Ensino Fundamental ⇒ Triângulo Tópico resolvido

[Flavio2020]

Dado o gráfico, S e T são pontos de tangencia e a medida do arco EM =120graus. Se AB=57 e BC=76, calcule OH.

IMG-20220523-WA0009.jpg (17.42 KiB) Exibido 643 vezes

a)18
b)12
c)24
d)25
e)28

Resposta

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c

[geobson]

Flavio2020, segue resolução ..

[petras]

Flavio2020, geobson,

Este livro está pior que o da racso;;Quase todos os gabaritos estão errados

[petras]

Flavio2020, geobson,

Aproveitando a ideia do geobson

[tex3]\mathrm{OH = x: \angle EOM = 120^0 \\
\triangle EOH: \angle =\frac{120^o}{2} = 60^0\implies OE= r = 2x\\
\triangle AA'F \sim ABC \implies \frac{x}{76} = \frac{AA'}{95} \therefore AA' =\frac{5x}{4}\implies A'F = \frac{3x}{4} \\
\triangle CC'D \sim ABC \implies \frac{x}{57} = \frac{CC'}{95} \therefore CC' =\frac{5x}{3}\implies C'D = \frac{4x}{3} \\
\triangle A'BC': r = \frac{b+c-a}{2} \implies 2r = (57+\frac{5x}{4}+(76+\frac{5x}{3})-(95+\frac{3x}{4}+\frac{4x}{3})\\
4x = 38+\frac{x}{2}+\frac{x}{3}\implies 24x =228+5x \therefore \boxed{x = 12}\color{green}\checkmark


}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathrm{OH = x: \angle EOM = 120^0 \\
\triangle EOH: \angle =\frac{120^o}{2} = 60^0\implies OE= r = 2x\\
\triangle AA'F \sim ABC \implies \frac{x}{76} = \frac{AA'}{95} \therefore AA' =\frac{5x}{4}\implies A'F = \frac{3x}{4} \\
\triangle CC'D \sim ABC \implies \frac{x}{57} = \frac{CC'}{95} \therefore CC' =\frac{5x}{3}\implies C'D = \frac{4x}{3} \\
\triangle A'BC': r = \frac{b+c-a}{2} \implies 2r = (57+\frac{5x}{4}+(76+\frac{5x}{3})-(95+\frac{3x}{4}+\frac{4x}{3})\\
4x = 38+\frac{x}{2}+\frac{x}{3}\implies 24x =228+5x \therefore \boxed{x = 12}\color{green}\checkmark


}[/tex3]

[geobson]

Flavio2020, geobson,

Aproveitando a ideia do geobson

[tex3]\mathrm{OH = x\implies r = 2x\\
\triangle AA'F \sim ABC \implies \frac{x}{76} = \frac{AA'}{95} \therefore AA' =\frac{5x}{4}\implies A'F = \frac{3x}{4} \\
\triangle CC'D \sim ABC \implies \frac{x}{57} = \frac{CC'}{95} \therefore CC' =\frac{5x}{3}\implies C'D = \frac{4x}{3} \\
\triangle A'BC': r = \frac{b+c-a}{2} \implies 2r = (57+\frac{5x}{4}+(76+\frac{5x}{3})-(95+\frac{3x}{4}+\frac{4x}{3})\\
4x = 38+\frac{x}{2}+\frac{x}{3}\implies 24x =228+5x \therefore \boxed{x = 12}\color{green}\checkmark


}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathrm{OH = x\implies r = 2x\\
\triangle AA'F \sim ABC \implies \frac{x}{76} = \frac{AA'}{95} \therefore AA' =\frac{5x}{4}\implies A'F = \frac{3x}{4} \\
\triangle CC'D \sim ABC \implies \frac{x}{57} = \frac{CC'}{95} \therefore CC' =\frac{5x}{3}\implies C'D = \frac{4x}{3} \\
\triangle A'BC': r = \frac{b+c-a}{2} \implies 2r = (57+\frac{5x}{4}+(76+\frac{5x}{3})-(95+\frac{3x}{4}+\frac{4x}{3})\\
4x = 38+\frac{x}{2}+\frac{x}{3}\implies 24x =228+5x \therefore \boxed{x = 12}\color{green}\checkmark


}[/tex3]

Obrigado, meu amigo!

[geobson]

Flavio2020, concorda com a solução apresentada? Se sim , por favor teria como validá- la?