Six Feet Less Than The Width

Six Feet Less Thanthe Width: Understanding and Solving This Common Math Phrase

When you encounter the expression “six feet less than the width” in a word problem, it signals a relationship between two measurements—typically the length and width of a rectangle, or sometimes the dimensions of a garden, a room, or a piece of fabric. Grasping how to translate this verbal clue into an algebraic expression is essential for solving geometry, algebra, and real‑world measurement tasks. This article breaks down the meaning, shows step‑by‑step methods for setting up and solving equations, highlights practical applications, warns of frequent pitfalls, and provides practice exercises to reinforce your confidence.

Introduction: What Does “Six Feet Less Than the Width” Mean?

The phrase “six feet less than the width” describes a quantity that is obtained by subtracting six feet from the width of an object. In algebraic terms, if we let W represent the width (in feet), then the expression becomes:

[ \text{Six feet less than the width} = W - 6 ]

Often this expression is used to define another dimension, such as the length (L) of a rectangle:

[L = W - 6 ]

Recognizing this pattern allows you to replace words with symbols, set up equations based on additional information (like perimeter or area), and solve for the unknowns.

Setting Up the Equation: From Words to Algebra

Step 1: Identify the Known and Unknown Quantities

Read the problem carefully and label each measurement. Typical unknowns include:

• W – width (feet)
• L – length (feet)
• P – perimeter (feet)
• A – area (square feet)

Step 2: Translate the Verbal Relationship

Replace “six feet less than the width” with W – 6. If the problem states that the length equals this quantity, write:

[ L = W - 6 ]

If instead the width is six feet less than the length, you would write:

[ W = L - 6 \quad \text{or} \quad L = W + 6]

Step 3: Incorporate Any Additional Conditions

Most problems give a second piece of information, such as:

• The perimeter is known: (P = 2L + 2W)
• The area is known: (A = L \times W)
• The sum of length and width is given: (L + W = \text{some number})

Substitute the expression for L (or W) from Step 2 into the second condition to obtain a single‑variable equation.

Step 4: Solve for the Unknown

Isolate the variable, perform arithmetic, and then back‑substitute to find the other dimension.

Worked Example: Finding Dimensions from Perimeter

Problem:
A rectangular garden has a width that is six feet less than its length. If the perimeter of the garden is 48 feet, what are the length and width?

Solution:

1. Define variables
   Let L = length (ft)
   Let W = width (ft)

2. Translate the relationship
   “Width is six feet less than the length” → (W = L - 6)

3. Write the perimeter formula
   (P = 2L + 2W)
   Given (P = 48), substitute:
   (48 = 2L + 2W)

4. Replace W with the expression from step 2
   (48 = 2L + 2(L - 6))

5. Simplify and solve [ \begin{aligned} 48 &= 2L + 2L - 12 \ 48 &= 4L - 12 \ 48 + 12 &= 4L \ 60 &= 4L \ L &= \frac{60}{4} = 15 \text{ ft} \end{aligned} ]

6. Find W
   (W = L - 6 = 15 - 6 = 9 \text{ ft})

Answer: Length = 15 ft, Width = 9 ft.

Worked Example: Finding Dimensions from Area

Problem:
A rectangular painting’s length is six feet less than twice its width. The area of the painting is 160 square feet. Determine the length and width.

Solution:

1. Variables
   Let W = width (ft)
   Let L = length (ft)

2. Translate the relationship
   “Length is six feet less than twice the width” → (L = 2W - 6)

3. Area formula
   (A = L \times W)
   Given (A = 160):
   (160 = (2W - 6) \times W)

4. Expand and rearrange
   [ 160 = 2W^{2} - 6W \ 0 = 2W^{2} - 6W - 160 ]

5. Divide by 2 to simplify
   (0 = W^{2} - 3W - 80)

6. Factor or use quadratic formula
   Factoring: ((W - 10)(W + 8) = 0)
   Thus, (W = 10) or (W = -8). Width cannot be negative, so (W = 10) ft.

7. Find L
   (L = 2W - 6 = 2(10) - 6 = 20 - 6 = 14) ft

Answer: Width = 10 ft, Length = 14 ft.

Real‑World Applications

Understanding the “six feet less than the width” relationship isn’t just an academic exercise; it appears in many practical scenarios:

of the court. | Dimensions for court construction or equipment placement. |

Conclusion

These examples demonstrate how seemingly simple word problems can be solved using algebraic principles. By carefully translating the given information into equations and applying basic algebraic techniques, such as substitution and factoring, we can determine unknown dimensions in a variety of real-world situations. The key is to identify the relationships between the variables and to formulate an equation that accurately represents the problem. While the specific phrases might vary – "six feet less than," "twice as much as," or "a certain proportion of" – the underlying methodology remains consistent. This skill is fundamental not only in mathematics but also in problem-solving across numerous disciplines, empowering us to analyze and understand the world around us with greater precision. Mastering these skills builds a strong foundation for more complex mathematical concepts and equips individuals with valuable analytical abilities applicable to everyday life and professional pursuits.