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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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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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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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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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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]
Agreed... It could be Rhombus or a square..

Hence C!
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
B could also generate a trapezoid!
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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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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]
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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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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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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Jeff is painting two murals on the front of an old apartment [#permalink]
Dear Bunuel VeritasKarishma GMATGuruNY EMPOWERgmatRichC,

What are the possible quadrilaterals according to Statement 2?

Are these the possibilities for Statement 2?
1. rectangle 2. isosceles trapezoid 3. kite 4. irregular quadrilateral 5. square

If possible, could you provide the full solution?
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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varotkorn wrote:
Dear Bunuel VeritasKarishma GMATGuruNY EMPOWERgmatRichC,

What are the possible quadrilaterals according to Statement 2?

Are these the possibilities for Statement 2?
1. rectangle 2. isosceles trapezoid 3. kite 4. irregular quadrilateral 5. square

If possible, could you provide the full solution?


With 2 equal diagonals, you can make a whole lot of different quadrilaterals. The most generic one will be an equidiagonal quadrilateral and the most specific will be a square.
Read about equidiagonal quads here: https://en.wikipedia.org/wiki/Equidiago ... drilateral

The square will have same area as circle while other figures will not so this statement alone will not be sufficient.
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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ST(1) : AB = BC = CD = DA and AB = XY√π
CASE I : let ,ABCD is a square, area of ABCD =(AB)^2 , area of Circle = π(XY)^2 =(AB)^2 ; =>EQUAL
CASE II : let ,ABCD is a rhombus ,
since area of rhombus < area of an square of same side
Area of ABCD ≠ area of Circle
so, ST (1) insufficient .

ST(2) : AC = BD and AC = XY√2π
AC =BD i.e. diagonals are equal .
CASE I : let ,ABCD is a square, area of ABCD =(AB)^2 =(AC/√2)^2=(AC)^2/2, area of Circle = π(XY)^2 =(AC)^2/2 ; =>EQUAL
CASE II : let ,ABCD is a rectangle ,
since area of rectangle < area of an square of same diagonal [ suppose , d= 5 , area of square = (5/√2)^2= 12.5 whereas area of rectangle = 3*4 = 12 ( 5^2 = 3^2+4^2) ]
Area of ABCD ≠ area of Circle
so, ST (2) insufficient .

Combine (1) & (2), ABCD has to be a square ( all sides are equal and diagonals are equal )
Area of ABCD = area of Circle. [b] sufficient[/b]

correct answer is C
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
Can a question of this difficulty level appear in GMAT?
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Re: Jeff is painting two murals on the front of an old apartment [#permalink]
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chiragr wrote:

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


We need to answer the question:

Is area_ABCD = pi(XY)^2 ?

Statement One Alone:

=> AB = BC = CD = DA and AB = XY√pi

Squaring the second equation, we have:

AB^2 = pi(XY)^2

Using the above information, we can rephrase the question:

Is area_ABCD = AB^2 ?

Since AB = BC = CD = DA, the sides of the quadrilateral are equal.

If quadrilateral ABCD is a square, then the answer is Yes. Whereas, if ABCD is a rhombus but not a square, then the answer is No since area_ABCD would be less than AB^2. [Think of the area formula for a rhombus, area = base × height. If the interior angles on the base of the rhombus are not right angles, then the height of the rhombus is less than its side length.]

Statement one is not sufficient. Eliminate answer choices A and D.

Statement Two Alone:

=> AC = BD and AC = XY√(2pi)

From the second equation, we have:

AC^2 = 2pi(XY)^2

(AC^2)/2 = pi(XY)^2

Using the above information, we can rephrase the question:

Is area_ABCD = (AC^2)/2 ?

Since AC = BD, the diagonals of the quadrilateral are equal.

If quadrilateral ABCD is a square, then the answer is Yes since a square’s area can be determined as (diagonal^2)/2. Whereas, if ABCD is a rectangle but not a square, then the answer is No since area_ABCD would be less than (AC^2)/2. [If a diagonal is given, then the square has the greatest area among all rectangles with the same diagonal.]

Statement two is not sufficient. Eliminate answer choice B.

Statements One And Two Together:

Since the sides of the quadrilateral are equal and the diagonals of the quadrilateral are equal, quadrilateral ABCD must be a square. In this case, we have a definite Yes answer to the rephrased questions.

The two statements together are sufficient.

Answer: C
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