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Question

The width of the path around a square field is 4.5 m and its area is 105.75 m 2. Find the cost of fencing the field at the rate of Rs. 100 per meter.

The correct answer is

Rs. 550

Finding the Cost of Fencing a Square Field

This problem involves a square field with a path around it. We are given the width of the path and the area of the path, and we need to find the cost of fencing the field itself.

Understanding the Geometry

We have an inner square which is the field and an outer square which includes the field and the path. The width of the path is uniform around the field.

  • Let the side length of the square field be \(s\) meters.
  • The path has a width of 4.5 meters.
  • The side length of the outer square (field + path) will be \(s + 2 \times \text{path width}\). This is because the path adds its width to both sides of the inner square.
  • So, the side length of the outer square is \(s + 2 \times 4.5 = s + 9\) meters.

Calculating the Area of the Path

The area of the path is the difference between the area of the outer square and the area of the inner square (the field).

  • Area of the inner square (field) = \(s^2\) square meters.
  • Area of the outer square = \((s + 9)^2\) square meters.
  • Area of the path = Area of outer square - Area of inner square
  • Area of the path = \((s + 9)^2 - s^2\)

We are given that the area of the path is 105.75 square meters.

So, we have the equation:

\[ (s + 9)^2 - s^2 = 105.75 \]

Solving for the Side Length of the Field

Let's expand the equation and solve for \(s\):

\[ (s^2 + 18s + 81) - s^2 = 105.75 \]

The \(s^2\) terms cancel out:

\[ 18s + 81 = 105.75 \]

Subtract 81 from both sides:

\[ 18s = 105.75 - 81 \] \[ 18s = 24.75 \]

Now, divide by 18 to find \(s\):

\[ s = \frac{24.75}{18} \] \[ s = 1.375 \]

So, the side length of the square field is 1.375 meters.

Calculating the Cost of Fencing

Fencing the field means fencing its perimeter. The field is a square with side length \(s = 1.375\) meters.

  • Perimeter of the square field = \(4 \times s\)
  • Perimeter = \(4 \times 1.375\) meters
  • Perimeter = 5.5 meters

The cost of fencing is Rs. 100 per meter.

  • Cost of fencing = Perimeter \(\times\) Rate per meter
  • Cost of fencing = \(5.5 \times 100\)
  • Cost of fencing = Rs. 550

Therefore, the cost of fencing the field is Rs. 550.

Summary of Calculation

Description Value Unit
Width of Path 4.5 m
Area of Path 105.75 m\(^2\)
Side of Inner Square (s) 1.375 m
Perimeter of Field (4s) 5.5 m
Fencing Rate 100 Rs./m
Cost of Fencing 550 Rs.

Revision Table - Square Field and Path Problem

Let's quickly review the key formulas and steps used in this problem:

  • Side of inner square = \(s\)
  • Width of path = \(w\)
  • Side of outer square = \(s + 2w\)
  • Area of inner square = \(s^2\)
  • Area of outer square = \((s + 2w)^2\)
  • Area of path = Area of outer square - Area of inner square = \((s + 2w)^2 - s^2\)
  • Perimeter of inner square (for fencing) = \(4s\)
  • Cost of fencing = Perimeter \(\times\) Rate

Additional Information - Areas and Perimeters

Understanding how areas and perimeters work for shapes with uniform borders is important. This problem is an example of finding the dimensions of an inner shape when the dimensions and area of a surrounding border are known.

  • Area: The amount of surface a 2D shape covers. Measured in square units (e.g., m\(^2\), cm\(^2\)).
  • Perimeter: The total distance around the boundary of a 2D shape. Measured in linear units (e.g., m, cm).
  • For a square with side length \(a\):
    • Area = \(a^2\)
    • Perimeter = \(4a\)
  • When a path of uniform width \(w\) surrounds a square of side \(s\), the outer shape is also a square. Its side length becomes \(s + 2w\). The area of the path is found by subtracting the inner area from the outer area.
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Important Questions from Plane Figures

  1. If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:

  2. What is the area of the square (in cm 2) whose vertices lie on a circle of radius 5 cm?

  3. The circumcentre of an equilateral triangle is at a distance of 3.2 cm from the base of the triangle. What is the length (in cm) of each of its altitudes?

  4. The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:

    Take \(\left(\pi=\frac{22}{7}\right)\)

  5. If the radius of a circle is decreased by 11% then the total decrease in the area of the circle is given as:

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