Question

the area of a rectangle is 77 ft², and the length of the rectangle is 3 ft more than twice 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 \((2x + 3)\) ft.

Step2: Use the area formula

The area formula for a rectangle is \(A=\text{length}\times\text{width}\). Given \(A = 77\) ft², we have the equation \(x(2x + 3)=77\).
Expand it: \(2x^{2}+3x-77 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 3\), \(c=-77\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-77)=9 + 616=625\).
Then \(x=\frac{-3\pm\sqrt{625}}{4}=\frac{-3\pm25}{4}\).
We get two solutions for \(x\):
\(x_{1}=\frac{-3 + 25}{4}=\frac{22}{4}=5.5\) and \(x_{2}=\frac{-3-25}{4}=\frac{-28}{4}=-7\).
Since the width cannot be negative, we take \(x = 5.5\) ft.

Step4: Find the length

Substitute \(x = 5.5\) into the length formula \(l=2x+3\).
\(l=2\times5.5+3=11 + 3=14\) ft.

Answer:

Length: \(14\) ft, Width: \(5.5\) ft