Tetrahedral pyramid

Calculate the volume and surface of the regular tetrahedral pyramid if content area of the base is 20 cm² and deviation angle of the side edges from the plane of the base is 60 degrees.

Correct result:

V =  36.51 cm³
S =  72.915 cm²

Solution:

$$\begin{gathered} S_1=20{\text{cm}}^2 \\ \mathrm{\Phi }=60^{\circ } \\ {S}_1=a^2 \\ a=\sqrt{S_1}=\sqrt{20}=2\sqrt{5}\text{ cm}\doteq 4.4721\text{ cm } \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 4.4721=2\sqrt{10}\text{ cm}\doteq 6.3246\text{ cm } \\ h={\displaystyle \frac{u}{2}}\cdot \tan {\mathrm{\Phi }}^{\circ }={\displaystyle \frac{u}{2}}\cdot \tan 60^{\circ }={\displaystyle \frac{6.3246}{2}}\cdot \tan 60^{\circ }={\displaystyle \frac{6.3246}{2}}\cdot 1.732051=5.47723 \\ V={\displaystyle \frac{1}{3}}\cdot S_1\cdot h={\displaystyle \frac{1}{3}}\cdot 20\cdot 5.4772=36.51{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} b=\sqrt{h^2+(a\mathrm{/}2{)}^2}=\sqrt{5.4772^2+(4.4721\mathrm{/}2{)}^2}=\sqrt{35}\text{ cm}\doteq 5.9161\text{ cm } \\ {S}_2=a\cdot b\mathrm{/}2=4.4721\cdot 5.9161\mathrm{/}2=5\sqrt{7}{\text{cm}}^2\doteq 13.2288{\text{cm}}^2 \\ S=4\cdot S_2+S_1=4\cdot 13.2288+20=72.915{\text{cm}}^2 \end{gathered}$$

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