题目内容
球放在墙角(两墙面,地面分别两两垂直),紧靠墙面和底面,球心到墙角顶点的距离是

【答案】分析:设球的半径为R,当球放在墙角时,同时与两墙面和地面相切可知球心与墙角顶点可构成边长为R的正方体,则正方体对角线即为球心到墙角顶点的距离,由此求出球的半径,可得球的体积.
解答:
解:根据题意可知球心与墙角顶点可构成边长为a的正方体
则球心到墙角顶点的距离为正方体的对角线即
R
即
R=
解得:R=1
故球的体积V=
=
故答案为:
点评:本题主要考查了空间两点的距离,以及利用构造正方体进行解题,属于基础题.
解答:

则球心到墙角顶点的距离为正方体的对角线即

即


解得:R=1
故球的体积V=


故答案为:

点评:本题主要考查了空间两点的距离,以及利用构造正方体进行解题,属于基础题.

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