Question

the area of a rectangle is 54 yd², and the length of the rectangle is 3 yd more than twice the width. find the dimensions of the rectangle. length : □ yd width : □ yd

Explanation:

Step1: Set up variables

Let the width of the rectangle be \(w\) yd. Then the length \(l=(2w + 3)\) yd.

Step2: Use the area formula

The area formula for a rectangle is \(A=l\times w\). Substitute \(A = 54\), \(l=(2w + 3)\) into the formula:

$$54=(2w + 3)\times w$$

$$54 = 2w^{2}+3w$$

$$2w^{2}+3w-54=0$$

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 3\), \(c=-54\)), use the quadratic formula \(w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-54)=9 + 432=441\).
Then \(w=\frac{-3\pm\sqrt{441}}{4}=\frac{-3\pm21}{4}\).
We have two solutions for \(w\):
\(w_{1}=\frac{-3 + 21}{4}=\frac{18}{4}=4.5\) and \(w_{2}=\frac{-3-21}{4}=\frac{-24}{4}=-6\).
Since width cannot be negative, \(w = 4.5\) yd.

Step4: Find the length

Substitute \(w = 4.5\) into \(l=(2w + 3)\):
\(l=2\times4.5+3=9 + 3=12\) yd.

Answer:

Length: \(12\) yd, Width: \(4.5\) yd