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Sagot :
Hi there!
[tex]\large\boxed{\text{ L = 6 ft, w = 5.5 ft}}[/tex]
Use the following equation for the area of a rectangle:
A = l × w
Since the length is 5 ft less than twice the width, we can write an expression for the length in terms of the width:
l = 2w - 5
Plug the given area and expression into the equation:
33 = (2w - 5) × w
Distribute:
33 = 2w² - 5w
Move everything over to one side:
0 = 2w² - 5w - 33
Factor the equation:
0 = (2w - 11)(w + 3)
Find each w value by setting each expression equal to 0:
0 = 2w - 11
11 = 2w
w = 5.5
0 = w + 3
-3 = w
Since dimensions must be positive, the width is 5.5 ft.
Plug this value into the initial equation to solve for length:
33 = 5.5l
33/5.5 = l
l = 6 ft.
The dimensions are length = 6ft, and width = 5.5ft.
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