Question 1210163
<pre>

The figure on the site is not to scale.  Here it is drawn to scale.
I will assume a kite ABCD on the left, although I agree with 
greenestamps that nothing tells us that's the case.

{{{drawing(400,3600/11,-8,14,-10,8,

locate(-.1,6.9,A),
locate(5.9,.9,B),
locate(12,-8,E),
locate(0,-8,C),
locate(-6.5,.65,D),
locate(.16,-.16,O),

line(-6,0,0,6),
line(0,6,6,0),
line(6,0,12,-8),
line(12,-8,0,-8),
line(0,-8,-6,0),
line(-6,0,6,0),
line(0,6,0,-8),
line(0,-8,6,0),
green(line(6,0,6,-8)),

locate(-3,0,6),
locate(-1.4,5,45^o),
locate(-3.9,-4,10),
locate(5.6,-8,12) 




)}}}

Triangle DOA is a 45-45-90 triangle so. DO = AO = 6. Its area
is {{{expr(1/2)*base*height=DO*OA=expr(1/2)*6*6=18}}}
Triangle BOA is congruent to triangle DOA so it also has area 18.

You may recognize triangle DOC as a 6-8-10 right triangle and see
right off that OC=8.  If not, use the Pythagorean theorem,

{{{OC = sqrt(DC^2-DO^2) = sqrt(10^2-6^2)=sqrt(100-36)=sqrt(64)=8}}}
Triangle DOC's area is {{{expr(1/2)*base*height=expr(1/2)*DO*OC=expr(1/2)*6*8=24}}}
Triangle BOC is congruent to triangle DOC so it also has area 24. 

So the area of kite ABCD is 18+18+24+24 = 84

Triangle BCE's height (the green line) is the same length as OC = 8.
so its area is {{{expr(1/2)*base*height=expr(1/2)*CE*8=expr(1/2)*12*8=48}}}

So the kite's area plus triangle BCE's area is 48.

So the area of the composite figure is 84 + 48 = 132 square units.

-----------------------------
For the other problem, they have added on an isosceles trapezoid on the right,
I use different lettering, as they didn't put any lettering on the figures.

{{{drawing(8000/9,3600/11,-8,32,-10,8,

locate(-.1,6.9,A),
locate(5.9,.9,B),
locate(12,-8,E),
locate(0,-8,C),
locate(-6.5,.65,D),
locate(.16,-.16,O),

line(-6,0,0,6),
line(0,6,6,0),
line(6,0,12,-8),
line(12,-8,0,-8),
line(0,-8,-6,0),
line(-6,0,6,0),
line(0,6,0,-8),
line(0,-8,6,0),
green(line(6,0,6,-8)),

locate(-3,0,6),
locate(-1.4,5,45^o),
locate(-3.9,-4,10),
locate(5.6,-8,12),
 
line(12,-8,24,-8), 
line(6,0,30,0),
line(30,0,24,-8),

locate(29.9,.9,G), locate(24,-8,H),

locate(18,-8,12), locate(18,.85,24)



)}}}

Let's draw in a couple of red lines and you will see what has been added:

{{{drawing(8000/9,3600/11,-8,32,-10,8,

locate(-.1,6.9,A),
locate(5.9,.9,B),
locate(12,-8,E),
locate(0,-8,C),
locate(-6.5,.65,D),
locate(.16,-.16,O),

line(-6,0,0,6),
line(0,6,6,0),
line(6,0,12,-8),
line(12,-8,0,-8),
line(0,-8,-6,0),
line(-6,0,6,0),
line(0,6,0,-8),
line(0,-8,6,0),
green(line(6,0,6,-8)),

locate(-3,0,6),
locate(-1.4,5,45^o),
locate(-3.9,-4,10),
locate(5.6,-8,12),
 
line(12,-8,24,-8), 
line(6,0,30,0),
line(30,0,24,-8),

locate(29.9,.9,G), locate(24,-8,H),

locate(18,-8,12), locate(18,.85,24),

red(line(12,-8,18,0), line(18,0,24,-8))



)}}}

Now you see that they have added on three more triangles all congruent to
triangle BCE.  We found triangle BCE's area to be 48, so we just add (3)(48)=144
to the area in the first problem, which was 132.

So the area of this new composite figure is 132+144 = 276 square units. 

Edwin</pre>