Problema 27 - Relaciones Métricas -Vol. 8
Enviado: 02 Jun 2024, 13:54
Na figura calcular "x".
A) 1m
B) 2m
C) 3m
D) 4m
E) 5m
Resposta:D
A) 1m
B) 2m
C) 3m
D) 4m
E) 5m
Resposta
Resposta:D
[tex3]\mathsf{
\triangle AMD \sim \triangle CMB \implies \frac{8}{2} = \frac{AM}{CM} \therefore AM = 4CM\\
\triangle ABC: BM^2= AM.CM \implies BM = \sqrt{AM.\frac{AM}{4}}=\frac{AM}{2}\\
\triangle AMB \sim \triangle BMC \implies \frac{x}{2}=\frac{AM}{BM} \therefore x = \frac{2AM}{BM}\\
Substituindo(BM): x = \frac{2AM}{\frac{AM}{2}} \implies \boxed{x=4}
}[/tex3]