Question

Find the dimensions of a rectangle whose area is 285 cm2 and whose perimeter is 68 cm.

Answer 1

$\bf \textit{area of a rectangle}\\\\ A=lw\quad \begin{cases} l=length\\ w=width\\ -----\\ A=285 \end{cases}\implies 285=lw\implies \boxed{\cfrac{285}{w}=l} \\\\\\ \textit{perimeter of a rectangle}\\\\ P=2(l+w)\quad \begin{cases} l=\frac{285}{w}\\\\ P=68 \end{cases}\implies 68=2\left( \boxed{\cfrac{285}{w}}+w \right) \\\\\\ \cfrac{68}{2}=\cfrac{285}{w}+w\implies 34=\cfrac{285+w^2}{w}\implies 34w=285+w^2$

$\bf 0=w^2-34w+285\implies 0=(w-19)(w-15)\implies w= \begin{cases} 19\\ 15 \end{cases} \\\\\\ \stackrel{lenght}{l}= \begin{cases} \frac{285}{19}\implies &15\\\\ \frac{285}{15}\implies &19 \end{cases}$