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 answer:

V =  314269.6805

Step-by-step explanation:

$$\begin{gathered} v=100 \\ \alpha =60^{\circ } \\ {\alpha }_1=\alpha =60^{\circ }=1.0472=\pi \mathrm{/}3 \\ n=8 \\ \tan \alpha =v\mathrm{/}r \\ r=v\mathrm{/}\tan ({\alpha }_1)=100\mathrm{/}\tan 1.0472\doteq 57.735 \\ \beta ={\displaystyle \frac{2\pi }{2\cdot n}}={\displaystyle \frac{2\cdot 3.1416}{2\cdot 8}}\doteq 0.3927\text{ rad } \\ \sin \beta =x\mathrm{/}r \\ x=r\cdot \sin (\beta )=57.735\cdot \sin 0.3927\doteq 22.0942 \\ {r}^2=x^2+w^2 \\ w=\sqrt{r^2-x^2}=\sqrt{57.735^2-22.0942^2}\doteq 53.3402 \\ {S}_1={\displaystyle \frac{w\cdot x}{2}}={\displaystyle \frac{53.3402\cdot 22.0942}{2}}\doteq 589.2557 \\ S=2\cdot n\cdot S_1=2\cdot 8\cdot 589.2557\doteq 9428.0904 \\ V={\displaystyle \frac{1}{3}}\cdot S\cdot v={\displaystyle \frac{1}{3}}\cdot 9428.0904\cdot 100=314269.6805=3.143\cdot 10^5 \end{gathered}$$

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