Cap. 2 - Segmentos de Recta ⇒ Solucionário:Racso - Cap II - Problemas de Geometria y Como Resolverlos - I Edição - Ex:11 Tópico resolvido

[petras]

Problema Proposto
11- Sobre uma linha reta se consideram os pontos consecutivos
A, B. C, D, E e F tal que: AC+BD+CE+DF=91 e BE= [tex3]\frac{5}{8}[/tex3]AF.
Calcular AF

Resposta

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A) 56

[petras]

[tex3]\mathsf{
AB = a\\
BC = b\\
CD = c\\
DE = d\\
EF=e\\
b+c+d=\frac{5}{8}(a+b+c+d+e)\implies a+b+c+d+e =\frac{8}{5}(b+c+d) \\
a+b+b+c+c+d+d+e=91\implies \frac{8}{5}(b+c+d)+(b+c+d)
=91\\ \therefore(b+c+d) = 35=\frac{5}{8}AF\\
\therefore (a+b) = \frac{3}{8}AF \implies a+b = 21\therefore \boxed{\color{red}AF=21+35=56 }









}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
AB = a\\
BC = b\\
CD = c\\
DE = d\\
EF=e\\
b+c+d=\frac{5}{8}(a+b+c+d+e)\implies a+b+c+d+e =\frac{8}{5}(b+c+d) \\
a+b+b+c+c+d+d+e=91\implies \frac{8}{5}(b+c+d)+(b+c+d)
=91\\ \therefore(b+c+d) = 35=\frac{5}{8}AF\\
\therefore (a+b) = \frac{3}{8}AF \implies a+b = 21\therefore \boxed{\color{red}AF=21+35=56 }









}[/tex3]

[LostWalker]

Definindo uma Reta

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Vamos estabelecer as variáveis a reta, de forma que:

[tex3]\begin{matrix}A&\leftarrow v\rightarrow&B&\leftarrow w\rightarrow&C&\leftarrow x\rightarrow&D&\leftarrow y\rightarrow&E&\leftarrow z\rightarrow&F\end{matrix}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\begin{matrix}A&\leftarrow v\rightarrow&B&\leftarrow w\rightarrow&C&\leftarrow x\rightarrow&D&\leftarrow y\rightarrow&E&\leftarrow z\rightarrow&F\end{matrix}[/tex3]

Transcrevendo as equações dadas temos:

[tex3]\mbox{I.}\,\,\,\,\,\,\,v+2(w+x+y)+z=91\\\mbox{II.}\,\,\,\,\,w+x+y=\frac{5}{8}\cdot(v+w+x+y+z)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{I.}\,\,\,\,\,\,\,v+2(w+x+y)+z=91\\\mbox{II.}\,\,\,\,\,w+x+y=\frac{5}{8}\cdot(v+w+x+y+z)[/tex3]

Mudança de Variável

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Olhando a estrutura das duas equações, convêm uma mudança de variável:


[tex3]\mbox{I.}\,\,\,\,\,\,\,{\color{PineGreen}v}+2{\color{Purple}(w+x+y)}+{\color{PineGreen}z}=91\\\mbox{II.}\,\,\,\,\,{\color{Purple}w+x+y}=\frac{5}{8}\cdot({\color{PineGreen}v}+{\color{Purple}w+x+y}+{\color{PineGreen}z})[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{I.}\,\,\,\,\,\,\,{\color{PineGreen}v}+2{\color{Purple}(w+x+y)}+{\color{PineGreen}z}=91\\\mbox{II.}\,\,\,\,\,{\color{Purple}w+x+y}=\frac{5}{8}\cdot({\color{PineGreen}v}+{\color{Purple}w+x+y}+{\color{PineGreen}z})[/tex3]

[tex3]\mbox{I.}\,\,\,\,\,\,\,{\color{PineGreen}a}+2{\color{Purple}b}=91\\\mbox{II.}\,\,\,\,\,{\color{Purple}b}=\frac{5}{8}\cdot({\color{PineGreen}a}+{\color{Purple}b})[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{I.}\,\,\,\,\,\,\,{\color{PineGreen}a}+2{\color{Purple}b}=91\\\mbox{II.}\,\,\,\,\,{\color{Purple}b}=\frac{5}{8}\cdot({\color{PineGreen}a}+{\color{Purple}b})[/tex3]

Reorganizando a segunda nessa ideia:

[tex3]\mbox{II.}\,\,\,\,\,3b=5a[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{II.}\,\,\,\,\,3b=5a[/tex3]

Resolvendo:

[tex3]\mbox{I.}\\a+2b=91\\{\color{PineGreen}5a}+10b=455\\{\color{PineGreen}3b}+10b=455\\\boxed{b=35}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{I.}\\a+2b=91\\{\color{PineGreen}5a}+10b=455\\{\color{PineGreen}3b}+10b=455\\\boxed{b=35}[/tex3]

[tex3]\mbox{II.}\\3b=5a\\3\cdot{\color{Red}\cancel{\color{Black}35}^7}={\color{Red}\cancel{\color{Black}5}}a\\\boxed{a=21}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mbox{II.}\\3b=5a\\3\cdot{\color{Red}\cancel{\color{Black}35}^7}={\color{Red}\cancel{\color{Black}5}}a\\\boxed{a=21}[/tex3]

[tex3]S=a+b\\S=35+21\\\color{MidNightBlue}\boxed{S=56}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]S=a+b\\S=35+21\\\color{MidNightBlue}\boxed{S=56}[/tex3]