The length of a rectangle is 2 m longer than its width. If the perimeter of the rectangle is 40 m , find its area.

Question

its an intermiediate algebra word problem and i need help with it

Answer 1

Let's start by calling the width W and the length L.  We know that since the length is 2m longer than the width, we can say L = W + 2.  Then we can substitute this into the equation for the perimeter.

 

In general, the perimeter of a rectangle is P = 2L + 2W, and so in this case we know that the perimeter is 40.  Thus,

40 = 2L + 2W

40 = 2(W+2) + 2W

40 = 2W + 4 + 2W

40 = 4W + 4

36 = 4W

  9 = W

 

Since the width is 9, we then know that the length is 11 (because L = W + 2).

Now that we know the length and the width, we can find the area by using A = LW

Answer 2

(contribution deleted) Misunderstood the problem. :-(