Question

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

Answer 1

Let the width of the path be \( x \) meters. The path runs along one side and one end of the rectangular plot, which has dimensions of 12 m by 5 m.

The dimensions of the larger rectangle, which includes the path, would then be:

- Length: \( 12 + x \) (adding the width on one side)
- Width: \( 5 + x \) (adding the width on the other side)

The area of the larger rectangle (the plot plus the path) can be represented as:

\[
\text{Area} = (12 + x)(5 + x)
\]

We know the area of the plot and the width of the path leads to a total area of 120 m²:

\[
(12 + x)(5 + x) = 120
\]

Expanding the left side, we get:

\[
60 + 12x + 5x + x^2 = 120
\]

Combining like terms gives:

\[
x^2 + 17x + 60 = 120
\]

Subtracting 120 from both sides leads to:

\[
x^2 + 17x - 60 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula, where \( a = 1 \), \( b = 17 \), and \( c = -60 \):

\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-17 \pm \sqrt{{17^2 - 4 \cdot 1 \cdot (-60)}}}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
17^2 = 289
\]
\[
4 \cdot 1 \cdot 60 = 240
\]

Thus,

\[
b^2 - 4ac = 289 + 240 = 529
\]

Taking the square root:

\[
\sqrt{529} = 23
\]

Plugging this back into the formula gives:

\[
x = \frac{{-17 \pm 23}}{2}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{{-17 + 23}}{2} = \frac{6}{2} = 3 \)
2. \( x = \frac{{-17 - 23}}{2} = \frac{-40}{2} = -20 \) (not a valid solution since width cannot be negative)

Therefore, the width of the path is:

\[
\boxed{3 \text{ meters}}
\]