To solve this problem, we need to find the value of $x$ such that the areas of the two rectangles are equal.

Step-by-step Solution:

1. Identify the dimensions and areas of each rectangle:

   • First Rectangle:

     • Width: $x + 9$ cm
     • Height: $4$ cm
     • Area: $(x + 9) \times 4$ cm²

   • Second Rectangle:

     • Width: $8$ cm
     • Height: $x + 1$ cm
     • Area: $8 \times (x + 1)$ cm²

2. Set the areas of the two rectangles equal to each other:

$(x + 9) \times 4 = 8 \times (x + 1)$

3. Expand both sides:

$4x + 36 = 8x + 8$

4. Rearrange the equation to isolate $x$:

$36 - 8 = 8x - 4x$

$28 = 4x$

5. Solve for $x$:

$x = \frac{28}{4} = 7$

Conclusion:

The value of $x$ is 7 cm.

Would you like further details or have any questions?

Related Questions:

1. What is the perimeter of each rectangle when $x = 7$ cm?
2. How would the areas change if the height of the first rectangle was 6 cm instead of 4 cm?
3. How would you solve this problem if the rectangles were triangles instead?
4. What if the rectangles have different lengths and widths, but you need them to have the same perimeter?
5. How can you use algebraic expressions to solve more complex geometric problems?

Tip:

When solving equations, always ensure all like terms are properly collected and simplified for clarity.