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Quadrilateral sides comparison

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Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 10:22
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Question Stats:

100% (01:32) correct 0% (00:00) wrong based on 3 sessions
I don't have the answer for it. But i have a solution not so sure though.
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Re: Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 13:05
I think the answer is C.

With the figure, I’d do this:

e^2= a^2+d^2 at the same time e^2=b^2+c^2 so, we can equal both equations:
a^2+d^2 = b^2+c^2
a^2+d^2 – c^2= b^2

Second, I'll compare the two options:
Quantity A
b^2 + 2ad

Quantity B
a^2+2ad+d^2-c^2

We can subtract “2ad” in both equations
Quantity A
b^2

Quantity B
a^2+d^2-c^2

We already know that “b^2= a^2+d^2 – c^2”

Please let me know if it is ok. It’s just my guess.
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Re: Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 13:16
Linamrs wrote:
I think the answer is C.

With the figure, I’d do this:

e^2= a^2+d^2 at the same time e^2=b^2+c^2 so, we can equal both equations:
a^2+d^2 = b^2+c^2
a^2+d^2 – c^2= b^2

Second, I'll compare the two options:
Quantity A
b^2 + 2ad

Quantity B
a^2+2ad+d^2-c^2

We can subtract “2ad” in both equations
Quantity A
b^2

Quantity B
a^2+d^2-c^2

We already know that “b^2= a^2+d^2 – c^2”

Please let me know if it is ok. It’s just my guess.


I dont think you can use pythagorean theorem, always look out for that. I used to fall for it many times.
Try without the theorem, your method is close to mine. I got B another friend D. Still not sure.
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Re: Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 13:47
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Tricky problem. There are a few hurdles here. The first is that the quantities look pretty complicated. I'd try to simplify if possible. Notice that Quantity B contains (a + d)^2, which can be rewritten as a^2 + 2ad + d^2. Then we can subtract the 2ad from both sides, and shuffle the leftovers till we have b^2 + c^2 and a^2 + d^2. See the first picture for the algebra.

Next, we should recognize b and c as the sides of the rightmost triangle and a and d as the sides of the leftmost. They have the same opposite side, but the rightmost triangle has a 92° angle, while the leftmost has an 88° angle. If the angle were 90° then we would of course know that b^2 + c^2 would be the same as e^2. But what does it mean when the angle is either bigger or smaller than 90°? Easiest to just draw a picture and exaggerate the difference to see what it means. See the 2nd picture. Doing so shows, without any math, that the two sides on either side of the smaller angle must be larger, so their squares must also be larger. Thus, B is the answer.
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Re: Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 14:30
SherpaPrep wrote:
Tricky problem. There are a few hurdles here. The first is that the quantities look pretty complicated. I'd try to simplify if possible. Notice that Quantity B contains (a + d)^2, which can be rewritten as a^2 + 2ad + d^2. Then we can subtract the 2ad from both sides, and shuffle the leftovers till we have b^2 + c^2 and a^2 + d^2. See the first picture for the algebra.

Next, we should recognize b and c as the sides of the rightmost triangle and a and d as the sides of the leftmost. They have the same opposite side, but the rightmost triangle has a 92° angle, while the leftmost has an 88° angle. If the angle were 90° then we would of course know that b^2 + c^2 would be the same as e^2. But what does it mean when the angle is either bigger or smaller than 90°? Easiest to just draw a picture and exaggerate the difference to see what it means. See the 2nd picture. Doing so shows, without any math, that the two sides on either side of the smaller angle must be larger, so their squares must also be larger. Thus, B is the answer.


Thanks I reached a similar conclusion but Im still not convinced. Can we have another method?


This is a friend method. I am more convinced of his method but not 100% sure.
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Re: Quadrilateral sides comparison [#permalink] New post 08 Feb 2018, 14:31
The problem with the above method is that we cant really tell from the manipulation.

Look


- decrease + increase 0 no change, these are the subscripts meaning.
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Re: Quadrilateral sides comparison [#permalink] New post 09 Feb 2018, 02:45
Expert's post
Please guys, could you edit your answers ??

Even though the discussion is quite interesting and useful per se, it is unuseful for the other students.
Who is trying to search for explanations will not find them due to the screenshots.

Whenever you do wanna a discussion unfold, is a good way to write all about as text.

The same is true when you post a new one question: the rules of the board forbid the use of screenshots.

Please refer to this https://greprepclub.com/forum/rules-for ... -1083.html
and this https://greprepclub.com/forum/qq-how-to ... -2357.html

Next time a similar topic will be locked.

Thank you for your collaboration.

regards
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Re: Quadrilateral sides comparison [#permalink] New post 09 Feb 2018, 12:32
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@Carcass-
Sorry about the pictures.

@trunks14-
Here's a simple way to imagine it: for any right triangle, the two legs squared will equal the hypotenuse squared. But what if you kept the two legs the same length and squeezed down the hypotenuse to something smaller? Then we know that the two legs squared will now be larger than the new hypotenuse squared, and we also know that the angle must now be less than 90°.

Similarly, if we widen the two legs, the hypotenuse would now have to be larger, but the angle now exceeds 90°.

This is actually a general rule:
a^2 + b^2 < c^2 when the angle is less than 90°
and
a^2 + b^2 > c^2 when the angle is greater than 90°

So from these thought experiments we know that in the original question the two sides of the triangle squared on either side of the 88° angle must be larger than the two sides of the triangle squared on either side of the 92° angle. So again the answer is B.

A picture would've been much easier and less wordy but that's not allowed, haha.
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Re: Quadrilateral sides comparison [#permalink] New post 12 Feb 2018, 03:17
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No need to say sorry :)

It is just a matter to keep the board efficient.

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Re: Quadrilateral sides comparison [#permalink] New post 16 Feb 2018, 11:50
SherpaPrep wrote:
@Carcass-
Sorry about the pictures.

@trunks14-
Here's a simple way to imagine it: for any right triangle, the two legs squared will equal the hypotenuse squared. But what if you kept the two legs the same length and squeezed down the hypotenuse to something smaller? Then we know that the two legs squared will now be larger than the new hypotenuse squared, and we also know that the angle must now be less than 90°.

Similarly, if we widen the two legs, the hypotenuse would now have to be larger, but the angle now exceeds 90°.

This is actually a general rule:
a^2 + b^2 < c^2 when the angle is less than 90°
and
a^2 + b^2 > c^2 when the angle is greater than 90°

So from these thought experiments we know that in the original question the two sides of the triangle squared on either side of the 88° angle must be larger than the two sides of the triangle squared on either side of the 92° angle. So again the answer is B.

A picture would've been much easier and less wordy but that's not allowed, haha.



I think need correction

For a right triangle: a2+b2=c2a2+b2=c2.
For an acute (a triangle that has all angles less than 90°) triangle: a2+b2>c2
For an obtuse (a triangle that has an angle greater than 90°) triangle: a2+b2<c2
Re: Quadrilateral sides comparison   [#permalink] 16 Feb 2018, 11:50
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