In a regular quadrangular pyramid SABCD, point O-center of the base, S-vertex, SO = 24, AC = 14. Find the lateral edge SD.

At the base of a regular quadrangular pyramid lies the square ABCD. The diagonals of the square are halved at the intersection. So AO = OC. In addition, the diagonals of the square are equal, that is, AC = BD. And, based on the first rule, – BO = OD. Find the length OD:

OD = 14/2 = 7.

Consider a triangle SOD. It is rectangular with SD hypotenuse. OD = 7, SO = 24. By the Pythagorean theorem:

SD = √ (OD² + SO²).

Let’s find the length of the side edge we need:

SD = √ (7² + 24²) = √ (49 + 576) = √625 = 25.

Answer: The length of the side edge SD of a regular quadrangular pyramid SABCD is 25.