dancing HEXAGON

First, join $\overline{FC}$ to form trapezoids $ABCF$ and $CDEF.$ The area of a trapezoid is $A = h\times w,$ where $h$ is the height and $w$ is the length of the median, that is $w = \frac{1}{2}(b_1 + b_2).$ Also, we aren't given the heights immediately, but we are given the angles in the trapezoids and slant lengths. So substitute $h = l\times\sin\theta,$ where $l$ is the slant length.

The area of $ABCF$ is $A_1 = 6\sin(60^\circ)\frac{3 + 9}{2} = 18\sqrt{3}$ The area of $CDEF$ is $A_2 = 3\sin(60^\circ)\frac{6 + 9}{2} = 45\frac{\sqrt{3}}{4}$

So the area of hexagon $ABCDEF$ is $A = A_1 + A_2 = 117\frac{\sqrt{3}}{4} \approx \boxed{50.66}$

angle [ABC]=[BCD]=120 degree and AB=CD so angle [BAD]=[CDA]=60 degree so BC||AD in same way FC||ED so quadrilateral ABCH is a Parrallelogram so CH=AB=CD=HD=3 CM and angle [HCD]=[HDC]=[DHC]=60 degree. so triangle (ABJ) congurrent to (CHD)and(FGE) and all are equilateral triangle. (√3*(3)^2)/4=(1/2)(3)(Height) height =(3√3)/2

so area of trapezium ABCD= 1/2(6+9) (3√3)2 (45 √3)/4=11.25√3 ----------------1

AREA OF parallelogram AJEF=6*(3√3)/2 =(18√3)/2---------------------2

TRIANGLE JED is a equilateral triangle of each side of 6 CM

area of Triangle(JED)= (√3(36))/4= --------3

SUM OF 1,2AND3=AREA OF HEXAGON = 50.66