Equilateral 81222

A sphere is inscribed in an equilateral cone with a base diameter of 12 cm. Calculate the volume of both bodies. What percentage of the volume of the cone is filled by the inscribed sphere?

Correct answer:

V₁ =  391.7807 cm³
V₂ =  174.1247 cm³
p =  44.4444 %

Step-by-step explanation:

$$\begin{gathered} d=12\text{cm} \\ r=d/2=12/2=6\text{cm} \\ s=d=12\text{cm} \\ h=\sqrt{s^2-r^2}=\sqrt{12^2-6^2}=6\sqrt{3}\text{cm}\doteq 10.3923\text{cm} \\ {S}_1=\pi \cdot r^2=3.1416\cdot 6^2\doteq 113.0973{\text{cm}}^2 \\ {V}_1=\frac{1}{3}\cdot S_1\cdot h=\frac{1}{3}\cdot 113.0973\cdot 10.3923=391.7807{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} R=\frac{\sqrt{3}}{6}\cdot d=\frac{\sqrt{3}}{6}\cdot 12=2\sqrt{3}\text{cm}\doteq 3.4641\text{cm} \\ {V}_2=\frac{4}{3}\cdot \pi \cdot R^3=\frac{4}{3}\cdot 3.1416\cdot 3.4641^3=174.1247{\text{cm}}^3 \end{gathered}$$

$p=100\cdot \frac{V_2}{V_1}=100\cdot \frac{174.1247}{391.7807}=44.4444\%$

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