Question

It's been a while since I am trying this ques. Not getting any result though.
PS: I'll upload the picture.

Answer 1

I assume there's a typo and the area of ABCD is 68 cm^2

Answer 2

Yes, my bad.

Answer 3

I think this problem is underspecified. Do you have any other data?

Answer 4

No that is it, it's one of the challengers.

Answer 5

For example, are the lines DE and BA parallel?

Answer 6

That's what should be the case, so ADEB and ADCF should be parallelograms.

Answer 7

"should be" or "is"

Answer 8

Okay is.

Answer 9

Anything else in the problem you haven't told us yet? This is a challenging problem for its sort.

Answer 10

That is it.

Answer 11

Ok.

So the way to get at this is the find the area of AND and AMD. Then the area of ANM is AND - AMD.

Add to your diagram two measurements. The first is the vertical distance between n and the line EF. Call it h. The second the is vertical distance between M and the line AD. Call it j.

We are going to solve for h and j and thereby find the areas of AND and AMD.

Notice first that as ABED and ADCE are parallelograms, they have area equal to their base times their perpendicular height. I will write ABED for both the figure ABED and its area.
ABED = 8x5 = 40. Likewise ADFC = 40.

Now look at those areas on the figure. They overlap in the triangle AND and they exclude from the total area of the figure, the triangle ENF. We know the total area of the figure ABCD is 68.

Hence ABED + ADCF + NEF - AND = 68 => NEF - AND = -12 --- (*)

Now the area of NEF = EF.h/2 = h. The area AND = AD(8-h)/2 = 5(8-h)/2.

Thus substituting these expressions into equation (*),

h + 5(h-8)/2 = -12
=> 7h/2 - 20 = -12
=> 7h/2 = 8
=> h = 16/7

Therefore the area of AND = 100/7.

Now for j and the area of AMD. We need to get at the area of the triangles AMD and EMC and use a similar method as we just did to find h.

AMD = 5j/2.

The base of EMC is EC = EF + FC = 2 + 5 = 7.
Hence the area of EMC = 7(8-j)/2

Note the triangle ADC is of area = 5x8/2 = 20.

Now
ABED + ACD + EMC - ADM = 68
=> 40 + 20 + 7(8-j)/2 - 5j/2 = 68
=> 28 - 6j = 8
=> j = 20/6 = 10/3

Hence area AMD = 5j/2 = 25/3

Therefore

ANM
= AND - AMD
= 100/7 - 25/3
= 125/21

Restoring units, the area of the triangle ANM is 125/21 cm^2.

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Hope that's right and I haven't made an error somewhere.

Answer 12

Thanks a lot man!