如图，在四边形ABCD中，对角线AC，BD交于点E，∠BAC＝90°，∠CED＝45°，∠DCE＝30°...
问题详情：
如图，在四边形ABCD中，对角线AC，BD交于点E，∠BAC＝90°，∠CED＝45°，∠DCE＝30°，DE＝
，BE＝2
．
(1)求CD的长：
(2)求四边形ABCD的面积
【回答】
解： (1)过点D作DH⊥AC，·········································································· 1分
∵∠CED=45°，
∴∠EDH=45°，
∴∠HED=∠EDH，
∴EH=DH，···································································································· 3分
∵EH2+DH2=DE2，DE=
，
∴EH2=1，
∴EH=DH=1，································································································· 5分
又∵∠DCE=30°，∠DHC=90°，
∴DC=2 ········································································································· 6分
(2)∵在Rt△DHC中，
························································· 7分
∴12+HC2=22，
∴HC=
，···································································································· 8分
∵∠AEB=∠CED=45°，∠BAC=90°，BE=2，
∴AB=AE=2，································································································· 9分
∴AC=2+1+
=3+，················································································ 10分
∴S四边形ABCD
=S△BAC+S△DAC································································································ 11分
=
×2×（3+）+
×1×（3+
）
=
······································································································· 12分
知识点：勾股定理
题型：解答题
