Quantitative Aptitude Ques 1383
Question: If $ x+y+z=12, $ $ x^{2}+y^{2}+z^{2}=44, $ then find $ xy+yz+zx. $
Options:
A) 26
B) 60
C) 36
D) 40
Show Answer
Answer:
Correct Answer: B
Solution:
- Given expression in the form of $ {{(a+b+c)}^{2}}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) $ $ 2(ab+bc+ca)={{(a+b+c)}^{2}}-(a^{2}+b^{2}+c^{2}) $ $ ab+bc+ca=\frac{{{(a+b+c)}^{2}}-(a^{2}+b^{2}+c^{2})}{2} $
$ \therefore $ $ xy+yz+zx=\frac{{{(x+y+z)}^{2}}-(x^{2}+y^{2}+z^{2})}{2} $ $ =\frac{{{(12)}^{2}}-44}{2}=\frac{144-44}{2}=50 $
BETA
