in the given triangle,
AB = BC = 6
As, the two sides are equal so, the given triangle is isosceles.
also, the angle < ABC = 120 degrees,
the area of the isosceles triangle is,
[tex]A=\frac{1}{2}\times a^2\times\sin \theta^{}[/tex]
here a = length of two congruent sides,
theta = the angle between congruent sides.
put the values,
A = 1/2 x 6^2 x sin 120
[tex]A=\frac{1}{2}\times6^2\times\sin 120[/tex][tex]\begin{gathered} A=\frac{36}{2}\times\frac{\sqrt[]{3}}{2} \\ A=9\sqrt[]{3} \end{gathered}[/tex]
thus, the correct answer is option D
