The area of a rectangle is 21 ft^2,and the length of the rectangle is
1 ft less than twice the width. Find the dimensions of the rectangle.
21=LW=(2W-1)W
solve for W, use the quadratic equation.
To find the dimensions of the rectangle, we need to first understand the relationship between the length and width, and then use that relationship to create an equation we can solve.
Let's assume the width of the rectangle is w ft.
According to the problem, we know that the area of the rectangle is 21 ft^2. The area of a rectangle is calculated by multiplying the length by the width. Therefore, we can write the equation:
Length × Width = Area
Since we are given the area as 21 ft^2 and the problem states that the length is 1 ft less than twice the width, we can write the length in terms of the width:
Length = 2w - 1
Now we can substitute this expression for the length into the equation for the area:
(2w - 1) × w = 21
Expanding and simplifying this equation:
2w^2 - w = 21
Rearranging the equation to set it equal to zero:
2w^2 - w - 21 = 0
At this point, we can solve the equation by factoring, using the quadratic formula, or by graphing. Let's solve it by factoring:
(2w + 7)(w - 3) = 0
Setting each factor equal to zero and solving for w:
2w + 7 = 0 or w - 3 = 0
Solving each of these equations gives us two possible values for the width:
w = -7/2 or w = 3
Since width cannot be negative, we discard the negative value. Therefore, the width of the rectangle is 3 ft.
Next, we can substitute this value of width back into the expression for the length:
Length = 2w - 1
Length = 2(3) - 1
Length = 6 - 1
Length = 5 ft
So, the dimensions of the rectangle are 3 ft by 5 ft.