Question

calcula a área total e o volume de um cone equilátero, sabendo que a área lateral é igual 48π cm^2
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Answer 1

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$\huge\boxed{\underline{\sf{Cone~equil\acute{a}tero}}}\\\tt{Dizemos~que~um~cone~\acute{e}~equil\acute{a}tero~quando~a~geratriz}\\\tt{\acute{e}~o~dobro~da~raio}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{G=2\cdot r}}}}}}$

$\rm{\underline{Dados}:}\\\sf{A_l=48\pi~cm^2}\\\tt{\underline{soluc_{\!\!,}\tilde{a}o}:}\\\sf{A_l=\pi\cdot r\cdot g}\\\sf{\diagdown\!\!\!\!\pi\cdot r\cdot 2r=48\diagdown\!\!\!\!\pi}\\\sf{r^2=\dfrac{48}{2}}\\\sf{r^2=24}\\\sf{r=\sqrt{24}~cm}\\\sf{A_t=\pi\cdot r(r+g)}\\\sf{A_t=\pi\cdot r(r+2r)}\\\sf{A_t=3\pi r^2}\\\sf{A_t=3\cdot \pi\cdot\left(\sqrt{24}\right)^2}\\\huge\boxed{\boxed{\boxed{\boxed{\rm{A_t=72\pi~cm^2}}}}}$

$\tt{\underline{C\acute{a}lculo~da~altura}:}\\\sf{h^2=g^2-r^2}\\\sf{h^2=(2\sqrt{24})^2-(\sqrt{24})^2}\\\sf{h^2=96-24}\\\sf{h^2=72}\\\sf{h=\sqrt{72}=6\sqrt{2}~cm}\\\sf{V=\dfrac{1}{3}\pi\cdot r^2\cdot h}\\\sf{V=\dfrac{1}{\diagup\!\!\!\!3}\pi\cdot\diagup\!\!\!\!\!24^8\cdot6\sqrt{2}}\\\sf{V=8\pi\cdot6\sqrt{6}}\\\huge\boxed{\boxed{\boxed{\boxed{\rm{V=48\pi\sqrt{6}~cm^3}}}}}$