[tex]\text{Given that,}\\\\\text{Area of the rectangle,}~ A = 344 ~ in^2\\\\\text{Width of the rectangle,}~w = 16 ~ in\\\\\text{Length of the rectangle,}~ l=?\\\\~~~~~~A = l \times w \\\\\implies l= \dfrac{A}w =\dfrac{344}{16}=21.5 ~in\\\\\text{Perimeter of the rectangle} = 2(l+w) = 2(21.5+16)=75~ in[/tex]
