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Jeff is painting two murals on the front of an old apartment [#permalink]
28 Apr 2006, 19:56

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

23% (02:01) correct
77% (01:51) wrong based on 88 sessions

Attachment:

Untitled.png [ 4.82 KiB | Viewed 1062 times ]

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.

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.

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. _________________

"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."

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!

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!

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."

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?

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
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."

Re: Jeff is painting two murals on the front of an old apartment [#permalink]
27 Apr 2014, 10:08

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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).

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