A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wie. If the path occupies `600 m^(2)`, then the width of the path is

Options

A. 30m
B. 5m
C. 18.75 m
D. 10m

Answer

To find the width of the path that runs around the inside of a rectangular field, we can follow these steps:

### Step 1: Define the dimensions of the rectangular field
The rectangular field has a length of 38 meters and a width of 32 meters.

### Step 2: Calculate the area of the rectangular field
The area of the rectangular field can be calculated using the formula:
\[ \text{Area} = \text{Length} \times \text{Width} \]
Substituting the values:
\[ \text{Area} = 38 \, \text{m} \times 32 \, \text{m} = 1216 \, \text{m}^2 \]

### Step 3: Define the width of the path
Let the width of the path be \( x \) meters. The path runs uniformly around the inside of the rectangular field.

### Step 4: Calculate the dimensions of the inner rectangle
The dimensions of the inner rectangle (after the path is removed) will be:
- Length: \( 38 - 2x \)
- Width: \( 32 - 2x \)

### Step 5: Calculate the area of the inner rectangle
The area of the inner rectangle can be calculated as:
\[ \text{Area of inner rectangle} = (38 - 2x)(32 - 2x) \]

### Step 6: Set up the equation for the area of the path
The area of the path is given as 600 m². Therefore, we can set up the equation:
\[ \text{Area of path} = \text{Area of outer rectangle} - \text{Area of inner rectangle} \]
This gives us:
\[ 600 = 1216 - (38 - 2x)(32 - 2x) \]

### Step 7: Expand the equation
Expanding the area of the inner rectangle:
\[ (38 - 2x)(32 - 2x) = 38 \times 32 - 2x \times 38 - 2x \times 32 + 4x^2 \]
Calculating \( 38 \times 32 \):
\[ 38 \times 32 = 1216 \]
So,
\[ (38 - 2x)(32 - 2x) = 1216 - 76x - 64x + 4x^2 = 1216 - 140x + 4x^2 \]

### Step 8: Substitute back into the equation
Substituting back into the area equation:
\[ 600 = 1216 - (1216 - 140x + 4x^2) \]
This simplifies to:
\[ 600 = 140x - 4x^2 \]

### Step 9: Rearranging the equation
Rearranging gives us:
\[ 4x^2 - 140x + 600 = 0 \]

### Step 10: Simplifying the equation
Dividing the entire equation by 4:
\[ x^2 - 35x + 150 = 0 \]

### Step 11: Factoring the quadratic equation
To factor the quadratic equation:
\[ (x - 30)(x - 5) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 30 \quad \text{or} \quad x = 5 \]

### Step 12: Determine the valid solution
Since the width of the path cannot exceed the dimensions of the field, we discard \( x = 30 \) and accept:
\[ x = 5 \]

### Conclusion
The width of the path is \( \boxed{5} \) meters.

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