In the title: ∵SA⊥ Aircraft ABC, and ∵AB⊥BC.∴SA, AB and BC are perpendicular to each other. At this time, B can be used as the parallel line BD of SA (D (D is the intersection of parallel line and circular surface). BD, AB and BC are perpendicular to each other. We can think that BD, AB and BC are three vertical sides of a cuboid, and the surface of this sphere is its circumscribed circle. According to the nature of a cuboid, the diameter of the circumscribed circle is the diagonal of the cuboid. ∴ The diameter of the circumscribed circle D = (AB*AB+BC*BC+BD*BD) 0.5 (i.e. the root number (AB*AB+BC*BC+BD*BD)).
It is easy to know that the quadrilateral ABDS is a cuboid. So SA=DB= 1, that is, BD=SA= 1, AB= 1, BC = 2 0.5 (that is, the root number 2). It can be deduced that the diameter of the circumscribed circle is d=2. So the radius of the ball r= 1. We can get the surface area O =4*π*r*r=4*π.