Exploring the Geometry of Rectangles: Solving for Length and Width

Rectangles are one of the most fundamental shapes in geometry, with a wide variety of applications from everyday life to complex engineering designs. In this article, we will delve into solving a specific problem related to the dimensions of a rectangle, where the area and one dimension are given.

The Problem

The area of a rectangle is 40 square centimeters (sq. cm). If the length is 3 centimeters (cm) more than its width, how can we find the dimensions of the rectangle?

Solving the Problem

Let's denote the width of the rectangle by w cm. According to the problem, the length l is 3 cm more than the width:

l w 3

The area A of a rectangle is given by the formula:

A l × w

Given that the area is 40 sq. cm, we can substitute for l:

40 (w 3) × w

Expanding this equation:

40 w2 3w

Rearranging the equation gives:

w2 - 3w - 40 0

Now we can solve this quadratic equation using the quadratic formula:

w (-b ± √(b2 - 4ac)) / 2a

where a 1, b -3, and c -40

Calculating the discriminant:

b2 - 4ac 32 - 4 × 1 × (-40) 9 160 169

Now substituting back into the quadratic formula:

w (-(-3) ± √169) / (2 × 1) (3 ± 13) / 2

Calculating the two possible values for w:

w (3 13) / 2 16 / 2 8 cm (not valid since width cannot be negative)

w (3 - 13) / 2 -10 / 2 -8 cm (not valid since width cannot be negative)

Therefore, the width w is 5 cm. Now we can find the length:

l w 3 5 3 8 cm

Conclusion

The length of the rectangle is 8 cm, and the width is 5 cm. This example demonstrates the application of quadratic equations in solving real-world geometric problems.