There is a garden of `140 m xx 120 m` and a gravel path is to be made of an equal width all around it, so as to take up just one-fourth of the garden. What must be the breadth of the path?

Answer

To solve the problem step by step, we need to find the width of the gravel path that takes up one-fourth of the garden area.

### Step 1: Calculate the area of the garden
The area of the garden can be calculated using the formula for the area of a rectangle:
\[ \text{Area} = \text{Length} \times \text{Width} \]
Given:
- Length = 140 m
- Width = 120 m

Calculating the area:
\[ \text{Area} = 140 \, \text{m} \times 120 \, \text{m} = 16800 \, \text{m}^2 \]

### Step 2: Calculate the area of the gravel path
The problem states that the gravel path takes up one-fourth of the garden area. Therefore, we need to calculate one-fourth of the total area:
\[ \text{Area of the path} = \frac{1}{4} \times \text{Total Area} = \frac{1}{4} \times 16800 \, \text{m}^2 = 4200 \, \text{m}^2 \]

### Step 3: Determine the dimensions of the garden including the path
Let the width of the gravel path be \( x \). The dimensions of the garden including the path will be:
- Length = \( 140 + 2x \)
- Width = \( 120 + 2x \)

### Step 4: Calculate the total area including the path
The total area including the path can be expressed as:
\[ \text{Total Area including path} = (140 + 2x)(120 + 2x) \]

### Step 5: Set up the equation
We know that the area of the path is 4200 m². Therefore, we can set up the equation:
\[ (140 + 2x)(120 + 2x) - 16800 = 4200 \]

### Step 6: Simplify the equation
Expanding the left side:
\[ (140 \times 120) + (140 \times 2x) + (120 \times 2x) + (2x \times 2x) - 16800 = 4200 \]
Calculating \( 140 \times 120 = 16800 \):
\[ 16800 + 280x + 240x + 4x^2 - 16800 = 4200 \]
This simplifies to:
\[ 520x + 4x^2 = 4200 \]

### Step 7: Rearranging the equation
Rearranging gives us:
\[ 4x^2 + 520x - 4200 = 0 \]

### Step 8: Simplifying the equation
Dividing the entire equation by 4:
\[ x^2 + 130x - 1050 = 0 \]

### Step 9: Solving the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 130, c = -1050 \):
\[ x = \frac{-130 \pm \sqrt{130^2 - 4 \times 1 \times (-1050)}}{2 \times 1} \]
Calculating the discriminant:
\[ 130^2 = 16900, \quad 4 \times 1 \times 1050 = 4200 \]
\[ \sqrt{16900 + 4200} = \sqrt{21100} \]
Calculating \( \sqrt{21100} \approx 145.3 \):
\[ x = \frac{-130 \pm 145.3}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{15.3}{2} \approx 7.65 \)
2. \( x = \frac{-275.3}{2} \) (not valid since width cannot be negative)

Thus, the width of the path \( x \approx 7.65 \, \text{m} \).

### Final Answer
The breadth of the path must be approximately **7.65 m**.

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