A) 4 cm
B) \[2+\sqrt{2}\,cm\]
C) \[4\sqrt{2}\,cm\]
D) \[2+2\sqrt{2}\,cm\]
Correct Answer: D
Solution :
Join OB, Then \[\angle OBA={{90}^{o}}\] \[A{{B}^{2}}+O{{B}^{2}}=A{{O}^{2}}.\] \[\therefore \]\[OB=AB=2\,cm\] Hence \[A{{O}^{2}}={{2}^{2}}+{{2}^{2}}\]or \[AO=2\sqrt{2}\,cm\] \[\therefore \]\[AC=AO+OC=\left( 2\sqrt{2}+2 \right)cm\]
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