Tetrahedral pyramid

Calculate the volume and surface area of a regular tetrahedral pyramid, its height is $b cm and the length of the edges of the base is 6 cm.

Correct result:

S =  196.1 cm²
V =  156 cm³

Solution:

$$\begin{gathered} a=6\text{ cm } \\ h=13\text{ cm } \\ s=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{13^2+(6\mathrm{/}2{)}^2}=\sqrt{178}\text{ cm}\doteq 13.3417\text{ cm } \\ {S}_1=a\cdot s\mathrm{/}2=6\cdot 13.3417\mathrm{/}2=3\sqrt{178}{\text{cm}}^2\doteq 40.025{\text{cm}}^2 \\ S=a^2+4\cdot S_1=6^2+4\cdot 40.025=196.1{\text{cm}}^2 \end{gathered}$$

$V={\displaystyle \frac{1}{3}}\cdot a^2\cdot h={\displaystyle \frac{1}{3}}\cdot 6^2\cdot 13=156{\text{cm}}^3$

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