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[几何] 征解

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张振华 Posted 2014-4-18 20:25 |Read mode
已知A,B,C,D四点在半径为(14^0.5)/2的球面上,且AC=BD=5^0.5,AD=BC=10^0.5,AB=CD,则三棱锥D-ABC的体积是______

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其妙 Posted 2014-4-18 20:49
对不对?已知$A,B,C,D$四点在半径为$\dfrac{\sqrt{14}}2$的球面上,且$AC=BD=\sqrt5,AD=BC=\sqrt{10},AB=CD,$则三棱锥$D-ABC$的体积是______

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kuing Posted 2014-4-18 20:51
看样子是构造长方体……

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其妙 Posted 2014-4-18 21:28
回复 3# kuing
,对棱都相等的四面体叫等腰四面体,其体积有公式的。外接球半径也是有公式的.甚至内切球半径也是有公式的.

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 Author| 张振华 Posted 2014-4-18 21:40
回复 2# 其妙
对的

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