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Finding the area of a parallelogram inside another

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Problem Statement
(The image is given in the attempt section,1st image.) In ABCD parallelogram E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively. Area of ABCD is 25 sq.unit. Find the area of PQRS.
Relevant Equations
area of parallelogram=height*base.
para.JPG


Through symmetry of parallelogram,I have come to this:

pa.JPG

Here 1,2,3,4 denotes the area of the particular regions.Then I am stuck.Please help what to do next or whether there is any other way.
 

BvU

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Hi,
the naming is a bit confusing, but you know that
1+2+4 = ...
2x1 + 3 =
4x1 + 2x2 + 2x4 + 3 = ...
manipulate these and see if you can end up with 3 by itself ...
 

Ray Vickson

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View attachment 242396

Through symmetry of parallelogram,I have come to this:

View attachment 242397
Here 1,2,3,4 denotes the area of the particular regions.Then I am stuck.Please help what to do next or whether there is any other way.
Your solution makes no sense: you have a small triangle with area 9 next to a much larger parallelogram with area 1, next to a smaller triangle with area 2, etc.

Anyway, you need to show your work. Just saying "Through symmetry of parallelogram,I have come to this:" is not an explanation.
 

BvU

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@Ray Vickson : I considered the numbers as names of regions; Akash used the word 'area' for region.

Anyway, you need to show your work
I fully agree. Apparently you have found that the two kinds of '1' have the same area.
If you now can show that area of '4' = area of '2', you're almost done !

(this also means my suggestion in #2 is not sufficient :frown: to crack this one !)
 

LCKurtz

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I think the answer is ##5## square units. :smile:
 

BvU

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Oops ! Here at PF we are not giving direct answers ! (Even if they are completely correct 😉)
 

LCKurtz

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Oops ! Here at PF we are not giving direct answers ! (Even if they are completely correct 😉)
It won't be helpful to the OP. That's why I didn't show how and now I'm expecting his next question to be "how did you do that?"
 

LCKurtz

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@BvU: I'm curious if you solved it by straight geometry messing with the given areas as suggested in above posts?
 

SammyS

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Hi,
the naming is a bit confusing, but you know that
1+2+4 = ...
2x1 + 3 =
4x1 + 2x2 + 2x4 + 3 = ...
manipulate these and see if you can end up with 3 by itself ...
I really like the very unusual arithmetic which results.

For a change, 1+2+4 ≠ 7 .

And certainly 4×1 ≠ 1×4 .
 
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BvU

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LCKurtz

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He's not the only thread abandoner. See my post here:
 
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No,I haven't abondoned the thread. BvU has given me some clues like these:
'1+2+4 = ...
2x1 + 3 =
4x1 + 2x2 + 2x4 + 3 = ...
manipulate these and see if you can end up with 3 by itself ... '
And:
'If you now can show that, 'area of '4' = area of '2' you're almost done !'
But still I can't solve the problem.Firstly, I have already tried those things which was suggested by BvU in his first post. But I didn't get anything useful.
Secondly,I couldn't prove that area of '4' = area of '2'.And even if I were able to prove that,I don't see how it would help me.The problem is from a math olympiad. And I am not so familiar with these problem.
I just remained silent because the clues you people were giving me wasn't sufficient for me and even BvU was so surprised with that!
 

LCKurtz

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If it was from a math olympiad, I'm not surprised it is tricky. If a person is taking a test where speed and the answer are what's important, here's a hint: If the problem can be solved with the only givens being the area and the fact that it is a parallelogram, then you must get the same answer for any parallelogram, and in particular, if the parallelogram is a square.
 
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BvU

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second @LCKurtz -- that's how I 'discovered' area of 2 = area of 4
 

LCKurtz

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Actually, that's how I did the whole problem. But technically that isn't a whole solution until you show the transformation from parallelogram to square preserves areas, which it does. I am curious if anyone solved it directly with just geometry and the given regions.
 
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If it was from a math olympiad, I'm not surprised it is tricky. If a person is taking a test where speed and the answer are what's important, here's a hint: If the problem can be solved with the only givens being the area and the fact that it is a parallelogram, then you must get the same answer for any parallelogram, and in particular, if the parallelogram is a square.
The following is an approach; not the solution: the fact that "E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively" is also among the givens, means that you can solve for the areas of the 2 congruent pairs of larger included triangles, sum them, then subtract the sum of the 4 smaller triangular areas that are common to the 2 pairs of larger triangles (thereby correcting for having counted the common smaller triangular areas twice when you were summing the areas of the 2 larger pairs of included triangles), and then subtracting the result of those operations from the area of the outer parallelogram, will give the area of the inner parallelogram (which is not part of either of the 2 pairs of included larger triangles).

[Edit: (for inclusion of illustration to clarify what I was hinting at)]:

If the fact that "E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively" were not also among the givens, you couldn't, without some further information, solve the problem:

242604


Did anyone else, other than me, Julius Caesar, and Benito Mussolini, notice that the inner parallelogram is designated by a counter-clockwise ##\text {SPQR}##?
 
