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

Correct result:

h =  6.532 cm

Solution:

$$\begin{gathered} a=8\text{ cm } \\ {a}^2=h_1^2+(a\mathrm{/}2{)}^2 \\ {h}_1=\sqrt{a^2-(a\mathrm{/}2{)}^2}=\sqrt{8^2-(8\mathrm{/}2{)}^2}=4\sqrt{3}\text{ cm}\doteq 6.9282\text{ cm } \\ {h}^2+(2\mathrm{/}3h_1{)}^2=a^2 \\ h=\sqrt{a^2-(2\mathrm{/}3\cdot h_1{)}^2}=\sqrt{8^2-(2\mathrm{/}3\cdot 6.9282{)}^2}=6.532\text{ cm} \end{gathered}$$

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