The diagonal AC of the base of the regular quadrangular pyramid SABCD is 6.

The diagonal AC of the base of the regular quadrangular pyramid SABCD is 6. The lateral edge SB is 5. Find the height of the pyramid SO.

Since a regular quadrangular pyramid SABCD is given, there is a square at its base, and equal triangles are its side faces. Consider an isosceles triangle SAC obtained in a diagonal section. From the condition of the problem it is known that the diagonal AC of the base is 6, the lateral edge SB = 5, then SA = 5. The height of the SO triangle will simultaneously be the height of the pyramid, we find it from the right-angled triangle ASO by the Pythagorean theorem: SA² = SO² + AO², where AO = AC / 2 = 6/2 = 3. We get the equation: 5² = SO² + 3²; SO = 4.
Answer: The height of the pyramid is 4.