Author 
Message 
TAGS:

Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 538
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
07 Feb 2012, 16:32
21
This post was BOOKMARKED
Question Stats:
23% (02:26) correct
77% (01:43) wrong based on 646 sessions
HideShow timer Statistics
Attachment:
Untitled.png [ 4.89 KiB  Viewed 22693 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
Last edited by Bunuel on 22 Jul 2013, 06:13, edited 1 time in total.
Edited the question.



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4128

Re: Area of a Parallelogram [#permalink]
Show Tags
07 Feb 2012, 22:54
14
This post received KUDOS
Expert's post
10
This post was BOOKMARKED
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. http://gmat.magoosh.com/questions/1023Please let me know if you have any questions on what I've said here. Mike
_________________
Mike McGarry Magoosh Test Prep



Manager
Status: Employed
Joined: 17 Nov 2011
Posts: 99
Location: Pakistan
Concentration: International Business, Marketing
GPA: 3.2
WE: Business Development (Internet and New Media)

Re: Area of a Parallelogram [#permalink]
Show Tags
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. 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. http://gmat.magoosh.com/questions/1023Please 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?
_________________
"Nowadays, people know the price of everything, and the value of nothing." Oscar Wilde



Manager
Joined: 31 Jan 2012
Posts: 74

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
11 Feb 2012, 02:20
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.
Just my Personal opinion.



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4128

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
12 Feb 2012, 13:44
Dear omerraufI 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
_________________
Mike McGarry Magoosh Test Prep



Intern
Joined: 15 Apr 2010
Posts: 49

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
07 Dec 2012, 02:19
11
This post received KUDOS
1
This post was BOOKMARKED
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.



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4128

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
07 Dec 2012, 11:25
Dear CatennacioThat was a brilliant approach. Thank you for sharing that. Mike
_________________
Mike McGarry Magoosh Test Prep



Intern
Joined: 08 Sep 2014
Posts: 18

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
09 Nov 2014, 17:37
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.



Director
Joined: 23 Jan 2013
Posts: 585

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
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)



Senior Manager
Joined: 01 Nov 2013
Posts: 345
WE: General Management (Energy and Utilities)

In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
26 Mar 2015, 06:06
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. 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.
_________________
Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.
I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.Mohammad Ali



SVP
Joined: 08 Jul 2010
Posts: 1736
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
20 Jul 2016, 10:12
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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 Answer: option A
Attachments
File comment: www.GMATinsight.com
Untitled1.jpg [ 145.42 KiB  Viewed 5085 times ]
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Intern
Joined: 18 Jul 2016
Posts: 9

Re: In the picture, quadrilateral ABCD is a parallelogram and [#permalink]
Show Tags
21 Sep 2016, 12:39
Awesome explanation Gmat Insight!!! Thank you so much..




Re: In the picture, quadrilateral ABCD is a parallelogram and
[#permalink]
21 Sep 2016, 12:39







