Ensino Médio ⇒ Geometria Espacial Tópico resolvido

[Gabi123]

Abaixo temos uma pirâmide quadrangular de altura igual a 1 m seccionada por dois planos α e β paralelos à sua base.

Capturar.JPG (23 KiB) Exibido 336 vezes

Considerando que a pirâmide é dividida em três sólidos de mesmo volume com alturas h1, h2 e h3, determine os valores dessas alturas.

[petras]

Gabi123,
Utilizando a semelhança:

[tex3]\mathtt{
h_1+h_2+h_3=1\\
\bullet(\frac{h_1}{1})^3=\frac{V}{3V}\implies h_1^3 = \frac{1}{3}\therefore h_1 = {\sqrt[3]{\frac{1}{3}}}=\frac{1}{\sqrt[3]{3}}=\boxed{\frac{\sqrt[3]{9}}{3}}\color{green}\checkmark\\

\bullet (\frac{(h_1+h_2)}{1})^3=\frac{2V}{3V}\implies (h_1+h_2)^3 =\frac{2}{3} \therefore h_1+h_2 = \sqrt[3]{\frac{2}{3}}=\frac{\sqrt[3]{2}.\sqrt[3]{9}}{3}\\
\therefore h_2 = \frac{\sqrt[3]{2}.\sqrt[3]{9}}{3}-\frac{\sqrt[3]{9}}{3}=\boxed{\frac{\sqrt[3]{9}}{3}({\sqrt[3]{2}}-1)}\color{green}\checkmark\\
\bullet h^3 = 1-(h_1+h_2) = 1 - [\frac{\sqrt[3]{9}}{3}+\frac{\sqrt[3]{9}}{3}(\sqrt[3]{2}-1)]=\\
1-(\frac{\sqrt[3]{9}}{3}.\sqrt[3]{2})=\boxed{1-\frac{\sqrt[3]{18}}{3}}\color{green}\checkmark

}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathtt{
h_1+h_2+h_3=1\\
\bullet(\frac{h_1}{1})^3=\frac{V}{3V}\implies h_1^3 = \frac{1}{3}\therefore h_1 = {\sqrt[3]{\frac{1}{3}}}=\frac{1}{\sqrt[3]{3}}=\boxed{\frac{\sqrt[3]{9}}{3}}\color{green}\checkmark\\

\bullet (\frac{(h_1+h_2)}{1})^3=\frac{2V}{3V}\implies (h_1+h_2)^3 =\frac{2}{3} \therefore h_1+h_2 = \sqrt[3]{\frac{2}{3}}=\frac{\sqrt[3]{2}.\sqrt[3]{9}}{3}\\
\therefore h_2 = \frac{\sqrt[3]{2}.\sqrt[3]{9}}{3}-\frac{\sqrt[3]{9}}{3}=\boxed{\frac{\sqrt[3]{9}}{3}({\sqrt[3]{2}}-1)}\color{green}\checkmark\\
\bullet h^3 = 1-(h_1+h_2) = 1 - [\frac{\sqrt[3]{9}}{3}+\frac{\sqrt[3]{9}}{3}(\sqrt[3]{2}-1)]=\\
1-(\frac{\sqrt[3]{9}}{3}.\sqrt[3]{2})=\boxed{1-\frac{\sqrt[3]{18}}{3}}\color{green}\checkmark

}[/tex3]