题目内容
正三棱锥V-ABC的底面边长为2a,E、F、G、H分别是VA、VB、BC、AC的中点,则四边形EFGH面积的取值范围是( )A.(0,+∞)
B.

C.
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D.
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【答案】分析:画出图形,求出EF,HG,说明EFHG是矩形,结合图形,说明V点在ABC平面时,面积最小,求出即可得到范围.
解答:
解:由条件可知:EF=HG=a,EFGH是平行四边形
因为正三棱锥V-ABC,所以EFGH是矩形而EH,FG,是变量,
当V点在ABC平面时,VA=VB=VC=
此时EH,FG有最小值,EH=FG=
VA=
EFGH的面积EF*•EH=a×
=
故选B.
点评:本题考查棱锥的结构特征,考查学生作图能力,分析问题解决问题的能力,是基础题.
解答:
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因为正三棱锥V-ABC,所以EFGH是矩形而EH,FG,是变量,
当V点在ABC平面时,VA=VB=VC=

此时EH,FG有最小值,EH=FG=


EFGH的面积EF*•EH=a×


故选B.
点评:本题考查棱锥的结构特征,考查学生作图能力,分析问题解决问题的能力,是基础题.

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