Question

na figura abaixo, o triangulo ABC é equilátero com AM =MB=4cm e CD=6 cm.​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm Observe\,a\,figura\,que\,eu\,anexei.}\\\boldsymbol{Pelo\,teorema\,de\,Menelaus\,temos:}\\\sf\dfrac{\diagdown\!\!\!\!4}{\diagdown\!\!\!\!\!4}\cdot\dfrac{8+6}{6}\cdot\dfrac{x}{y}=1\\\\\sf\dfrac{14x}{6y}=1\\\sf 6y=14x\\\sf y=\dfrac{14x\div2}{6\div2}\\\\\sf y=\dfrac{7x}{3}\\\\\sf AE+EC=8\\\sf x+y=8\\\sf x+\dfrac{7x}{3}=8\cdot(3)\\\sf 3x+7x=24\\\sf 10x=24\\\sf x=\dfrac{24\div2}{10\div2}\\\\\sf x=\dfrac{12}{5}\end{array}}$

$\large\boxed{\begin{array}{l}\sf A\,\acute area\,do\,tri\hat angulo\,CDE\,\acute e\,dado\,por\\\sf A=\dfrac{1}{2}\cdot 6\cdot\dfrac{12}{5}\cdot sen(120^\circ)\\\\\sf A=\dfrac{72}{10}\cdot\dfrac{\sqrt{3}}{2}\\\\\sf A=\dfrac{72\sqrt{3}\div4}{20\div4}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{18\sqrt{3}}{5}\,cm^2}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\red{\maltese}~\blue{alternativa~a}}}}}\end{array}}$