Question

the area of a rectangle is 21 ft², and the length of the rectangle is 1 ft less than double the width. find the dimensions of the rectangle. length : ft width : ft

Explanation:

Step1: Set up variables

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

Step2: Use the area formula

The area formula for a rectangle is \(A=\text{length}\times\text{width}\). Given \(A = 21\) ft², we have the equation \(x(2x - 1)=21\).
Expanding gives \(2x^{2}-x - 21=0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 2\), \(b=-1\), \(c = - 21\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-1)^{2}-4\times2\times(-21)=1 + 168=169\).
Then \(x=\frac{1\pm\sqrt{169}}{4}=\frac{1\pm13}{4}\).
We get two solutions: \(x_{1}=\frac{1 + 13}{4}=\frac{14}{4}=3.5\) and \(x_{2}=\frac{1-13}{4}=\frac{-12}{4}=-3\).
Since the width cannot be negative, we take \(x = 3.5\) ft.

Step4: Find the length

Substitute \(x = 3.5\) into the length formula \(l=2x-1\). So \(l=2\times3.5-1=7 - 1=6\) ft.

Answer:

Length: \(6\) ft
Width: \(3.5\) ft