Respostas

2013-09-29T02:38:40-03:00
O problema nos diz que:
c=\ell+6

Calculando o valor da largura a partir da área dada, temos:

c\times\ell=47,25\\
(\ell+6)\cdot\ell=47,25\\
\ell^2+6\ell-47,25=0\\\\
\Delta=b^2-4\cdot a\cdot c\\
\Delta=6^2-4\cdot1\cdot(-47,25)\\
\Delta=36+189\\
\Delta=225\\\\
\ell=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-6\pm\sqrt{225}}{2\cdot1}=\dfrac{-6\pm15}{2}\Longrightarrow\begin{cases}\ell_1=\frac{-6+15}{2}=\frac{9}{2}=4,5\\\ell_2=\frac{-6-15}{2}=\frac{-21}{2}=-10.5\end{cases}

Como a largura deve possuir medida positiva, temos que:
\boxed{\ell=4,5\;m}\\
\Longrightarrow c=\ell+6\Longrightarrow c=4,5+6\Longrightarrow\boxed{c=10,5\;m}