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#3**+1 **

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

CPhill Nov 5, 2018