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