PLEASE HELP ME ;-;

Assuming that ABCD is a trapezoid.

Drop a perpendicular from A to BC; call this point X.

Triangle(BAX) is a 30^o - 60^o - 90^o right triangle with AB the hypotenuse.

To find AX, divide AB by the sqrt(3)   --->   AX  = 6·sqrt(3)/sqrt(3)  =  6  [this is the height of the trapezoid]

BX  =  6·sqrt(3)/2  =  3·sqrt(3).

Drop a perpendicular from D to BC; call this point Y.

XY  =  AD  =  16.

Triangle(DCY) is a 30^o - 60^o - 90^o right triangle with CD the hypotenuse.

DY  =  AX  = 6.

YC  =  DY/sqrt(3)   --->   YC  =  6/sqrt(3)  =  2·sqrt(3).

BC  =  BX + XY + YC.