In the regular quadrangular pyramid SABCD, point O is the center of the base

In the regular quadrangular pyramid SABCD, point O is the center of the base, S is the apex, SO = 8, SB = 10. Find the length of the line segment BD.

At the base of a regular quadrangular pyramid lies a square, the center of the base O is the intersection point of the diagonals of this square, dividing them in half. Consider a right-angled triangle SOB. SO and OB – legs, SB – hypotenuse. The sum of the squares of the legs is equal to the square of the hypotenuse: SO ^ 2 + OB ^ 2 = SB ^ 2, hence
OB ^ 2 = SB ^ 2-SO ^ 2 = 10 ^ 2-8 ^ 2 = 100-64 = 36;
OB = √36 = 6.
The OB segment is equal to half of the BD segment, therefore BD = 2 * OB = 2 * 6 = 12.