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In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
07 Feb 2012, 15:32

5

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

24% (02:23) correct
76% (01:40) wrong based on 211 sessions

Attachment:

Untitled.png [ 4.89 KiB | Viewed 5837 times ]

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.

Re: Area of a Parallelogram [#permalink]
07 Feb 2012, 21:54

11

This post received KUDOS

Expert's post

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.

Answer = A.

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.

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
07 Dec 2012, 01:19

9

This post received KUDOS

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.

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
11 Feb 2012, 01:20

1

This post received KUDOS

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.

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
12 Feb 2012, 12:44

1

This post received KUDOS

Expert's post

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.

Re: Area of a Parallelogram [#permalink]
10 Feb 2012, 22:58

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.

Answer = A.

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.

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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