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In a triangle $ABC$,if $\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}$,$a = 2$,$c = 5$ and $b$ is an integer,then the area (in sq. units) of triangle $ABC$ is

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$(D) \text{locus of point } A \text{ is a pair of straight lines}$

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