Question 1148863: In the diagram AD=5cm, EF=2cm, and parallel segments are indicated. If the total area of the trapezoid is 68cm2, what is the area, in cm2, of triangle AMN
Diagram: https://imgur.com/a/JhZZ9W0
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Cool problem. Solve by determining the height of the trapezoid from the given information and then looking at some similar triangles....
Given: AD=5, EF=2; area of ABCD is 68.
BE=5 because ABED is a parallelogram; FC=5 because AFCD is a parallelogram.
BC=5+2+5 = 12.
Area of ABCD = 68 = (1/2)(5+12)(h) --> the height of the trapezoid is 8.
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.
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.
The area of triangle AMN is the difference between the areas of triangles AND and AMD:
ANSWER: The area of triangle AMN is 125/21 cm^2.
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