Correct Answer: $25\sqrt{3}$ $cm^2$

Solution :

In $\triangle$ BGC,
BG = CG, $\angle$BGC = $120^{\circ}$, BC = 10 cm
$\triangle$ BGC is an isosceles triangle.
⇒ $\angle$GBC = $\angle$GCB
Sum of angles of a triangle = $180^{\circ}$
$\angle$GBC + $\angle$GCB + $\angle$BGC = $180^{\circ}$
$2\angle$GBC = $180^{\circ}-120^{\circ}$ = $60^{\circ}$
$\angle$GBC = $\frac{60^{\circ}}{2}$ = $30^{\circ}$
GD $\perp$ BC
$\angle$BGD = $\angle$CGD = $60^{\circ}$
$\cos 30^{\circ} = \frac{BD}{BG}$
⇒ BG = $5 × \frac{2}{\sqrt3}$ = $\frac{10}{\sqrt3}$
⇒ GD = BG × $\sin 30^{\circ}$ = $\frac{10}{\sqrt3}$ × $\frac{1}{2}$ = $\frac{5}{\sqrt3}$
Since G is the centroid of $\triangle$ABC,
AG : GD = 2 :1
AD : GD = 3 : 1
AD = 3 ×$\frac{5}{\sqrt3}$ = $\frac{15}{\sqrt3}$
Since AD is the altitude of $\triangle$ABC,
Area of $\triangle$ABC = $\frac{1}{2} × BC × AD = \frac{1}{2}×10×\frac{15}{\sqrt3} = \frac{75}{\sqrt3} = 25\sqrt3 \;cm^2$
Hence, the correct answer is $25\sqrt3$ $cm^2$.