Question : If $a^2+b^2+c^2=62$ and $a + b + c = 12$, then what is the value of $4ab + 4bc + 4ca$?
Option 1: 164
Option 2: 148
Option 3: 152
Option 4: 160
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Correct Answer: 164
Solution : We can solve this problem using the identity, $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ Given that $a^2 + b^2 + c^2 = 62$ and $a + b + c = 12$, $⇒12^2 = 62 + 2(ab + bc + ca)$ $⇒144 = 62 + 2(ab + bc + ca)$ $⇒82 = 2(ab + bc + ca)$ $⇒41 = ab + bc + ca$ $⇒4(ab + bc + ca) = 4 \times 41$ $\therefore 4ab + 4bc + 4ca=164$ Hence, the correct answer is 164.
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