A) \[1\]
B) \[4\]
C) \[2\]
D) \[2\sqrt{2}\]
Correct Answer: B
Solution :
\[1.02\times 1.02+1.96\times 1.02+0.98\times 0.98\] Let \[a=1.02\] and \[b=0.98\] \[\because \,\,\,\,\,\,\,{{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}\] \[\therefore \,\,\,\,\,\,{{(1.02+0.98)}^{2}}=\underline{\mathbf{4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
