Questões Perdidas ⇒ Solucionário:Racso - Cap XXI - Problemas de Geometria y Como Resolverlos - I Edição - Ex:04 Tópico resolvido

[petras]

Problema Proposto
4 - Na figura se : Área (triângulo ABC) = 9 m²
Calcular SX(AO=2, AB = 3)

Resposta

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C) 5m²

[petras]

[tex3]\mathsf{OG=OH=r\\
BE \perp AC (E \in AC)\\
△ABE∼△AOG \implies \frac{S_{AOG}}{S_{ABE}} = \frac{4}{9} \therefore S_{ABE} =\frac{9S_{AOG}}{4} \\
△COH \sim △CBE \implies \frac{S_{COH}}{S_{BCE}}=\frac{4}{9}\therefore S_{BCE} = \frac{9S_{COH}}{4}\\
S_{ABC}=S_{ABE}+S_{BCE}=9 \\
\frac{9S_{AOG}}{4}+\frac{9S_{COH}}{4}=9 \therefore S_{AOG}+S_{COH} = 4 \implies
\boxed{\color{red}S_X=9 - 4 = 5}
}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{OG=OH=r\\
BE \perp AC (E \in AC)\\
△ABE∼△AOG \implies \frac{S_{AOG}}{S_{ABE}} = \frac{4}{9} \therefore S_{ABE} =\frac{9S_{AOG}}{4} \\
△COH \sim △CBE \implies \frac{S_{COH}}{S_{BCE}}=\frac{4}{9}\therefore S_{BCE} = \frac{9S_{COH}}{4}\\
S_{ABC}=S_{ABE}+S_{BCE}=9 \\
\frac{9S_{AOG}}{4}+\frac{9S_{COH}}{4}=9 \therefore S_{AOG}+S_{COH} = 4 \implies
\boxed{\color{red}S_X=9 - 4 = 5}
}[/tex3]

(Solução:MathLover)