Let the width of the path be \( w \) meters.
The original rectangle has dimensions:
- Length = 12 m
- Width = 5 m
The area of the original rectangle is: \[ A_{\text{rectangle}} = 12 \times 5 = 60 , \text{m}^2 \]
When the path of width \( w \) is added to one side and one end of the rectangle, the new dimensions become:
- New length = \( 12 + w \)
- New width = \( 5 + w \)
The area of the plot including the path is: \[ A_{\text{total}} = (12 + w)(5 + w) \]
We know the total area of the plot and the path is 120 m², so we can set up the equation: \[ (12 + w)(5 + w) = 120 \]
Expanding the left side: \[ 12 \times 5 + 12w + 5w + w^2 = 120 \] \[ 60 + 17w + w^2 = 120 \]
Rearranging gives us: \[ w^2 + 17w + 60 - 120 = 0 \] \[ w^2 + 17w - 60 = 0 \]
Now we can solve this quadratic equation using the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 17 \), and \( c = -60 \).
Calculating the discriminant: \[ b^2 - 4ac = 17^2 - 4 \times 1 \times (-60) = 289 + 240 = 529 \]
Taking the square root: \[ \sqrt{529} = 23 \]
Now substituting into the quadratic formula: \[ w = \frac{-17 \pm 23}{2} \]
Calculating the two possible solutions: 1. \[ w = \frac{-17 + 23}{2} = \frac{6}{2} = 3 \] 2. \[ w = \frac{-17 - 23}{2} = \frac{-40}{2} = -20 \]
Since a width cannot be negative, we take the positive solution: \[ w = 3 \]
Thus, the width of the path is: \[ \boxed{3 , \text{m}} \]