2025-04-12
设正四面体棱长为aaa,相关公式如下:
高:h=63ah = \frac{\sqrt{6}}{3}ah=36
a
表面积:S=3a2S = \sqrt{3}a^{2}S=3
a2
体积:V=212a3V = \frac{\sqrt{2}}{12}a^{3}V=122
a3
外接球半径:R=64aR = \frac{\sqrt{6}}{4}aR=46
内切球半径:r=612ar = \frac{\sqrt{6}}{12}ar=126