A) $ \frac{25}{4}cm $
B) $ \frac{25}{\sqrt{3}}cm $
C) $ \frac{25\sqrt{3}}{4}cm $
D) $ 25\sqrt{3},cm $
Correct Answer: C
$ \therefore $ $ \angle APQ=\angle AQP=60{}^\circ $
$ \therefore $ $ \Delta APQ $ is an equilateral triangle.
So, $ \Delta APQ,\sim ,\Delta ABC $
$ \therefore ,Area,of,(\Delta APQ) $ $ =\frac{\sqrt{3}}{4}{{(PQ)}^{2}}=\frac{\sqrt{3}}{4}\times 25=\frac{25\sqrt{3}}{4}cm^{2} $