Olá
Flavio2020, boa noite. Tive a mesma ideia do colega
pedrolopes...
Solução:
- fig.png (9.58 KiB) Exibido 329 vezes
Da figura, segue que:
[tex3]\begin{cases}
De \ \overline{CD}\parallel OX, vem \ que \ os \ ângulos \ vermelhos \ são \ iguais \ , assim \ \beta=30^{\circ}\\
\angle BAO=60^{\circ} \ (ângulo \ externo \ de \ um \ hexágono)\\
\#ABCD \ é \ um \ trapézio \ isósceles \implies \angle BAD=60^{\circ}\\
Assim, \ \angle DAF=60^{\circ}
\end{cases}[/tex3]
Do [tex3]\Delta BOA[/tex3]
:
1°)
[tex3]\begin{cases}
OA=\ell\cos(60^{\circ})\implies OA=\frac{\ell}{2}\iff A\(\frac{\ell}{2},0\)\\
OB=\ell\sen(60^{\circ})\implies OB=\frac{\ell\sqrt3}{2}\iff B\(0,\frac{\ell\sqrt3}{2}\)\\
Como \ x_C=\frac{\ell}{2} \ e \ y_C=2\cdot y_B=\ell\sqrt3\iff C\(\frac{\ell}{2},\ell\sqrt3\)
\end{cases}
[/tex3]
Equação da reta [tex3]\overline{AD}[/tex3]
:
[tex3]y-y_A=\tg(60^{\circ})\cdot(x-x_A)\\
y-0=\sqrt3\cdot\(x-\frac{\ell}{2}\)\\
\boxed{\overline{AD}:\ y=x\sqrt3-\frac{\ell\sqrt3}{2}}[/tex3]
Equação da reta [tex3]\overline{CE}[/tex3]
:
[tex3]y-y_C=-\tg(30^{\circ})\cdot(x-x_C)\\
y-\ell\sqrt3=-\frac{\sqrt3}{3}\cdot\(x-\frac{\ell}{2}\)\\
\boxed{\overline{CE}: \ y=\ell\sqrt3-\frac{x\sqrt3}{3}+\frac{\ell\sqrt{3}}{6}}\\
[/tex3]
[tex3]K=\overline{AD}\cap \overline{CE}[/tex3]
:
[tex3]x\sqrt3-\frac{\ell\sqrt3}{2}=\ell\sqrt3-\frac{x\sqrt3}{3}+\frac{\ell\sqrt{3}}{6}\\
x\sqrt3+\frac{x\sqrt3}{3}=\frac{\ell\sqrt3}{2}+\ell\sqrt3+\frac{\ell\sqrt3}{6}\\
\frac{4\sqrt3x}{3}=\frac{5\ell\sqrt{3}}{3}\\
\boxed{x=\frac{5\ell}{4}}\iff\boxed{y=\frac{3\ell\sqrt3}{4}}[/tex3]
Logo, [tex3]K\(\frac{5\ell}{4},\frac{3\ell\sqrt3}{4}\)[/tex3]
.
Portanto, a inclinação da reta [tex3]\mathscr{L}[/tex3]
, é:
[tex3]m_{\mathscr{L}}=\frac{y_{B}-y_{K}}{x_{B}-x_{K}}\\
m_{\mathscr{L}}=\frac{\frac{\ell\sqrt3}{2}-\frac{3\ell\sqrt3}{4}}{0-\frac{5\ell}{4}}\\
m_{\mathscr{L}}=\frac{-\frac{\ell\sqrt3}{4}}{-\frac{5\ell}{4}}\\
\boxed{\boxed{m_{\mathscr{L}}=\frac{\sqrt3}{5}}}[/tex3]
att>>rodBR
"Uma vida sem questionamentos não merece ser vivida".