Question

the length of a rectangle is 5 ft more than double the width, and the area of the rectangle is 63 ft². find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be \(w\) ft. Then the length \(l=(2w + 5)\) ft.

Step2: Use the area formula

The area of a rectangle \(A=l\times w\). Given \(A = 63\) \(ft^{2}\), so \((2w + 5)\times w=63\).
Expanding gives \(2w^{2}+5w-63 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 5\), \(c=-63\)), use the quadratic formula \(w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(5)^{2}-4\times2\times(-63)=25 + 504=529\).
Then \(w=\frac{-5\pm\sqrt{529}}{4}=\frac{-5\pm23}{4}\).
We have two solutions for \(w\):

• \(w_{1}=\frac{-5 + 23}{4}=\frac{18}{4}=4.5\)
• \(w_{2}=\frac{-5-23}{4}=\frac{-28}{4}=-7\) (reject since width cannot be negative)

Step4: Find the length

If \(w = 4.5\) ft, then \(l=2w+5=2\times4.5+5=9 + 5=14\) ft.

Answer:

Length: \(14\) ft, Width: \(4.5\) ft