又∵在直角△ECN中,由勾股定理得到:NE2=CN2+CE2, ∴(BE+DN)2=CN2+CE2,即(2√33+DN)2=(6-DN)2+(6-2√33)2, 解得DN=12-6√33. ∴S△AEN=S□ABCD-S△ABE-S△ECN-S△ADN, =6×6-1212×6×2√33-1212×(6-2√33)×(6-DN)-1212×6×DN, ...