已知等腰梯形ABCD中,AB∥CD,對(duì)角線AC、BD相交于O,∠ABD = 30°,AC⊥BC,AB =" 8" cm,則△COD的面積為( ).
A.
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cm
2 B.
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cm
2 C.
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cm
2 D.
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cm
2
解:∵梯形ABCD是等腰梯形,CD∥AB,
由SAS可證△DAB≌△CBA,
∴∠CAB=∠DCA=30°,
∵∠CAB=30°,又因?yàn)锳C⊥BC,
∴∠DAB=∠CBA=60°,
∴∠DAC=∠DCA=30°,
∴CD=AD=BC=4cm,
∴AC
2=AB
2-BC
2,
∴AC=4
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cm,
∵梯形ABCD是等腰梯形,
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∴AC=BD=4
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cm,
∴S
△ABC=
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×4×4
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=8
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cm,
設(shè)DO為x,則CO=x,則AO=BO=(4
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-x)cm,
在Rt△COB中,CO
2+BC
2=BO
2,
即:x
2+4
2=(4
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-x)
2∴D0=
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cm,
∴S
△ADO=
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×
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×4=
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,
∴S
△AOB=S
△ABC-S
△ADO=
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∵AB∥CD,
∴△AOB∽△DOC,
∴(
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)
2=
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∴S
△DOC=
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,故選A
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