Inradius of circle which is inscribed in a isoceles triangle one of whose angle is 2pi/3 is \( \sqrt{3} \), then area of triangle is

- \( 4 \sqrt{3} \)
- \( 12 - 7 \sqrt{3} \)
- \( 12 + 7 \sqrt{3} \)
- None of the above.

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