In a regular quadrangular pyramid, the height is 7 dm, and the lateral rib is 9 dm. Find the perimeter of the base of the pyramid.

Let’s draw the diagonals AC and BD at the base of the pyramid. In a right-angled triangle KOC, according to the Pythagorean theorem, determine the length of the leg OS.

OC ^ 2 = KC ^ 2 – KO ^ 2 = 81 – 49 = 32.

OC = √32 = 4 * √2 dm.

Since at the base of the pyramid is a square, then its diagonals, at the point of intersection, are divided in half and intersect at right angles, then the triangle COD is rectangular, with a right angle at point O, and OD = OC = 4 * √2 dm.

By the Pythagorean theorem, CD ^ 2 = OD ^ 2 = OC ^ 2 = 3 ^ 2 + 3 ^ 2 = 64.

CD = 8 dm.

Then the perimeter of the base is: Rosn = 4 * CD = 4 * 8 = 32 dm.

Answer: The perimeter of the base is 32 dm.