Question 1188319
<br>
Here is a re-drawing of the given figure:<br>
{{{drawing(400,400,-1,4,-1,4
,line(0,0,3,0),line(0,0,1.2,1.731),line(3,0,1.2,1.731)
,line(1,0,1.8,1.154),line(1,0,0.4,0.577)
,locate(1.1,2,A),locate(-.2,-.1,B),locate(2.8,-.1,C)
,locate(0.2,0.7,E),locate(0.9,-.1,D),locate(1.9,1.3,F)
)}}}<br>
Given the parallel segments, we know triangles BED and DFC are similar; and knowing that the area of triangle DFC is 4 times the area of triangle BED, we know DC is twice BD.<br>
Then we can draw several other segments parallel to the sides of triangle ABC...<br>
{{{drawing(400,400,-1,4,-1,4
,line(0,0,3,0),line(0,0,1.2,1.731),line(3,0,1.2,1.731)
,line(1,0,1.8,1.154),line(1,0,0.4,0.577)
,locate(1.1,2,A),locate(-.2,-.1,B),locate(2.8,-.1,C)
,locate(0.2,0.7,E),locate(0.9,-.1,D),locate(1.9,1.3,F)
,line(0.8,1.154,1.8,1.154),line(0.4,0.577, 2.4, 0.577)
,line(2,0,2.4,0.577),line(0.8,1.154,2,0)
)}}}<br>
... to see that triangle ABC can be divided into 9 congruent triangles.<br>
Parallelogram AEDF is composed of 4 of those triangles, so its area is 4/9 of the area of triangle ABC.<br>
84(4/9) = 28(4/3) = 112/3 = 37 1/3<br>
ANSWER: D<br>