Cap. 13 - Relaciones Métricas en Triângulos Rectângulos ⇒ Solucionário:Racso - Cap XIII - Problemas de Geometria y Como Resolverlos - I Edição - Ex:28 Tópico resolvido

[petrasMOD]

Problema Proposto
28 - Na figura: BC || AD. Se AP=4 , DL=a , CL=b
[tex3]\sqrt{2}a[/tex3]+[tex3]\sqrt{2}a^2[/tex3]- b²=6
Calcular: 3a + b[tex3]\sqrt{2}[/tex3].

Resposta

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D) 14

[petrasMOD]

[tex3]\mathsf{\triangle PCL(retângulo-isósceles) \implies \boxed{PC=LP=\frac{b\sqrt2}{2}}(I)\\
\triangle ADP: AL (bissetriz)\implies \frac{LP}{AP}=\frac{DL}{AD}\\
\therefore \boxed{AD =\frac{4a\sqrt2}{b}}(II)\\
\triangle ABP \sim \triangle APD \implies \frac{AP}{CP} = \frac{DP}{BP} = \frac{AD}{BC}\implies\\
\frac{4}{\frac{b\sqrt2}{2}}=\frac{8}{b\sqrt2}=\frac{\frac{4a\sqrt2}{b}}{BC}\therefore \boxed{BC = a} \\
LH \perp AD (H\in AD) \implies \triangle BCP \sim \triangle DHL\implies \frac{CP}{LH}=\frac{\underbrace{BC}_{=a=DL}}{DL}=\frac{BP}{DH}\\\therefore BP = DH ~e ~
CP=LH =LP= \frac{b\sqrt2}{2}\implies \triangle APL \cong\triangle ALH \therefore AH = AP = 4\\
Em(II)AD = AH+HD = 4+HD = \frac{4a\sqrt2}{b}\implies \boxed{HD = \frac{4a\sqrt2-b}{b}}(III)\\
\\\triangle DHL: a^2 = HD^2 + (\frac{b\sqrt2}{2})^2 \implies \boxed{4\sqrt{2}(a\sqrt{2}-b)=b\sqrt{2a^2-b^2}}(IV)\\
mas~ \sqrt{2}a+\sqrt{2a^2-b^2}=6\sqrt{2} \implies \sqrt{2a^2-b^2}=\sqrt2(6-a)\\
Substituindo ~em ~(IV):\boxed{ 8a-10b\sqrt{2}=-ab\sqrt{2}}(V)\\
HD = BP=AD\frac{b\sqrt{2}}{8}\rightarrow BP=\frac{b\sqrt{2}(b\sqrt{2}+2a)}{16}=\frac{b(b+a\sqrt{2})}{8}(VI)\\
Utilizando~(VI)~novamente ~\triangle DHL: \\
a^2 = LH^2+HD^2=LD^2 \implies -ab\sqrt{2}=-48+8a+b^2(VII)\\
(VII)em(V) \implies b=2\sqrt2\\
Substituindo(V) \implies a = \frac{10}{3}\\
\therefore \boxed{\color{red}{3a+2\sqrt2 b = 3*\frac{10}{3}+2\sqrt{2}*\sqrt{2}=14}}

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(Solução - jvmago - viewtopic.php?f=2&t=63006&p=168231&hili ... Da#p168231)