Vol. 09 - Geometria Plana FME 2013 ⇒ 082 - FME 09 -Teste de Vestibulares 2013 Tópico resolvido

[petrasMOD]

(FEI-SP) Num triângulo ABC, os lados medem AB = 5 cm, AC = 7 cm e BC = 8 cm. Se M é o ponto médio do lado BC, então a medida do segmento AM é:

a) [tex3]\sqrt{29}cm[/tex3]
b) 2[tex3]\sqrt{3}cm[/tex3]
c) [tex3]\sqrt{21}cm[/tex3]
d) 6cm
e) [tex3]\sqrt{19}cm[/tex3]

Resposta

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Gabarito: c

[rcompany]

[tex3]S(ABC)=\sqrt{\frac{AB+BC+AC}{2}\left(\frac{AB+BC+AC}{2}-AB\right)\left(\frac{AB+BC+AC}{2}-BC\right)\left(\frac{AB+BC+AC}{2}-AC\right)}=\sqrt{300}=10\sqrt{3}\\
S(ABM)=\frac{1}{2}S(ABC)=5\sqrt{3}\\
\implies 75=\frac{AB+BM+AM}{2}\left(\frac{AB+BM+AM}{2}-AB\right)\left(\frac{AB+BM+AM}{2}-BM\right)\left(\frac{AB+BM+AM}{2}-AM\right)\\
\implies 1200=(AM+9)(AM-1)(AM+1)(9-AM)\\
\implies 1200=(81-AM^2)(AM^2-1)\\
\implies AM^4-82AM^2+1281=0\\
\implies AM=\sqrt{61}\text{ ou }AM=\sqrt{21}\\
\implies AM=\sqrt{21}\text{ já que }5<AM<7\\
\\\fbox{$\quad$resposta c$\quad$}


[/tex3]