In a right-angled triangle ABC, right angled at B, AB = `(x)/(2)`, BC = `x+2` and AC = `x+3`. Find value of x.

Answer

To solve the problem, we will use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

### Step-by-Step Solution:

1. **Identify the sides of the triangle**:
- Let \( AB = \frac{x}{2} \) (one leg)
- Let \( BC = x + 2 \) (the other leg)
- Let \( AC = x + 3 \) (the hypotenuse)

2. **Apply the Pythagorean theorem**:
According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values we have:
\[ (x + 3)^2 = \left(\frac{x}{2}\right)^2 + (x + 2)^2 \]

3. **Expand both sides**:
- Left side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
- Right side:
\[ \left(\frac{x}{2}\right)^2 = \frac{x^2}{4} \]
\[ (x + 2)^2 = x^2 + 4x + 4 \]
Combining these:
\[ \frac{x^2}{4} + x^2 + 4x + 4 = \frac{x^2}{4} + \frac{4x^2}{4} + \frac{16x}{4} + \frac{16}{4} = \frac{5x^2 + 16x + 16}{4} \]

4. **Set the equation**:
Now we have:
\[ x^2 + 6x + 9 = \frac{5x^2 + 16x + 16}{4} \]

5. **Eliminate the fraction**:
Multiply through by 4 to eliminate the fraction:
\[ 4(x^2 + 6x + 9) = 5x^2 + 16x + 16 \]
This simplifies to:
\[ 4x^2 + 24x + 36 = 5x^2 + 16x + 16 \]

6. **Rearrange the equation**:
Move all terms to one side:
\[ 4x^2 + 24x + 36 - 5x^2 - 16x - 16 = 0 \]
Simplifying gives:
\[ -x^2 + 8x + 20 = 0 \]
or
\[ x^2 - 8x - 20 = 0 \]

7. **Factor the quadratic equation**:
We can factor this equation:
\[ (x - 10)(x + 2) = 0 \]

8. **Solve for x**:
Setting each factor to zero gives:
\[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

9. **Determine the valid solution**:
Since \( x \) represents a length, it cannot be negative. Therefore, we discard \( x = -2 \).

Thus, the value of \( x \) is:
\[ \boxed{10} \]