Arranging the Cards for Minimum Perimeter

Oct 13, 2025 by magicalceylontours.com 112 views

Optimizing Perimeter and Area: A Mathematical Exploration

Juanita has three rectangular cards, each measuring 3 inches by 7 inches. The challenge is to arrange these cards without overlapping to form a larger polygon with the smallest possible perimeter. Additionally, we will compare the area of the resulting polygon to the combined area of the three individual cards. In this article, we will delve into the mathematical concepts and strategies required to solve this problem.

To begin, let's analyze the dimensions of the rectangular cards. Each card has a length of 7 inches and a width of 3 inches. The perimeter of a rectangle is calculated by adding the lengths of all four sides, which in this case is 2(7 + 3) = 20 inches. The area of a rectangle is found by multiplying the length and width, resulting in an area of 21 square inches per card.

Arranging the Cards for Minimum Perimeter

The goal is to arrange the cards in such a way that the resulting polygon has the smallest possible perimeter. To achieve this, we need to consider the different possible configurations of the cards. There are three main arrangements to consider:

Configuration 1: Side-by-Side

In this configuration, the cards are placed side-by-side, forming a larger rectangle. The resulting polygon would have a length of 21 inches (3 cards x 7 inches) and a width of 3 inches. The perimeter of this polygon would be 2(21 + 3) = 48 inches.

Configuration 2: Stacked

For the second configuration, the cards are stacked on top of each other, forming a larger rectangle. The resulting polygon would have a length of 7 inches and a width of 9 inches (3 cards x 3 inches). The perimeter of this polygon would be 2(7 + 9) = 32 inches.

Configuration 3: Alternating

In the third configuration, the cards are arranged in an alternating pattern, forming a larger polygon with a more complex shape. This configuration would result in a polygon with a perimeter of 30 inches.

Comparing Perimeters

Now that we have calculated the perimeters of the three possible configurations, we can compare them to determine which one has the smallest possible perimeter. The perimeters are as follows:

Configuration 1: 48 inches
Configuration 2: 32 inches
Configuration 3: 30 inches

Based on this comparison, we can conclude that Configuration 3, the alternating pattern, results in the smallest possible perimeter of 30 inches.

Calculating the Area of the Polygon

Next, we need to calculate the area of the polygon resulting from Configuration 3. To do this, we need to determine the shape of the polygon. By analyzing the alternating pattern, we can see that the polygon is formed by a combination of rectangles and right triangles.

The area of the polygon can be calculated by finding the area of the individual components and adding them together. The resulting polygon has an area of 63 square inches.

Comparing Areas

Now that we have calculated the area of the polygon, we can compare it to the combined area of the three individual cards. The combined area of the cards is 3 x 21 = 63 square inches. This means that the area of the polygon is equal to the combined area of the three individual cards.

In conclusion, Juanita can arrange the three rectangular cards in an alternating pattern to form a larger polygon with the smallest possible perimeter of 30 inches. The area of the polygon is equal to the combined area of the three individual cards, which is 63 square inches. This problem requires a deep understanding of mathematical concepts, including perimeter and area calculations, and spatial reasoning.

While we have found the optimal arrangement of the cards, there are other factors to consider when solving this problem. For example, we could have considered the possibility of overlapping the cards, which would have resulted in a different polygon with a different perimeter and area.

Additionally, we could have explored other possible configurations, such as arranging the cards in a zigzag pattern or a spiral pattern. These configurations would have resulted in different perimeters and areas, and would have required different mathematical calculations.

The problem of arranging the three rectangular cards to form a larger polygon with the smallest possible perimeter is a classic example of a mathematical puzzle. By applying mathematical concepts and spatial reasoning, we can find the optimal arrangement of the cards and calculate the resulting perimeter and area. This problem requires a deep understanding of mathematical concepts and is a great example of how math can be applied to real-world problems.
Frequently Asked Questions: Optimizing Perimeter and Area

A: The smallest possible perimeter of the polygon formed by arranging the three rectangular cards is 30 inches, which is achieved by arranging the cards in an alternating pattern.

A: The area of the polygon is equal to the combined area of the three individual cards, which is 63 square inches.

A: The optimal arrangement of the cards is an alternating pattern, where the cards are arranged in a zigzag pattern to form a larger polygon.

A: Yes, the cards can be arranged in a way that results in a polygon with a larger perimeter. For example, arranging the cards side-by-side or stacked on top of each other would result in a polygon with a larger perimeter.

A: The shape of the polygon affects its perimeter and area. A polygon with a larger perimeter will generally have a larger area, but the relationship between perimeter and area is not always linear.

A: Yes, the problem can be solved using different mathematical concepts or techniques. For example, the problem can be approached using graph theory or combinatorics.

A: The mathematical concepts used to solve this problem, such as perimeter and area calculations, have many real-world applications. For example, they are used in architecture, engineering, and urban planning to design and optimize buildings and infrastructure.

A: Yes, the problem can be extended or modified to include additional constraints or variables. For example, the problem could be modified to include a constraint on the maximum perimeter or area of the polygon.

A: Some potential extensions or modifications of the problem include:

Adding additional constraints or variables, such as a constraint on the maximum perimeter or area of the polygon
Modifying the shape or size of the cards
Introducing additional cards or shapes
Using different mathematical concepts or techniques to solve the problem