| If \[a+b+c=13,\]\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=69,\]then find b \[ab+bc+ca.\] |
A) \[-\,\,50\]
B) \[50\]
C) \[69\]
D) \[75\]
Correct Answer: B
Solution :
| We know that, |
| \[{{(a+b+c)}^{2}}=({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+2\,\,(ab+bc+ca)\] |
| Now substituting the given values |
| \[{{(13)}^{2}}=69+2\,\,(ab+bc+ca)\] |
| \[\Rightarrow \] \[169=69+2\,\,(ab+bc+ca)\] |
| \[\Rightarrow \]\[\frac{169-69}{2}=ab+bc+ca\] |
| \[\Rightarrow \]\[ab+bc+ca=50\] |
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