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

• A

  $\frac{\sqrt{297}}{4}$
• Correct answer
  B

  $\frac{\sqrt{231}}{4}$
• C

  $\frac{\sqrt{385}}{4}$
• D

  $\frac{\sqrt{185}}{4}$

Hence, the correct answer is option: 'B'.