Question

The length of a rectangle is twice the width. The perimeter is 48in. Find the dimensions of the rectangle.

Answer 1

w = 6in

l = 12in

Explanation:

To solve this problem we propose 2 equations

2w + 2l = 48in

l = 2w

We replace the value of l = (2w) into the first equation and solve

2w + 2(2w) = 48in

2w + 4w = 48in

6w = 48in

w = 48in/6

w = 8in

we replace the value of w and solve

l = 2(8in)

l = 16in

Answer 2

perimeter = p
length = l
width = w

p = 2l + 2w

p = 48
l = 2w
w = ?

now plug those values into the base equation (p = 2l + 2w)

48 = 2(2w) + 2w
then solve for w
48 = 4w + 2w
48 = 6w
w = 8

because l is 2w, find the length using the width value (8)

l = 2w = 2(8) = 16

SO

length = 16 in
width = 8 in
perimeter = 48 in