Algebra.Com's Answer #819207 by ikleyn(52781)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "In the diagram line AD = 2 cm, EF = 1 cm, and parallel segments are indicated.
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document.write( "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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document.write( "From parallelogram ABED, we have BE = AD = 2 cm.\r\n" );
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document.write( "From parallelogram AFCD, we have FC = AD = 2 cm.\r\n" );
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document.write( "So, the base BC of the trapezoid ABCD is 2 + 1 + 2 = 5 cm long.\r\n" );
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document.write( "Hence, the altitude of this trapezoid is 105 cm^2 divided by  =  =  = 3.5 cm, \r\n" );
document.write( "which gives for the altitude  = 30 cm.\r\n" );
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document.write( "Next, triangles AND and FNE are similar (OBVIOUSLY) with the similarity coefficient  = 2;\r\n" );
document.write( "So, their altitudes, drawn from their common vertex N to their parallel bases are 20 cm and 10 cm, in this order;\r\n" );
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document.write( "hence, the areas of these triangles are  = 5 cm^2 for triangle FNE and  = 20 cm^2 for triangle AND.\r\n" );
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document.write( "Further, triangles AMD and CME are similar with similarity coefficient  (as their bases AD and EC are in this ratio);\r\n" );
document.write( "so, their altitudes, drawn from their common vertex M to their parallel bases AD and EC are  = 6*2 = 12 cm and 30-12 = 18 cm.\r\n" );
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document.write( "hence, the areas of these triangles are  = 12 cm^2 for triangle AMD and  = 3*9 = 27 cm^2 for triangle CME.\r\n" );
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document.write( "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.\r\n" );
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document.write( "ANSWER. The area of the triangle AMN is 8 cm^2.\r\n" );
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document.write( "Solved.\r
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