Tetrahedral pyramid 8

Let’s all side edges of the tetrahedral pyramid ABCDV be equally long and its base let’s be a rectangle. Determine its volume if you know the deviations A=40° B=70° of the planes of adjacent sidewalls and the plane of the base and the height h=16 of the pyramid.

Result

V =  2368.923

Solution:

A=40 ∘ B=70 ∘ h=16  tan⁡A=hb/2 tan⁡B=ha/2  b=2⋅ h/tan⁡(A)=2⋅ 16/tan⁡(40∘)≐38.1361 a=2⋅ h/tan⁡(B)=2⋅ 16/tan⁡(70∘)≐11.647  S=a⋅ b=11.647⋅ 38.1361≐444.1731  V=13⋅ S⋅ h=13⋅ 444.1731⋅ 16≐2368.9234≐2368.923A=40 \ ^\circ \ \\ B=70 \ ^\circ \ \\ h=16 \ \\ \ \\ \tan A=\dfrac{ h }{ b/2 } \ \\ \tan B=\dfrac{ h }{ a/2 } \ \\ \ \\ b=2 \cdot \ h / \tan( A)=2 \cdot \ 16 / \tan( 40^\circ ) \doteq 38.1361 \ \\ a=2 \cdot \ h / \tan( B)=2 \cdot \ 16 / \tan( 70^\circ ) \doteq 11.647 \ \\ \ \\ S=a \cdot \ b=11.647 \cdot \ 38.1361 \doteq 444.1731 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 444.1731 \cdot \ 16 \doteq 2368.9234 \doteq 2368.923

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