In a regular quadrangular pyramid SABCD, point O is the center of the base. SO = 10.BD = 48. Find the side rib SA.

Given: SABCD is a regular quadrangular pyramid,

Point O is the center of the base,

SO = 10,

BD = 48.

Find: SA.

Solution: since the pyramid is correct, so at its base there is a square and SO perpendicular to the ABSD plane (SO is perpendicular to AC and BD) and is the height of the pyramid, and point O is the point of intersection of its diagonals (AC = BD).

Means: ASO – right-angled triangle (where SOA angle = 90 °),

then:

SA ^ 2 = AO ^ 2 + SO ^ 2,

AO = AC / 2 = 48/2 = 12,

SA ^ 2 = 122 + 102,

SA = √ (144 + 100),

SA = √244.

Answer: SA = √244.