Respuesta :

[tex]\bf \textit{perimeter of a rectangle}=p=side+side+side+side \\\\ or\implies p=w+w+l+l\implies p=2w+2l \\\\ p=2(w+l)\qquad \begin{cases} w=width\\ l=length\\ ------\\ p=48 \end{cases}\implies 40=2(w+l)\\\\ -----------------------------\\\\ \textit{area of it}=A=w\cdot l\qquad \begin{cases} w=width\\ l=length\\ ------\\ A=40 \end{cases}\implies 40=wl\\\\ -----------------------------\\\\ [/tex]

[tex]\bf thus\qquad \begin{cases} 40=2(w+l)\to \frac{40}{2}=w+l\to\frac{40}{2}-l=\boxed{w} \\\\ 40=wl\\ --------------\\ 40=\left( \boxed{\frac{40}{2}-l} \right)\cdot l \end{cases}[/tex]

solve for "l" to find its length

Otras preguntas