In `DeltaABC, angleB=90^(@)`, find the sides of the triangle, if :
(i) `AB=(x-3)cm, BC=(x+4)cm and AC=(x+6)cm`
(ii) `AB=xcm, BC=(4x+4)cm and AC=(4x+5)cm`

Answer

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

### Part (i): Given AB = (x - 3) cm, BC = (x + 4) cm, AC = (x + 6) cm

1. **Set up the equation using the Pythagorean theorem**:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the values:
\[ (x - 3)^2 + (x + 4)^2 = (x + 6)^2 \]

2. **Expand each term**:
\[ (x - 3)^2 = x^2 - 6x + 9 \]
\[ (x + 4)^2 = x^2 + 8x + 16 \]
\[ (x + 6)^2 = x^2 + 12x + 36 \]
Now substitute these expansions into the equation:
\[ (x^2 - 6x + 9) + (x^2 + 8x + 16) = (x^2 + 12x + 36) \]

3. **Combine like terms**:
\[ 2x^2 + 2x + 25 = x^2 + 12x + 36 \]

4. **Rearrange the equation**:
\[ 2x^2 + 2x + 25 - x^2 - 12x - 36 = 0 \]
Simplifying gives:
\[ x^2 - 10x - 11 = 0 \]

5. **Factor the quadratic equation**:
\[ (x - 11)(x + 1) = 0 \]
This gives us two solutions:
\[ x = 11 \quad \text{or} \quad x = -1 \]
Since lengths cannot be negative, we take \( x = 11 \).

6. **Calculate the lengths of the sides**:
\[ AB = x - 3 = 11 - 3 = 8 \text{ cm} \]
\[ BC = x + 4 = 11 + 4 = 15 \text{ cm} \]
\[ AC = x + 6 = 11 + 6 = 17 \text{ cm} \]

### Final Answer for Part (i):
- \( AB = 8 \text{ cm} \)
- \( BC = 15 \text{ cm} \)
- \( AC = 17 \text{ cm} \)

---

### Part (ii): Given AB = x cm, BC = (4x + 4) cm, AC = (4x + 5) cm

1. **Set up the equation using the Pythagorean theorem**:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the values:
\[ x^2 + (4x + 4)^2 = (4x + 5)^2 \]

2. **Expand each term**:
\[ (4x + 4)^2 = 16x^2 + 32x + 16 \]
\[ (4x + 5)^2 = 16x^2 + 40x + 25 \]
Now substitute these expansions into the equation:
\[ x^2 + (16x^2 + 32x + 16) = (16x^2 + 40x + 25) \]

3. **Combine like terms**:
\[ 17x^2 + 32x + 16 = 16x^2 + 40x + 25 \]

4. **Rearrange the equation**:
\[ 17x^2 + 32x + 16 - 16x^2 - 40x - 25 = 0 \]
Simplifying gives:
\[ x^2 - 8x - 9 = 0 \]

5. **Factor the quadratic equation**:
\[ (x - 9)(x + 1) = 0 \]
This gives us two solutions:
\[ x = 9 \quad \text{or} \quad x = -1 \]
Since lengths cannot be negative, we take \( x = 9 \).

6. **Calculate the lengths of the sides**:
\[ AB = x = 9 \text{ cm} \]
\[ BC = 4x + 4 = 4(9) + 4 = 36 + 4 = 40 \text{ cm} \]
\[ AC = 4x + 5 = 4(9) + 5 = 36 + 5 = 41 \text{ cm} \]

### Final Answer for Part (ii):
- \( AB = 9 \text{ cm} \)
- \( BC = 40 \text{ cm} \)
- \( AC = 41 \text{ cm} \)

---