If \(\frac{(a+ b + c)}{2}\) = 16 and 2ab + 2bc + 2ca = 120, then what is the value of $4a^2 +4b^2 + 4c^2$?

3616

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