已知斜三棱柱ABC-A1B1C1,侧面A1ACC1与底面垂直,角ABC=90度,BC=2,AC=2√3,且AA1⊥A1C,AA1=A1C.

题目

已知斜三棱柱ABC-A1B1C1,侧面A1ACC1与底面垂直,角ABC=90度,BC=2,AC=2√3,且AA1⊥A1C,AA1=A1C.
求顶点C到侧面A1ABB1的距离

答案

解法一：由点C作平面A1ABB1的垂线,垂足为H则CH的长是顶点C到平面A1ABB1的距离连接HB,由于AB⊥BC,得AB⊥HB又A1E⊥AB知HB∥A1E,且BC∥ED∴∠HBC＝∠A1ED＝60°∴CH＝BC•sin60°＝√3为所求解法二：连接A1B根据定...