A
分析:由△AMC為等邊三角形,取CM中點,則AD⊥CM,AD交BC于E,證明AE⊥平面BCM,利用等體積法,即可求得結論.
解答:由已知得AB=2,AM=MB=MC=1,BC=
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,
由△AMC為等邊三角形,取CM中點,則AD⊥CM,AD交BC于E,則AD=
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,DE=
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,CE=
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.
折起后,由BC
2=AC
2+AB
2,知∠BAC=90°,
又cos∠ECA=
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,∴AE
2=CA
2+CE
2-2CA•CEcos∠ECA=
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,于是AC
2=AE
2+CE
2.
∴∠AEC=90°.
∵AD
2=AE
2+ED
2,∴AE⊥平面BCM,即AE是三棱錐A-BCM的高,AE=
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設點M到面ABC的距離為h,則
∵S
△BCM=
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∴由V
A-BCM=V
M-ABC,可得
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,∴h=
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故選A.
點評:本題考查由平面圖形折成空間圖形求其體積,考查點到平面距離的計算,求此三棱錐的高是解決問題的關鍵.