Sure, let's break this problem down step by step.
1. Understand the Problem:
- We are given the area of a rectangular piece of carpet: [tex]\( 150 \, \text{yd}^2 \)[/tex].
- We know that the width ([tex]\(w\)[/tex]) is 5 yards less than the length ([tex]\(l\)[/tex]).
- We need to find the perimeter ([tex]\(P\)[/tex]) of the rectangle.
2. Establish Relations and Equations:
- Given that the width is 5 yards less than the length, we can write:
[tex]\[ w = l - 5 \][/tex]
- The area of a rectangle is given by the product of its length and width. Hence:
[tex]\[ l \times w = 150 \][/tex]
- Substituting [tex]\( w \)[/tex] from the above relationship into the area formula:
[tex]\[ l \times (l - 5) = 150 \][/tex]
3. Form and Solve the Quadratic Equation:
- This expands to:
[tex]\[ l^2 - 5l - 150 = 0 \][/tex]
- This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex] with:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = -5 \][/tex]
[tex]\[ c = -150 \][/tex]
4. Solve the Quadratic Equation:
- Applying the quadratic formula [tex]\( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ l = \frac{5 \pm \sqrt{25 + 600}}{2} \][/tex]
[tex]\[ l = \frac{5 \pm \sqrt{625}}{2} \][/tex]
[tex]\[ l = \frac{5 \pm 25}{2} \][/tex]
- This gives us two potential solutions:
[tex]\[ l = \frac{30}{2} = 15 \][/tex]
[tex]\[ l = \frac{-20}{2} = -10 \][/tex]
- Since a length cannot be negative, we discard [tex]\( l = -10 \)[/tex] and take:
[tex]\[ l = 15 \][/tex]
5. Calculate the Width:
- Using [tex]\( w = l - 5 \)[/tex]:
[tex]\[ w = 15 - 5 = 10 \][/tex]
6. Find the Perimeter of the Rectangle:
- The perimeter of a rectangle is given by:
[tex]\[ P = 2(l + w) \][/tex]
- Substituting the values of the length and width:
[tex]\[ P = 2(15 + 10) \][/tex]
[tex]\[ P = 2 \times 25 \][/tex]
[tex]\[ P = 50 \][/tex]
Conclusively, the length of the carpet is 15 yards, the width is 10 yards, and the perimeter is 50 yards.
