The area of a rectangle is 33 ft2 , and the length of the rectangle is 5 ft less than double the width. Find the dimensions of the rectangle.

Question

The area of a rectangle is
33 ft2
, and the length of the rectangle is
5 ft
less than double the width. Find the dimensions of the rectangle.

Answer 1

Hi Patty, 

Step One:

Let width=w

Let length= 2w-5

Step Two: Since the area =33ft² and length*width=area, we can create the equation:

(2w-5)(w)=33

2w²-5w-33=0

w²-[(5w)/2]-16.5=0

Use the Quadratic Equation

x=5.5 and x=-3 

We disregard the -3 because we cannot have a negative dimension.

∴The width of the rectangle is 5.5 feet while the length is 6 feet because 2(5.5)-5 is 6.

Best,

Star 

 

 

 

Answer 2

Patty,

We know that the area of a rectangle is found by multiplying the length by the width. So we set up an equation using the given info as follows:

 

l * w = A

l = 2w - 5

A = 33

 

using substitution:

 

(2w - 5)w = 33

 

2w² - 5w = 33

 

We can't solve this type of equation unless we set it equal to zero and apply the zero product rule.

 

2w² -5w - 33 = 0

 

Now factor if possible:

 

2 * -33 = -66

 

-66 can factor to -11 * 6

 

and -11 + 6 = -5  so

 

(2w -11) and (2w+6)

 

any set with a common factor must be reduced, so (2w + 6)/2 gives:

 

(2w - 11)(w + 3) = 0

 

If you're not convinced, foil it out to see it gives you the original equation.

 

Now we apply the zero product rule:  if any of the factors =0, then the product must =0.

 

2w - 11 = 0  ⇒  2w = 11  ⇒  w = 11/2

 

w + 3 = 0   ⇒  w = -3

 

Since we can't have a negative physical dimension, the width must be 11/2 ft.

 

Plugging this back into l = 2w - 5 gives l = 2(11/2) - 5 = 6 ft.