設(shè)棱長(zhǎng)為a=2的正四棱錐A-BCDE與正三棱錐A-CDF,取AC中點(diǎn)G,連BG,DG,FG.易知AC⊥BG,AC⊥DG,AC⊥FG,
∴∠BGD、∠DGF分別是二面角B-AC-D、D-AC-F的平面角.
BG=DG=FG=√3.BD=2√2.
cosBGD=(BG^2+DG^2-BD^2)/(2BG*DG)=-1/3,
cosDGF=(DG^2+FG^2-DF^2)/(2DG*FG)=1/3.
∴∠BGD+∠DGF=180°,
∴A,B,C,F四點(diǎn)共面；
同理,A,E,D,F四點(diǎn)共面.
∴多面體ABCDEF是五面體.