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 will use the properties of triangles, particularly focusing on the relationships between the sides and the semi-perimeter.

### Given:
- Triangle ABC is right-angled at B.
- \( s - a = 3 \)
- \( s - c = 2 \)

### Step 1: Define the semi-perimeter
The semi-perimeter \( s \) of triangle ABC is given by:
\[ s = \frac{a + b + c}{2} \]

### Step 2: Express \( s \) in terms of \( a \) and \( c \)
From the equations given:
1. \( s - a = 3 \) implies \( s = a + 3 \)
2. \( s - c = 2 \) implies \( s = c + 2 \)

### Step 3: Set the two expressions for \( s \) equal to each other
Since both expressions represent \( s \):
\[ a + 3 = c + 2 \]

### Step 4: Rearrange to find a relationship between \( a \) and \( c \)
Rearranging gives:
\[ a - c = -1 \quad \text{or} \quad c - a = 1 \]
This means:
\[ c = a + 1 \]

### Step 5: Use the Pythagorean theorem
Since triangle ABC is right-angled at B, we can apply the Pythagorean theorem:
\[ a^2 + b^2 = c^2 \]

### Step 6: Substitute \( c \) in the Pythagorean theorem
Substituting \( c = a + 1 \) into the equation gives:
\[ a^2 + b^2 = (a + 1)^2 \]
Expanding the right side:
\[ a^2 + b^2 = a^2 + 2a + 1 \]

### Step 7: Simplify the equation
Subtract \( a^2 \) from both sides:
\[ b^2 = 2a + 1 \]

### Step 8: Find the semi-perimeter \( s \)
Now, we can express \( s \) using \( a \) and \( c \):
\[ s = a + 3 = (a + 1) + 2 = c + 2 \]

### Step 9: Substitute \( b^2 \) into the semi-perimeter equation
Using the value of \( b^2 \):
\[ s = \frac{a + b + c}{2} = \frac{a + \sqrt{2a + 1} + (a + 1)}{2} \]

### Step 10: Solve for \( a \) and \( c \)
We already have \( c = a + 1 \). Now we need to find \( a \):
Using the equation \( s - a = 3 \):
\[ \frac{(a + \sqrt{2a + 1} + (a + 1))}{2} - a = 3 \]
This simplifies to:
\[ \frac{2a + \sqrt{2a + 1} + 1 - 2a}{2} = 3 \]
\[ \frac{\sqrt{2a + 1} + 1}{2} = 3 \]
Multiplying through by 2:
\[ \sqrt{2a + 1} + 1 = 6 \]
\[ \sqrt{2a + 1} = 5 \]
Squaring both sides:
\[ 2a + 1 = 25 \]
\[ 2a = 24 \quad \Rightarrow \quad a = 12 \]

### Step 11: Find \( c \)
Now substituting \( a \) back to find \( c \):
\[ c = a + 1 = 12 + 1 = 13 \]

### Final Answer:
The values of \( a \) and \( c \) are:
\[ a = 12, \quad c = 13 \]