解∶∵SA⊥AB,SA⊥AC,又∵AB,AC均在平面AC内,∴SA⊥平面AC,在正方形ABCD中,AB=BC=2√2,∵AC＝√﹙AB＋BC﹚＝4,设S到平面AC的距离为d,由勾股定理得d＋AC＝sc,∴d＝√﹙SC－AC﹚＝√﹙5－4﹚＝3,∴S到平面AC的距离为3