The length of a rectangle is 1 ft less than two times its width. The perimeter of the rectangle is 34 ft. What is the length of the rectangle?

Question

a)6

b)11
c)36
d)equation
e)None of these

Answer 1

let w = width;  then length = 2w - 1.

 

Perimeter = 34 = 2lengths + 2 widths = 2[2w-1 + w]

 

34 = 6w -2;  36 = 6w; width w = 6. You now solve for the length.

Answer 2

L = 2W - 1

 

34 = 2L + 2W = 2(2W - 1) + 2W

 

34 = 6W - 2

 

36 = 6W

 

W = 6

 

L = 2W - 1= 2(6) - 1 = 11

 

The answer is (b).

Answer 3

Hi Alexis!

 

This problem can be set up as a system of two equations, using the two facts you know about the rectangle: first, that its length in feet (L) is 1 ft less than 2 times its width in feet (W), which can be represented in equation form as:

 

L=2W-1

 

and, second, that its perimeter is 34 ft.  The perimeter of a rectangle is given by twice the length plus twice the width, so we can write this in equation form as:

 

2L+2W=34

 

This is two equations with two unknowns, a system of equations we can solve.  The solution can be done in multiple ways.  I will choose substitution, using the first equation to transform the second equation into an equation with only one unknown, like this:

 

2(2W-1) + 2W = 34   (I replaced L with 2W-1, as the first equation allows me to do)

 

Now, I can simplify the second equation and solve for W:

 

4W-2 + 2W = 34 --> 6W-2 = 34 --> 6W = 36 --> W = 6 ft.

 

Knowing what the width is 6 ft., we can now go back to the first equation and solve for L.  If you do that, you should get one of the answers you listed.  

 

Hope this helps!

Steve