Question

the area of a rectangle is 27 yd², and the length of the rectangle is 3 yd less than double the width. find the dimensions of the rectangle. length : yd width : yd

Explanation:

Step1: Set up variables

Let the width of the rectangle be \(x\) yd. Then the length is \((2x - 3)\) yd.

Step2: Use the area formula

The area of a rectangle is \(A=\text{length}\times\text{width}\). So \(x(2x - 3)=27\).
Expand to get \(2x^{2}-3x - 27=0\).

Step3: Solve the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 2\), \(b=-3\), \(c=-27\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-3)^{2}-4\times2\times(-27)=9 + 216=225\).
Then \(x=\frac{3\pm\sqrt{225}}{4}=\frac{3\pm15}{4}\).
We have two solutions: \(x_1=\frac{3 + 15}{4}=\frac{18}{4}=4.5\) and \(x_2=\frac{3-15}{4}=\frac{-12}{4}=-3\).
Since the width cannot be negative, \(x = 4.5\).

Step4: Find the length

Substitute \(x = 4.5\) into the length formula \(2x-3\). So the length is \(2\times4.5-3=9 - 3=6\).

Answer:

Length: \(6\) yd
Width: \(4.5\) yd