7. The Sides Of A Rectangle Are In The Ratio 5:2. If The Perimeter Of The Rectangle Is 42 Cm, Find The
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Introduction
Understanding geometric concepts such as the properties of rectangles is essential in mathematics, particularly in solving real-world problems involving measurements and ratios. In this article, we'll explore a specific problem involving rectangles where the sides are in a ratio of 5:2, and the perimeter is given as 42 centimeters. The goal is to find the lengths of the sides based on these conditions. By breaking down the problem step-by-step, applying algebraic methods, and understanding the concepts involved, readers can develop a clear approach to solving similar problems in geometry.
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Understanding the Problem
Before diving into the calculations, it's important to clearly understand what the problem states:
- The rectangle's sides are in a ratio of 5:2.
- The perimeter of the rectangle is 42 centimeters.
- The task is to find the lengths of the sides.
Key Concepts Involved
- Ratio of sides: The ratio between the length and width of the rectangle.
- Perimeter of a rectangle: The total distance around the rectangle, which is calculated as 2 times the sum of length and width.
- Algebraic representation: Using variables to represent the sides and ratios for setting up equations.
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Step-by-Step Approach to Solving the Problem
To find the sides of the rectangle, follow these steps:
Step 1: Define Variables
Let’s denote:
- L as the length of the rectangle.
- W as the width of the rectangle.
Since the sides are in the ratio 5:2, we can express:
- L = 5x
- W = 2x
where x is the common multiplying factor.
Step 2: Write the Perimeter Equation
The perimeter P of a rectangle is given by:
\[ P = 2 (L + W) \]
Given that the perimeter is 42 cm, substitute the expressions for L and W:
\[ 42 = 2 (5x + 2x) \]
Simplify inside the parentheses:
\[ 42 = 2 (7x) \]
Which simplifies to:
\[ 42 = 14x \]
Step 3: Solve for the Common Factor \( x \)
Divide both sides of the equation by 14:
\[ x = \frac{42}{14} = 3 \]
Step 4: Calculate the Actual Length and Width
Now, substitute \( x = 3 \) into the expressions for L and W:
- L = 5x = 5 \times 3 = 15 \text{ cm}
- W = 2x = 2 \times 3 = 6 \text{ cm}
Final Answer:
- Length = 15 cm
- Width = 6 cm
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Additional Insights and Related Concepts
Understanding Ratios in Geometry
Ratios are fundamental in comparing quantities, especially when dealing with similar figures or proportional relationships. In this problem, the sides' ratio helps set up an algebraic framework to solve for actual lengths.
Significance of Perimeter in Geometry
The perimeter provides a measure of the boundary length of a shape, which is crucial in various applications such as fencing, framing, or material estimation.
Applications of the Problem
- Design and Architecture: Calculating dimensions based on ratios and total boundary length.
- Manufacturing: Determining material dimensions for rectangular components.
- Education: Teaching students to translate word problems into algebraic equations.
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Common Mistakes to Avoid
When solving similar problems, keep in mind:
- Incorrectly setting up the ratio: Ensure the ratio is expressed correctly in terms of a common variable.
- Miscalculating the perimeter: Remember that the perimeter formula involves doubling the sum of the sides.
- Forgetting to simplify: Always simplify equations step-by-step to avoid errors.
- Ignoring units: Maintain consistent units throughout calculations.
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Practice Problems
To reinforce understanding, try solving these problems:
- If the sides of a rectangle are in the ratio 3:4 and the perimeter is 56 cm, find the lengths of the sides.
- A rectangle has sides in the ratio 7:3, and its perimeter is 50 cm. Determine the side lengths.
- The length of a rectangle is twice its width, and its perimeter is 48 cm. Find the length and width.
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Tips for Solving Ratio and Perimeter Problems
- Always start by defining variables for the sides.
- Express the sides in terms of a common variable based on the given ratio.
- Write the perimeter equation carefully and substitute the expressed sides.
- Simplify step-by-step to isolate the variable.
- Verify your answer by plugging back into the perimeter formula.
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Conclusion
This problem exemplifies how ratios and basic algebra can be combined to solve geometric problems involving rectangles. By understanding the relationships between sides and perimeter, and translating the problem into algebraic equations, solutions become more straightforward. Remember to carefully define variables, set up your equations correctly, and solve systematically. With practice, these types of problems will become more intuitive, enhancing both your algebraic and geometric problem-solving skills.
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Summary
| Step | Description | Example/Formula |
|---|---|---|
| 1 | Define variables for sides based on ratio | \( L = 5x, W = 2x \) |
| 2 | Write perimeter equation | \( P = 2 (L + W) \) |
| 3 | Substitute expressions and solve for \( x \) | \( 42 = 14x \) \(\Rightarrow x=3\) |
| 4 | Calculate actual side lengths | \( L=15 \text{ cm}, W=6 \text{ cm} \) |
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Final Remarks
Mastering problems involving ratios and perimeters equips students with critical problem-solving skills applicable in many areas of mathematics and real-world applications. By following structured approaches and practicing different scenarios, learners can develop confidence and competence in geometry.
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Disclaimer: This article is for educational purposes and aims to provide a comprehensive understanding of solving ratio-based rectangle perimeter problems.
