Let's solve the problem.

The problem states that the sum of the areas of the three rectangles is $189 \, \text{cm}^2$, and we need to find the value of $x$.

Step 1: Determine the area of each rectangle.

1. First rectangle:

   • Dimensions: $x$ by $x + 2$
   • Area: $\text{Area}_1 = x \times (x + 2) = x^2 + 2x$

2. Second rectangle:

   • Dimensions: $x$ by $x + 1$
   • Area: $\text{Area}_2 = x \times (x + 1) = x^2 + x$

3. Third rectangle:

   • Dimensions: $x$ by $x$
   • Area: $\text{Area}_3 = x \times x = x^2$

Step 2: Set up the equation for the total area.

The total area is the sum of the areas of the three rectangles: $\text{Total Area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3$ Substitute the areas: $x^2 + 2x + x^2 + x + x^2 = 189$ Combine like terms: $3x^2 + 3x = 189$

Step 3: Simplify and solve the quadratic equation.

Divide the entire equation by 3 to simplify: $x^2 + x - 63 = 0$ This is a standard quadratic equation. We can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 1$, and $c = -63$.

Step 4: Calculate the discriminant and the roots.

$\text{Discriminant} = b^2 - 4ac = 1^2 - 4(1)(-63) = 1 + 252 = 253$

So, $x = \frac{-1 \pm \sqrt{253}}{2}$

Since $\sqrt{253}$ is approximately $15.9$, we get: $x = \frac{-1 + 15.9}{2} \quad \text{or} \quad x = \frac{-1 - 15.9}{2}$ $x = \frac{14.9}{2} = 7.45 \quad \text{or} \quad x = \frac{-16.9}{2} \approx -8.45$

Since $x$ represents a length, it must be positive, so $x = 7.45$.

Final Answer:

The value of $x$ is approximately $\boxed{7.45 \, \text{cm}}$.

Would you like more details on any step or have any questions? Here are a few questions related to the problem:

1. How do you derive the quadratic formula?
2. What is the significance of the discriminant in solving quadratic equations?
3. How do you simplify and combine like terms in polynomial expressions?
4. How can you check if the calculated value of $x$ is correct?
5. What is the geometric significance of the problem setup?

Tip: When solving quadratic equations, always check both roots to ensure the correct context, such as positive lengths for geometric problems.