Finding the Dimensions of a Rectangle Given Its Length and Area

Imagine you are dealing with a rectangle where the width is 7 cm less than the length, and the area is 78 cm2. How would you find the dimensions of this rectangle? This article will guide you through the process, offering multiple methods to arrive at the solution. We will also discuss the mathematical background that supports these methods.

Method 1: Factorization

One simple and intuitive way to find the dimensions is by factorizing the area. The area of the rectangle is 78 square centimeters, and one dimension is 7 cm less than the other. By factorizing 78, we can find the suitable dimensions.

The factors of 78 are:

Among these, the pair 13 and 6 fits the condition that one dimension is 7 cm less than the other. Hence, the dimensions of the rectangle are 13 cm and 6 cm.

Method 2: Quadratic Equation

Another method involves setting up and solving a quadratic equation. Let's denote the length of the rectangle as n cm. Therefore, the width is n-7 cm. The area of the rectangle is given by the product of length and width, which equals 78 square centimeters.

We can write the equation as:

nn-7 78

Expanding this, we get the quadratic equation:

n2-7n-780

To solve this quadratic equation, we can factorize it or use the quadratic formula. Here, we choose the factorization method for simplicity.

The factors of -78 that add up to -7 are -13 and 6. Therefore, the equation can be factorized as:

n-13 x (n 6) 0

This gives us two solutions: n -6 and n 13. Since the length cannot be negative, we select the positive root n 13.

Substituting n 13 into the width equation:

Width n-7 13-7 6 cm

Thus, the length is 13 cm and the width is 6 cm.

Method 3: Contemporary Algebraic Approach

Using a more conventional algebraic approach, let l be the length of the rectangle and l-7 be the width. The area is given by:

ll-7 78

Simplifying, we get:

l2-7l-780

This is a standard quadratic equation. We can solve it using the quadratic formula:

L (-b ± √(b^2 - 4ac)) / 2a

For our equation: a 1, b -7, and c -78. Plugging these values into the formula, we get:

L (7 ± √((-7)^2 - 4 x 1 x -78)) / 2 x 1

L (7 ± √(49 312)) / 2

L (7 ± √361) / 2

L (7 ± 19) / 2

This gives us two solutions:

L 13 and L -6

Since the length cannot be negative, we select L 13 cm. Substituting this into the width equation:

Width L-7 13-7 6 cm

Thus, the length is 13 cm and the width is 6 cm.

Conclusion

Through the methods of factorization, quadratic equation solving, and contemporary algebraic techniques, we have determined that the dimensions of the rectangle are 13 cm and 6 cm. This problem not only reinforces basic algebraic skills but also demonstrates how to apply different problem-solving strategies to reach a solution.

Understanding these methods is crucial not only for solving geometric problems but also for developing a broader mathematical perspective. By practicing and applying different approaches to the same problem, you can enhance your problem-solving skills and deepen your comprehension of mathematical concepts.

References

Books on Algebra and Geometry: These provide a comprehensive understanding of the algebraic methods used in solving geometric problems.