[tex]\large\text{$O ~volume ~do ~cone ~ \Rightarrow ~V = 48,042 ~cm ^3 $}[/tex]
A geratriz do cone equilátero é o dobro do raio da base
Encontrar a altura:
[tex]h^2 = g^2 - r^2\\\\h^2 = 6^2 - 3^2\\\\h^2 = 36 - 9\\\\h^2 = 27\\\\h = \sqrt{27} \\\\h = 3\sqrt{3} ~cm[/tex]
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Volume do cone:
[tex]V = \dfrac{1}{3} ~\pi~. ~r^2~. ~h\\\\\\V = \dfrac{1}{3} ~\pi~. ~3^2~. ~3\sqrt{3} \\\\\\V = \dfrac{1}{3} ~\pi~. ~9~. ~3\sqrt{3} \\\\\\V = \dfrac{1}{3} ~\pi~. 27\sqrt{3} \\\\\\V = \dfrac{27\sqrt{3}}{3} ~\pi\\\\\\V =9\sqrt{3} ~\pi\\\\\\V = 48,042 ~cm ^3[/tex]
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