Tetrahedral pyramid

Determine the surface of a regular tetrahedral pyramid when its volume is V = 120 and the angle of the sidewall with the base plane is α = 42° 30´.

Result

S =  200.643

Solution:

V=120 A=42+3060=852=42.5 ∘ A1=A∘→rad=A∘⋅ π180 =42.5∘⋅ 3.1415926180 =0.74176 =17π/72  tan⁡A=h/(a/2)  h=a/2⋅ tan⁡A V=13a2 h V=16a3 tan⁡A  a=6⋅ V/tan⁡(A1)3=6⋅ 120/tan⁡(0.7418)3≐9.2277  h=a/2⋅ tan⁡(A1)=9.2277/2⋅ tan⁡(0.7418)≐4.2278 sin⁡A1=h:w  w=h/sin⁡(A1)=4.2278/sin⁡(0.7418)≐6.258  S1=a⋅ w/2=9.2277⋅ 6.258/2≐28.8733  S=a2+4⋅ S1=9.22772+4⋅ 28.8733≐200.6434=200.643V = 120 \ \\ A = 42 + \dfrac{ 30 }{ 60 } = \dfrac{ 85 }{ 2 } = 42.5 \ ^\circ \ \\ A_{ 1 } = A ^\circ \rightarrow rad = A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ = 42.5 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ = 0.74176 \ = 17π/72 \ \\ \ \\ \tan A = h /(a/2) \ \\ \ \\ h = a/2 \cdot \ \tan A \ \\ V = \dfrac{ 1 }{ 3 } a^2 \ h \ \\ V = \dfrac{ 1 }{ 6 } a^3 \ \tan A \ \\ \ \\ a = \sqrt[3]{ 6 \cdot \ V / \tan(A_{ 1 }) } = \sqrt[3]{ 6 \cdot \ 120 / \tan(0.7418) } \doteq 9.2277 \ \\ \ \\ h = a/2 \cdot \ \tan(A_{ 1 }) = 9.2277/2 \cdot \ \tan(0.7418) \doteq 4.2278 \ \\ \sin A_{ 1 } = h : w \ \\ \ \\ w = h / \sin(A_{ 1 }) = 4.2278 / \sin(0.7418) \doteq 6.258 \ \\ \ \\ S_{ 1 } = a \cdot \ w/2 = 9.2277 \cdot \ 6.258/2 \doteq 28.8733 \ \\ \ \\ S = a^2+4 \cdot \ S_{ 1 } = 9.2277^2+4 \cdot \ 28.8733 \doteq 200.6434 = 200.643

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