Question

The length of fence required to enclose a rectangular field is 3000 meters. What are the dimensions of the field if it is known that the difference between its length and width is 50 meters?

Short answer

The length is 775 meters and the width is 725 meters.

Step-by-step solution

Define Variables

Let the length of the field be denoted by \(L\) and the width be denoted by \(W\). According to the problem, the difference between the length and width is given by: \[ L - W = 50 \]

Use the Perimeter Formula

The perimeter of a rectangle is given by the formula: \[ 2L + 2W = 3000 \] Simplify this to: \[ L + W = 1500 \]

Set Up a System of Equations

From Step 1 and Step 2, we have the system of equations: \[ L - W = 50 \] \[ L + W = 1500 \]

Solve for Length and Width

Add the two equations: \[ (L - W) + (L + W) = 50 + 1500 \] This simplifies to: \[ 2L = 1550 \] Then, solve for \(L\): \[ L = 775 \]

Calculate the Width

Substitute the value of \(L\) back into \(L + W = 1500\): \[ 775 + W = 1500 \] Subtract 775 from both sides to solve for \(W\): \[ W = 725 \]

Verify the Solution

Check if \(L - W = 50\) with the calculated values: \[ 775 - 725 = 50 \] Both conditions are satisfied.

Key concepts

perimeter of a rectangle

The perimeter of a rectangle is the total distance around the outside of the rectangle. Imagine you are placing a fence around a field, the perimeter would be the length of that fence. For a rectangular field, the perimeter can be calculated using the formula: \[ P = 2L + 2W \] where:

• \(P\) is the perimeter
• \(L\) is the length
• \(W\) is the width

In the given problem, the perimeter is 3000 meters. This tells us the total length of all four sides combined.

system of equations

When you have more than one equation working together, it is called a system of equations. The solution must satisfy all equations in the system. In our exercise, we receive two important pieces of information:

• The perimeter: \[ L + W = 1500 \]
• The difference between length and width: \[ L - W = 50 \]

These two linear equations form a system. We need to solve them together to find the values of the length and width that satisfy both conditions.

algebraic expressions

Algebraic expressions use variables and constants to represent different values. In the exercise, expressions like\( L - W = 50 \) and \( L + W = 1500 \)are examples. These expressions help us model real-life problems using algebra. By solving them, we can find unknown values. In our problem:

• \(L\) represents the length of the field
• \(W\) represents the width of the field

By manipulating these expressions using basic algebra operations (addition, subtraction), we can find the actual measurements of the field.

problem solving

Problem solving involves applying various strategies to find a solution. Here's a simple plan we followed:

• Define Variables: Let \(L\) be the length and \(W\) be the width.
• Set Up Equations: Use given information to create equations.
• Solve the System: Combine and solve the equations to find \(L\) and \(W\).
• Check the Solution: Verify if the solution meets all conditions.

In the given problem, we first set up our variables and equations. Then, by adding the equations \(L - W = 50\) and \(L + W = 1500\), we eliminated \(W\) and solved for \(L\). Finally, we found \(W\) and checked our work. This process showcased how systematic problem solving can tackle complex problems efficiently.