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# In the picture, quadrilateral ABCD is a parallelogram and

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Updated on: 22 Jul 2013, 06:13
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Difficulty:

95% (hard)

Question Stats:

17% (02:12) correct 83% (02:14) wrong based on 918 sessions

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In the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle. What is the area of parallelogram ABCD (figure not drawn to scale)?

(1) The area of rectangle DEFG is 8√5.
(2) Line AH, the altitude of parallelogram ABCD, is 5.

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Originally posted by enigma123 on 07 Feb 2012, 16:32.
Last edited by Bunuel on 22 Jul 2013, 06:13, edited 1 time in total.
Edited the question.
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Re: Area of a Parallelogram  [#permalink]

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07 Feb 2012, 22:54
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Hi, there. I'm happy to help with this.

As a geometry geek myself, I found this a very cool geometry problem, but I will say --- it is WAY harder than anything you would be expected to figure out for yourself on the real GMAT.

Statement #1: The area of rectangle DEFG is 8√5.

Well, to cut to the chase, this statement is sufficient because the rectangle and the parallelogram must have equal area. Why do the rectangle and parallelogram have equal area? You will see the complete geometric argument in the pdf attachment to this post. Leaving those details aside for the moment, Statement #1 is sufficient.

Statement #2: Line AH, the altitude of parallelogram ABCD, is 5.
Area of a parallelogram = (base)*(altitude). If we know the altitude and not the base, that's not enough. Therefore, Statement #2 is insufficient.

Does all this (including everything in the pdf) make sense?

Here's another geometry DS, a little closer to the actual level of difficulty of the GMAT itself.

http://gmat.magoosh.com/questions/1023

Please let me know if you have any questions on what I've said here.

Mike
Attachments

rectangle & parallelogram with equal area.pdf [205.82 KiB]

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07 Dec 2012, 02:19
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Hi, mikemcgarry's is good but it uses similar triangles to prove. I think it's doesn't need to be that complicated. I use the same diagram that mikemcgarry provided.

First, we all agree that by considering DC as base and EQ as height, Area DEC = 1/2 * EQ * DC (1). It also equals 1/2 Area ABCD (area of parallelogram is base * height). This is just normal formula, no problem.

The tricky part is how to link it with the rectangle DEFG. Now, from C, draw a line CP that is perpendicular with DE with P is on DE. Now, for triangle DEC, consider ED as base and CP as height, we have Area of DEC = 1/2 CP * DE (2)

From (1) and (2), the 2 area is the same, we have EQ * DC = CP * DE (3). But in rectangle DEFG, CP = EF (since DEFG is rectangle, CP perpendicular with DE, so CP must = EF)

So (3) can be rewritten as EQ * DC = EF * DE. LHS is area of ABCD. RHS is area of DEFG. So (1) Suff.

(2) obviously NS.

So A is correct. I must admit I couldn't get this right, but after reading the explanation of mikemcgarry, I think this way is simpler as you don't have to think and prove similars. You just need to substitute side for side.
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Re: Area of a Parallelogram  [#permalink]

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10 Feb 2012, 23:58
mikemcgarry wrote:
Hi, there. I'm happy to help with this.

As a geometry geek myself, I found this a very cool geometry problem, but I will say --- it is WAY harder than anything you would be expected to figure out for yourself on the real GMAT.

Statement #1: The area of rectangle DEFG is 8√5.

Well, to cut to the chase, this statement is sufficient because the rectangle and the parallelogram must have equal area. Why do the rectangle and parallelogram have equal area? You will see the complete geometric argument in the pdf attachment to this post. Leaving those details aside for the moment, Statement #1 is sufficient.

Statement #2: Line AH, the altitude of parallelogram ABCD, is 5.
Area of a parallelogram = (base)*(altitude). If we know the altitude and not the base, that's not enough. Therefore, Statement #2 is insufficient.

Does all this (including everything in the pdf) make sense?

Here's another geometry DS, a little closer to the actual level of difficulty of the GMAT itself.

http://gmat.magoosh.com/questions/1023

Please let me know if you have any questions on what I've said here.

Mike

Dear Mike.. What is the likelihood of such a question on the GMAT. The more I see Kaplan questions, the more I feel the questions can be extremely hard. Whereas the questions on GMATPREP seem to be much simpler than this, No?
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11 Feb 2012, 02:20
2
I personal think it would be on GMAT, but will be a 700 or 800 question. Calculation is straight forward. The only thing you need to recognize is that they both share the same triangle and if a triangle has the same height and width as a parallelogram thats not a trapezoid; then the triangle will always be 1/2 the area of the parallelogram. This is due to the simple mathematical equation to calculate the both of them.

