A rectangular park 60 metre long and 40 mete wide has concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is `2109 "metre"^(2)` then the width of the road is

Options

A. 3 metre
B. 5 metre
C. 6 metre
D. 2 metre

Answer

To solve the problem, we need to find the width of the concrete road running through the middle of the rectangular park.

### Step-by-Step Solution:

1. **Calculate the Area of the Park:**
The area of a rectangle is given by the formula:
\[ \text{Area} = \text{Length} \times \text{Width} \]
For the park:
\[ \text{Area of the park} = 60 \, \text{m} \times 40 \, \text{m} = 2400 \, \text{m}^2 \]

2. **Determine the Area of the Lawn:**
We are given that the area of the lawn is \( 2109 \, \text{m}^2 \).

3. **Calculate the Area of the Road:**
The area of the road can be found by subtracting the area of the lawn from the area of the park:
\[ \text{Area of the road} = \text{Area of the park} - \text{Area of the lawn} \]
\[ \text{Area of the road} = 2400 \, \text{m}^2 - 2109 \, \text{m}^2 = 291 \, \text{m}^2 \]

4. **Set Up the Equation for the Width of the Road:**
Let the width of the road be \( x \, \text{m} \). The road runs through the middle of the park, creating a crossroad. The dimensions of the road will be:
- Length of the road = 60 m (the length of the park)
- Width of the road = \( x \, \text{m} \) (the width we need to find)

The area of the road can be expressed as:
\[ \text{Area of the road} = \text{Length} \times \text{Width} + \text{Width} \times \text{Length} \]
Since there are two roads (one horizontal and one vertical), the area becomes:
\[ \text{Area of the road} = 60x + 40x - x^2 \]
Here, \( x^2 \) is subtracted because the intersection of the two roads has been counted twice.

5. **Set Up the Equation:**
Now we equate the area of the road to the calculated area:
\[ 60x + 40x - x^2 = 291 \]
Simplifying this gives:
\[ 100x - x^2 = 291 \]
Rearranging gives:
\[ x^2 - 100x + 291 = 0 \]

6. **Solve the Quadratic Equation:**
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -100, c = 291 \).
\[ x = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 1 \cdot 291}}{2 \cdot 1} \]
\[ x = \frac{100 \pm \sqrt{10000 - 1164}}{2} \]
\[ x = \frac{100 \pm \sqrt{8836}}{2} \]
\[ x = \frac{100 \pm 94}{2} \]
This gives us two potential solutions:
\[ x = \frac{194}{2} = 97 \quad \text{(not possible, as it exceeds the park's width)} \]
\[ x = \frac{6}{2} = 3 \]

7. **Conclusion:**
The width of the road is \( 3 \, \text{m} \).