Question

the area of a rectangle is 35 m², and the length of the rectangle is 3 m more than twice the width. find the dimensions of the rectangle. length: m width: m

Explanation:

Step1: Set up variables

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

Step2: Use the area formula

The area of a rectangle is \(A=\text{length}\times\text{width}\). Given \(A = 35\space m^{2}\), we have the equation \(x(2x + 3)=35\).
Expand it: \(2x^{2}+3x-35 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 2\), \(b = 3\), \(c=-35\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-35)=9 + 280=289\).
Then \(x=\frac{-3\pm\sqrt{289}}{4}=\frac{-3\pm17}{4}\).
We get two solutions for \(x\):
\(x_{1}=\frac{-3 + 17}{4}=\frac{14}{4}=\frac{7}{2}=3.5\) and \(x_{2}=\frac{-3-17}{4}=\frac{-20}{4}=- 5\).
Since the width cannot be negative, we take \(x = 3.5\).

Step4: Find the length

Substitute \(x = 3.5\) into the length formula \(l=2x + 3\).
\(l=2\times3.5+3=7 + 3=10\).

Answer:

Length: \(10\space m\)
Width: \(3.5\space m\)