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The image shows a square with side lengths of 12 meters, and an inner triangle where one of the sides (marked as "x") is unknown. There is another right-angled triangle with the dimensions 7 meters and 5 meters.

Here's how to solve for \( x \):

1. **Right-Angle Triangle in the Square**: 
   The larger right-angled triangle is formed by the diagonal of the square (which acts as the hypotenuse) and the two sides of the square, both of which are 12 meters.

   We use the Pythagorean theorem to find the diagonal \( d \) of the square:
   \[
   d = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} \approx 16.97 \, \text{meters}
   \]
   
2. **Second Triangle Inside the Square**:
   The second triangle has sides of 7 meters and 5 meters, and \( x \) is the third side. Again, we can use the Pythagorean theorem here:
   \[
   x = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.6 \, \text{meters}
   \]

Thus, the value of \( x \) is approximately **8.6 meters**.

Would you like further clarification or more details?

Here are some related questions to consider:
1. What is the formula for the Pythagorean theorem, and when is it applied?
2. How can we verify that a triangle is a right triangle?
3. How do diagonals behave in other types of polygons, like rectangles or parallelograms?
4. How does the Pythagorean theorem extend to three-dimensional shapes?
5. What are some applications of right-angled triangles in real-world scenarios?

**Tip:** Always double-check if you are dealing with a right triangle before applying the Pythagorean theorem!

Learn how to calculate the unknown dimension x by applying the Pythagorean theorem. This geometry problem involves a square with side lengths of 12m and an internal triangle with dimensions 7m and 5m. Suitable for students in Grades 9-10.

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