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 $
BETA
