Question

A rectangle measures 12m by 5m a path of constant width runs along one side and one end. If the total area of the plot and the path is 120m² find the width of the path

Answer 1

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:
\[
\boxed{3 \, \text{m}}
\]