Vol. 09 - Geometria Plana FME 2004 ⇒ 100 - FME 09 -Teste de Vestibulares Tópico resolvido

[petrasMOD]

(U.E.BA-84) Na figura ao lado são dados [tex3]\frac{AE}{EC}=\frac{1}{3}[/tex3], BE = 8 cm e ED = 6 cm. O comprimento de AC, em cm, é:

Resposta

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Gabarito: c) 16

[rcompany]

[tex3]ABCD\text{ cíclico}\implies \angle BDC=\angle BAC\text{ e }\angle DCA=\angle DBA\\
\therefore\triangle CED\sim\triangle AEB,\text{ razão }r=\frac{AE}{ED}=\frac{EB}{EC}=\frac{CD}{AB}\\
\frac{AE}{EC}=\frac{1}{3}\implies EC=3\cdot AE\\
\therefore \frac{AE}{ED}=\frac{EB}{EC}\implies \frac{AE}{ED}=\frac{EB}{3AE}\implies AE^2=\frac{EB\cdot ED}{3}=\frac{8\cdot 6}{3}=16\implies AE=4\\
EC=3\cdot AE\implies EC=3\cdot 4=12\\
AC=AE+EC=4+12=16[/tex3]