Tetrahedron

Calculate height and volume of a regular tetrahedron whose edge has a length 18 cm.

Result

h =  14.7 cm
V =  687.31 cm3

Solution:

h2=a2(a/2)2 h=63a=6318=14.7 cmh^2 = a^2 - (a/2)^2 \ \\ h = \dfrac{ \sqrt{6}}{3} a = \dfrac{ \sqrt{6}}{3} \cdot 18 = 14.7 \ \text{cm}
V=13Sh=1334a263a V=212a3=212183=687.31 cm3V = \dfrac13 S h = \dfrac{1}{3} \cdot \dfrac{ \sqrt{3}}{4} \cdot a^2 \cdot \dfrac{ \sqrt{6}}{3} a \ \\ V = \dfrac{ \sqrt{2}}{12} \cdot a^3 = \dfrac{ \sqrt{2}}{12} \cdot 18^3 = 687.31 \ \text{cm}^3



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