# 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:

$h^2 = a^2 - (a/2)^2 \ \\ h = \dfrac{ \sqrt{6}}{3} a = \dfrac{ \sqrt{6}}{3} \cdot 18 = 14.7 \ \text{cm}$
$V = \dfrac13 S h = \dfrac{1}{3} \cdot \dfrac{ \sqrt{3}}{4} \cdot a^2 \cdot \dfrac{ \sqrt{6}}{3} a \ \\ V = \dfrac{ \sqrt{2}}{12} \cdot a^3 = \dfrac{ \sqrt{2}}{12} \cdot 18^3 = 687.31 \ \text{cm}^3$

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