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The image shows a geometric figure consisting of three squares arranged around a right triangle. The squares are labeled with their areas:

1. The largest square has an area of \( 1369 \, \text{cm}^2 \).
2. The smallest square has an area of \( 144 \, \text{cm}^2 \).
3. The shaded square (whose area we need to find) is adjacent to these two squares.

Since the squares are arranged around a right triangle, this setup follows the Pythagorean theorem. The area of the shaded square corresponds to the square on the remaining side of the right triangle.

### Solution:
Let's denote:
- The side length of the largest square as \( c \).
- The side length of the smallest square as \( a \).
- The side length of the shaded square as \( b \).

According to the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]

Given:
- \( c^2 = 1369 \, \text{cm}^2 \Rightarrow c = \sqrt{1369} = 37 \, \text{cm} \)
- \( a^2 = 144 \, \text{cm}^2 \Rightarrow a = \sqrt{144} = 12 \, \text{cm} \)

Now we can find \( b^2 \):
\[
b^2 = c^2 - a^2
\]
\[
b^2 = 1369 - 144 = 1225
\]

Therefore:
\[
b = \sqrt{1225} = 35 \, \text{cm}
\]

The area of the shaded square is \( b^2 \), which is:
\[
1225 \, \text{cm}^2
\]

### Answer:
The area of the shaded square is \( \mathbf{1225 \, cm^2} \).

Would you like more details or have any questions?

Here are some related questions for further exploration:
1. How does the Pythagorean theorem apply to other shapes beyond squares?
2. What are some real-world applications of the Pythagorean theorem?
3. How can you use the side lengths to find the perimeter of the triangle?
4. Can the Pythagorean theorem be applied to non-right triangles?
5. How do you calculate the area of a square given its diagonal length?

**Tip:** Remember that the Pythagorean theorem only applies to right-angled triangles, so the relationship \( a^2 + b^2 = c^2 \) only holds in these cases.

Learn how to calculate the area of the shaded square using the Pythagorean theorem. This problem demonstrates how to work with square areas and apply the formula a^2 + b^2 = c^2. Suitable for students in Grades 8-10.

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