Math Problem Statement

The floor of a shed has an area of 90 square feet. The floor is in the shape of a rectangle whose length is 3 feet less than twice the width. Find the length and the width of the floor of the shed. Question content area bottom Part 1 The width of the floor of the shed is

enter your response here ft.

Solution

Let the width of the floor be ww feet. According to the problem, the length of the floor is 2w32w - 3 feet. The area of the rectangular floor is given as 90 square feet.

The formula for the area of a rectangle is: Area=length×width\text{Area} = \text{length} \times \text{width}

Substitute the given expressions for length and width: 90=w(2w3)90 = w \cdot (2w - 3)

Step 1: Expand and simplify the equation

90=2w23w90 = 2w^2 - 3w

Rearrange the equation to standard quadratic form: 2w23w90=02w^2 - 3w - 90 = 0

Step 2: Solve the quadratic equation

We can solve this using the quadratic formula: w=b±b24ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here:

  • a=2a = 2,
  • b=3b = -3,
  • c=90c = -90.

Substitute these values into the quadratic formula: w=(3)±(3)24(2)(90)2(2)w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-90)}}{2(2)} w=3±9+7204w = \frac{3 \pm \sqrt{9 + 720}}{4} w=3±7294w = \frac{3 \pm \sqrt{729}}{4} w=3±274w = \frac{3 \pm 27}{4}

Step 3: Evaluate the two possible solutions

  1. w=3+274=304=7.5w = \frac{3 + 27}{4} = \frac{30}{4} = 7.5
  2. w=3274=244=6w = \frac{3 - 27}{4} = \frac{-24}{4} = -6

Since width cannot be negative, the width is: w=7.5ftw = 7.5 \, \text{ft}

Step 4: Find the length

The length is: Length=2w3=2(7.5)3=153=12ft\text{Length} = 2w - 3 = 2(7.5) - 3 = 15 - 3 = 12 \, \text{ft}

Final Answer:

  • Width: 7.5ft7.5 \, \text{ft}
  • Length: 12ft12 \, \text{ft}

Would you like a detailed verification of the solution? Here are related questions you might explore:

  1. What are the steps to verify the solution by substituting the values back into the area formula?
  2. How can the quadratic equation be solved using factoring instead of the quadratic formula?
  3. What happens if the area or dimensions are scaled by a factor of 2?
  4. How would this problem change if the area was a fraction or irrational number?
  5. Can we generalize this approach for any rectangle where the length is a linear function of the width?

Tip: Always check for realistic constraints (like positive dimensions) when solving real-world geometry problems.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Quadratic Equations

Formulas

Area of a rectangle: Area = length × width
Quadratic formula: w = (-b ± √(b² - 4ac)) / 2a

Theorems

Quadratic formula

Suitable Grade Level

Grades 8-10