Geometry help

Need to see a pic, bambam.....  

Mmmmmm.......this is an interesting one  !!!!

Maybe easier ways to do this....but......

Perpendicular to   AB.....draw IJ   through E  so that I is a lies on AB  and J  lies on  DC

This divides the gray area on the bottom into 3 areas....

Triangle AIE, Triangle HEJ and Quadrilateral IJCB

We can find the area of  Tiriangle AIE as follows

AE = AB  = 12

Angle IAE = 30

Angle AIE  = 90

So  IE  = 6  and AI = 6sqrt(3)

So...the area of Triangle AIE  = (1/2)(6)(6sqrt(3))  =   18sqrt (3)   

So....IB  =   12 - AI =  12 - 6qrt(3)

So....the area of Quadrilateral IJBC = BC * IB =  18 * [ 12 - 6sqrt(3) ]  =  216 - 108 sqrt (3)  

And we can find the area of Triangle HEJ as follows

Angle HJE = 90

Angle JHE  = 60

EJ  = BC - IE  = 18 - 6  = 12

And HJ is opposite the 30 ° angle in Triangle HEJ  =   12 / sqrt (3) =  4sqrt (3)

So...the area of triangle HEJ = (1/2) (12)(4sqrt(3) )  = 24sqrt(3)

So...the "bottom" Grey Area is the sum of the areas in red =  216 - 66sqrt(3)  units^2

So....the "white" area  =  Area of  Rectangle ABCD  -  bottom Grey Area =

18 * 12  - [ 216 - 66sqrt(3) ]  =

216 - 216 + 66sqrt(3)  =

66sqrt(3)  units^2

And the "top" Grey Area is just the area of Rectangle AEFG - white area  =

Area of Rectangle ABCD - white area  =

216 - 66sqrt (3)    units^2

So.....the shaded area is just the bottom grey area + top grey area  

[ 216 - 66sqrt (3) ] + [ 216 - 66sqrt (3) ]  =

 [432 - 132sqrt(3) ]   units^2