正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r (2)用R表正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R
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![正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r (2)用R表正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R](/uploads/image/z/8297896-40-6.jpg?t=%E6%AD%A3%E5%9B%9B%E9%9D%A2%E4%BD%93ABCD%E5%86%85%E6%8E%A5%E4%BA%8E%E5%8D%8A%E5%BE%84%E4%B8%BAR%E7%9A%84%E7%90%83O%EF%BC%88%E5%8D%B3%E5%9B%9B%E4%B8%AA%E9%A1%B6%E7%82%B9%E5%9C%A8%E7%90%83%E9%9D%A2%E4%B8%8A%EF%BC%89%2C%E5%85%B6%E5%86%85%E5%88%87%E7%90%83%E5%8D%8A%E5%BE%84%E4%B8%BAr%2C%EF%BC%881%EF%BC%89.%E8%AF%81%E6%98%8ER%EF%BC%9D3r+%EF%BC%882%EF%BC%89%E7%94%A8R%E8%A1%A8%E6%AD%A3%E5%9B%9B%E9%9D%A2%E4%BD%93ABCD%E5%86%85%E6%8E%A5%E4%BA%8E%E5%8D%8A%E5%BE%84%E4%B8%BAR%E7%9A%84%E7%90%83O%EF%BC%88%E5%8D%B3%E5%9B%9B%E4%B8%AA%E9%A1%B6%E7%82%B9%E5%9C%A8%E7%90%83%E9%9D%A2%E4%B8%8A%EF%BC%89%2C%E5%85%B6%E5%86%85%E5%88%87%E7%90%83%E5%8D%8A%E5%BE%84%E4%B8%BAr%2C%EF%BC%881%EF%BC%89.%E8%AF%81%E6%98%8ER)
正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r (2)用R表正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R
正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r (2)用R表
正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r
正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R=3r (2)用R表正四面体ABCD内接于半径为R的球O(即四个顶点在球面上),其内切球半径为r,(1).证明R
如图,正四面体P-ABC内接于球O,O的半径为R
过点P作面ABC的垂线,垂足为O'
则,O'为等边△ABC所在圆面的圆心,且球心O在PO'上
设正四面体P-ABC的棱长为a,OO'=x
那么,BO'=a*(√3/2)*(2/3)=(√3/3)a
PO=BO=R,PB=a
那么,由勾股定理有:
BO^2=BO'^2+OO'^2 ===> R^2=(a^2/3)+x^2
===> R^2-x^2=a^2/3
===> (R+x)*(R-x)=a^2/3……………………………………(1)
PB^2=PO'^2+BO'^2 ===> a^2=(R+x)^2+(a^2/3)
===> (R+x)^2=(2/3)a^2
===> R+x=(√6/3)a
代入(1)有:(√6/3)a*(R-x)=a^2/3
===> R-x=(√6/6)a
所以:2R=(√6/3)a+(√6/6)a=(√6/2)a
所以,a=(2√6/3)R.