A) \[12\,\,c{{m}^{2}}\]
B) \[6\sqrt{2}\,c{{m}^{2}}\]
C) \[\frac{12}{\sqrt{3}}\,c{{m}^{2}}\]
D) \[6\,\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Area of the triangle \[=\frac{1}{2}\,\,ab\,\,\sin \,C\] \[=\frac{1}{2}\times 4\times 6\times \sin {{30}^{o}}=6\,c{{m}^{2}}\]
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