A) \[\pi \]
B) \[2\,\pi \]
C) \[4\,\pi \]
D) \[6\,\pi \]
Correct Answer: C
Solution :
Here, \[b=2,B={{30}^{o}}\] \[\therefore \] \[R=\frac{b}{2\sin B}=\frac{2}{2\sin {{30}^{o}}}=\frac{2}{1}\] \[R=2\] Area of circumcircle \[=\pi \,{{R}^{2}}\] \[=\pi \times {{(2)}^{2}}\] \[=1=4\pi \] isq. UnitsYou need to login to perform this action.
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