If `(a +b) : (b +c) : (c +a) ::6:7:8 & a+b+c= 14` then find c=?

Options

A. 4
B. 6
C. 8
D. 7

Answer

To solve the problem step by step, we will use the given ratios and the total sum of \(a\), \(b\), and \(c\).

### Step 1: Set up the equations based on the ratios
Given the ratios:
\[ (a + b) : (b + c) : (c + a) = 6 : 7 : 8 \]
We can express \(a + b\), \(b + c\), and \(c + a\) in terms of a common variable \(k\):
\[ a + b = 6k \]
\[ b + c = 7k \]
\[ c + a = 8k \]

### Step 2: Add the equations
Now, we will add all three equations:
\[ (a + b) + (b + c) + (c + a) = 6k + 7k + 8k \]
This simplifies to:
\[ 2a + 2b + 2c = 21k \]
Dividing the entire equation by 2 gives:
\[ a + b + c = \frac{21k}{2} \]

### Step 3: Use the given total of \(a + b + c\)
We know from the problem statement that:
\[ a + b + c = 14 \]
Setting the two expressions for \(a + b + c\) equal to each other:
\[ \frac{21k}{2} = 14 \]

### Step 4: Solve for \(k\)
To find \(k\), multiply both sides by 2:
\[ 21k = 28 \]
Now, divide by 21:
\[ k = \frac{28}{21} = \frac{4}{3} \]

### Step 5: Substitute \(k\) back to find \(a + b\), \(b + c\), and \(c + a\)
Now we can find the values of \(a + b\), \(b + c\), and \(c + a\):
\[ a + b = 6k = 6 \times \frac{4}{3} = 8 \]
\[ b + c = 7k = 7 \times \frac{4}{3} = \frac{28}{3} \]
\[ c + a = 8k = 8 \times \frac{4}{3} = \frac{32}{3} \]

### Step 6: Express \(c\) in terms of \(a + b\) and \(b + c\)
We know:
\[ a + b + c = 14 \]
From \(a + b = 8\), we can substitute:
\[ 8 + c = 14 \]
This gives:
\[ c = 14 - 8 = 6 \]

### Final Answer
Thus, the value of \(c\) is:
\[ \boxed{6} \]