Rectangular Park Path Width Calculation

We need to find the width of a path inside a rectangular park. Key information provided is:

• Outer Dimensions of the park: Length ($L$) = $37$ m, Width ($B$) = $30$ m.
• Area of the path: $A_{path} = 570 \text{ m}^2$.
• The path runs along the inner side.

Inner Park Dimensions Calculation

Let the uniform width of the path be $w$ meters. Since the path is on the inside, the dimensions of the inner rectangular area (the park excluding the path) are:

• Inner Length ($L'$): $L - 2w = 37 - 2w$ m.
• Inner Width ($B'$): $B - 2w = 30 - 2w$ m.

Path Area Equation Formulation

The area of the path is the difference between the total area of the park (outer rectangle) and the area of the inner rectangle.

• Area of Outer Rectangle: $A_{outer} = L \times B = 37 \times 30 = 1110 \text{ m}^2$.
• Area of Inner Rectangle: $A_{inner} = L' \times B' = (37 - 2w)(30 - 2w) \text{ m}^2$.

Using the given area of the path: $A_{path} = A_{outer} - A_{inner}$ $570 = 1110 - (37 - 2w)(30 - 2w)$

Quadratic Equation Solution

Expand the expression for the inner area and solve the equation for $w$:

$570 = 1110 - (1110 - 74w - 60w + 4w^2)$ $570 = 1110 - 1110 + 134w - 4w^2$ $570 = 134w - 4w^2$

Rearrange this into a standard quadratic equation ($ax^2 + bx + c = 0$): $4w^2 - 134w + 570 = 0$

Simplify by dividing by 2: $2w^2 - 67w + 285 = 0$

Use the quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $w = \frac{-(-67) \pm \sqrt{(-67)^2 - 4(2)(285)}}{2(2)}$ $w = \frac{67 \pm \sqrt{4489 - 2280}}{4}$ $w = \frac{67 \pm \sqrt{2209}}{4}$ $w = \frac{67 \pm 47}{4}$

Valid Path Width Determination

This gives two possible values for the width $w$:

• $w_1 = \frac{67 + 47}{4} = \frac{114}{4} = 28.5$ m
• $w_2 = \frac{67 - 47}{4} = \frac{20}{4} = 5$ m

The width of the path ($w$) must be physically possible within the park's dimensions. Specifically, $2w$ must be less than the park's width ($30$ m), meaning $w$ must be less than $15$ m.

The value $w_1 = 28.5$ m is greater than $15$ m, making it an invalid solution.

The value $w_2 = 5$ m is less than $15$ m and is therefore the valid width of the path.