The length of a rectangle is 1ft. more than twice the width, and the area of the rectangle is 66 ft^2. Find the dimensions of the rectangle.
length:
width:
L=1+2w
area=lw=(1+2w)w=66
2w^2+w-66=0
use the quadratic equation to solve for w.
To find the dimensions of the rectangle, we'll solve the problem step by step.
Let's assume the width of the rectangle is 'w' ft. According to the given information, the length of the rectangle is 1 ft more than twice the width. So, the length can be calculated as follows:
Length = 1 + 2w
Next, we know that the area of a rectangle is given by the formula:
Area = Length * Width
Given that the area is 66 ft^2, we can substitute the length and width into the equation and solve for 'w'.
66 = (1 + 2w) * w
Expanding the equation, we get:
66 = w + 2w^2
2w^2 + w - 66 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. For simplicity, we'll use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = 1, and c = -66. Substituting these values into the quadratic formula, we have:
w = (-(1) ± √((1)^2 - 4(2)(-66))) / (2(2))
Simplifying further:
w = (-1 ± √(1 + 528)) / 4
w = (-1 ± √529) / 4
Since √529 = 23, we have two possible solutions for 'w':
w1 = (-1 + 23) / 4 = 22 / 4 = 5.5
w2 = (-1 - 23) / 4 = -24 / 4 = -6
Since the width cannot be negative, we discard w2 = -6.
Therefore, the width of the rectangle is w = 5.5 ft.
Now, to find the length, we substitute this value back into the equation:
Length = 1 + 2w = 1 + 2(5.5) = 1 + 11 = 12 ft
So, the dimensions of the rectangle are:
Length = 12 ft
Width = 5.5 ft