Question

the area of a rectangle is 99 m², and the length of the rectangle is 7 m more than double 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 \((2x + 7)\) meters.

Step2: Use the area formula

The area of a rectangle is \(A=\text{length}\times\text{width}\). So, \(x(2x + 7)=99\).
Expand to get \(2x^{2}+7x - 99 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 2\), \(b = 7\), \(c=-99\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(7)^{2}-4\times2\times(-99)=49 + 792=841\).
Then \(x=\frac{-7\pm\sqrt{841}}{4}=\frac{-7\pm29}{4}\).
We have two solutions for \(x\):
\(x_{1}=\frac{-7 + 29}{4}=\frac{22}{4}=5.5\) and \(x_{2}=\frac{-7-29}{4}=\frac{-36}{4}=-9\).
Since the width cannot be negative, we take \(x = 5.5\).

Step4: Find the length

Substitute \(x = 5.5\) into the length formula \(l=2x + 7\).
\(l=2\times5.5+7=11 + 7=18\).

Answer:

Length: \(18\) m, Width: \(5.5\) m