Question 1210155
<pre>
{{{drawing(400,480,-2,28,-18,18, 


green( triangle(10,0,10,sqrt(264),10,0),triangle(21,0,21,sqrt(264),21,0),

locate(9.5,17.5,H),locate(21,17.5,I)), 

line(0,0,21,0),
line(0,0,5,sqrt(264)),
line(5,sqrt(264),26,sqrt(264)),
line(26,sqrt(264),21,0),
line(5,sqrt(264),5,-sqrt(264)),
line(5,sqrt(264),10,0),
line(0,0,5,-sqrt(264)),
line(5,-sqrt(269),10,0),

locate(4.5,17.5,A),
locate(26,17.5,B),
locate(21,0,C),
locate(10.3,0,D),
locate(-.9,.5,F),
locate(2.5,1.5,5),
locate(.5,-7,17),
locate(15.5,0,11),
locate(5,-sqrt(264),E),
locate(5.3,0,G)


)}}}

We calculate GE by using the Pythagorean theorem on right triangle FGE.

{{{GE=sqrt(17^2-5^2)=sqrt(264)=sqrt(4*66)=2sqrt(66)}}}
AG is also {{{2sqrt(66)}}}, so diagonal of the rhombus, AE = {{{4sqrt(66)}}}
FG = GD = 5 so the other diagonal of the rhombus, FD is 10.

The area of a rhombus is {{{1/2}}} the product of the diagonals, or,
Area of rhombus ADEF is {{{expr(1/2)*AE*FD=expr(1/2)*4sqrt(66)*10=20sqrt(66)}}}.


Since ABCF is a parallellogram, 
ABCD is an isosceles trapezoid (or trapezium if you live in the UK).
We draw two green lines at D and C perpendicular to AB and CD. They partition
the trapezoid into two congruent right triangles, and a rectangle in the middle.
Since AH = GD = 5, the area of right triangle AHD is
{{{expr(1/2)*base*height = expr(1/2)*AH*HD = expr(1/2)*5*2sqrt(66)=5sqrt(66)}}}
The area of right triangle BIC is also {{{5sqrt(66)}}}
All that's left of isosceles trapezoid ABCD is the area of rectangle HICD which
is {{{CD*HD= 11*2sqrt(66) = 22sqrt(66).

So the area of isosceles trapezoid ABCD is 

Area of right triangle AHD = {{{5sqrt(66)}}}
Area of rectangle = {{{22sqrt(66)}}}
Area of right triangle BIC = {{{5sqrt(66)}}}
Area of isosceles trapezoid ABCD = {{{32sqrt(66)}}}

Area of rhombus ADEF = {{{20sqrt(66)}}}
Area of isosceles trapezoid ABCD = {{{32sqrt(66)}}}

Area of composite figure = {{{52sqrt(66)}}}

That's approximately 422.4 sqruare units.

Edwin</pre>