Question

No triângulo abaixo, calcule a medida de X.

Answer 1

Veja figura em anexo:

$A1 = \dfrac{x.y}{2} \\ \\ \\ A2 = \dfrac{4x.4x}{2} . sen 120^o \\ \\ A1 + A2 = \dfrac{x(x + 4)}{2} * sen 120^0 \\ \\ \\ \dfrac{x(4 + 4)}{2} = \dfrac{4x}{2} + \dfrac{4*4}{2} *sen 120^0 \\ \\ \\ \dfrac{x(4 + 4)}{\not 2} = \dfrac{4x}{\not 2} + \dfrac{16}{\not 2} * sen 120^o$

$x(y + 4 ) = 4x + 16 * \dfrac{ \sqrt{3} }{2} \\ \\ \\ xy + 4x = 4x + 16 * \dfrac{ \sqrt{3} }{2} \\ \\ \\ 4x = 16 .\dfrac{ \sqrt{3} }{2} \\ \\ \\ 4x = 8 \sqrt{3} \\ \\ \\ x = \dfrac{8 \sqrt{3}}{4} \\ \\ \\ \fbox{ \ 2 \sqrt{3} \ }$