A) \[4\sqrt{53}\,cm\]
B) \[36\,cm\]
C) \[2\sqrt{53}\,cm\]
D) 52 cm
Correct Answer: A
Solution :
Let \[AC=x\,cm\]and \[BD=4\,cm\](Given) Given area of ABCD \[ABCD=28\,c{{m}^{2}}\] \[\Rightarrow \]\[\frac{1}{2}\times x\times 4=28\Rightarrow x=14\,cm\] Clearly, \[AO=\frac{14}{2}=7cm.\] By Pythagoras? theorem, \[A{{O}^{2}}+B{{O}^{2}}=A{{B}^{2}}\] or \[{{7}^{2}}+{{2}^{2}}=53\]or \[AB=\sqrt{53}\] \[\therefore \]Perimeter \[=4AB=4\sqrt{53}\]
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