We are given the problem where the total shaded area is $75 \, \text{m}^2$, and we need to find the area of the smaller rectangle. Here is the approach:

Step-by-step solution:

1. Outer rectangle dimensions:
   The dimensions of the outer green rectangle are:

   • Length = $x + 4 \, \text{m}$
   • Width = $2x + 3 \, \text{m}$

   So, the area of the outer rectangle is: $A_{\text{outer}} = (x + 4)(2x + 3)$

2. Inner (smaller) rectangle dimensions:
   The dimensions of the smaller rectangle (which is white in the image) are:

   • Length = $x + 2 \, \text{m}$
   • Width = $x + 1 \, \text{m}$

   So, the area of the inner rectangle is: $A_{\text{inner}} = (x + 2)(x + 1)$

3. Shaded area:
   The shaded area is the difference between the outer rectangle and the inner rectangle, and we know it equals $75 \, \text{m}^2$: $A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} = 75$

4. Set up the equation:
   Expanding both expressions:

   • $A_{\text{outer}} = (x + 4)(2x + 3) = 2x^2 + 3x + 8x + 12 = 2x^2 + 11x + 12$
   • $A_{\text{inner}} = (x + 2)(x + 1) = x^2 + x + 2x + 2 = x^2 + 3x + 2$

   Subtracting these: $A_{\text{shaded}} = (2x^2 + 11x + 12) - (x^2 + 3x + 2) = x^2 + 8x + 10$

   Setting it equal to 75: $x^2 + 8x + 10 = 75$

5. Solve the quadratic equation:
   Rearrange the equation: $x^2 + 8x - 65 = 0$

   Solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 8$, and $c = -65$.

   Substituting the values: $x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-65)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 260}}{2} = \frac{-8 \pm \sqrt{324}}{2} = \frac{-8 \pm 18}{2}$

   So, $x = \frac{-8 + 18}{2} = \frac{10}{2} = 5$.

6. Area of the smaller rectangle:
   Now that $x = 5$, we substitute this into the expression for the area of the smaller rectangle: $A_{\text{inner}} = (x + 2)(x + 1) = (5 + 2)(5 + 1) = 7 \times 6 = 42 \, \text{m}^2$

Final Answer:

The area of the smaller rectangle is $42 \, \text{m}^2$.

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

Here are some related questions you might consider:

1. How would the problem change if the shaded area was different?
2. What happens if we change the dimensions of the inner rectangle?
3. Can we solve for $x$ if the outer rectangle has different dimensions?
4. How does the quadratic formula handle cases where there are no real roots?
5. Can the shaded area ever be negative in a similar setup?

Tip: Always check that your calculated areas make sense in the context of the problem to avoid logical errors!