Question

The length of a rectangle is 1 ft more than double the width, and the area of the rectangle is 28ft^2. Find the dimensions of the rectangle

Answer 1

Answer: The length is 8 ft and the  width is 3.5  ft  and those are the dimensions.

Step-by-step explanation:

The length of the rectangle is 1 more than double the width so we can represent it by the equation  L = 2w + 1   where l is the length and w is the width. We know that to find the area of a rectangle, you will have to multiply the length by the width and the area is 28 ft.  so we can represent it by the equation,

w(2w+1) = 28  Now solve for w to find the width

2$w^{2}$ + w = 28   Subtract 28 from both sides for the whole equation to equal zero.

$2w^{2}$ + w -28 = 0   Now solve for w using the quadratic formula

  x= -b ± $\sqrt{b^2-4ac} /2a$  In this case, x is representing w(the width)

a = 2      b = 1     c = -28  

 x = -1 ± $\sqrt{225} / 4$  

x= -1 ± 15 / 4

x=  14/4

x =  3.5  

The width is 3.5 ft  and the length is 1 more than double the width so

L =  2(3.5)  + 1

L = 7 + 1

l = 8  

The length is 8  

Check  

8 * 3.5 =  28  

 28 = 28