Question

The perimeter of a rectangle is 48 in. If the length is twice

Answer 1

Answer:

$\boxed{\sf Length \ of \ the \ rectangle = 16 \ in}$

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

$\sf \implies Perimeter \ of \ rectangle = 2(Length + Width \\ \\ \sf \implies 48 = 2(2w + w) \\ \\ \sf \implies 48 = 2(3w) \\ \\ \sf \implies 48 = 6w \\ \\ \sf \implies 6w = 48 \\ \\ \sf \implies \frac{ \cancel{6}w}{ \cancel{6}} = \frac{48}{6} \\ \\ \sf \implies w = \frac{48}{6} \\ \\ \sf \implies w = \frac{8 \times \cancel{6}}{ \cancel{6}} \\ \\ \sf \implies w = 8 \: in$

Width of the rectangle (w) = 8 in

Length of the rectangle = 2w

= 2 × 8

= 16 in