Question

A rectangular park 60 m long and 40 m wide has two 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 sq. m, then what is the width of the road?

• A. 2.91 m
• B. 3 m
• C. 5.82 m
• D. None of these

Answer 1

The Correct Option is B

The area of the rectangular park is calculated by multiplying its length and width, which gives:
Area of park = \(60 \, \text{m} \times 40 \, \text{m} = 2400 \, \text{sq. m}\). 

The area of the lawn is given as 2109 sq. m. The difference between the total area of the park and the area of the lawn equals the area occupied by the roads.
Area of roads = \(2400 \, \text{sq. m} - 2109 \, \text{sq. m} = 291 \, \text{sq. m}\).

Let the width of the roads be \(x \, \text{m}\). The roads form a plus pattern, crossing each other. The total area covered by the roads consists of two strips: one horizontal and one vertical, both sharing a square at their intersection.
The length of the horizontal road is equal to the width of the park, so it covers an area of \(40 \, \text{m} \times x \, \text{m}\).
The length of the vertical road is equal to the length of the park, covering an area of \(60 \, \text{m} \times x \, \text{m}\).
However, the \(x \times x\) square at the intersection has been counted twice, so we need to subtract its area once.
Thus, the total area of the roads is \(40x + 60x - x^2 = 100x - x^2\).

Setting the area of the roads equal to 291 sq. m., we derive the equation:
\(100x - x^2 = 291\).

Rearrange to form a quadratic equation:
\(x^2 - 100x + 291 = 0\).

We solve this by using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-100\), and \(c=291\):
\(x = \frac{100 \pm \sqrt{100^2 - 4 \times 1 \times 291}}{2}\).

\(x = \frac{100 \pm \sqrt{10000 - 1164}}{2}\)
\(x = \frac{100 \pm \sqrt{8836}}{2}\)
\(x = \frac{100 \pm 94}{2}\).

This yields solutions \(x = \frac{194}{2} = 97\) and \(x = \frac{6}{2} = 3\).

Since 97 m is too wide for the context of park roads, the only reasonable solution is \(x = 3 \, \text{m}\).

The width of the road is therefore 3 m.