Dados os valores para o raio ( r ) e a geratriz ( g ) do cone, chegamos ao resultado que o volume do cone:
[tex]\large \text {$\Rightarrow ~ 10\sqrt{10} ~\pi ~cm^3$}[/tex]
[tex]\Large\text{$ S\acute{o}lidos ~Geom\acute{e}tricos - Cone $}[/tex]
[tex]\large\begin{cases}\sf Altura = 3 ~. ~r = 9 ~cm \\\sf Geratriz = g= 10 ~cm\end{cases}[/tex]
Por Pitágoras: Encontrar o valor do raio do cone.
[tex]\large \text {$h^2 = g^2 - r^2$}\\\\\large \text {$(3r)^2 = 10^2 - r^2$}\\\\\large \text {$9r^2 = 100 - r^2$}\\\\\large \text {$9r^2 + r^2 = 100$}\\\\\large \text {$10r^2 = 100$}\\\\\\\large \text {$r^2 = \dfrac{100}{10}$}\\\\\\\large \text {$r^2 = 10$}\\\\\\\large \text {$r = \sqrt{10} ~cm$}\\\\[/tex]
Com o valor do raio, o valor da altura - triplo do raio - 3√10 cm
[tex]\large \text {$V = \dfrac{1}{3} ~\pi~. ~r^2~. ~h$}\\\\\\\large \text {$V = \dfrac{~\pi~. ~(\sqrt{10})^2~. ~3\sqrt{10}}{3} $}\\\\\\\large \text {$V = \dfrac{~\pi~. ~10~. ~3\sqrt{10}}{3} $}\\\\\\\large \text {$V = \dfrac{30~\sqrt{10}\pi}{3} $}\\\\\\\large \text {$V = 10\sqrt{10}~\pi ~cm^3$}[/tex]
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