Cap. 3 - Relações Métricas na Circunferência ⇒ Problema 092 - Relaciones Métricas -Vol. 8 Tópico resolvido

[petrasMOD]

Na figura se BC = 3m e CD = 1m.
Calcular AE.
A) 2[tex3]\sqrt{6}[/tex3]m
B) 2[tex3]\sqrt{3}[/tex3]m
C) 3[tex3]\sqrt{6}[/tex3]m
D) [tex3]\sqrt{3}[/tex3]m
E) [tex3]\sqrt{6}[/tex3]m

Resposta

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Resposta:E

[petrasMOD]

[tex3]\mathsf{

\angle BFA = \alpha \implies \angle OO_1A =2\alpha\\
\angle EAH = \theta=\frac{\angle AOC}{2} \implies \angle AOC = 2\theta\\
\therefore 2\theta +\theta = 90^o \implies \theta = 30^o \\
\triangle AOO_1: 2\alpha+ 90^o + 2\theta = 180^0 \implies 2\alpha = 30^o \therefore \alpha = 15^o \\
\angle AHE \cong \angle BFA= \alpha=15^o \implies \angle EH = 180^o -(15^o +30^o ) = 135^o \\
\therefore \angle O_1EH = 45^ o \implies O_1H \perp EF\\
\triangle ECF: \angle DEC = 180^o -90^o - 45^o-15^o = 30^o \\
\triangle ECD: tg 30^o = \frac{1}{EC} \implies EC = \sqrt3 \\
\triangle EBC: tg \angle CEB = \frac{3}{\sqrt3} = \sqrt3 \therefore \angle CEB = 60^o \\
sen 60^o =\frac{BC}{BE }=\frac{\sqrt3}{2} \implies \frac{3}{BE} =\frac{\sqrt3}{2}\implies BE = 2\sqrt3\\
\therefore \angle BEA = 180^o -(45^o +30^o +60^o ) = 45^o \implies \triangle EAB_{(ret.)}\\
cos 45^o =\frac{AE}{BE} \implies \frac{\sqrt2}{2} = \frac{AE}{2\sqrt3} \\
\therefore \boxed{AE = \sqrt6}












}[/tex3]