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
