Finding Rectangular Field Dimensions

Let the original length of the rectangular field be L meters and the original width be W meters.

The original area is given as 600 square meters. This can be written as:

$ L \times W = 600 \quad (1) $

According to the problem, if the length were increased by 10 meters (new length = \( L + 10 \)) and the width decreased by 2 meters (new width = \( W - 2 \)), the area remains the same (600 square meters). This gives us the second equation:

$ (L + 10) \times (W - 2) = 600 \quad (2) $

Algebraic Manipulation of Equations

Expand the second equation:

$ LW - 2L + 10W - 20 = 600 $

Substitute the value of \( LW \) from equation (1) into the expanded equation:

$ 600 - 2L + 10W - 20 = 600 $

Simplify the equation by subtracting 600 from both sides and rearranging:

$ -2L + 10W - 20 = 0 $

$ -2L + 10W = 20 $

Divide the entire equation by -2 to simplify further:

$ L - 5W = -10 $

Now, express L in terms of W:

$ L = 5W - 10 \quad (3) $

Solving for Width

Substitute the expression for L from equation (3) into the original area equation (1):

$ (5W - 10) \times W = 600 $

Expand this equation to form a quadratic equation:

$ 5W^2 - 10W = 600 $

Rearrange it into the standard quadratic form \( aW^2 + bW + c = 0 \):

$ 5W^2 - 10W - 600 = 0 $

Divide the equation by 5:

$ W^2 - 2W - 120 = 0 $

Factor the quadratic equation:

$ (W - 12)(W + 10) = 0 $

This gives two possible values for W:

• \( W - 12 = 0 \Rightarrow W = 12 \)
• \( W + 10 = 0 \Rightarrow W = -10 \)

Since the width of a field cannot be negative, we take the positive value:

$ W = 12 \text{ meters} $

Calculating Length

Now, use equation (3) to find the original length L:

$ L = 5W - 10 $

Substitute \( W = 12 \):

$ L = 5(12) - 10 $

$ L = 60 - 10 $

$ L = 50 \text{ meters} $

Verification

Let's check if these dimensions satisfy the problem conditions:

• Original Area: \( 50 \text{ m} \times 12 \text{ m} = 600 \text{ m}^2 \) (Correct)
• New Length: \( 50 + 10 = 60 \text{ m} \)
• New Width: \( 12 - 2 = 10 \text{ m} \)
• New Area: \( 60 \text{ m} \times 10 \text{ m} = 600 \text{ m}^2 \) (Correct)

The original dimensions of the field are Length = 50 meters and Width = 12 meters.