Question

the area of a rectangular wall of a barn is 171 square feet. its length is 10 feet longer than the width. find the length and width of the wall of the barn. the width is □ feet.

Explanation:

Step1: Set up the equation

Let the width of the wall be \(x\) feet. Then the length is \((x + 10)\) feet. The area of a rectangle is \(A=\text{length}\times\text{width}\), so \(x(x + 10)=171\).
Expanding gives \(x^{2}+10x-171 = 0\).

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 1\), \(b = 10\), \(c=-171\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(10)^{2}-4\times1\times(-171)=100 + 684=784\).
Then \(x=\frac{-10\pm\sqrt{784}}{2}=\frac{-10\pm28}{2}\).
We have two solutions: \(x_{1}=\frac{-10 + 28}{2}=\frac{18}{2}=9\) and \(x_{2}=\frac{-10-28}{2}=\frac{-38}{2}=-19\). Since width cannot be negative, we take \(x = 9\).

Step3: Find the length

The length is \(x + 10\). Substituting \(x = 9\), we get length \(=9+10 = 19\) feet.

Answer:

The width is \(9\) feet and the length is \(19\) feet.