If \(\frac{(a+ b + c)}{2}\) = 16 and 2ab + 2bc + 2ca = 120, then what is the value of $4a^2 +4b^2 + 4c^2$?
Answer & explanation
Correct answer: option 3
Given, If \(\frac{(a+ b + c)}{2}\) = 16
then, (a + b + c) = 2 × 16 = 32
2ab + 2bc + 2ca = 120
Now, let's square the first equation:
(a+b+c)2 = (32)2
a2 + b2 + c2 + 2ab + 2bc + 2ac = 1024
Put the value of 2ab + 2bc + 2ca in the above equation,
a2 + b2 + c2 + 120 = 1024
a2 + b2 + c2 = 1024 – 120 = 904
4a2 + 4b2 + 4c2 = 4 × [ a2 + b2 + c2 ]
= 4 × 904 = 3616