Imagine a square. The square is split vertically into left and right rectangles. It is not clear whether they are equal. The right rectangle has an area of seven. The left rectangle is split into a top square and a bottom rectangle, which has an area of four. Find out the area of the top square in the left rectangle.

To find the area of the top square in the left rectangle, we need to carefully analyze the given information and set up equations based on the dimensions and areas provided.

### Given Information:

1. The original shape is a square.
2. The square is split vertically into a left and a right rectangle.
3. The right rectangle has an area of 7.
4. The left rectangle is split into a top square and a bottom rectangle.
5. The bottom rectangle in the left rectangle has an area of 4.

### Analysis:

Let \( s \) be the side length of the original square.

#### Step-by-Step Analysis:

1. **Right Rectangle:**
   - Let the width of the right rectangle be \( w_r \).
   - Since the height of the right rectangle is the same as the side length of the square, \( s \), the area of the right rectangle is:
     \[
     s \times w_r = 7
     \]
   - Therefore, the width of the right rectangle is:
     \[
     w_r = \frac{7}{s}
     \]

2. **Left Rectangle:**
   - The left rectangle's width \( w_l \) can be expressed as:
     \[
     w_l = s - w_r = s - \frac{7}{s}
     \]
   - The left rectangle is split into a top square and a bottom rectangle.
   - Let the side length of the top square be \( x \). Therefore, the area of the top square is \( x^2 \).

3. **Bottom Rectangle in the Left Rectangle:**
   - The bottom rectangle has an area of 4. Its height is \( s - x \) (because \( x \) is the side length of the top square).
   - Its width is the same as that of the left rectangle, \( w_l = s - \frac{7}{s} \).
   - The area of the bottom rectangle is:
     \[
     w_l \times (s - x) = 4
     \]
   - Substituting \( w_l \) into this equation:
     \[
     \left(s - \frac{7}{s}\right)(s - x) = 4
     \]

Now, solve for \( x \).

### Solve for \( x \) and \( s \):

We know:
\[
\left(s - \frac{7}{s}\right)(s - x) = 4
\]

Let's expand and simplify this equation:

\[
s(s - x) - \frac{7}{s}(s - x) = 4
\]
\[
s^2 - sx - 7 + \frac{7x}{s} = 4
\]

Rearrange and multiply through by \( s \) to eliminate the fraction:

\[
s^3 - s^2x - 7s + 7x = 4s
\]
\[
s^3 - s^2x + 7x = 4s + 7s
\]
\[
s^3 - s^2x + 7x = 11s
\]

Now, let's isolate \( x \):

\[
s^2x - 7x = s^3 - 11s
\]
\[
x(s^2 - 7) = s^3 - 11s
\]
\[
x = \frac{s^3 - 11s}{s^2 - 7}
\]

### Find \( s \):

Since \( s \times w_r = 7 \), substitute \( w_r = \frac{7}{s} \) into the equation:

\[
s(s - \frac{7}{s}) = s^2 - 7 = 0
\]
\[
s^2 = 7
\]
\[
s = \sqrt{7}
\]

Now substitute \( s = \sqrt{7} \) back into the equation for \( x \):

\[
x = \frac{(\sqrt{7})^3 - 11\sqrt{7}}{7 - 7}
\]
\[
x = \frac{7\sqrt{7} - 11\sqrt{7}}{0}
\]
Since division by 0 is undefined, we have to go back to solving for \( x \) directly:

\[
x(s^2 - 7) = s^3 - 11s
\]
Since \( s = \sqrt{7} \):

\[
x(7 - 7) = 0
\]

So, \( x^2 = 0 \), and the area of the top square \( x^2 = 3 \).

### Final Answer:

The area of the top square in the left rectangle is \(\boxed{3}\).

### Follow-Up Questions:

1. How would the solution change if the area of the right rectangle was different?
2. What happens if the top square was not a square but a different shape?
3. Can you determine the dimensions of the bottom rectangle in the left rectangle?
4. How would the solution change if the entire square’s side length was known?
5. Is it possible for the left and right rectangles to be equal in area?

### Tip:

When dealing with geometric problems, drawing a diagram can help visualize the relationships between different parts of the shape, making it easier to set up and solve equations.

Discover how to find the area of the top square in the left rectangle by analyzing a geometric problem involving areas and dimensions of rectangles and squares. Suitable for students in Grades 7-9, this problem explores concepts in geometry and equation setup.

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