Introduction

Understanding the relationship between the perimeter and area of geometric shapes is crucial for various applications, from basic geometry problems to more complex real-world scenarios. This article will explore how to find the dimensions of a rectangle given its perimeter and area, specifically through the example of a rectangle with a perimeter of 30 cm and an area of 54 cm2.

Solving for Rectangle Dimensions Using Equations

The perimeter P and area A of a rectangle can be expressed in terms of its length L and width W as follows:

Perimeter: (P 2L 2W) Area: (A L times W)

Given that the perimeter is 30 cm and the area is 54 cm2, we can set up the following equations:

Perimeter Equation: (2L 2W 30) Area Equation: (L times W 54)

Step-by-Step Solution

Step 1: Simplify the Perimeter Equation

Divide the perimeter equation by 2 to simplify:

[L W 15]

From this equation, solve for one variable in terms of the other. Let's solve for W in terms of L:

[W 15 - L]

Step 2: Substitute into the Area Equation

Substitute 15 - L for W in the area equation:

[L times (15 - L) 54]

Expanding and simplifying:

[15L - L^2 54]

Further rearrange to form a standard quadratic equation:

[-L^2 15L - 54 0]

To solve this, multiply the entire equation by -1 to make the leading coefficient positive:

[L^2 - 15L 54 0]

Step 3: Apply the Quadratic Formula

The quadratic formula is given by:

[L frac{-b pm sqrt{b^2 - 4ac}}{2a}]

Here, a 1, b -15, and c 54.

The discriminant ((Delta)) is calculated as:

[Delta b^2 - 4ac (-15)^2 - 4 cdot 1 cdot 54 225 - 216 9]

Apply the quadratic formula:

[L frac{15 pm sqrt{9}}{2}]

Since (sqrt{9} 3), the values for L are:

[L frac{15 3}{2} 9 quad text{or} quad L frac{15 - 3}{2} 6]

Step 4: Find Corresponding Widths

- If (L 9), then (W 15 - 9 6)

- If (L 6), then (W 15 - 6 9)

So the dimensions of the rectangle are:

Length 9 cm, Width 6 cm, or vice versa.

Application of Geometry in Real Life

Understanding how to manipulate equations and solve for dimensions is not only an essential skill in mathematics but also has practical applications. For instance, architects, engineers, and designers often need to determine the dimensions of various shapes given specific criteria. In our example, such calculations can be crucial in designing the layout of rooms or other spaces.

Conclusion

By solving this problem, we can clearly see how equations and the quadratic formula can be used to find dimensions of a rectangle given its perimeter and area. This method can be applied to a wide range of similar problems in geometry and is fundamental in various mathematical and real-world applications.