Question 1208980
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I'll recreate the diagram just in case the link might expire (at some point in the future).
{{{
drawing(400,400,-1,12,-1,19,
line(2.46,18,0,0),line(0,0,11,0),line(11,0,6.46,18),line(6.46,18,2.46,18),line(2.46,18,7,0),line(6.46,18,4,0),line(2.46,18,11,0),

line(2.46,18,1.66190558,12.160284732),line(1.66190558,12.160284732,1.43900558,12.705884732),line(1.66190558,12.160284732,2.02300558,12.626084732),line(2.46,18,4.87,18),line(4.87,18,4.6613,17.8795),line(4.87,18,4.6613,18.1205),line(2.46,18,4.0969258628,11.5099855662),line(4.0969258628,11.5099855662,3.6306258628,11.9901855662),line(4.0969258628,11.5099855662,4.2796258628,12.1538855662),line(6.46,18,5.8493574143,13.531883519),line(5.8493574143,13.531883519,5.6788574143,13.949383519),line(5.8493574143,13.531883519,6.1256574143,13.888283519),line(6.46,18,8.1243641745,11.4011993081),line(8.1243641745,11.4011993081,7.6502641745,11.8894993081),line(8.1243641745,11.4011993081,8.3101641745,12.0558993081),line(7,0,9.19,0),line(9.19,0,9.0003,-0.1095),line(9.19,0,9.0003,0.1095),

locate(2.46-0.5,18,"A"),locate(0,0,"B"),locate(11,0,"C"),locate(6.46+0.5,18,"D"),locate(4,0,"E"),locate(7,0,"F"),locate(5.5654545455+0.5,11.4545454545,"M"),locate(5.0542857143+0.5,7.7142857143,"N"),
locate(5.2,0,"3 cm"),locate(4.2,18,"4 cm"),
locate(9,18,matrix(1,2,"Diagram","is")),
locate(9,17.2,matrix(1,2,"to","scale"))

)
}}}
Note that I'm copying the arrow style (to the best of my ability) shown in the diagram link. Your teacher/textbook should use different arrow styles for a different set of parallel segments. For instance AB & ED could have single arrow, while AF & DC have double arrows.


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The arrows on the segments tell us the following 
AB is parallel to DE
BC is parallel to AD
AF is parallel to DC


Consequently it means ADEB and ADCF are parallelograms.
The opposite sides of a parallelogram are congruent, which leads to AD = BE = 4 and AD = FC = 4.
BC = BE+EF+FC = 4+3+4 = 11


h = height of trapezoid ABCD
area of trapezoid = 0.5*height*(base1+base2)
area of trapezoid ABCD = 0.5*h*(AD+BC)
135 = 0.5*h*(AD+BC)
0.5*h*(4+11) = 135
7.5h = 135
h = 135/7.5
h = 18
Trapezoid ABCD has height 18 cm.
This is also the height of triangle ABF when working with base BF.


Focus on triangle ABF. 
To avoid clutter, you can create a quick sketch of triangle ABF off to the side somewhere. 
area = 0.5*base*height
area = 0.5*BF*h
area = 0.5*(BE+EF)*h
area = 0.5*(4+3)*18
area = 63 square cm


Notice that triangles ABF and NEF are similar triangles (use the Angle Angle Theorem; I will leave this as an exercise for the student).


Let's compare the horizontal portions of each triangle.
The ratio of segment EF to BF is 3:7
Square both parts to get 9:49 which is the ratio of their areas.
This new ratio means that if triangle NEF had area 9 square cm, then triangle ABF would have an area of 49 square cm.


In other words, triangle NEF takes up 9/49 of triangle ABF's area.
The remaining portion is 1 - (9/49) = 40/49 which is what trapezoid ANEB takes up.


area of ANEB = (40/49)*(area of triangle ABF) 
= (40/49)*(63)
= (360/7) square cm
= 51.42857 square cm approximately


Now focus on triangle ABC.
area = 0.5*base*height
area = 0.5*BC*h
area = 0.5*11*18
area = 99 square cm


Triangles MEC and ABC are similar.
The ratio of the horizontal pieces EC to BC is 7:11
Both parts square to 49:121 which is the ratio of the areas MEC to ABC.
Triangle MEC takes up 49/121 of triangle ABC.
Trapezoid AMEB takes up 1 - (49/121) = 72/121 of triangle ABC's area.


area of AMEB = (72/121)*(area of triangle ABC)
= (72/121)*(99)
= (648/11) square cm
= 58.90909 square cm approximately


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There are a lot of things to keep track of. 
But the two key important things are: 
area of trapezoid AMEB = (648/11) square cm
area of trapezoid ANEB = (360/7) square cm


Subtracting those areas will get us the area of triangle AMN.
area AMN = (area AMEB) - (area ANEB)
area AMN = (648/11) - (360/7)
area AMN = (648/11)*(7/7) - (360/7)*(11/11)
area AMN = (648*7)/(11*7) - (360*11)/(7*11)
area AMN = 4536/77 - 3960/77
area AMN = (4536 - 3960)/77
area AMN = <font color=red>(576/77) square cm exactly</font>
area AMN = <font color=red>7.480519480519 square cm approximately</font>
Round this approximate value however needed.


I have confirmed this answer is correct using <a href="https://www.geogebra.org/calculator">GeoGebra</a> 


A similar problem is found <a href="https://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.1187703.html">here</a>
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