Tetrahedral pyramid

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

Correct answer:

V =  36.51 cm³
S =  72.915 cm²

Step-by-step explanation:

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

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

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