The length of a rectangle is 5 ft longer than its width. if the perimeter of the rectangle is 34 ft, find its area

Length = 11 ft
Width = 6 ft
11 times 6 = 66.
The area of the rectangle is 66 feet.
Hope this helps!
I would appreciate it if I got brainliest! Thanks! =D

First, write an equation based on the problem
"The length of a rectangle is 5 ft longer than its width" could be written as
⇒ l = w + 5  (first equation)
"the perimeter of the rectangle is 34 ft" could be written as
⇒ p = 34
⇒ 2l + 2w = 34  (second equation)

Second, solve the first and second equation by substitution method
Substitute l with (w + 5) in the second equation to find the value of w
2l + 2w = 34
2(w + 5) + 2w = 34

with distribution property, remove the parenthesis
2w + 2×5 + 2w = 34
2w + 10 + 2w = 34

add like terms, the one with variable with the one with variable, and the constant with the constant
2w + 10 + 2w = 34
2w + 2w = 34 - 10
4w = 24
  w = 24/4
  w = 6

Find its length
l = w + 5
l = 6 + 5
l = 11

Third, find its area
The dimension of the rectangle is 11 ft long and 6 ft wide. Plug the numbers to area formula
a = l × w
a = 11 × 6
a = 66
The area of the rectangle is 66 square feet

Ithink if ur looking for a ratio it would be 12: 3