C. Solve the given problem using Polya’s problem-solving strategy.
A rectangular wall has a width that measures 2 meters more than its height. If the area of the wall is 15 square meters, what are its dimensions?

To solve the problem using **Polya's problem-solving strategy**, we'll follow these four steps: **Understand the problem, Devise a plan, Carry out the plan, and Look back**.

### 1. Understand the Problem:
- **Given:**
  - The wall is rectangular.
  - The width of the wall is 2 meters more than its height.
  - The area of the wall is 15 square meters.

- **Find:** The dimensions of the wall (height and width).

### 2. Devise a Plan:
Let \( h \) be the height of the wall in meters.

- **Width:** Since the width is 2 meters more than the height, we can express the width as \( w = h + 2 \).

- **Area of the rectangle:** The area of the wall is given by \( \text{Area} = \text{height} \times \text{width} \). Therefore,
  \[
  h \times (h + 2) = 15
  \]
  
This equation is a quadratic equation in terms of \( h \). We can solve this equation to find the height, and then determine the width.

### 3. Carry Out the Plan:
Let's solve the quadratic equation:

\[
h \times (h + 2) = 15
\]
\[
h^2 + 2h = 15
\]
\[
h^2 + 2h - 15 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 2 \), and \( c = -15 \).

Substituting these values:
\[
h = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-15)}}{2(1)}
\]
\[
h = \frac{-2 \pm \sqrt{4 + 60}}{2}
\]
\[
h = \frac{-2 \pm \sqrt{64}}{2}
\]
\[
h = \frac{-2 \pm 8}{2}
\]

This gives two possible solutions for \( h \):
\[
h = \frac{-2 + 8}{2} = \frac{6}{2} = 3
\]
\[
h = \frac{-2 - 8}{2} = \frac{-10}{2} = -5
\]

Since height cannot be negative, we discard \( h = -5 \). Thus, \( h = 3 \) meters.

- **Width:** \( w = h + 2 = 3 + 2 = 5 \) meters.

### 4. Look Back:
- **Check:** The dimensions of the wall are 3 meters by 5 meters. The area is:
  \[
  3 \times 5 = 15 \text{ square meters}
  \]
This confirms that our solution is correct.

### Final Answer:
The dimensions of the wall are **3 meters (height)** and **5 meters (width)**.

---

Do you have any questions or would you like further details?

### Related Questions:
1. How would the problem change if the width was 3 meters more than the height?
2. What if the area was given as 20 square meters instead of 15?
3. How would you solve the problem if the height was described as 3 meters less than the width?
4. What are the steps to solve a quadratic equation using factorization instead of the quadratic formula?
5. What if the width was described using a percentage more than the height instead of a fixed value?

### Tip:
When solving quadratic equations, always check for the feasibility of the solutions in the context of the problem, especially when dealing with physical dimensions.

Learn how to solve a problem involving a rectangular wall where the width is 2 meters more than its height and the area is 15 square meters. This problem covers geometric concepts and quadratic equations. Suitable for Grades 7-9.

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