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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 (\( w \)) is smaller than the length (\( l \)).

### Step-by-Step Solution:

1. **Set up the equations**:

   For a rectangle:
   - The perimeter (\( P \)) formula is:
     \[
     P = 2(l + w)
     \]
     Given \( P = 44 \):
     \[
     2(l + w) = 44 \Rightarrow l + w = 22
     \]

   - The area (\( A \)) formula is:
     \[
     A = l \cdot w
     \]
     Given \( A = 105 \):
     \[
     l \cdot w = 105
     \]

2. **Express \( l \) in terms of \( w \)**:

   From \( l + w = 22 \):
   \[
   l = 22 - w
   \]

3. **Substitute \( l \) in the area equation**:

   Substitute \( l = 22 - w \) into \( l \cdot w = 105 \):
   \[
   (22 - w) \cdot w = 105
   \]
   Expanding this equation:
   \[
   22w - w^2 = 105
   \]
   Rearrange it to form a standard quadratic equation:
   \[
   w^2 - 22w + 105 = 0
   \]

4. **Solve the quadratic equation**:

   The equation \( w^2 - 22w + 105 = 0 \) can be solved using the quadratic formula:
   \[
   w = \frac{22 \pm \sqrt{22^2 - 4 \cdot 1 \cdot 105}}{2 \cdot 1}
   \]
   Calculate the discriminant:
   \[
   22^2 - 4 \cdot 1 \cdot 105 = 484 - 420 = 64
   \]
   So:
   \[
   w = \frac{22 \pm \sqrt{64}}{2}
   \]
   \[
   w = \frac{22 \pm 8}{2}
   \]
   This gives two solutions for \( w \):
   \[
   w = \frac{22 + 8}{2} = 15 \quad \text{and} \quad w = \frac{22 - 8}{2} = 7
   \]

5. **Determine the dimensions**:

   Since the width is smaller than the length:
   - Let \( w = 7 \) and \( l = 22 - 7 = 15 \).

### Answer:
- **Width**: \( 7 \) feet
- **Length**: \( 15 \) feet

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

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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.

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.

Solve for the dimensions of a rectangle with a perimeter of 44 feet and an area of 105 square feet. This geometry problem explores algebra and quadratic equations to determine the length and width, assuming the width is smaller than the length. Suitable for Grade 10 students.

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