Question

qual o volume de um cone equilátero cujo o raio da base é r=3cm.(dado TT =3,14e
$\sqrt{3}$
3=1,7) .
$\sqrt[3]{?}$
​

Answer 1

$\largeO ~volume ~do ~cone ~ \Rightarrow ~V = 48,042 ~cm ^3$

A geratriz do cone equilátero é o dobro do raio da base

Encontrar a altura:

$h^2 = g^2 - r^2\\\\h^2 = 6^2 - 3^2\\\\h^2 = 36 - 9\\\\h^2 = 27\\\\h = \sqrt{27} \\\\h = 3\sqrt{3} ~cm$

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Volume do cone:

$V = \dfrac{1}{3} ~\pi~. ~r^2~. ~h\\\\\\V = \dfrac{1}{3} ~\pi~. ~3^2~. ~3\sqrt{3} \\\\\\V = \dfrac{1}{3} ~\pi~. ~9~. ~3\sqrt{3} \\\\\\V = \dfrac{1}{3} ~\pi~. 27\sqrt{3} \\\\\\V = \dfrac{27\sqrt{3}}{3} ~\pi\\\\\\V =9\sqrt{3} ~\pi\\\\\\V = 48,042 ~cm ^3$

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