Question 1148863
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Cool problem.  Solve by determining the height of the trapezoid from the given information and then looking at some similar triangles....<br>
Given: AD=5, EF=2; area of ABCD is 68.<br>
BE=5 because ABED is a parallelogram; FC=5 because AFCD is a parallelogram.<br>
BC=5+2+5 = 12.<br>
Area of ABCD = 68 = (1/2)(5+12)(h)  -->  the height of the trapezoid is 8.<br>
Triangles AND and FNE are similar, in the ratio 5:2.  In particular, the heights of those two triangles (measured perpendicular to the bases of the trapezoid) are in the ratio 5:2.  Since the height of the trapezoid is 8, the height of triangle AND is (5/7)*8 = 40/7.<br>
Triangles ADM and CME are similar, in the ratio 5:7.  In particular, the heights of those two triangles (measured perpendicular to the bases of the trapezoid) are in the ratio 5:7.  Since the height of the trapezoid is 8, the height of triangle ADM is (5/12)*8 = 40/12 = 10/3.<br>
The area of triangle AMN is the difference between the areas of triangles AND and AMD:<br>
{{{(1/2)(5)(40/7-10/3) = (5/2)((120-70)/21) = 125/21}}}<br>
ANSWER: The area of triangle AMN is 125/21 cm^2.<br>