A) 60
B) 50
C) 75
D) 80
Correct Answer: B
Solution :
We have \[{{(a+b+c)}^{2}}=({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+2\,\,(ab+bc+ca)\] \[\Rightarrow \] \[{{(13)}^{2}}=69+2\,\,(ab+bc+ca)\] \[\Rightarrow \] \[2\,\,(ab+bc+ca)=(169-69)=100\] \[\Rightarrow \] \[(ab+bc+ca)=50\]You need to login to perform this action.
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