Cap. 4 - Problemas Adicionais ⇒ Problema 109 - Relaciones Métricas -Vol. 8 Tópico resolvido

[petrasMOD]

Na figura mostrada calcular "r", se
"R"= 17m
A) 1m
B) 2m
C) 3m
D) 4m
E) 5m

Resposta

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Resposta:D

[petrasMOD]

[tex3] \mathsf{ ON = x\\
s = \frac{34}{4} = 8,5\\
\triangle MO_1N: (s+t)^2 = (s+x)^2+t^2\\
s^2+2st+t^2 = s^2+2sx+x^2+t^2 \implies 2st = 2sx+x^2\\
17t = 17x + x^2(I)\\
\triangle OO_1N: (R-t)^2 = t^2+x^2 \\
R^2-2RT+t^2 = t^2+x^2 \implies R^2-2Rt = x^2\\
289-34t = x^2 \implies 289- 2(17t) =x^2 \\
De(I): 289 - 2(17x+x^2) = x^2 \implies 3x^2+34x-289=0\\
{x = -17} \vee x = \frac{17}{3}
de(I): 17t = 17.\frac{17}{3}+ (\frac{17}{3})^2\\
t = \frac{17}{3}+\frac{17}{9} = \frac{68}{9}\\
h = O_2Y : w = MY\\
\triangle MOO_2:p = \frac{MO+MO_2+OO_2}{2} =\frac{8,5+(8,5+r)+(17-r)}{2}=17\\
S_\triangle MOO_2 = \frac{8,5.h}{2} = \sqrt{17(17-8,5)(17-(8,5+r))(17-(17-r))}\\
\therefore h^2 = 4(17)t-8r^2\\
\triangle MYO_2: w^2 = (8,5+r)^2-h^2 \implies w = 3r - 8,5\\
XO_1=YN=MN−w=8.5+\frac{17}{3}−(3r−8.5)=−3r+\frac{4(17)}{3}\\
O_ 2X=O_2Y−O_ 1N=h−\frac{68}{9}\\
O_1O_2=r+\frac{68}{9}\\
△O_1XO_2:
(-3r + \frac{4(17)}{3})^2 + (h - \frac{68}{9})^2 = (r + \frac{68}{9})^2 \\
9r^2 - 8(17)r + \frac{16(289)}{9} + h^2 - \frac{136h}{9} + (\frac{68}{9})^2 = r^2 + \frac{136r}{9} + (\frac{68}{9})^2 \\
8r^2 - 8(17)r + \frac{16(289)}{9} + (4(17)r - 8r^2) - \frac{136h}{9} = \frac{136r}{9} \\
- 4(17)r + \frac{16(289)}{9} - \frac{136h}{9} = \frac{136r}{9} \\
-9r + 4(17) - 2h = 2r \\
-11r + 4(17) = 2h \\
121r^2 - 8(11)(17)r + (4(17))^2 = 16(17)r - 32r^2 \\
153r^2 - 104(17)r + (4(17))^2 = 0 \\
9r^2 - 104r + 16(17) = 0
\therefore \cancel{r = \frac{68}{9} }(r < \frac{68}{9})\vee \boxed{r=4}
}
[/tex3]
(Solução:JohnOmielan-adaptada)