Solving Rectangle Area Problems Using Ratios and Perimeter Information

In this article, we will explore several methods to solve a specific problem involving the area of a rectangle given its dimensions as a ratio and its perimeter. The problem is stated as follows: the length and width of a rectangle are in the ratio 4:2, and the perimeter is 30 cm. We will provide detailed solutions to this problem using various strategies to ensure a deep understanding of the concepts involved.

Solution 1: Direct Calculation

The first method involves breaking down the ratio and using the given perimeter to find the actual dimensions of the rectangle. Here’s the step-by-step process:

The ratio of the length to the width is 4:2. When we sum the parts of the ratio, we get 4 2 6. The perimeter of the rectangle is given as 30 cm. To find the length of each part, we divide the perimeter by the sum of the parts: 30 cm / 6 5 cm. Therefore, the length of the rectangle is 4 parts, which is 4 * 5 cm 20 cm, and the width is 2 parts, which is 2 * 5 cm 10 cm. The area of the rectangle can be calculated by multiplying the length and the width: 20 cm * 10 cm 200 cm2.

Conclusion: The area of the rectangle is 200 cm2.

Solution 2: Variable Substitution

We can also use variable substitution to solve the problem. Here’s how:

Assume the width of the rectangle is 2 units and the length is 3 units. The perimeter with these values would be 10 units, but it is given as 30 units, which is 3 times the assumed value. Multiplying the assumed width and length by 3 gives the actual dimensions: width 6 cm and length 9 cm. The area is then calculated as: 6 cm * 9 cm 54 cm2.

Solution 3: Algebraic Method

Another approach is using algebraic equations:

Let the width be represented by w and the length by l. The perimeter equation is 2l 2w 30, and the ratio equation is l 2w. Substituting l 2w into the perimeter equation gives: 2(2w) 2w 30, which simplifies to 4w 2w 30, or 6w 30. Solving for w gives w 5 cm, and l 2w 10 cm. The area is then l * w 10 * 5 50 cm2.

Solution 4: J Programming Language Approach

The J programming language can also be used to solve this problem through a brute force approach:

Using the J programming environment and the NuVoc, we find that the solution is: width 5 cm and length 10 cm. The area is then 5 cm * 10 cm 50 cm2.

Conclusion

Regardless of the method used, the area of the rectangle is consistently found to be 50 cm2. This article demonstrates several approaches to solving the same problem, emphasizing the importance of using ratios and perimeters to find unknown dimensions in geometric shapes. Understanding these methods can be particularly useful in advancing your skills in solving geometric problems in mathematics and related fields.