Quantitative Aptitude Ques 1278

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:

  • Given, $ (a+b+c)=2 $ and $ ab+bc+ca=1 $ Now, $ {{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}} $ $ =a^{2}+b^{2}+2ab+b^{2}+c^{2}+2bc+c^{2}+a^{2}+2ca $ $ =a^{2}+b^{2}+c^{2}+2ab+2bc+2ca+a^{2}+b^{2}+c^{2} $ $ ={{(a+b+c)}^{2}}a^{2}+b^{2}+c^{2} $ $ ={{(a+b+c)}^{2}}+{{(a+b+c)}^{2}}-2ab-2bc-2ca $ $ =2{{(a+b+c)}^{2}}-2(ab+bc+ca) $ $ =2\times {{(2)}^{2}}-2\times (1)=8-2=6 $
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