题目内容
顶点为P的圆锥的轴截面积是等腰直角三角形,A是底面圆周上的点,O为底面圆的圆心,AB⊥OB,垂足为B,OH⊥PB,垂足为H,且PA=4,C为PA的中点,则当三棱锥O-HPC的体积最大时,OB的长是( )A.
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B.
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C.
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D.
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【答案】分析:画出图形,说明PC是三棱锥P-OCH的高,△OCH的面积在OH=HC=
时取得最大值,求出OB即可.
解答:
解:AB⊥OB,可得PB⊥AB,即AB⊥面POB,所以面PAB⊥面POB.
OH⊥PB,则OH⊥面PAB,OH⊥HC,OH⊥PC,
又,PC⊥OC,所以PC⊥面OCH.即PC是三棱锥P-OCH的高.PC=OC=2.
而△OCH的面积在OH=HC=
时取得最大值(斜边=2的直角三角形).
当OH=
时,由PO=2
,知∠OPB=30°,OB=POtan30°=
.
故选D.
点评:本题考查圆锥的结构特征,棱锥的体积等知识,考查空间想象能力,是中档题.
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解答:
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OH⊥PB,则OH⊥面PAB,OH⊥HC,OH⊥PC,
又,PC⊥OC,所以PC⊥面OCH.即PC是三棱锥P-OCH的高.PC=OC=2.
而△OCH的面积在OH=HC=
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当OH=
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故选D.
点评:本题考查圆锥的结构特征,棱锥的体积等知识,考查空间想象能力,是中档题.
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