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