摘要：在长方体ABCD-A1B1C1D1中.底面是边长为2的正方形.高为4.则点A1到截面AB1D1的距离是( ) A. B. C. D. 解析:如图.设A1C1∩B1D1=O1.∵B1D1⊥A1O1.B1D1⊥AA1.∴B1D1⊥平面AA1O1.故平面AA1O1⊥AB1D1.交线为AO1.在面AA1O1内过A1作A1H⊥AO1于H.则易知A1H长即是点A1到平面AB1D1的距离.在Rt△A1O1A中.A1O1=.AO1=3.由A1O1·A1A=h·AO1,可得A1H=. 答案:C

网址：http://m.1010jiajiao.com/timu3_id_510032[举报]