SOLUTION: Can you please help me on the following problem: "A rectangle whose perimeter is 200 is 4 times as long as it is wide. Find the dimensions and area." CAn you please help me.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Can you please help me on the following problem: "A rectangle whose perimeter is 200 is 4 times as long as it is wide. Find the dimensions and area." CAn you please help me.      Log On


   

Question 57012: Can you please help me on the following problem: "A rectangle whose perimeter is 200 is 4 times as long as it is wide. Find the dimensions and area." CAn you please help me.
Answer by rcmcc(152) About Me  (Show Source):
You can put this solution on YOUR website!
You can solve this question by solving for X.
Let X be the width of the rectangle.
Since the length is 4 times greater than the width it can be stated as 4X
Because we are calculating perimeter all the values must be multiplied by 2, as there are two identical sides.
Now we can form a simple equation.
(2)4X+(2)X=200
Simplify the equation
8X+2X=200
10X=200
divide 10 out to isolate x
X=20
Hence the width is 20 and the length is 80
The area is LxW
20X80=1600