Dimension Problem

Question

The length of a rectangle is
11 ft
more than twice the width, and the area of the rectangle is
63 ft2
. Find the dimensions of the rectangle.

Answer 1

Let's start with what we know already:

We know that the area is Length times Width, thus  L x W.

We also know that L x W = 63

We know that L = 11 ft more than twice the width  OR L= 2W + 11

Since W is unknown, we will leave it as W at this point.

 

So now we set up our equations:

       L      x   W  = 63

(2W + 11) ( W) = 63

2W² + 11W = 63                    Combining by multiplication

2W² + 11W - 63 = 0               Set the equation to zero

(2W-7) (W+9)                        Using the quadratic equation

                                             Now we need to set each equation to zero

 

2W - 7 = 0                                                                     

       +7  +7     add 7 to both sides                                   

2W        =  7                                                                   

2               2    divide by 2 on both sides

           

          W = 7

                2

 

 

 

W+9=0

    -9 -9             subtract 9 from both sides

W = -9              Dimensions can't be negative so we can't use this answer.

 

 

 

Therefore we now know that our Width is 7/2.

 

To figure out the Length, we substitute 7/2 in our Length equation from above.   

 

2   W    +11= L

2 (7/2) + 11,    thus the Length is 18.

  

 

Final result,  The width is 7/2 feet.  The Length is 18 feet.

 

To check, multiply the answers that we received for the length and the width and they should equal 63, which they do.