如圖，在四邊形ABCD中，對角線AC，BD交於點E，∠BAC＝90°，∠CED＝45°，∠DCE＝30°...

問題詳情：

    如圖，在四邊形ABCD中，對角線AC，BD交於點E，∠BAC＝90°，∠CED＝45°，∠DCE＝30°，DE＝，BE＝2．

  (1)求CD的長：

(2)求四邊形ABCD的面積

【回答】

解： (1)過點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四邊形ABCD

=S△BAC+S△DAC································································································ 11分

=×2×（3+）+×1×（3+）

=······································································································· 12分

知識點：勾股定理

題型：解答題