Question

the area of a rectangle is $66 yd^{2}$, and the length of the rectangle is 1 yd more than double the width. find the dimensions of the rectangle.

Explanation:

Step1: Set up variables

Let the width of the rectangle be \(x\) yd. Then the length is \((2x + 1)\) yd.

Step2: Use the area formula

The area formula for a rectangle is \(A=\text{length}\times\text{width}\). So \(66=(2x + 1)\times x\), which simplifies to \(2x^{2}+x-66 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 1\), \(c=-66\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(1)^{2}-4\times2\times(-66)=1 + 528=529\).
Then \(x=\frac{-1\pm\sqrt{529}}{2\times2}=\frac{-1\pm23}{4}\).
We get two solutions: \(x_{1}=\frac{-1 + 23}{4}=\frac{22}{4}=5.5\) and \(x_{2}=\frac{-1-23}{4}=\frac{-24}{4}=-6\). Since the width cannot be negative, we take \(x = 5.5\).

Step4: Find the length

If \(x = 5.5\), then the length \(l=2x+1=2\times5.5 + 1=11 + 1=12\).

Answer:

Length: \(12\) yd, Width: \(5.5\) yd