Thread: Area of triangle in parallelogram

The questions asks for finding the area of triangle AMN. Also, ADEB and ADCF are parallelograms.

PS: there's a typo in the diagram, area of ADCB = 68cm^2

Looks like a longwinded problem.

Use the fact that the area of this trapezium is equal to the average length of the parallel sides - ( and ) multiplied by the height
This will give you an unknown length and you can use this to deduce some areas.
I can see a very large amount of right triangles.
Also use the fact that paralellograms have 2 pairs of equal sides with equal angles.
You will also need to use angle theorems. What can you say about angles and for example? What about and ? Start with what you know, and see what you can deduce. Show us your deductions so far if you need further help.

angle AME = angle DMC, and angle PAR + angle MRQ = 180 degrees ?

That was just an example; what have you managed to work out so far?

area of triangle ABC = 48cm^2, and BE = FC = 5cm, I need to figure out how to find AB. Also area of parallelograms = 40cm^2
Area of triangle ABF = 28cm^2 , so Area of triangle AFC should be 48 - 28 = 20cm^2

I'm stuck at this point

I have to be honest, I'm progressing extremely slowly!

To speed things up for other people, we know that and because otherwise the parallelograms are false. and in order for the rectangle to stand true.

We know that the area of triangle going by the fact that the area of a triangle is and we know that through the area of the trapezium. This makes the area of

As , and which means the area of . By the same logic,

This is where you got up to. We know the area of the parallelograms through various bits of logic.

So where do we proceed from here?
Well, we also know that and are similar triangles, and we know that the sides are in the ratio of and the same can be said of and , although the ratio will be different. I'm not entirely sure how this helps yet. is also similar to both and , in a ratio of

Not sure whether this helps, but I think you can use it to work out some of the other unknown sides/areas (be careful when using ratios to work out the areas though - it isn't as straightforward as sides, remember). I have to go for a bit so I can't finish, but this problem has taken an unusual amount of time for me to make any progress. Hopefully someone else will be able to give you better support than I could.

Got the answer.
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 the 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.