Question 1187703
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There is probably an easier way to solve this using ordinary geometry, but I'm not seeing it....<br>
With the two sets of parallel lines, we know BE and FC are each 2cm.<br>
With that, and knowing the area of ABCD is 105 cm^3, we can easily determine that the height of the trapezoid is 30cm.<br>
Now we can use a strategy for solving the problem that is applicable to this and many similar problems.  Since no angles are specified, the result must be independent of the angles, so we can make them any angles we want.<br>
In particular, we can re-draw the figure with angle B a right angle, making the figure like this:<br>
{{{drawing(500,500, -2,6,-10,40
,line(0,0,5,0),line(5,0,2,30),line(2,30,0,30),line(0,30,0,0)
,line(0,30,3,0),line(0,30,5,0),line(2,0,2,30)
,locate(0,33,"A(0,30)"),locate(0,-1,"B(5,0)"),locate(5,-1,"C(5,0)"),locate(2,33,"D(2,30)")
,locate(1.8,-1,"E(2,0)"),locate(3,-1,"F(3,0)"),locate(2.1,12,N),locate(2.1,20,M)
)}}}<br>
Now we can use coordinate geometry to find the equations of segments DE, AF, and AC and use their intersections to find that the coordinates of N and M are (2,10) and (2,18).  (I leave it to you to do that much of the work.)<br>
Then we see that triangle AMN has base length 8 and height 2, making its area (one-half base times height) 8.<br>
ANSWER: The area of triangle AMN is 8 cm^3<br>
NOTE: I worked the problem with another arbitrary angle B and obtained the same result; choosing the right angle made the computations far easier....<br><hr>
Thanks to tutor @ikleyn for showing a purely geometrical solution.  In looking for that kind of solution, I wasn't seeing enough pairs of similar triangles.<br>
On the other hand, note that my solution is useful to the student in demonstrating the powerful problem-solving technique of being able to re-draw the figure in a way that makes the problem easier to solve.<br>