Question : If $a^2+b^2+c^2=14$ and $a+b+c=6$, then the value of $(ab+bc+ca)$ is:
Option 1: 11
Option 2: 12
Option 3: 13
Option 4: 14
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Correct Answer: 11
Solution :
Given: $a^2+b^2+c^2=14$ and $a+b+c=6$.
On squaring the given equation $a+b+c=6$, we get,
$(a+b+c)^2=6^2$
⇒ $a^2+b^2+c^2+2(ab+bc+ca)=36$
Substitute the value of $a^2+b^2+c^2=14$ in the above equation and we get,
⇒ $14+2(ab+bc+ca)=36$
⇒ $2(ab+bc+ca)=22$
⇒ $(ab+bc+ca)=11$
Hence, the correct answer is 11.
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