Question**: A rectangle's length is three times its width. If the perimeter of the rectangle is 48 meters, what is the area of the rectangle? - Aurero
Solving for the Area of a Rectangle: A Practical Math Problem
Solving for the Area of a Rectangle: A Practical Math Problem
When faced with a geometry question like "A rectangle’s length is three times its width, and the perimeter is 48 meters. What is the area?", it’s easy to break the problem down step by step—especially with SEO in mind. This article will guide you through solving this common math challenge while optimizing for search engines to help readers understand both the solution and key terms like rectangle area, perimeter formula, and algebraic modeling.
Understanding the Context
Understanding the Problem
Before solving, it’s important to clearly define the relationship between the rectangle’s dimensions. According to the question:
- The length (L) is three times the width (W)
- That is, ( L = 3W )
- The perimeter (P) is given as 48 meters
- We are asked to find the Area (A) of the rectangle
This setup is ideal for teaching algebraic problem-solving, making it highly relevant for students, educators, and DIY home project planners.
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Key Insights
Step 1: Use the Perimeter Formula
The perimeter of a rectangle is calculated using the formula:
[
P = 2(L + W)
]
Plugging in the known perimeter:
[
48 = 2(L + W)
]
Divide both sides by 2:
[
24 = L + W
]
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Step 2: Substitute the Length in Terms of Width
Since ( L = 3W ), substitute into the sum:
[
24 = 3W + W = 4W
]
Solve for ( W ):
[
W = \frac{24}{4} = 6 \ ext{ meters}
]
Step 3: Find the Length
Using ( L = 3W ):
[
L = 3 \ imes 6 = 18 \ ext{ meters}
]
Step 4: Calculate the Area
The area ( A ) of a rectangle is:
[
A = L \ imes W
]
Substitute the values:
[
A = 18 \ imes 6 = 108 \ ext{ square meters}
]
