The area of a rectangle is 66 yd^2, and the length of the rectangle is 1 yd more than twice the width. find the dimensions of the rectangle.

The length of the rectangle is 11 yards, the width of the rectangle is 6 yards.

Step-by-step explanation:

1. Let's check the information given to resolve the question:

Width of the rectangle = x

Length of the rectangle = 2x - 1 (Length of the rectangle is 1 yd more than twice the width)

Area of the rectangle = 66 square yards

2. Let's find the value of the length and the width

Area of the rectangle = Length of the rectangle * Width of the rectangle

66 = x (2x - 1)

66 = 2x² - x

0 = 2x² - x - 66 (Subtracting 66 at both sides)

This is a quadratic equation and the formula to solve it is this:

x = [ - b +/- √ (b² - 4ac)/2a]

x = - [- (- 1) +/- √(1 ² - 4 (2) (-66))/2(2)]

x = - [1 +/- √(1 + 528)/4]

x = - [1 +/- √529)/4]

x = - [1 +/- 23/4]

x₁ = 24/4 = 6

x₂ = -22/4 = -5.5

We will take the value of x₁  because x₂ is negative and the width can't be negative.

Width of the rectangle = 6 yards

Length of the rectangle = 2x - 1 = 2(6) -1 = 12 -1 = 11 yards

The length of the rectangle is 11 yards, the width of the rectangle is 6 yards.

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