Rectangle Perimeter: Max Vs Min (Area 60 Cm²)

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Rectangle Perimeter: Max vs Min (Area 60 cm²)

Let's dive into a fun math problem involving rectangles, areas, and perimeters! This question explores how the shape of a rectangle, while maintaining the same area, can drastically change its perimeter. We're given a rectangle with an area of 60 cm², and the lengths of its sides are whole numbers (integers). The challenge is to figure out the biggest possible difference between the largest and smallest perimeters this rectangle could have. Sounds intriguing, right? Let's break it down step by step.

Understanding the Basics: Area and Perimeter

Before we jump into solving the problem, let's quickly recap the formulas for the area and perimeter of a rectangle. These are fundamental concepts that will guide our approach.

  • Area of a rectangle: The area is the space enclosed within the rectangle. It's calculated by multiplying the length and the width: Area = Length × Width
  • Perimeter of a rectangle: The perimeter is the total distance around the outside of the rectangle. It's calculated by adding up the lengths of all four sides. Since a rectangle has two pairs of equal sides, the formula is: Perimeter = 2 × (Length + Width)

In our case, we know the area is fixed at 60 cm². This means that no matter what the length and width are, their product must always be 60. However, the length and width can vary, and this variation directly affects the perimeter.

Finding Possible Dimensions

The key to solving this problem is recognizing that different pairs of integer side lengths can result in the same area. Since the area is 60 cm², we need to find all the pairs of whole numbers that multiply to give 60. Let's list them out systematically:

  • 1 cm × 60 cm = 60 cm²
  • 2 cm × 30 cm = 60 cm²
  • 3 cm × 20 cm = 60 cm²
  • 4 cm × 15 cm = 60 cm²
  • 5 cm × 12 cm = 60 cm²
  • 6 cm × 10 cm = 60 cm²

These are all the possible integer side lengths for our rectangle. Notice how we've covered all the factors of 60. Now, for each of these pairs, we need to calculate the perimeter.

Calculating the Perimeters

Using the formula Perimeter = 2 × (Length + Width), let's calculate the perimeter for each of the side length pairs we found:

  • 1 cm × 60 cm: Perimeter = 2 × (1 + 60) = 2 × 61 = 122 cm
  • 2 cm × 30 cm: Perimeter = 2 × (2 + 30) = 2 × 32 = 64 cm
  • 3 cm × 20 cm: Perimeter = 2 × (3 + 20) = 2 × 23 = 46 cm
  • 4 cm × 15 cm: Perimeter = 2 × (4 + 15) = 2 × 19 = 38 cm
  • 5 cm × 12 cm: Perimeter = 2 × (5 + 12) = 2 × 17 = 34 cm
  • 6 cm × 10 cm: Perimeter = 2 × (6 + 10) = 2 × 16 = 32 cm

We now have the perimeters for all possible rectangular shapes with an area of 60 cm² and integer side lengths. It's pretty cool how much the perimeter changes just by altering the shape, even though the area stays the same!

Finding the Maximum and Minimum Perimeters

Looking at the perimeters we calculated, it's clear that:

  • The maximum perimeter is 122 cm (for the 1 cm × 60 cm rectangle).
  • The minimum perimeter is 32 cm (for the 6 cm × 10 cm rectangle).

This highlights a key concept: for a fixed area, a more elongated rectangle (one with very different side lengths) will have a larger perimeter, while a rectangle closer to a square shape will have a smaller perimeter. Think about it – stretching out the sides increases the total distance around the shape.

Calculating the Difference

The final step is to find the difference between the maximum and minimum perimeters:

Difference = Maximum Perimeter - Minimum Perimeter Difference = 122 cm - 32 cm Difference = 90 cm

So, the difference between the maximum and minimum possible perimeters for a rectangle with an area of 60 cm² and integer side lengths is 90 cm. That's a significant difference!

Key Takeaways

  • For a fixed area, the perimeter of a rectangle varies depending on its shape.
  • Elongated rectangles have larger perimeters compared to rectangles closer to a square shape.
  • Finding all possible integer side length pairs is crucial for solving this type of problem.

Let's Summarize Guys!

To put it simply, the problem asked us to find the difference between the largest and smallest possible perimeters of a rectangle. The area was fixed at 60 cm², and the side lengths had to be integers. We figured out all the possible side length combinations. Then, we calculated the perimeter for each combination and found the maximum and minimum values. Finally, we subtracted the minimum from the maximum to get the answer, which is 90 cm. We nailed it!

Why This Matters: Real-World Connections

This type of problem isn't just about math for the sake of math. It actually has connections to real-world situations. Imagine you're fencing off a garden with a limited amount of fencing material (representing the perimeter). You want to enclose a certain area to grow your plants. This problem helps you understand how to choose the dimensions of your garden to maximize the area you enclose with your limited fencing. Or, think about packaging design – companies often want to minimize the amount of material used (perimeter) while still enclosing a certain volume (related to area). Understanding the relationship between perimeter and area is super important in many practical applications.

Extra Practice

Want to practice this concept further? Try these variations:

  1. Change the Area: What if the area was 48 cm² instead of 60 cm²? How would the difference between the maximum and minimum perimeters change?
  2. Different Shapes: Can you explore a similar problem with other shapes, like triangles or parallelograms? How does the relationship between area and perimeter differ for these shapes?
  3. Non-Integer Sides: What if the side lengths didn't have to be integers? How would that affect the possible perimeters?

By playing with these variations, you'll deepen your understanding of the relationship between area and perimeter. Keep exploring, keep questioning, and keep having fun with math!

Conclusion

This problem beautifully illustrates how mathematical concepts can be both challenging and engaging. By understanding the formulas for area and perimeter and by systematically exploring the possibilities, we were able to find the solution. Remember, math isn't just about numbers; it's about problem-solving, logical thinking, and making connections to the world around us. Keep practicing, and you'll become a math whiz in no time! Hope you guys enjoyed this problem-solving journey! Let's tackle more math challenges together in the future!