Questões Perdidas ⇒ Solucionário:Racso - Cap II - Problemas de Geometria y Como Resolverlos - I Edição - Ex:24 Tópico resolvido

[petras]

Problema Proposto
24 - A, B, C e D são pontos colineares e consecutivos.
Se AC é a média proporcional entre AD e BD.
Calcular o valor de C se [tex3]C = 2\frac{AD}{AC}(\frac{AB}{CD}-1)[/tex3]

Resposta

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E)2

[petras]

[tex3]\mathsf{
AC = a\\
AB = b\\
BD = C\\
Média~Geométrica: AC^2 = AD.BD\implies a^2 = bc+c^2\\
C= 2.(\frac{AD.AB}{AC.CD}-\frac{AD}{AC}) \implies 2.(\frac{AD.AB-AD.CD}{AC.CD})\\
2[\frac{(b+c).(b-(b+c-a))}{a(b+c-a)}]=2.[\frac{(b+c)(a-c)}{ab+ac-a^2}]=2.(\frac{ab-bc+ac-c^2}{ab+ac-a^2})\\
2.(\frac{ab+ac-(\underbrace{bc+c^2}_{a^2})}{ab+ac-a^2)})=2.(\frac{\cancel{ab+ac-a^2}}{\cancel{ab+ac-a^2}})\\
\therefore \boxed{\color{red}C=2}









}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
AC = a\\
AB = b\\
BD = C\\
Média~Geométrica: AC^2 = AD.BD\implies a^2 = bc+c^2\\
C= 2.(\frac{AD.AB}{AC.CD}-\frac{AD}{AC}) \implies 2.(\frac{AD.AB-AD.CD}{AC.CD})\\
2[\frac{(b+c).(b-(b+c-a))}{a(b+c-a)}]=2.[\frac{(b+c)(a-c)}{ab+ac-a^2}]=2.(\frac{ab-bc+ac-c^2}{ab+ac-a^2})\\
2.(\frac{ab+ac-(\underbrace{bc+c^2}_{a^2})}{ab+ac-a^2)})=2.(\frac{\cancel{ab+ac-a^2}}{\cancel{ab+ac-a^2}})\\
\therefore \boxed{\color{red}C=2}









}[/tex3]