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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