If `a^(2)+b^(2)+c^(2)=20` and `a+b+c=9` , find ab+bc+ca .

Answer

To solve the problem, we need to find the value of \( ab + bc + ca \) given the equations:

1. \( a^2 + b^2 + c^2 = 20 \)
2. \( a + b + c = 9 \)

We can use the identity that relates these quantities:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

### Step 1: Calculate \( (a + b + c)^2 \)

From the second equation, we have:

\[ a + b + c = 9 \]

Now, squaring both sides:

\[ (a + b + c)^2 = 9^2 = 81 \]

### Step 2: Substitute into the identity

Now we can substitute the values we have into the identity:

\[ 81 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

Substituting \( a^2 + b^2 + c^2 = 20 \):

\[ 81 = 20 + 2(ab + bc + ca) \]

### Step 3: Rearrange the equation

Now, we will isolate \( 2(ab + bc + ca) \):

\[ 81 - 20 = 2(ab + bc + ca) \]

\[ 61 = 2(ab + bc + ca) \]

### Step 4: Solve for \( ab + bc + ca \)

Now, divide both sides by 2:

\[ ab + bc + ca = \frac{61}{2} = 30.5 \]

### Final Answer

Thus, the value of \( ab + bc + ca \) is:

\[ \boxed{30.5} \]