Question 1210610
To solve for the area of one of the seven identical rectangles, we need to determine the relationship between the rectangle's height ($h$) and its width ($w$), and how they relate to the side of the large square.

### 1. Define the Dimensions
Let each of the seven identical rectangles have:
* **Height** = $h$
* **Width** = $w$
* **Area** = $hw$

Since the large shape is a **square**, its total width must equal its total height. 

### 2. Analyze the Layout
In these types of geometric puzzles (commonly seen in Math Kangaroo or AMC competitions), the seven rectangles are typically arranged such that:
* A group of them are stacked vertically (sharing their widths).
* A group of them are placed horizontally (sharing their heights).

A common configuration for 7 identical rectangles forming a square is having **3 horizontal** rectangles stacked on top of **4 vertical** ones (or vice versa).
* **The width of the square** would be the sum of the widths of the vertical rectangles: $4w$.
* **The height of the square** would be the height of a vertical rectangle ($h$) plus the width of a horizontal rectangle ($w$): $h + w$.

However, there is a more standard layout for "7 identical rectangles in a square" where:
* Two rectangles are placed side-by-side ($2w$).
* Five rectangles are placed perpendicular to them ($5h$).

In the most frequent version of this problem:
1.  The total side of the square can be expressed as **$5h$** (if five rectangles are lined up by their heights).
2.  The same side can be expressed as **$2w$** (if two are lined up by their widths).
3.  Therefore, $2w = 5h$, which means $w = \frac{5}{2}h$.

### 3. Calculate the Area
Once we have the width in terms of $h$, we can find the area ($A$) of one rectangle:

$$A = \text{height} \times \text{width}$$
$$A = h \times \left(\frac{5}{2}h\right)$$
$$A = \frac{5}{2}h^2$$

### Summary of Result
Depending on the specific visual arrangement of your puzzle (which usually dictates whether $2w=5h$ or $3w=4h$):
* If the layout implies **$2$ widths = $5$ heights**: The area is **$2.5h^2$**.
* If the layout implies **$3$ widths = $4$ heights**: The area is **$\frac{4}{3}h^2$** (approx. $1.33h^2$).

Given the common "identical pieces" geometry standards, the most likely intended answer is **$\frac{5}{2}h^2$** or **$2.5h^2$**.