A) 53
B) 52
C) 51
D) 54
Correct Answer: C
Solution :
\[a+b+c=16\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=154\] \[\therefore \,\,{{(a+b+c)}^{2}}={{(16)}^{2}}\] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)=256\] \[2(ab+bc+ca)=256-154\] \[ab+bc+ac=\underline{\mathbf{51}}\]You need to login to perform this action.
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