(2014?合肥三模)如图,多面体ABCDEF中,面ABCD为边长为a的菱形,且∠DAB=60°,DF=2BE=2a,DF∥BE,DF

2025-05-16 21:46:25
推荐回答(1个)
回答1:

(Ⅰ)当点G是AF中点时,EG∥平面ABCD.
取AD中点H,连接GH,GE,BH,则
∵GH∥DF,GH=

1
2
DF,
∴GH∥BE且GH=BE,
∴四边形BEGH为平行四边形,
∴EG∥BH,
∵BH?平面ABCD,EG?平面ABCD,
∴EG∥平面ABCD;
(Ⅱ)连接BD,由V=VA-BDFE+V C-BDFE=2VA-BDFE=2?
1
3
?
1
2
(a+2a)?a?
3
2
a=
3
2
a3