Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 05 Jan 2006
Posts: 374

Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
Updated on: 08 Feb 2018, 05:29
Question Stats:
27% (02:07) correct 73% (02:32) wrong based on 564 sessions
HideShow timer Statistics
Jeff is painting two murals on the front of an old apartment building that he is renovating. One mural will cover the quadrilateral face ABCD while the other will cover the circular face (shown to the right, with radius XY). Assuming that the thickness of the coats of paint is negligible, will each mural require the same amount of paint? Note: Figures are not drawn to scale. (1) AB = BC = CD = DA and \(AB=XY\sqrt{\pi}\) (2) AC = BD and \(AC=XY\sqrt{2\pi}\) Attachment:
Untitled.png [ 4.82 KiB  Viewed 9095 times ]
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by chiragr on 28 Apr 2006, 20:56.
Last edited by Bunuel on 08 Feb 2018, 05:29, edited 5 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.



VP
Joined: 29 Apr 2003
Posts: 1373

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:22
I wud Say A
From I we know that the quad is a square.
Hence the Area = AB^2 == Area of the circle (pi*XY^2).
From B. You could have multiple quadirlaterals!
If the two diagnals are perpendicular, then the given equation could be used to determine (the quad will then be a square or a kite). But since nothing is provided abt their angle of intersection, we cannot assume that.
Hence, A remains!
Whats the OA?



VP
Joined: 21 Sep 2003
Posts: 1034
Location: USA

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:31
Is it C?
1. ABCD could be square or a rhombus.
If ABCD is a square, then area of Circle = area of ABCD
If ABCD is a rhombus with sat 6012060120 degress,
Area of ABCD < AB^2 = XY^*pi
INSUFF
2. AC = BD = XY*SQRT(2pi).
ABCD could be a rectangle or a square or a rhombus.
IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle
If ABCD is a square, both areas are equal.
Rhombus.. unequal.
Combine (1) & (2), ABCD has to be a square with the same area as that of the circle.
_________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
 Bernard Edmonds



Senior Manager
Joined: 05 Jan 2006
Posts: 374

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:33
sm176811 wrote: I wud Say A
From I we know that the quad is a square.
Hence the Area = AB^2 == Area of the circle (pi*XY^2).
From B. You could have multiple quadirlaterals!
If the two diagnals are perpendicular, then the given equation could be used to determine (the quad will then be a square or a kite). But since nothing is provided abt their angle of intersection, we cannot assume that.
Hence, A remains!
Whats the OA?
it's quadirlaterals not square... it can be rhombus!



Senior Manager
Joined: 05 Jan 2006
Posts: 374

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:34
giddi77 wrote: Is it C?
1. ABCD could be square or a rhombus. If ABCD is a square, then area of Circle = area of ABCD If ABCD is a rhombus with sat 6012060120 degress, Area of ABCD < AB^2 = XY^*pi INSUFF
2. AC = BD = XY*SQRT(2pi). ABCD could be a rectangle or a square or a rhombus. IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle If ABCD is a square, both areas are equal. Rhombus.. unequal.
Combine (1) & (2), ABCD has to be a square with the same area as that of the circle.
for 2 AC=BD can not be rhombus, rhombus has equal sides not diagonal!



VP
Joined: 29 Apr 2003
Posts: 1373

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:35
Agreed... It could be Rhombus or a square..
Hence C!



VP
Joined: 29 Apr 2003
Posts: 1373

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:36
B could also generate a trapezoid!



VP
Joined: 21 Sep 2003
Posts: 1034
Location: USA

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 21:36
chiragr wrote: for 2 AC=BD can not be rhombus, rhombus has equal sides not diagonal!
Oops... take out rhombus. Still it can be a rectangle or square. Hence B is INSUFF?
_________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
 Bernard Edmonds



Senior Manager
Joined: 05 Jan 2006
Posts: 374

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 22:02
C is the correct answer, the only reason I have this question is following explaination
"
Statement 2 tells us that the diagonals are equalthus telling us that ABCD has right angle corners (The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees.) Statement 2 also gives us a numerical relationship between the diagonal of ABCD and the radius of the circle. If we assume that ABCD is a square, this relationship would allow us to determine that the area of the square and the area of the circle are equal. However, once again, we cannot assume that ABCD is a square."
The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees.
I can not really visualize this! Can any one show me the light?



VP
Joined: 21 Sep 2003
Posts: 1034
Location: USA

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 22:34
chiragr wrote: The only way for a quadrilateral to have equal diagonals is if its corners are 90 degrees. I can not really visualize this! Can any one show me the light?
chiragrbuddy,
I think there is no light to show here.. There was a blackout in Manhattan when this question was created and explained by 99%tile students
They probably didn't know that a regular trapezium also has equal diagonals. If the diagonals are equal and also bisect each other, then the corners have to be 90 degrees.
Still it's a very good question. We should remember that Sides equal => the quadrilateral could be a nice Square or an ugly Rhombus!
_________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
 Bernard Edmonds



Senior Manager
Joined: 05 Jan 2006
Posts: 374

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
28 Apr 2006, 23:18
now I can sleep without nightmares



Senior Manager
Joined: 19 Oct 2012
Posts: 327
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35 GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
29 Aug 2016, 21:05
giddi77 wrote: Is it C?
1. ABCD could be square or a rhombus. If ABCD is a square, then area of Circle = area of ABCD If ABCD is a rhombus with sat 6012060120 degress, Area of ABCD < AB^2 = XY^*pi INSUFF
2. AC = BD = XY*SQRT(2pi). ABCD could be a rectangle or a square or a rhombus. IF ABCD is a rectangle, Area of ABCD is smaller than the area of the Circle If ABCD is a square, both areas are equal. Rhombus.. unequal.
Combine (1) & (2), ABCD has to be a square with the same area as that of the circle. If the diagonals are equal it cant be a rhombus. It would be either a square or a rectangle. Combining 1 and 2, we get C.
_________________
Citius, Altius, Fortius



Director
Joined: 26 Oct 2016
Posts: 642
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
01 Jan 2017, 04:09
St1  ABCD is a rhombus. If it were a square, the condition would have implied paint is equal coz area would be equal. However, we dont know that. All we know is its a rhombus, and the area of the rhombus could be anything relative to the circle depending on the ratio of its diagonal to its side > INSUFF St2  ABCD is a m. Again, a parallelogram has area = base x height. But we only know info about diagonals. In other words, for the same length of diagonals, you could have infinitely many areas of ms  so not enough info  INSUFF Both combined  we know sides of ABCD are equal and so are the diagonals. Ratio of diagonal to side is sqrt(2) > This means that ABCD is now a square. In this case area = piXY^2 = area of circle Thus answer is C
_________________
Thanks & Regards, Anaira Mitch



NonHuman User
Joined: 09 Sep 2013
Posts: 8496

Re: Jeff is painting two murals on the front of an old apartment
[#permalink]
Show Tags
08 Feb 2018, 05:27
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Jeff is painting two murals on the front of an old apartment &nbs
[#permalink]
08 Feb 2018, 05:27






