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Jeff is painting two murals on the front of an old apartment

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Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post Updated on: 08 Feb 2018, 05:29
2
31
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

30% (01:14) correct 70% (01:32) wrong based on 554 sessions

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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 \(AC=XY\sqrt{2\pi}\)


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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.
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 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?
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 28 Apr 2006, 21:31
4
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 60-120-60-120 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.
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 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!
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New post 28 Apr 2006, 21:34
1
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 60-120-60-120 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!
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 28 Apr 2006, 21:35
Agreed... It could be Rhombus or a square..

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

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New post 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 equal--thus 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? :)
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 28 Apr 2006, 22:34
1
1
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? :)


chiragr-buddy,
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 :P
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!
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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

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New post 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 60-120-60-120 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.
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Re: Jeff is painting two murals on the front of an old apartment  [#permalink]

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New post 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
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