Vol. 09 - Geometria Plana FME 2004 ⇒ 109 - FME 09 -Teste de Vestibulares Tópico resolvido

[petrasMOD]

(U.MACK.-79) No triângulo retângulo ABC da figura, b = 1 e c = 2. Então, x vale:

Resposta

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Gabarito: e) [tex3]\frac{2\sqrt{2}}{3}[/tex3]

[petrasMOD]

Seja PC = a
PH = h = altura do triângulo APB


[tex3]\triangle ABC: BC^2 = AB^2+AC^2 = 2^2+1^2 \implies BC = \sqrt5\\
T.Bissetriz: \frac{PC}{AC} = \frac{AB}{PB} \implies\frac{1}{a} = \frac{2}{\sqrt5-a} \implies \therefore a =\frac{\sqrt5}{3}\\
\therefore PB = \sqrt5 - \frac{\sqrt5}{3} = \frac{2\sqrt5}{3}\\
\triangle BHP \sim \triangle BAC: \frac{h}{AC} = \frac{PB}{BC} \implies \frac{h}{1} = \frac{\frac{2\sqrt5}{3}}{\sqrt5}\\
\therefore h = \frac{2}{3}\\
\triangle APH: sen 45^o = \frac{h}{x} \implies \frac{\sqrt2}{2} = \frac{\frac{2}{3}}{x} \implies x = \frac{4}{3\sqrt2} = \boxed{\frac{2\sqrt2}{3}}
[/tex3]

[rcompany]

[tex3]\text{Seja $Q$ a projeção ortogonal de $P$ sobre $(AB)$}\\
(AP) \text{ bissetriz}\implies \frac{CP}{AB}=\frac{BP}{AB}\implies CP=\frac{1}{2}\cdot BP\implies AQ=\frac{1}{2}\cdot QB\implies AQ=\frac{1}{3}\cdot AB=\frac{2}{3}\\
\triangle APQ:\, \angle APQ=180-\angle QAP-\angle PQA=180-45-90=45°=\angle QAP\implies AQ=QP=\frac{2}{3}\\
AP^2=2\cdot AQ^2\implies AP^2=\frac{8}{9}\implies AP=\frac{2}{3}\cdot\sqrt{3}[/tex3]