A) \[6\sqrt{3}\]
B) \[12\sqrt{3}\]
C) \[4\sqrt{3}\]
D) \[8\sqrt{3}\]
Correct Answer: A
Solution :
In \[\Delta AOB,\] \[\tan 30{}^\circ =\frac{OB}{OA}\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}=\frac{OB}{6}\] \[\Rightarrow \] \[OB=\frac{6\sqrt{3}}{3}\] \[=2\sqrt{3}\] \[\therefore \]Area of \[\Delta AOB=\frac{1}{2}\times OA\times OB\] \[=\frac{1}{2}\times 6\times 2\sqrt{3}\] \[=6\sqrt{3}\]sq unitsYou need to login to perform this action.
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