题目内容
如图a,在直角梯形ABCD中,AB⊥AD,AD∥BC,F为AD的中点,E在BC上,且EF∥AB.已知AB=AD=CE=2,沿线EF把四边形CDFE折起如图b,使平面CDFE⊥平面ABEF.
(1)求证:AB⊥平面BCE;
(2)求三棱锥C ADE体积.
(1)见解析 (2)
解析(1)证明:在题图a中,EF∥AB,AB⊥AD,
∴EF⊥AD,在题图b中,CE⊥EF,又平面CDFE⊥平面ABEF,且平面CDFE∩平面ABEF=EF,
CE⊥平面ABEF,AB?平面ABEF,∴CE⊥AB,又∵AB⊥BE,BE∩CE=E,∴AB⊥平面BCE;
(2)解:∵平面CDFE⊥平面ABEF,且平面CDFE∩平面ABEF=EF,AF⊥FE,AF?平面ABEF,∴AF⊥平面CDEF,∴AF为三棱锥A CDE的高,且AF=1,又∵AB=CE=2,∴S△CDE=×2×2=2,
∴VC ADE=·S△CDE·AF=×2×1=.
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