Question

已知正△ABC的邊長為3,D、E分別是邊BC上的三等分點(diǎn)〔如圖(1)所示〕.沿AD、AE把△ABC折成三棱錐A—DEF,使B、C兩點(diǎn)重合于點(diǎn)F〔如圖(2)〕,且G是DE的中點(diǎn).

(1)求證:DE⊥平面AGF;

(2)求二面角A-DE-F的大小;

(3)求點(diǎn)F到平面ADE的距離.

Answer 1

(1)證明:在△ADE中,AD=AE,G為DE中點(diǎn),∴AG⊥DE.?

同理,在△FDE中,FG⊥DE.?

又∵FG∩AG=G,∴DE⊥面AGF.?

(2)解析：由(1)知同∠AGF即為所求二面角的平面角.AF=3,AG=,DG=,FD=1,FG=.?

cos∠AGF=.?

∴∠AGF=π-arccos.?

(3)解析：過F作FH⊥AG于H.?

∵DE⊥面AGF,DE面ADE,?

∴面AFG⊥面ADE.?

又∵交線為AG,?

∴FH⊥面ADE,AF=3,FG=,AG=.?

設(shè)HG=x,則FH²=AF²-AH²=FG²-GH²,即9-(+x)²=-x² x=.?

FH=.