Math Problem Statement

A rectangle has perimeter 44 feet and an area of 105 square feet. Find the dimensions of the rectangle assuming the width is smaller than the length.

Solution

To solve this problem, let’s use the information provided.

Given Information:

  1. The perimeter of the rectangle is 44 feet.
  2. The area of the rectangle is 105 square feet.
  3. We assume that the width (ww) is smaller than the length (ll).

Step-by-Step Solution:

  1. Set up the equations:

    For a rectangle:

    • The perimeter (PP) formula is: P=2(l+w)P = 2(l + w) Given P=44P = 44: 2(l+w)=44l+w=222(l + w) = 44 \Rightarrow l + w = 22

    • The area (AA) formula is: A=lwA = l \cdot w Given A=105A = 105: lw=105l \cdot w = 105

  2. Express ll in terms of ww:

    From l+w=22l + w = 22: l=22wl = 22 - w

  3. Substitute ll in the area equation:

    Substitute l=22wl = 22 - w into lw=105l \cdot w = 105: (22w)w=105(22 - w) \cdot w = 105 Expanding this equation: 22ww2=10522w - w^2 = 105 Rearrange it to form a standard quadratic equation: w222w+105=0w^2 - 22w + 105 = 0

  4. Solve the quadratic equation:

    The equation w222w+105=0w^2 - 22w + 105 = 0 can be solved using the quadratic formula: w=22±2224110521w = \frac{22 \pm \sqrt{22^2 - 4 \cdot 1 \cdot 105}}{2 \cdot 1} Calculate the discriminant: 22241105=484420=6422^2 - 4 \cdot 1 \cdot 105 = 484 - 420 = 64 So: w=22±642w = \frac{22 \pm \sqrt{64}}{2} w=22±82w = \frac{22 \pm 8}{2} This gives two solutions for ww: w=22+82=15andw=2282=7w = \frac{22 + 8}{2} = 15 \quad \text{and} \quad w = \frac{22 - 8}{2} = 7

  5. Determine the dimensions:

    Since the width is smaller than the length:

    • Let w=7w = 7 and l=227=15l = 22 - 7 = 15.

Answer:

  • Width: 77 feet
  • Length: 1515 feet

Would you like further clarification on any steps, or have any additional questions?


Here are some related questions to expand your understanding:

  1. How would the solution change if the width were larger than the length?
  2. Can you solve this problem using a different approach, like completing the square?
  3. What would the dimensions be if the area was different but the perimeter remained the same?
  4. How does the quadratic formula work in finding the dimensions here?
  5. Can you find the dimensions using trial and error instead of algebra?

Tip: For geometry problems involving rectangles, remember that the perimeter and area equations are powerful tools to set up and solve for unknowns.

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Math Problem Analysis

Mathematical Concepts

Geometry
Algebra
Quadratic Equations

Formulas

Perimeter of rectangle: P = 2(l + w)
Area of rectangle: A = l * w
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grade 10