Question

Kaurav and Pandav have a rectangular field of area 20,000 sq. \(\mathrm{m}\). They decided to divide it into two equal parts by dividing it with a single straight line. Kaurav wanted to fene their land immediately, so they incurred total expenses for the fencing all the four sides alone at Rs. 2 per metre. Whatis the minimum cost that Kaurav had to incur? (a) Rs. 800 (b) Rs. 1600 (c) Rs. 1200 (d) Rs. 600

Short answer

Answer: (a) Rs. 800

Step-by-step solution

Calculate dimensions of the entire field.

Given area of the rectangular field = 20,000 sq. m Let the length of the field be 'L' and the width of the field be 'W'. So, LW = 20,000. We need to find out the dimensions of this rectangular field. Since they are splitting the field into two equal parts, the area of half-field = (20,000 / 2) = 10,000 sq. m.

Finding dimensions for minimum perimeter of half field.

To minimize the fencing cost, Kaurav needs to minimize the perimeter of his half field. As we know that L and W are integers, we can find the dimensions by finding integer factors of 10,000. Here we should note that for a rectangle with a fixed area, the perimeter will be minimized when its length and width are as close to each other as possible. Factors of 10,000: 1 x 10,000, 2 x 5,000, 4 x 2,500, 5 x 2,000, 8 x 1,250, 10 x 1,000, 20 x 500, 25 x 400, 40 x 250, 50 x 200, 100 x 100 As we can see, 100 x 100 are the dimensions of the half field that minimizes the perimeter as the difference between them is the least.

Calculating the minimum fencing cost of Kaurav's half field.

Now that we know the dimensions of Kaurav's half field are 100 m x 100 m, we can calculate the perimeter. Perimeter = 2(L + W) = 2(100 + 100) = 2(200) = 400 m Now we need to find the minimum cost incurred to put fence all the four sides of Kaurav's half field. Minimum cost = Perimeter * cost per metre Minimum cost = 400 m * Rs. 2/m = Rs. 800 So, the minimum cost Kaurav has to incur for fencing his half of the field is (a) Rs. 800.

Key concepts

Perimeter Minimization

The main goal when minimizing the perimeter is to find dimensions where the sum of lengths around the shape is as small as possible. For a given area, a rectangle will have the minimal perimeter when its length and width are equal or as close as possible.

In this example, the area of each half of the field is 10,000 square meters. We look for two integers that multiply to give 10,000, and that are as close to equal as possible. Starting from 100 x 100, which is a square, we confirm these dimensions minimize the perimeter. The reasoning behind this is that a square has the minimum perimeter for a given area compared to other rectangular configurations. This means fewer resources are used for the same enclosed area, leading to lower cost when fencing or materials are involved.

Fence Cost Calculation

The cost to fence a field depends largely on the required perimeter of the fence and the cost per unit length of the material used. Calculating the total cost involves multiplying the perimeter by the price per meter of fencing material. Here, Kaurav's field perimeter was found to be 400 meters after determining optimal dimensions.

Given a cost of Rs. 2 per meter for fencing, we can easily find the total cost by:- Calculating the perimeter first: \[\text{Perimeter} = 2(L + W) = 2(100\, \text{m} + 100\, \text{m}) = 400\, \text{m}\] - Then multiplying this by the fence cost per meter: \[\text{Total Cost} = 400\, \text{m} \times Rs. 2/m = Rs. 800\]

• Ensure accurate perimeter calculation.
• Multiply by the cost per meter accurately.
• Double-check numbers to avoid errors.

Thus, the total fence cost sums up to Rs. 800 for the half field, assuming the fence costs are consistent all around.

Rectangular Dimensions

Understanding the dimensions of a rectangle involves considering both its length and width in relation to its total area. For the problem, dividing a 20,000 sq. m field into two equal parts results in each half having an area of 10,000 sq. m.

To determine potential dimensions, consider: - The fact that any length and width combination that multiplies to given area makes a valid rectangle. - In this problem, integer factor pairs of 10,000 are explored to find where length and width are closest. By trial and error, 100 x 100 was found to be most optimal, ensuring minimized perimeter. This optimizes material usage.

• Start by calculating possible integer pairs.
• Compare closeness of length and width.
• Verify that these dimensions give the needed area.

This approach not only proves efficient in cost but also demonstrates a practical method in geometry to find compact, cost-effective shapes for a given land division.