Let’s derive a few formulae. The octagonal base has radius R and vertical slant angle θ. A regular octagon can be split into 8 isosceles triangles with a common vertex at the centre O of the octagon and vertex angle 45° (π/4).
Each isosceles triangle can be split into two congruent right triangles, back to back. The hypotenuses are all length R, base Rcos(π/8), height Rsin(π/8), area=½(Rcos(π/8))(Rsin(π/8))=½R²sin(π/8)cos(π/8)=¼R²sin(π/4)=R²√2/8.
There are 16 right triangles making up the area of the octagon, so its area is 2R²√2.
The height of the pyramid is Rtanθ and so its volume is ⅓(2R²√2)(Rtanθ)=⅔R³tanθ√2. When the pyramid is truncated the volume of the pyramid removed from the top is ⅔r³tanθ√2, where r is the base radius of the removed pyramid (the radius of the top of the truncated pyramid). The difference between these volumes is the volume of the truncated octagonal pyramid (frustum), V=⅔tanθ√2(R³-r³). The height h of the frustum is (R-r)tanθ.
We now have two formulae involving the radii of the top and base and the slant angle.
From these we can derive a quadratic in r:
Since R=r+hcotθ, R³-r³=3r²hcotθ+3rh²cot²θ+h³cot³θ,
V=⅔tanθ√2(3r²hcotθ+3rh²cot²θ+h³cot³θ),
V=2√2(r²h+rh²cotθ+⅓h³cot²θ),
V√2/4=r²h+rh²cotθ+⅓h³cot²θ,
r²h+rh²cotθ+⅓h³cot²θ-V√2/4=0, which has the solution:
r=[√(Vh√2-⅓h⁴cot³θ)-h²cotθ]/(2h), and R=r+hcotθ. cotθ can be replaced by tan(90-θ). This shows how r and R are related to V, h and θ, variables which in this question have been given values.
There are two extreme conditions: when r=0, so that the frustum becomes a pyramid; when r=R, so that the solid becomes a prism. By plugging in these conditions we can see their effect.
r=0: V=⅔R³tanθ√2. So for a constant volume V, tanθ=3V√2/(4R³);
r=R⇒θ=90°, and V=⅓2R²h√2. Therefore, R²=3V√2/(4h). In this case, if h and V are constants, R=√(3V√2/h)/2.
Now we can refer to the conditions given in the question.
For V=12m³ and h=2m, and r=0:
tanθ=9√2/R³. But h=Rtanθ, so R²h=9√2. When h=2m, R²=9√2/2=6.3640 approx, making R=2.5227m; tan²θ=4√2/9, tanθ=0.7928, θ=38.41° approx. This angle is the minimum slope of the pyramid. The base area is 18m².
For V=12m³ and h=2m, and r=R:
V=(base area)×h, base area=V/h=12/2=6m² and 2R²√2=6, making R=r=1.4565m approx.
V=14m³, θ=80°:
Top area=5.4927m², base area=8.6245m²;
To complete the table, plug in V=14m³, θ=75°:
Top area=4.7624m², base area=9.5084m².
