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The problem asks to find the value of \( x \) in the right-angled triangle \( \triangle ABC \), where:

- \( AC = 3 \) (the base),
- \( BC = 1 \) (the height),
- \( AB = x \) (the hypotenuse).

We can apply the **Pythagorean theorem** for a right triangle, which is:

\[
AB^2 = AC^2 + BC^2
\]

Substituting the known values:

\[
x^2 = 3^2 + 1^2
\]
\[
x^2 = 9 + 1 = 10
\]
\[
x = \sqrt{10}
\]

Thus, \( x = \sqrt{10} \approx 3.16 \).

Would you like more details on this, or have any further questions?

Here are 5 questions that build on this concept:
1. How would the Pythagorean theorem change if this were not a right triangle?
2. What are the trigonometric functions of angle \( A \) in this triangle?
3. Can you calculate the area of triangle \( ABC \)?
4. How would the value of \( x \) change if \( AC \) and \( BC \) were doubled?
5. What would the value of \( x \) be if the triangle was isosceles with the same base and height?

**Tip:** Always check if a triangle is right-angled before applying the Pythagorean theorem!

Find the value of x in the right-angled triangle ABC where AC = 3, BC = 1, and AB = x (the hypotenuse).

This problem involves finding the hypotenuse x of a right-angled triangle where AC = 3 and BC = 1. Using the Pythagorean theorem, we calculate that x = √10 ≈ 3.16. Ideal for students in Grades 7-9, this example reinforces key concepts in geometry and the application of the Pythagorean theorem.

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