If in a triangle ABC, right angled at B, s-a=3, s-c=2, then the values of a and c are respectively

Options

A. 2,3
B. 3,4
C. 4,3
D. 6,8

Answer

To solve the problem, we need to find the values of sides \( a \) and \( c \) in triangle \( ABC \), which is right-angled at \( B \). We are given the conditions \( s - a = 3 \) and \( s - c = 2 \), where \( s \) is the semi-perimeter of the triangle.

### Step-by-step Solution:

1. **Express \( s \) in terms of \( a \) and \( c \)**:
From the given conditions, we can express \( s \) as:
\[ s = a + 3 \quad \text{(1)} \]
\[ s = c + 2 \quad \text{(2)} \]

2. **Set the two expressions for \( s \) equal**:
Since both expressions equal \( s \), we can set them equal to each other:
\[ a + 3 = c + 2 \]

3. **Rearrange to find a relationship between \( a \) and \( c \)**:
Rearranging the equation gives us:
\[ a - c = -1 \quad \Rightarrow \quad a = c - 1 \quad \text{(3)} \]

4. **Use the semi-perimeter formula**:
The semi-perimeter \( s \) is also given by:
\[ s = \frac{a + b + c}{2} \]
Substituting \( s \) from equation (1):
\[ a + 3 = \frac{a + b + c}{2} \]
Multiplying through by 2:
\[ 2a + 6 = a + b + c \]
Rearranging gives:
\[ b = a + 6 - c \quad \text{(4)} \]

5. **Use the relationship from step 3 in equation (4)**:
Substitute \( a = c - 1 \) into equation (4):
\[ b = (c - 1) + 6 - c \]
Simplifying gives:
\[ b = 5 \]

6. **Use the Pythagorean theorem**:
Since triangle \( ABC \) is right-angled at \( B \), we apply the Pythagorean theorem:
\[ a^2 + c^2 = b^2 \]
Substituting \( b = 5 \) gives:
\[ a^2 + c^2 = 25 \quad \text{(5)} \]

7. **Substitute \( a \) from equation (3) into equation (5)**:
Substitute \( a = c - 1 \):
\[ (c - 1)^2 + c^2 = 25 \]
Expanding this:
\[ c^2 - 2c + 1 + c^2 = 25 \]
Combining like terms:
\[ 2c^2 - 2c + 1 = 25 \]
Rearranging gives:
\[ 2c^2 - 2c - 24 = 0 \]
Dividing through by 2:
\[ c^2 - c - 12 = 0 \]

8. **Factor the quadratic equation**:
Factoring gives:
\[ (c - 4)(c + 3) = 0 \]
Thus, \( c = 4 \) or \( c = -3 \) (we discard \( c = -3 \) since lengths cannot be negative).

9. **Find \( a \)**:
Using equation (3) \( a = c - 1 \):
\[ a = 4 - 1 = 3 \]

### Final Values:
The values of \( a \) and \( c \) are:
\[ a = 3, \quad c = 4 \]