Just my Personal opinion.
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12 Feb 2012, 13:44
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Dear omerrauf

I would say a question like this ---- a question that hinges on a relatively obscure geometry theorem that one probably would have to prove from scratch to answer the question ---- is something far harder than what they would put on the GMAT. Any GMAT math question, no matter how challenging, is something that someone facile with math would be able to solve in under a minute. If you've never seen this theorem, there's virtually no way that you will derive the full geometry proof in under a minute, unless you operate at Isaac Newton level. The GMAT doesn't expect that, even on 800 level questions. You don't have to have be Isaac Newton to answer the hardest questions.

That's my take on it. I am not as familiar with Kaplan questions overall, I am not qualified to make a statement about them. I know that Magoosh has a few hundred math questions, all appropriate difficulty for the GMAT, and each followed but its own video solution. The link above will give you a sample.

Mike
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07 Dec 2012, 11:25
Dear Catennacio
That was a brilliant approach. Thank you for sharing that.
Mike
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09 Nov 2014, 17:37
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enigma123 wrote:
Attachment:
Untitled.png
In the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle. What is the area of parallelogram ABCD (figure not drawn to scale)?

(1) The area of rectangle DEFG is 8√5.
(2) Line AH, the altitude of parallelogram ABCD, is 5.

Just providing my 2 cents on the problem...

Theorem: Triangles between two parallel lines with same base have equal areas.

Even if you're not familiar with the above theorem, it's pretty intuitive from the area formula of the triangle.

In the figure, join EC.

Then from the above theorem, it's clear that area (tri DEC) = area (tri DBC) = $$\frac{1}{2}$$* area (ABCD) ---- (*)

Similarly, area (tri DEC) = area (tri DEF) = $$\frac{1}{2}$$* area (DEFG) ---- (**)

So, from (*) and (**), area (ABCD) = area (DEFG)

Clearly, (1) is sufficient and (2) is not, so answer is A.

This approach takes less than 2 minutes to solve. I think it is quite possible that similar questions are likely to be seen in GMAT at 700 level.
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25 Mar 2015, 23:08
Got it intuitevly + elimination in 2 mins.

Stat.2 is clearly insuff, so eliminate B and D

Stat.1. Rectangle is a part of parallelogram or may be even equal

A,
sorry for my absence of discipline)
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Updated on: 19 Oct 2018, 06:34
1
enigma123 wrote:
Attachment:
The attachment Untitled.png is no longer available
In the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle. What is the area of parallelogram ABCD (figure not drawn to scale)?

(1) The area of rectangle DEFG is 8√5.
(2) Line AH, the altitude of parallelogram ABCD, is 5.

hi guys

here is another solution to this hard problem based on a different approach.

i hope you find the solution interesting and easy.

the diagram in the pdf is self explanatory.
Attachment:
Samichange.pdf [141.14 KiB]

Once we understand the diagram, the solution looks so easy.

Took me 15 minutes to figure out the approach.

Press kudos if you like the solution.

Attachments

Samusa_simple solution.jpg [ 98.66 KiB | Viewed 4263 times ]

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Originally posted by samusa on 26 Mar 2015, 06:06.
Last edited by samusa on 19 Oct 2018, 06:34, edited 1 time in total.
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20 Jul 2016, 10:12
4
1
enigma123 wrote:
Attachment:
The attachment Untitled.png is no longer available
In the picture, quadrilateral ABCD is a parallelogram and quadrilateral DEFG is a rectangle. What is the area of parallelogram ABCD (figure not drawn to scale)?

(1) The area of rectangle DEFG is 8√5.
(2) Line AH, the altitude of parallelogram ABCD, is 5.

Please find the attached file for explanation

Attachments

File comment: www.GMATinsight.com

Untitled1.jpg [ 145.42 KiB | Viewed 18626 times ]

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21 Sep 2016, 12:39
Awesome explanation Gmat Insight!!! Thank you so much..
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14 Nov 2017, 19:16
Somehow I came to know that in any parallelogram, if a triangle is formed joining two vertexes of the parallelogram and a point of opposite side of the vertexes, then the area of resulting triangle is half of the area of the parallelogram. Need confirmation from expert.

Draw a line EC.

Now in DEFG rectangle, Area of DEC = Area of $$\frac{1}{2}$$ DEFG
Again in ABCD parallelogram, Area of DEC = Area of $$\frac{1}{2}$$ ABCD

∴ Area of $$\frac{1}{2}$$ DEFG = Area of $$\frac{1}{2}$$ ABCD => Area of DEFG = Area of ABCD.

So, A is sufficient.
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29 Dec 2018, 04:44
Re: In the picture, quadrilateral ABCD is a parallelogram and   [#permalink] 29 Dec 2018, 04:44
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