Quadriláteros - 2003 - Vol 4 ⇒ Problema 06 - Quadriláteros -Vol. 4 Tópico resolvido

[petrasMOD]

Se ABCD é um quadrado e "M' e "N" são pontos médios, calcular QC sendo BQ = 1m.

Resposta

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Gabarito: B) [tex3]\sqrt{2}[/tex3]

[petrasMOD]

[tex3]\mathsf{
\triangle BMQ_{(notável)}: R:2R: \sqrt{5}R\\
2R=1 \therefore R = \frac{1}{2} = MQ \implies BM=MC =CN= \frac{\sqrt{5}}{2}\\
\angle MAB \cong \angle BM \therefore \triangle ABM \sim \triangle BQM \implies \angle MQB = 90^o \\
\therefore MQNC_{(inscritivel)} \implies CQ.MN = MC.QN+MQ.CN(I)\\
\triangle BCN: BN^2 = ( \sqrt {5} )^2+(\frac{\sqrt5}{2})^2 \implies BN = \frac{5}{2}\\
\therefore QN = \frac{5}{2} - 1 = \frac{3}{2}\\
\triangle MQN: MN^2 = (\frac{1}{2})^2+(\frac{3}{2})^2 \implies MN = \frac{\sqrt{10}}{2}\\
De(I): QC.\frac{\sqrt{10}}{2} = \frac{\sqrt5}{2}.\frac{3}{2}+\frac{1}{2}.\frac{\sqrt5}{2} \implies \boxed{QC = \sqrt2}







}[/tex3]