Cap. 18 - Areas de Regiones Triangulares ⇒ Solucionário:Racso - Cap XVIII - Problemas de Geometria y Como Resolverlos - I Edição - Ex:38 Tópico resolvido

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Problema Proposto
38 - Calcular a área da região sombreada.
se AD = 12 ; DC= 5.

Resposta

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D) [tex3]\frac{1225}{128}[/tex3]

[petrasMOD]

[tex3]\mathsf{EL \perp CD (L \in CD )\implies CL=CI = 5-2 = 3\\
\triangle ABC \sim \triangle GIC: \frac{GI}{AB}=\frac{CI}{BC}=\frac{GC}{AC}\implies \\
\frac{GI}{5}=\frac{3}{12}=\frac{GC}{13}\\
\therefore GI = \frac{15}{12} ~e~CG=\frac{39}{12}\\EG = EI+GI = 2+\frac{15}{12} =\frac{39}{12} \\
BG = 12 - CG=\frac{105}{12}\\
\triangle BGH \sim \triangle ICG: \frac{GH}{CG}=\frac{BG}{CI}\implies\frac{12.GH}{39}=\frac{105}{12.3}\therefore GH = \frac{455}{48}\\
S_{HGE} = \frac{1}{2}.\frac{455}{48}.\frac{39}{12}=\frac{5915}{384}\\
\triangle EIG\sim \triangle EGH> \frac{EG}{EI}=\frac{GH}{IM}\implies \frac{39}{12.2}=\frac{455}{48.IM}\therefore IM = \frac{455}{78}\\
S_{EIM}=\frac{455.2}{78.2}=\frac{455}{78} \\
S_{HMGI} = S_{HGE}-S_{EIM}=\frac{5915}{384}-\frac{455}{78} =\boxed{\color{red}\frac{1225}{128}}
}[/tex3]