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LCKurtz

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@sysprog: I'm not following that. I have included a better picture with the areas labeled. Since this is given not to be a homework problem I don't think anyone will object to more detail and I haven't found a "simple" solution of that type myself. Here's a picture for reference:
parallelogram.jpg

What would your equations be? Call the area of the parallelogram ##A##.
 
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@LCKurtz Does the illustration and comment edited into my post #17 help to clarify what I was trying to hint at? Your illustration makes the corner-to-side line segments that define the inner parallelogram look like they terminate at non-midpoints on the sides. If the fact the point are side midpoints isn't given, then unless there's some other information, the problem isn't solvable.
 

LCKurtz

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@LCKurtz Does the illustration and comment edited into my post #17 help to clarify what I was trying to hint at?
No. It's hard to figure out which triangles you refer to. There are more than 2 "larger" triangles. I understand adding the triangles around the outside of the inner parallelogram and subtracting the duplicates, but unless I'm missing something simple, it doesn't help much.
 
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No. It's hard to figure out which triangles you refer to. There are more than 2 "larger" triangles. I understand adding the triangles around the outside of the inner parallelogram and subtracting the duplicates, but unless I'm missing something simple, it doesn't help much.
I said 2 pairs of congruent (for this problem, the fact that the 2 members of the pair have the same area is the part of congruency that matters -- it means you only have to figure the area for one of each pair, then double each of them to get the sums of the areas for each pair) larger triangles, not just 2 larger triangles.

HDC and AFB are one such pair, and ADE and GCB are the other. Summing their areas means counting their common areas twice, so you have to subtract out their common areas once. Their common areas are the 2 congruent pairs of smaller triangles: APE and GCR, and HDS and QFB.

You can figure all the areas given the area of the outer parallelogram, the fact that both pairs of line segments, DE and GB, and HC and AF, are parallel, and the fact that "E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively".
 
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LCKurtz

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@LCKurtz Your illustration makes the corner-to-side line segments that define the inner parallelogram look like they terminate at non-midpoints on the sides. If the fact the point are side midpoints isn't given, then unless there's some other information, the problem isn't solvable.
My picture was drawn exactly with a good drawing program and the lines do go to the mid points of the sides. Maybe it's an optical illusion. It measures correctly on my screen.

I said 2 pairs of congruent (for this problem, the fact that the 2 members of the pair have the same area is the part of congruency that matters -- it means you only have to figure the area for one of each pair, then double each of them to get the sums of the areas for each pair) larger triangles, not just 2 larger triangles.

HDC and AFB are one such pair, and ADE and GCB are the other. Summing their areas means counting their common areas twice, so you have to subtract out their common areas once. Their common areas are the 2 congruent pairs of smaller triangles: APE and GCR, and HDS and QFB.

You can figure all the areas given the area of the outer parallelogram, the fact that both pairs of line segments, DE and GB, and HC and AF, are parallel, and the fact that "E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively".
Yes, that's what I tried to describe in post #20. The only thing I'm missing here is how you get the areas of the smaller triangles, for example APE, in terms of the area of the parallelogram.
 

ehild

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As @sysprog pointed out, ABF and HCD are congruent and so are AED and BCG.
As the lines DE and GB are parallel, and HC and AF also, the following pairs of triangles are similar: AEP and ABQ; BFQ and BCR; CGR and CDS; DHS and DAP. The bigger ones are twice the size of the smaller ones, so their areas are 4 times the smaller areas.
242610

The following figure shows the areas:

242612

The area of the triangle ABF is ab/4 sin (B)=4s+t
The area of triangle AED is ab/4 sin(A) = 4t+s.
As sin(A) = sin(B), the areas are equal.
The area of the parallelogram is absin(A), and it can be written in terms of s and t plus the area of the small parallelogram SPQR.
 
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My picture was drawn exactly with a good drawing program and the lines do go to the mid points of the sides. Maybe it's an optical illusion. It measures correctly on my screen.
In response to your saying that, I checked to 1 pixel of accuracy, and I now agree that you're right about that -- sorry for incorrectly stating otherwise without checking first -- I think you're right about the 'optical illusion', too -- I'm a bit miffed at myself about that, because I usually get those right if they're posed as questions, but I let that one get me, probably at least partly because it wasn't posed as a question ...
Yes, that's what I tried to describe in post #20. The only thing I'm missing here is how you get the areas of the smaller triangles, for example APE, in terms of the area of the parallelogram.
Well, I was going to answer that, but @ehild mostly beat me to it, and I have some pressing other stuff I have to attend to at present, so I'll get back with my answer on that later.
 

LCKurtz

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OK, so here's a summary using ehild's diagram, the fact (which I was missing) that the quadrilaterals are 3 times the area of the triangles, and no trig, just geometry.
Call the area of the given parallelogram ##A##.
From triangle DHC ##t + 4s = \frac A 4##.
And from triangle GCB ##s+ 4t = \frac A 4##. So ##t = s = \frac A {20}##.
The area of the smaller parallelogram is ##A-16t = A -16\frac A {20} = \frac A 5##.
 

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