Question

08. Seja A = área do polígono (ABCDE)
B = área do polígono (GHIJF)
C = área do polígono (KLMNP), então podemos afirmar:
a) A + B = C
b) C-A = 2B
c) A-C = B
d) B+C = 10A
e) B-C = A
09. Qual o volume da pirâmide VGHIJF se o volume da pirâmide VABCDE é 1 m??

10. A área do polígono ABCDE é d unidades de área. Qual o volume da pirâmide VKLMNP?

Answer 1

$\mathbf{08.}$

$\mathrm{O\ plano\ \gamma\ cont\acute{e}m\ a\ base\ pentagonal\ da\ pir\hat{a}mide\ VKLMNP.}$

$\mathrm{Os\ planos\ \beta\ e\ \alpha\ seccionam\ a\ pir\hat{a}mide\ VKLMNP\ em\ duas\ pir\hat{a}mides}\\ \mathrm{menores}\text{: VGHIJF e VABCDE.}$

$\mathrm{A\ altura\ dos\ planos\ \alpha,\ \beta\ e\ \gamma\ em\ rela\c{c}\tilde{a}o\ ao\ v\acute{e}rtice\ V\ \acute{e},}\\ \mathrm{respectivamente,}\ d,\ 2d\ \text{e}\ 3d.$

$\mathrm{Seja}\ A\ \mathrm{a\ \acute{a}rea\ do\ pol\acute{\i}gono\ ABCDE},\ B\ \mathrm{a\ \acute{a}rea\ do\ pol\acute{\i}gono\ GHIJF}\\ \mathrm{e}\ C\ \mathrm{a\ \acute{a}rea\ do\ pol\acute{\i}gono\ KLMNP.}$

$\mathrm{Seja}\ V_A\ \mathrm{o\ volume\ da\ pir\hat{a}mide\ VABCDE},\ V_B\ \mathrm{o\ volume\ da\ pir\hat{a}mide}\\ \mathrm{VFGHIJF\ e}\ V_C\ \mathrm{o\ volume\ da\ pir\hat{a}mide\ VKLMNP.}$

$V_A=\dfrac{1}{3}A(d)\Longrightarrow V_A=\dfrac{1}{3}Ad$

$V_B=\dfrac{1}{3}B(2d)\Longrightarrow V_B=\dfrac{2}{3}Bd$

$V_C=\dfrac{1}{3}C(3d)\Longrightarrow V_C=Cd$

$\mathrm{A\ rela\c{c}\tilde{a}o\ entre\ os\ volumes\ de\ uma\ pir\hat{a}mide\ e\ uma\ pir\hat{a}mide}\\ \mathrm{obtida\ pela\ sec\c{c}\tilde{a}o\ de\ um\ plano\ \acute{e}}\text{:}$

$\dfrac{V_s}{V_p}=\bigg(\dfrac{h_s}{h_p}\bigg)^3$

$\mathrm{Rela\c{c}\tilde{a}o\ entre}\ A\ \text{e}\ B\text{:}$

$\dfrac{V_A}{V_B}=\bigg(\dfrac{d}{2d}\bigg)^3\Longrightarrow \dfrac{\frac{1}{3}Ad}{\frac{2}{3}Bd}=\bigg(\dfrac{1}{2}\bigg)^3\Longrightarrow \dfrac{1}{2}\dfrac{A}{B}=\dfrac{1}{8}\ \therefore\ \boxed{B=4A}$

$\mathrm{Rela\c{c}\tilde{a}o\ entre}\ A\ \text{e}\ C\text{:}$

$\dfrac{V_A}{V_C}=\bigg(\dfrac{d}{3d}\bigg)^3\Longrightarrow \dfrac{\frac{1}{3}Ad}{Cd}=\bigg(\dfrac{1}{3}\bigg)^3\Longrightarrow \dfrac{1}{3}\dfrac{A}{C}=\dfrac{1}{27}\ \therefore\ \boxed{C=9A}$

$\mathrm{Rela\c{c}\tilde{a}o\ entre}\ B\ \text{e}\ C\text{:}$

$\dfrac{V_B}{V_C}=\bigg(\dfrac{2d}{3d}\bigg)^3\Longrightarrow \dfrac{\frac{2}{3}Bd}{Cd}=\bigg(\dfrac{2}{3}\bigg)^3\Longrightarrow \dfrac{2}{3}\dfrac{B}{C}=\dfrac{8}{27}\ \therefore\ \boxed{C=\dfrac{9}{4}B}$

$\mathrm{Logo}\text{:}$

$C-A=9A-A=8A=2(4A)\ \therefore\ \boxed{C-A=2B}$

$\mathbf{Alternativa\ B.}$

$\mathbf{09.}$

$\mathrm{Rela\c{c}\tilde{a}o\ entre}\ V_A\ \text{e}\ V_B\text{:}$

$\dfrac{V_A}{V_B}=\bigg(\dfrac{d}{2d}\bigg)^3\Longrightarrow \dfrac{V_A}{V_B}=\bigg(\dfrac{1}{2}\bigg)^3\Longrightarrow \dfrac{V_A}{V_B}=\dfrac{1}{8}\ \therefore\ \boxed{V_B=8V_A}$

$\mathrm{Se}\ V_A=1\ \mathrm{u.v.}\Longrightarrow V_B=8(1)\ \therefore\ \boxed{V_B=8\ \mathrm{u.v.}}$

$\mathbf{10.}$

$\mathrm{Seja}\ A=d\ \mathrm{u.a.}$

$\mathrm{Rela\c{c}\tilde{a}o\ entre}\ A\ \text{e}\ C\Longrightarrow \boxed{C=9A}$

$\mathrm{Logo}\text{:}$

$V_C=Cd=(9A)d=(9d)d\ \therefore\ \boxed{V_C=9d^2\ \mathrm{u.v.}}$