Question : If $(a+b+c)=16$ and $\left(a^2+b^2+c^2\right)=90$, find the value of $(a b+b c+c a)$.
Option 1: 84
Option 2: 83
Option 3: 82
Option 4: 81
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Answer (1)
7th Jan, 2024
Correct Answer: 83
Solution :
We have, $(a+b+c) = 16$ and $a^2 + b^2 + c^2 = 90$
We know that $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$
So, $16^2 = 90 + 2(ab+bc+ca)$
⇒ $ab+bc+ca = \frac{16^2 - 90}{2}=83$
Hence, the correct answer is 83.
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