Question

Find the missing measurements for the rectangle when the area equals 216 and the perimeter equals 66

Answer 1

`\bf \stackrel{\textit{perimeter of a rectangle}}{P=2(L+w)}~~ \begin{cases} L=length\\ w=width\\ \cline{1-1} P=66 \end{cases}\implies 66=2(L+w) \\\\\\ 33=L+w\implies \boxed{33-w=L} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a rectangle}}{A=Lw}\qquad \implies 216=Lw\implies 216=(33-w)w \\\\\\ 216=33w-w^2\implies w^2-33w+216=0 \\\\\\ (w-24)(w-9)=0\implies w= \begin{cases} 24\\ 9 \end{cases}`

now, both values are valid, so if "w" is either one, "L" is the other.