:
V = 1/3 * Sosn * h = 42 * 2/3 = 28.