Octagonal pyramid

Find the volume of a regular octagonal pyramid with height v = 100 and the angle of the side edge with the plane of the base is α = 60°.

Correct result:

V =  314269.681

Solution:

v=100 α=60 ∘ α1=α=60∘=1.0472=π/3 n=8  tan⁡α=v/r  r=v/tan⁡(α1)=100/tan⁡1.0472≐57.735  β=2π2⋅ n=2⋅ 3.14162⋅ 8≐0.3927 rad  sin⁡β=x/r  x=r⋅ sin⁡(β)=57.735⋅ sin⁡0.3927≐22.0942  r2=x2+w2 w=r2−x2=57.7352−22.09422≐53.3402  S1=w⋅ x2=53.3402⋅ 22.09422≐589.2557  S=2⋅ n⋅ S1=2⋅ 8⋅ 589.2557≐9428.0904  V=13⋅ S⋅ v=13⋅ 9428.0904⋅ 100=314269.681v=100 \ \\ α=60 \ ^\circ \ \\ α_{1}=α=60^\circ =1.0472=π/3 \ \\ n=8 \ \\ \ \\ \tan α=v/r \ \\ \ \\ r=v / \tan (α_{1})=100 / \tan 1.0472 \doteq 57.735 \ \\ \ \\ β=\dfrac{ 2 \pi }{ 2 \cdot \ n }=\dfrac{ 2 \cdot \ 3.1416 }{ 2 \cdot \ 8 } \doteq 0.3927 \ \text{rad} \ \\ \ \\ \sin β=x / r \ \\ \ \\ x=r \cdot \ \sin (β)=57.735 \cdot \ \sin 0.3927 \doteq 22.0942 \ \\ \ \\ r^2=x^2 + w^2 \ \\ w=\sqrt{ r^2-x^2 }=\sqrt{ 57.735^2-22.0942^2 } \doteq 53.3402 \ \\ \ \\ S_{1}=\dfrac{ w \cdot \ x }{ 2 }=\dfrac{ 53.3402 \cdot \ 22.0942 }{ 2 } \doteq 589.2557 \ \\ \ \\ S=2 \cdot \ n \cdot \ S_{1}=2 \cdot \ 8 \cdot \ 589.2557 \doteq 9428.0904 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S \cdot \ v=\dfrac{ 1 }{ 3 } \cdot \ 9428.0904 \cdot \ 100=314269.681

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