Question 1208980: In the diagram AD = 4 cm, EF = 3 cm, and parallel segments are indicated. If the total area of the trapezoid is 135 cm2, what is the area, in cm2, of triangle AMN?
https://ibb.co/ggBgSQB
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'll recreate the diagram just in case the link might expire (at some point in the future).
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 = (576/77) square cm exactly
area AMN = 7.480519480519 square cm approximately
Round this approximate value however needed.
I have confirmed this answer is correct using GeoGebra
A similar problem is found here
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
I saw several similar problems, solved at this forum years before, where the input data
were carefully given/selected by the problems creators in a very nice way giving integer numbers along the solution.
These problems were created by reasonable/professional math composers.
To solve these problems was a pleasure. I myself solved one of such problems here at the forum
https://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.1187703.html
In this version, the input data are monstrously ugly and lead to monstrous fractions.
The lowest possible score to the current math composer for his monstrous irresponsible work.
Instead of fostering a love of mathematics, such problems and their creators generate hatred in readers.
Such "math composers" need to be driven out of their positions with flamethrowers, immediately,
unconditionally and forever, and their creatures should be prohibited for spreading in the Internet.
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