how else can i say this? the way i did it he doesnt want.
$\displaystyle FB = DB $
$\displaystyle \sqrt{(FB)^2+(DB)^2} = \sqrt{8^2 \cdot 2} $
$\displaystyle \sqrt{(DB)^2+(DB)^2} = 8 \sqrt{2} $
$\displaystyle \sqrt{2(DB)^2} = 8 \sqrt{2} $
$\displaystyle \sqrt{2}(DB)^2 = 8 \sqrt{2} $
$\displaystyle DB = 8 = FB $
for the total area of segment AB we have 11+8 = 19 , therefore the area of the rectangle ANDB is 19*8 = 152 and the area of the triangle FDB is (8*8)/2 = 32 = ANG
so subtracting 152 by 2(32) = 88 which is the area of the parallelogram
now for the rectangle we have 11 * 8 = 88