# Triangular pyramid

A regular tetrahedron is a triangular pyramid whose base and walls are identical equilateral triangles. Calculate the height of this body if the edge length is a = 8 cm

Result

h =  6.532 cm

#### Solution:

$a=8 \ \text{cm} \ \\ \ \\ a^2=h_{1}^2 + (a/2)^2 \ \\ h_{1}=\sqrt{ a^2 - (a/2)^2 }=\sqrt{ 8^2 - (8/2)^2 } \doteq 4 \ \sqrt{ 3 } \ \text{cm} \doteq 6.9282 \ \text{cm} \ \\ \ \\ h^2 + (2/3 \ h_{1})^2=a^2 \ \\ \ \\ h=\sqrt{ a^2 - (2/3 \cdot \ h_{1})^2 }=\sqrt{ 8^2 - (2/3 \cdot \ 6.9282)^2 } \doteq 6.532 \doteq 6.532 \ \text{cm}$

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