Find ratio of areas of parallelograms: parallelogram ABCD; P midpoint of AB, Q of BC,

Azazello

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ABCD is a parallelogram. P is the midpoint of AB, Q is the midpoint of BC, R is the midpoint of CD and S is the midpoint of AD. Lines AQ, CS, BR and DP, when intersecting with each other, make a new parallelogram. Find the ratio of the area of that parallelogram and the area of ABCD.


P.S. I'm assuming I need to introduce variables for two sides of the parallelogram (e.g. a for AB and DC, and b for BC and AD) and maybe α for the angle BAD, and thus say that the area of ABCD is ab*sin(α), but I'm not sure how to get the other parallelogram's area (fyi, I named it A1B1C1D1).
 
ABCD is a parallelogram. P is the midpoint of AB, Q is the midpoint of BC, R is the midpoint of CD and S is the midpoint of AD. Lines AQ, CS, BR and DP, when intersecting with each other, make a new parallelogram. Find the ratio of the area of that parallelogram and the area of ABCD.

P.S. I'm assuming I need to introduce variables for two sides of the parallelogram (e.g. a for AB and DC, and b for BC and AD) and maybe α for the angle BAD, and thus say that the area of ABCD is ab*sin(α), but I'm not sure how to get the other parallelogram's area (fyi, I named it A1B1C1D1).

Variables aren't necessarily required; I solved it by a purely visual or geometric method.

What is the context of the problem? Are there any expectations as to what methods you should use? Is there a reason to think trig should be needed?

My method was to sketch the result of rearranging the 9 pieces, fitting pairs of pieces together to make parallelograms congruent to the one in the middle. This changes the figure into one in which the question is very easy to answer.

You may find this easiest if you draw a very special parallelogram, namely a square. What works for that will work for any.
 
Variables aren't necessarily required; I solved it by a purely visual or geometric method.

What is the context of the problem? Are there any expectations as to what methods you should use? Is there a reason to think trig should be needed?

My method was to sketch the result of rearranging the 9 pieces, fitting pairs of pieces together to make parallelograms congruent to the one in the middle. This changes the figure into one in which the question is very easy to answer.

You may find this easiest if you draw a very special parallelogram, namely a square. What works for that will work for any.

Thanks! I'll get on solving it now. And no, there was no context or anything, I just thought that if you need the ratio, you have to do it with the algebraic method :D


P.S. I solved it, but kind of in a different way, I just found the ratios of the area of ABCD and the areas of all the shapes except for A1B1C1D1​ and the answer I got is 1/5. But thank you anyway!
 
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Thanks! I'll get on solving it now. And no, there was no context or anything, I just thought that if you need the ratio, you have to do it with the algebraic method :D


P.S. I solved it, but kind of in a different way, I just found the ratios of the area of ABCD and the areas of all the shapes except for A1B1C1D1​ and the answer I got is 1/5. But thank you anyway!

Okay, here's the quick visual way:

FMH113021.png

The two pieces of each color combine to make a square identical to the one in the middle. Thus the whole square has the same area as five squares of that size.

As I said, you can do the same with any parallelogram, but it's easier to draw with a square.
 
Okay, here's the quick visual way:

View attachment 10370

The two pieces of each color combine to make a square identical to the one in the middle. Thus the whole square has the same area as five squares of that size.

As I said, you can do the same with any parallelogram, but it's easier to draw with a square.

That's a pretty awesome way to solve it! Thanks, I'll keep it mind :D
 
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