Question

1) Na figura seguinte, temos que AB = (x+2), BC = x, MN = (x-2) e NP= 3cm. Se A//B//C, determine as medidas dos seguimentos AB, BC, AC.

Answer 1

$\large\boxed{\begin{array}{l}\sf\dfrac{x}{x+2}=\dfrac{3}{x-2}\\\\\sf x\cdot (x-2)=3\cdot(x+2)\\\sf x^2-2x=3x+6\\\sf x^2-2x-3x-6=0\\\sf x^2-5x-6=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-5)^2-4\cdot1\cdot(-6)\\\sf\Delta=25+24\\\sf\Delta=49\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-5)\pm\sqrt{49}}{2\cdot1}\\\\\sf x=\dfrac{5\pm7}{2}\\\sf como~x > 0\,devemos\,pegar\,a\,parte\,positiva\,somente\\\sf x=\dfrac{5+7}{2}=\dfrac{12}{2}=6\end{array}}$

$\large\boxed{\begin{array}{l}\sf AB=x+2=6+2=8\\\sf BC=6\\\sf AC=6+3=9\end{array}}$