9-gon pyramid

Calculate the volume and the surface of a nine-sided pyramid, the base of which can be inscribed with a circle with radius ρ = 7.2 cm and whose side edge s = 10.9 cm.

Correct answer:

V =  409.0382 cm³
S =  299.899 cm²

Step-by-step explanation:

$$\begin{gathered} \rho =7.2\text{cm} \\ s=10.9\text{cm} \\ n=9 \\ {h}^2+{\rho }^2=s^2 \\ h=\sqrt{s^2-{\rho }^2}=\sqrt{10.9^2-7.2^2}\doteq 8.1835\text{cm} \\ \alpha =\frac{2\pi }{2\cdot n}=\frac{2\cdot 3.1416}{2\cdot 9}\doteq 0.3491\text{rad} \\ \sin \alpha =a/2:\rho \\ a=2\cdot \rho \cdot \sin (\alpha )=2\cdot 7.2\cdot \sin (0.3491)\doteq 4.9251\text{cm} \\ {h}_2=\sqrt{{\rho }^2-(a/2{)}^2}=\sqrt{7.2^2-(4.9251/2{)}^2}\doteq 6.7658\text{cm} \\ {S}_1=\frac{a\cdot h_2}{2}=\frac{4.9251\cdot 6.7658}{2}\doteq 16.6611\text{cm} \\ {S}_9=n\cdot S_1=9\cdot 16.6611\doteq 149.9495{\text{cm}}^2 \\ V=\frac{1}{3}\cdot S_9\cdot h=\frac{1}{3}\cdot 149.9495\cdot 8.1835=409.0382{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} S_2=\frac{a\cdot h_2}{2}=\frac{4.9251\cdot 10.6182}{2}\doteq 16.6611{\text{cm}}^2 \\ {h}_2=\sqrt{h^2+h_2^2}=\sqrt{8.1835^2+10.6182^2}\doteq 10.6182\text{cm} \\ S=S_9+n\cdot S_2=149.9495+9\cdot 16.6611=299.899{\text{cm}}^2 \end{gathered}$$

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