Let’s analyze the problem. Here's what we gather from the image:

1. Rectangle dimensions:

   • Length: $2x - 7$ cm
   • Width: $x - 6$ cm
   • Area of the rectangle: $(2x - 7)(x - 6)$.

2. Square dimensions:

   • Side: $x - 5$ cm
   • Area of the square: $(x - 5)^2$.

3. Given: The area of the rectangle is 53 cm² more than the area of the square.

Step-by-Step Solution:

Step 1: Write the equation for the given condition.

$\text{Area of Rectangle} = \text{Area of Square} + 53$ Substitute the formulas for the areas: $(2x - 7)(x - 6) = (x - 5)^2 + 53$

Step 2: Expand both sides.

Expand $(2x - 7)(x - 6)$: $(2x - 7)(x - 6) = 2x^2 - 12x - 7x + 42 = 2x^2 - 19x + 42$

Expand $(x - 5)^2$: $(x - 5)^2 = x^2 - 10x + 25$

Substitute these into the equation: $2x^2 - 19x + 42 = x^2 - 10x + 25 + 53$

Step 3: Simplify the equation.

Combine like terms: $2x^2 - 19x + 42 = x^2 - 10x + 78$

Move all terms to one side: $2x^2 - 19x + 42 - x^2 + 10x - 78 = 0$

Simplify further: $x^2 - 9x - 36 = 0$

Step 4: Solve the quadratic equation.

Factorize $x^2 - 9x - 36 = 0$: $(x - 12)(x + 3) = 0$

Thus: $x = 12 \quad \text{or} \quad x = -3$

Step 5: Verify the solution.

Since dimensions cannot be negative, $x = -3$ is not valid. So: $x = 12$

Final Answer:

$\boxed{x = 12}$

Would you like a breakdown of the verification process?

Follow-Up Questions:

1. How can we verify the solution for $x = 12$ with the given dimensions?
2. What would happen if the problem allowed for negative dimensions?
3. How do we simplify quadratic equations when factoring is not possible?
4. Why must the area difference between the shapes equal 53?
5. Can this problem be solved graphically?

Tip:

When solving quadratic equations, always check the feasibility of your solutions within the given context (e.g., negative dimensions are usually invalid in geometry problems).