Answer:
[tex] \boxed{\sf Length \ of \ the \ rectangle = 16 \ in} [/tex]
Given:
Perimeter of rectangle = 48 in
Length = Twice the width
To Find:
Length of the rectangle
Step-by-step explanation:
Let width of the rectangle be 'w'.
So,
Length of the rectangle = 2w
[tex]\sf \implies Perimeter \ of \ rectangle = 2(Length + Width \\ \\ \sf \implies 48 = 2(2w + w) \\ \\ \sf \implies 48 = 2(3w) \\ \\ \sf \implies 48 = 6w \\ \\ \sf \implies 6w = 48 \\ \\ \sf \implies \frac{ \cancel{6}w}{ \cancel{6}} = \frac{48}{6} \\ \\ \sf \implies w = \frac{48}{6} \\ \\ \sf \implies w = \frac{8 \times \cancel{6}}{ \cancel{6}} \\ \\ \sf \implies w = 8 \: in[/tex]
Width of the rectangle (w) = 8 in
Length of the rectangle = 2w
= 2 × 8
= 16 in
