Respuesta :

pretty sure this is the answer, hope this helps :')
Ver imagen taehyungtheguccigod
1rstar
◆ AREA OF 2-D FIGURES ◆

[tex]let \: the \: length \: of \: rectangle \: be \: l \: yd\: \\ \\ and \: width \: of \: rectangle \: be \: b \: yd\\ \\ now \: , \: its \: given \: that \: , \\ \\ length \: of \: rectangle \: is \: 11 \: yd \\ \: less \: than \: 3 \: times \: the \: width \: , \\ \\ l = 3b - 11 \: \: \: \\ \\ area \: of \: rectangle \: = length \times width \\ \\ 70 = (3b - 11) \times b \\ \\ 3 {b}^{2} - 11b - 70 = 0 \\ \\ 3 {b}^{2} - 21b + 10b - 70 = 0 \\ \\ 3b(b - 7) + 10(b - 7) = 0 \\ \\ (3b + 10)(b - 7) = 0 \\ \\ solving \: , \: we \: get \: , \\ b = \frac{ - 10}{3} \: \: \: \: or \: \: \: b = 7 \\ \\ side \: of \: any \: shape \: can \: not \: be \: negative \: , \\ \\ therefore \: , \\ \\ width \: = 7 \: yd \: \: \: \: ans.\\ \\ length \: = 10 \: yd \: \: \: ans.[/tex]