The area of a rectangle is 66ft ^2, and the length of the rectangle is 1ftmore than double the width. Find the dimensions of the rectangle.

Question

find length and width

Answer 1

So, Eden, let's let w = the width.  If the length is one foot more than double the width, then we can write it as 2w+1.

 

How do we find the area?

 

       w(2w+1)=66.

          2w²+w=66

      2w²+w-66=0

(2w-11)(w+6)=0

 

2w-11=0

       2w=11

         w=5.5

 

w+6=0

    w=-6

 

Since we can't have a negative width, -6 won't work, so our width is 5.5. 

 

Our length is 5.5(2)+1 =12

 

Check our work. 5.5 x12 = 66, so we're good.

 

 

Answer 2

Here's what we know:

A _rectangle = l * w

l = 2w + 1

A = 66

 

So we start by putting in what we know:

66 = (2w + 1) w

What we do next is multiply the l and w values together and also move the 66 to the other side. That way, we get a quadratic equation:

0 = 2w² + w - 66

The factors of 66 are: 11, 3, 2

And what we need is to find the proper factor combination that will give us a +1 (for the w)

 

If we combine the 2 and 3, the product will be 6, and if we combine that with the 2 in the 2w, that product will give us 12. So:

0 = (2w - 11)(w + 6)

 

We can throw out the w+6 part because to make that equal 0, we need a -6, and we can't have a length that's negative, so that leaves us to solve:

0= 2w - 11

11 = 2w

¹¹/₂ = w

 

And so, we have our width. Now to solve for length:

l = 2w + 1

l = 2 * 11 / 2 + 1

l = 11 + 1

l = 12

 

Now to check our answer:

11 / 2 * 12 = 66

11 * 6 = 66

66 = 66