Quantitative Aptitude Ques 580

Question: Given that $ a+b+c=2 $ and $ ab+bc+ca=1, $ then the value of $ {{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}} $ is

Options:

A) 10

B) 16

C) 6

D) 8

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ {{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}} $ $ =a^{2}+b^{2}+c^{2}+b^{2}+a^{2}+c^{2}+2,(ab+bc+ca) $ $ =2,(a^{2}+b^{2}+c^{2})+2,(ab+bc+ca) $ $ =2,(a^{2}+b^{2}+c^{2})+2\times 1 $ Now, $ (a+b+c)=2 $ On squaring both sides, we get $ {{(a+b+c)}^{2}}=a^{2}+b^{2}+c^{2}+2,(ab+bc+ca) $

$ \Rightarrow $ $ a^{2}+b^{2}+c^{2}={{(2)}^{2}}-2\times 1=4-2=2 $

$ \therefore $ $ {{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}} $ $ =2\times 2+2=4+2=6 $

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