[tex3]\mathsf{
\text{os lados do triangulo são iguais à R+r}\\
\therefore h=\frac{\sqrt3}{2}(R+r)h\\
x=R−\frac{R+r}{2}=\frac{R−r}{2}\\
x^2+h^2=BP^2\\
(\frac{R−r}{2})^2+(\frac{\sqrt3}{2}.\frac{R+r}{2})^2=2^2\\
R^2+Rr+r^2=4\\
S_{AMNC} = S_{AMD }+S_{CNE}+S_{MNED}\\
=\frac{R\sqrt3}{2}.\frac{R}{2}.\frac{1}{2}+\frac{r\sqrt3}{2}.\frac{r}{2}.\frac{1}{2}+\frac{R+r}{2}.\frac{(\frac{R\sqrt3}{2}+\frac{r\sqrt3}{2})}{2}=\\
=\sqrt3(\frac{R^2+Rr+r^2}{4})=\\
\therefore \boxed{\color{red}S_{AMNC} = =\sqrt3.\frac{4}{4}=\sqrt3}
}[/tex3]
(Solução: jedi -
viewtopic.php?f=4&t=58891&p=155467&hili ... C.#p155467)