Solving Geometry Problems: Rectangle Perimeter and Area Calculations

Solving geometry problems is an essential skill, particularly in fields such as engineering, architecture, and interior design. One common task involves calculating the perimeter and area of a rectangle given certain conditions or constraints. In this article, we will detail the steps and methods to solve a specific problem that involves determining the area of a rectangle given its length and breadth relationship and perimeter.

Problem Statement

We are given a rectangle where the difference between its length and breadth is 23 meters. The perimeter of the rectangle is 206 meters. The task is to find the area of the rectangle in square meters.

Solution 1

Let's denote the length of the rectangle as L meters and the breadth as B meters. We are given two key pieces of information:

The length and breadth relationship: (L - B 23) The perimeter: (2L 2B 206)

Simplifying the perimeter equation, we get:

[2(L B) 206]

[L B 103]

Now, we have a system of linear equations:

[L - B 23] [L B 103]

To solve for L and B, we can add these two equations:

[2L 126]

[L 63]

Substituting (L 63) into (L B 103), we find:

[63 B 103]

[B 40]

Now that we have the length (L 63 meters) and the breadth (B 40 meters), we can calculate the area (A) of the rectangle using the formula:

[A L times B]

[A 63 times 40]

[A 2520 text{ square meters}]

Solution 2

Another approach involves using simplified steps:

1. Let the length be x meters and the breadth be y meters.

2. The length and breadth relationship is:

[x - y 25]

3. The perimeter is given by:

[2(xy) 210]

4. Dividing both sides by 2:

[xy 105]

5. We can solve these two equations to find x and y:

[2x 130]

[x 65text{ meters}]

[y 40text{ meters}]

6. Using the area formula for a rectangle:

[A x times y]

[A 65 times 40]

[A 2600 text{ square meters}]

7. Another method directly calculates the area without explicitly solving for length and breadth:

[2L 126]

[L 63text{ meters}]

[B 103 - 63 40text{ meters}]

[A L times B]

[A 63 times 40]

[A 2520 text{ square meters}]

Solution 3

Yet another approach is as follows:

1. Let L and B be the length and breadth of the rectangle in centimeters.

2. Given conditions:

[L - B 23text{ cm}]

[2(L B) 206text{ cm}]

3. Simplifying the perimeter condition:

[L B 103text{ cm}]

4. Solving the system of equations:

[2L 126text{ cm}]

[L 63text{ cm}]

[B 103 - 63 40text{ cm}]

5. Calculating the area:

[A L times B]

[A 63 times 40]

[A 2520text{ square centimeters}]

Conclusion

Understanding and solving geometry problems is valuable for various practical applications. By carefully analyzing the given conditions and systematically applying algebraic methods, we can determine the desired measurements such as length, breadth, and area of geometric shapes. The example provided demonstrates multiple approaches to solving a problem involving the perimeter and area of a rectangle. These strategies can be adapted and applied to similar geometry problems in real-world scenarios.