C
分析:可設DE=3k,則AE=4k,AD=5K,BE=k,從而求出邊長及高,計算面積;連接BD、AC,根據(jù)勾股定理可求對角線的長度;
作DH⊥BC于H點,則DH=DE,比較DH與DF的大�。�
解答:
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解:設DE=3k,則AE=4k,AD=5K,BE=k=1,
∴AB=5,DE=3.故①正確;
S
梯形DEBC=
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×(1+5)×3=9,故②正確;
∵DE=3,EB=1,∴DB=
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.
又∵S
ABCD=AB×DE=5×3=15,
S
ABCD=
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×BD×AC,
∴15=
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×
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×AC,
AC=3
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.
(AC+BD)(AC-BD)
=AC
2-BD
2=(3
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)
2-
2=90-10=80.故③正確;
作DH⊥BC于H點.
∵DE⊥AB,DH⊥BC,∠ABD=∠CBD,
∴DE=DH.
又DH<DF,
∴DE<DF.故④錯誤.
所以①②③正確.
故選C.
點評:此題主要考查了菱形的性質及面積計算,有一定的綜合性,屬中等難度.