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
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