And ∵ quadrilateral ABCD is a diamond, ∴AC⊥BD, ∵BD is on the curved surface ABCD,
∴BD⊥ surface ACFE
∴BD⊥EG,BD⊥FG
∵EG and FG are on BDE and BDF respectively, and BD is the intersection of BDE and BDF.
∴∠EGF is the included angle between BDE and BDF, that is, dihedral angle.
∵BD⊥EG, in RT△BEG, EG? = Yes? -BG？ ，BG= 1/2BD= 1，BE？ =3
∴EG？ =2
∫ The quadrilateral ABCD is a diamond.
RT△ABG, ∴ in AG? =AB？ -BG？ =AB？ -( 1/2BD)？ =3
∵ Quadrilateral ACFE is a parallelogram.
∴EF？ =AC？ =(2AG)？ = 12
∵FG？ = 10
∴EF？ = 12=FG？ +EG？
∴△EFG is RT△, ∠ EGF = 90,
* EGF is the angle between BDE and BDF.
∴ BDE⊥ BDF
(2) Do FO⊥AC at point F and cross the extension line of AC at point O. ..
Then F0 is the height of the parallelogram ACFE.
∫AC(AO)∑EF
∴∠EFG=∠FGC(∠FGO)
∴sin∠efg=sin∠fgc(sin∠fgo)=eg/fe=fo/fg
∵EG？ =2，EF？ = 12，FG？ = 10
∴FO= fifteen under one third of the root sign.
S parallelogram ACFE=EF*FO= twice the root sign of the next five.
∫ In diamond ABCD, DG=BG.
∴V polygon FEABCD=2V quadrangular cone B-ACFE=2* 1/3*S parallelogram ACFE*BG= three quarters of the root sign, and the next five.