Question 1187703
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https://imgur.com/mLTjbMU 
In the diagram line AD = 2 cm, EF = 1 cm, and parallel segments are indicated. 
If the total area of the trapezoid is 105 cm^2, what is the area, in cm^2, of triangle AMN?
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From parallelogram ABED, we have BE = AD = 2 cm.

From parallelogram AFCD, we have FC = AD = 2 cm.

So, the base BC of the trapezoid ABCD is 2 + 1 + 2 = 5 cm long.

Hence, the altitude of this trapezoid is 105 cm^2 divided by {{{(AD+BC)/2}}} = {{{(2+5)/2}}} = {{{7/2}}} = 3.5 cm, 
which gives for the altitude {{{105/3.5}}} = 30 cm.



Next, triangles AND and FNE are similar (OBVIOUSLY) with the similarity coefficient {{{2/1}}} = 2;
So, their altitudes, drawn from their common vertex N to their parallel bases are 20 cm and 10 cm, in this order;

hence, the areas of these triangles are  {{{(1*10)/2}}} = 5 cm^2 for triangle FNE and {{{(2*20)/2}}} = 20 cm^2 for triangle AND.



Further, triangles AMD and CME are similar with similarity coefficient {{{2/3}}} (as their bases AD and EC are in this ratio);
so, their altitudes, drawn from their common vertex M to their parallel bases AD and EC are {{{(30/5)*2}}} = 6*2 = 12 cm and 30-12 = 18 cm.

hence, the areas of these triangles are {{{(2*12)/2}}} = 12 cm^2 for triangle AMD and {{{(3*18)/2}}} = 3*9 = 27 cm^2 for triangle CME.



Finally, the area of the triangle AMN is the difference of the area AND (20 cm^2) and the area AMD (12 cm^2), i.e. 20 - 12 = 8 cm^2.


<U>ANSWER</U>.  The area of the triangle AMN is 8 cm^2.
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Solved.