Question : If $\frac{(a+b+c)}{2}=16$ and $2ab+2 bc+2ca=120$ ,then what is the value of $4a^2+4b^2+4c^2$?
Option 1: 2828
Option 2: 3368
Option 3: 3616
Option 4: 3056
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Correct Answer: 3616
Solution :
Given: $\frac{(a+b+c)}{2}=16$
$⇒ a+b+c=32$
Also given $2ab+2 bc+2ca=120$
We know that, $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Putting the values we get:
⇒ $(32)^2=a^2+b^2+c^2+120$
⇒ $1024-120=a^2+b^2+c^2$
⇒ $a^2+b^2+c^2=904$
$\therefore 4(a^2+b^2+c^2)=4\times904=3616$
Hence, the correct answer is 3616.
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