In $\triangle ABC$,with usual notations,if $a \cos B = b \cos A$ and $a \cos C \neq c \cos A$,then the area of $\triangle ABC$ is . . . . . . sq. units.

  • A
    $\frac{c}{2} \sqrt{4a^2 - b^2}$
  • B
    $\frac{c}{4} \sqrt{4a^2 - c^2}$
  • C
    $\frac{b}{2} \sqrt{4b^2 - c^2}$
  • D
    $\frac{b}{4} \sqrt{4b^2 - c^2}$

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