Area of triangle in parallelogram

• Nov 24th 2011, 07:07 AM
Kanwar245
Area of triangle in parallelogram

PS: there's a typo in the diagram, area of ADCB = 68cm^2
• Nov 24th 2011, 07:45 AM
Quacky
Re: It's been a while since I am trying this question, but without any results.
Looks like a longwinded problem.

Use the fact that the area of this trapezium $(68cm^2)$ is equal to the average length of the parallel sides - ( $BC$ and $AD$) multiplied by the height $(8cm)$
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 $AME$ and $DMC$ for example? What about $PAR$ and $MRQ$? Start with what you know, and see what you can deduce. Show us your deductions so far if you need further help.
• Nov 24th 2011, 07:57 AM
Kanwar245
Re: It's been a while since I am trying this question, but without any results.
angle AME = angle DMC, and angle PAR + angle MRQ = 180 degrees ?
• Nov 24th 2011, 07:57 AM
Quacky
Re: It's been a while since I am trying this question, but without any results.
That was just an example; what have you managed to work out so far?
• Nov 24th 2011, 08:00 AM
Kanwar245
Re: It's been a while since I am trying this question, but without any results.
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
• Nov 24th 2011, 08:14 AM
Kanwar245
Re: It's been a while since I am trying this question, but without any results.
I'm stuck at this point
• Nov 24th 2011, 09:10 AM
Quacky
Re: It's been a while since I am trying this question, but without any results.
I have to be honest, I'm progressing extremely slowly!

To speed things up for other people, we know that $BE=5$ and $FC=5$ because otherwise the parallelograms are false. $PQ=5$ and $DQ=8$ in order for the rectangle to stand true.

We know that the area of triangle $ABC=48$ going by the fact that the area of a triangle is $\frac{base\times height}{2}$ and we know that $BC=12$ through the area of the trapezium. This makes the area of $ADC=20$

As $FC=5$, $EC=7$ and $DQ=8$ which means the area of $EDC=28$. By the same logic, $ABF=28$

This is where you got up to. We know the area of the parallelograms $=40cm$ through various bits of logic.

So where do we proceed from here?
Well, we also know that $ABC$ and $MEC$ are similar triangles, and we know that the sides are in the ratio of $12:7$ and the same can be said of $APC$ and $RQC$, although the ratio will be different. I'm not entirely sure how this helps yet. $NEF$ is also similar to both $ABF$ and $EDC$, in a ratio of $2:7$

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.
• Nov 24th 2011, 11:42 AM
Kanwar245
Re: It's been a while since I am trying this question, but without any results.
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.