In a `DeltaABC`, right-angled at B, if `AB = x/2, BC = x + 2 and AC = x + 3`, then the quadratic equation, formed in x, is  :

Options

A. `x^2 - 8x - 20 = 0`
B. `x^2 - 2x + 5 = 0 `
C. `x^2 + 8x + 20 = 0`
D. `x^2 + 2x + 5 = 0`

Answer

To solve the problem, we will use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AB and BC).

### Step-by-Step Solution:

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

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

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

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

5. **Eliminate the fraction**:
- Multiply the entire equation 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:
\[ 4x^2 + 24x + 36 - 5x^2 - 16x - 16 = 0 \]
- Combine like terms:
\[ -x^2 + 8x + 20 = 0 \]
- Multiply through by -1 to make the leading coefficient positive:
\[ x^2 - 8x - 20 = 0 \]

7. **Final quadratic equation**:
- The quadratic equation formed in \( x \) is:
\[ x^2 - 8x - 20 = 0 \]