题目内容
如下的三个图中,上面的是一个长方体截去一个角所得多面体的直观图,它的正视图和左视图在下面画出(单位:cm).
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(1)在正视图下面,按照画三视图的要求画出该多面体的俯视图;
(2)按照给出的尺寸,求该多面体的体积;
(3)在所给直观图中连接BC′,证明:BC′∥平面EFG.
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(1)在正视图下面,按照画三视图的要求画出该多面体的俯视图;
(2)按照给出的尺寸,求该多面体的体积;
(3)在所给直观图中连接BC′,证明:BC′∥平面EFG.
(1)
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(2)
cm(3)证明略
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(2)
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如图(1)所示.
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图(1)
(2)解 所求多面体体积
V=V长方体-V正三棱锥
=4×4×6-
×(
×2×2)×2=
(cm3).
(3)证明 如图,在长方体ABCD—A′B′C′D′中,
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连接AD′,则AD′∥BC′.
因为E,G分别为AA′,A′D′的中点,
所以AD′∥EG,从而EG∥BC′.
又BC′
平面EFG,
所以BC′∥面EFG.

图(1)
(2)解 所求多面体体积
V=V长方体-V正三棱锥
=4×4×6-
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(3)证明 如图,在长方体ABCD—A′B′C′D′中,
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连接AD′,则AD′∥BC′.
因为E,G分别为AA′,A′D′的中点,
所以AD′∥EG,从而EG∥BC′.
又BC′

所以BC′∥面EFG.
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