高中数学立体几何题 高中数学题，立体几何19题
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2.根据正四面体的高可以求出其棱长为2sqrt（6），棱长和高是1：sqrt（6）/3的关系，然后过顶点和球心相连，得到一个直角三角形（顶点包括一个顶点、球心、底面中心），设球半径为R,列方程：R^2=(R-4)^2+(2sqrt（2）)^2，解得半径R=3，则表面积S=4*pi*R^2=36pi1.A
2.设正四面体的棱长为a，球的半径为R，球的体积为M，则：
有方程组：a^2=(4 - R)^2 - (sqrt(3) * /2) * 2a/3
a^2=4^2 + (sqrt(3) * /2) * 2a/3
M= 4*pi*R^2
解得：M=36pi 其中pi = 3.1415926
1.A2.256/9派
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