連結BD交EF于O,
EF∥AC,AC⊥BD ,EF⊥BD
BB1⊥ABCD,BB1⊥EF
EF⊥BB1O
∠BOB1為AC交BD于P,
BO/BP=1/2
BP=√2 * a/2
BOtg∠B0B1=BB1/BO=a/(√2 * a/4)=2√2
二面角B1-EF-B的平面角的正切值為2√2
SB-B1EF=1/3 * EF * BO/2 * BB1
=[1/3] * [(√2)*a/2] * [√2 * a/4]/2 * a
=(a^3)/24