(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
时间:2024-12-09 21:17:50
答案

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

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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?
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?
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(a+2a)?a?

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