Ensino Médio ⇒ Circunferência Tópico resolvido

[AngelitaB]

Se ON=4[tex3]\sqrt{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{2}[/tex3]

, UH=4, HP=5 é CP=7, calcular X.

IMG-20240422-WA0008.jpg (61.55 KiB) Exibido 109 vezes

a)5
b)6
c)7
d)8
e)9

Resposta

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b

[petras]

AngelitaB,

[tex3]\triangle UHN: c=NU : a=NH : b=UH=4 \\
p=\frac{(a+b+c)}{2}\\

\triangle ODN:ON^2=OD^2+ND^2⇔32=(p−c)^2+(p−b)^2\\

p−c=\frac{a+b−c}{2}=\frac{a−(c−b)}{2}\\
p−b=\frac{a+c−b}{2}=\frac{a+(c−b)}{2}
\\
32=\frac{1}{4}[({a−(c−b))}^2+(a+(c−b))^2]=\frac{1}{2}(a^2+b^2+c^2−2bc)=\frac{1}{2}(c^2+c^2−8c)=c2−4c
\\
\therefore c^2−4c-32=0 \implies c = 8 \vee \cancel{c = - 4}.
\\
\therefore NU=c=8 \implies T.Pitot: X=NU+CP−UP=7+8−9=\boxed{6}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\triangle UHN: c=NU : a=NH : b=UH=4 \\
p=\frac{(a+b+c)}{2}\\

\triangle ODN:ON^2=OD^2+ND^2⇔32=(p−c)^2+(p−b)^2\\

p−c=\frac{a+b−c}{2}=\frac{a−(c−b)}{2}\\
p−b=\frac{a+c−b}{2}=\frac{a+(c−b)}{2}
\\
32=\frac{1}{4}[({a−(c−b))}^2+(a+(c−b))^2]=\frac{1}{2}(a^2+b^2+c^2−2bc)=\frac{1}{2}(c^2+c^2−8c)=c2−4c
\\
\therefore c^2−4c-32=0 \implies c = 8 \vee \cancel{c = - 4}.
\\
\therefore NU=c=8 \implies T.Pitot: X=NU+CP−UP=7+8−9=\boxed{6}[/tex3]

(Solução:LuisFuentes-adaptada)