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!