Determine o volume de um cubo inscrito em uma esfera cujo volume mede 2,304πcm³.... Pergunta de ideia deNabouvier

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December 2019 1 19 Report

Determine o volume de um cubo inscrito em uma esfera cujo volume mede 2,304πcm³

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FibonacciTH

Verified answer

O volume de uma esfera de raio (R) é determinado pela formula:

$\mathsf{V_{esfera}=\dfrac{4}{3}\pi R^3}$

= = = = =

Foi informado que o volume da esfera é $\mathsf{2,304\pi
\:cm^3}$ . Logo:

$\mathsf{V_{esfera}=2,304\pi
\:cm^3}\\\\\\\mathsf{2,304\diagup\!\!\!\!\!\pi
=\dfrac{4}{3}\diagup\!\!\!\!\!\pi
R^3}\\\\\\\mathsf{2,304=\dfrac{4R^3}{3}}\\\\\\\boxed{\mathsf{R^3=\dfrac{2,304\cdot
3}{4}}}\:\:\:\:\:\:\:\:\mathsf{(i)}$

= = = = =

Aplicando o teorema de Pitágoras na Figura 2 determinaremos que o valor da diagonal da base do quadrado:

$\mathsf{d^2=l^2+l^2}\\\\\mathsf{d^2=2l^2}\\\\\mathsf{d=\sqrt{2l^2}}\\\\\mathsf{d=l\sqrt{2}}$

= = = = =

A diagonal (D) do cubo equivale ao valor da diagonal (2R) da esfera. Aplicando o teorema de Pitágoras na Figura 1:

$\mathsf{D^2=l^2+d^2}\\\\\mathsf{\left(2R\right)^2=l^2+\left(l\sqrt{2}\right)^2}\\\\\mathsf{4R^2=l^2+2l^2}\\\\\mathsf{4R^2=3l^2}\\\\\mathsf{R^2=\dfrac{3l^2}{4}}\\\\\\\mathsf{R=\sqrt{\dfrac{3l^2}{4}}}\\\\\\\mathsf{R=\dfrac{l\sqrt{3}}{2}}$

Elevando os dois termos ao cubo teremos:

$\mathsf{R^3=\left(\dfrac{l\sqrt{3}}{2}\right)^3}\\\\\\\mathsf{R^3=\dfrac{l^3\left(\sqrt{3}\right)^3}{2^3}}\\\\\\\boxed{\mathsf{R^3=\dfrac{3l^3\sqrt{3}}{8}}}\:\:\:\:\:\:\:\:\mathsf{(ii)}$

= = = = =

O volume do cubo é determinado pela formula:

$\mathsf{V_{cubo}=l^3}$

= = = = =

Igualando i e ii:

$\mathsf{\dfrac{2,304\cdot
3}{4}=\dfrac{3l^3\sqrt{3}}{8}}\\\\\\\mathsf{\dfrac{2,304\cdot
\diagup\!\!\!\!3}{\diagup\!\!\!\!4}=\dfrac{\diagup\!\!\!\!3\cdot
l^3\sqrt{3}}{2\cdot
\diagup\!\!\!\!4}}\\\\\\\mathsf{2,304=\dfrac{l^3\sqrt{3}}{2}}\\\\\\\mathsf{2,304\cdot
2=l^3\sqrt{3}}\\\\\\\mathsf{l^3=\dfrac{2,304\cdot
2}{\sqrt{3}}}\\\\\\\mathsf{l^3=\dfrac{4,608}{\sqrt{3}}}\\\\\\\boxed{\mathsf{V_{cubo}=\dfrac{4,608}{\sqrt{3}}}}$

= = = = =

Racionalizando teremos:

$\mathsf{V_{cubo}=\dfrac{4,608}{\sqrt{3}}\cdot
\dfrac{\sqrt{3}}{\sqrt{3}}}\\\\\\\mathsf{V_{cubo}=\dfrac{4,608\sqrt{3}}{3}}\\\\\\\boxed{\mathsf{V_{cubo}=1,536\sqrt{3}}}$