SOLUTION: suppose that the width of a certain rectangle is 2 inches more than one fourth of its length. the perimeter is 54 inches. find the length and the width.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: suppose that the width of a certain rectangle is 2 inches more than one fourth of its length. the perimeter is 54 inches. find the length and the width.      Log On


   

Question 262207: suppose that the width of a certain rectangle is 2 inches more than one fourth of its length. the perimeter is 54 inches. find the length and the width.
Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length = x
Given, the width of a certain rectangle is 2 inches more than one fourth of its length.
So, width = x/4 +2
Perimeter is 2*(x+x/4+2) = 54
2x+x/2+4 = 54
(4x+x)/2 = 54-4
5x = 2*50
5x = 100
x = 100/5
x = 20 inches
Therefore the length = 20
width = x/4+2 = 20/4+2 = 5+2 = 7 inches