Question : If $a^2+b^2+c^2=160$ and $a+b+c=16$, find $a b+b c+c a$.
Option 1: 96
Option 2: 84
Option 3: 48
Option 4: 42
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Correct Answer: 48
Solution :
Given, $a^2+b^2+c^2=160$ and $a+b+c=16$
We know, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
⇒ $16^2 = 160 + 2(a b+b c+c a)$
⇒ $2(a b+b c+c a)=256-160$
⇒ $ab+bc+ca=48$
Hence, the correct answer is 48.
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