A) \[2\sqrt{3}\]sq units
B) 4 sq units
C) 3 sq units
D) \[4\sqrt{3}\]sq units
Correct Answer: A
Solution :
(a): \[AB\parallel DC\] and \[AD\parallel BC\] In\[\Delta ABE\], \[\angle EAB=\angle ABE={{60}^{{}^\circ }}\] \[\Rightarrow \] \[\angle AE={{60}^{{}^\circ }}\] \[\Rightarrow \] \[\Delta ABE\] is an equilateral triangle. Now, Perimeter of \[\Delta ABE=6\] \[\Rightarrow \]\[AB+BE+EA=2\] And in \[\Delta ADE,\text{ }A{{E}^{2}}=A{{D}^{2}}+E{{D}^{2}}\] \[\Rightarrow \]\[4=A{{D}^{2}}+1\] (Since. E is mid-point of CD) \[\Rightarrow \]\[AD=\sqrt{3}\]units Hence, area of quadrilateral \[ABCD=AB\times AD\] \[=2\times \sqrt{3}=2\sqrt{3}\]sq units
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