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Bunuel
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1: The quadrilateral can either be a square or a quadrilateral with sides X, X, X root 2, and 2X. In the first case, the perimeter is 4X -> X is 6, so area is 36, and in the second case, the perimeter is 4X + X root 2 -> the area is not 36. So 1 is not sufficient.

2: All angles are equal -> quadrilateral is a rectangle, and the width is 25% of the length, so the sides are X, X, .25X and .25X -> perimeter is 2.5X and so the area is calculate-able.

Therefore, 2 is sufficient, so the answer is B.
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Quick approach

Statement 1: not possible as 24 needs to be divided into 2 pairs of equal numbers. Multiple possibilities. So out.

Statement 2: sufficient
concept - sum of interior angles of a convex quadrilateral is always 360 degrees (assuming we have only convex polygons in gmat)
Therefore quadrilateral is a rectangle. Its not a square as the relation between length and breadth is given. So solvable.
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Quick approach

Statement 1: not possible as 24 needs to be divided into 2 pairs of equal numbers. Multiple possibilities. So out.

Statement 2: sufficient
concept - sum of interior angles of a convex quadrilateral is always 360 degrees (assuming we have only convex polygons in gmat)
Therefore quadrilateral is a rectangle. Its not a square as the relation between length and breadth is given. So solvable.

This question is badly written.

1. How can perimeter be in square centimeters ?
2. Answers are contradictory through both the statements! A rectangle and a square. How is that possible ?
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anewbeginning
Jackal
Quick approach

Statement 1: not possible as 24 needs to be divided into 2 pairs of equal numbers. Multiple possibilities. So out.

Statement 2: sufficient
concept - sum of interior angles of a convex quadrilateral is always 360 degrees (assuming we have only convex polygons in gmat)
Therefore quadrilateral is a rectangle. Its not a square as the relation between length and breadth is given. So solvable.

This question is badly written.

1. How can perimeter be in square centimeters ?
2. Answers are contradictory through both the statements! A rectangle and a square. How is that possible ?

totally agree on the framing of the question
1. Ignore square part in the units :)
2. But Square is a rectangle with all sides equal. please refer gmatclub math book.
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Jackal
anewbeginning
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Quick approach

Statement 1: not possible as 24 needs to be divided into 2 pairs of equal numbers. Multiple possibilities. So out.

Statement 2: sufficient
concept - sum of interior angles of a convex quadrilateral is always 360 degrees (assuming we have only convex polygons in gmat)
Therefore quadrilateral is a rectangle. Its not a square as the relation between length and breadth is given. So solvable.

This question is badly written.

1. How can perimeter be in square centimeters ?
2. Answers are contradictory through both the statements! A rectangle and a square. How is that possible ?

totally agree on the framing of the question
1. Ignore square part in the units :)
2. But Square is a rectangle with all sides equal. please refer gmatclub math book.

According to me the answer should be D.

You can easily see from statement 1 that it is a square.

a + b + c + d= 24
4a =6
a=1.5

Area = 2.25 square centimeters

Where as the second statement yields a different value of the area.

Please check. I was saying it in that terms :-D
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Quote:


According to me the answer should be D.

You can easily see from statement 1 that it is a square.

a + b + c + d= 24
4a =6
a=1.5

Area = 2.25 square centimeters

Where as the second statement yields a different value of the area.

Please check. I was saying it in that terms :-D


Hello. Good evening.

The statement 1 can be a square but always is not a square. Please refer attached diagram. As you rightly said this is not a well written problem. But in the Gmat world I think this is a smart problem.
Attachments

Untitledmmm.png
Untitledmmm.png [ 32.92 KiB | Viewed 8491 times ]

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Jackal
Quote:


According to me the answer should be D.

You can easily see from statement 1 that it is a square.

a + b + c + d= 24
4a =6
a=1.5

Area = 2.25 square centimeters

Where as the second statement yields a different value of the area.

Please check. I was saying it in that terms :-D


Hello. Good evening.

The statement 1 can be a square but always is not a square. Please refer attached diagram. As you rightly said this is not a well written problem. But in the Gmat world I think this is a smart problem.

Oh Yes. Thanks a lot. :)

I tried forming the quadrilateral only by joining the two hypotenuses. But you are right in saying that it can be a square and cannot be a square. Well, you illustrated your point with a diagram that clears my doubt by joining the other triangle through the base of the first triangle.

Yes, this is a smart problem and well an add on to my error log.

Thank You Bunuel for the question :-D
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What is the area of a quadrilateral with a perimeter of 24 square centimeters?

Stmt (1) : The quadrilateral is formed by combining two isosceles right triangles

if both the triangles have same sides say (1,1,2) & (1,1,2) then it can be a rectangle.
but if one triangle is (1,1,2) & (2, 2, 2\(\sqrt{2}\)) , then we can make one diagonal as 2 (2 as common side) and make a quadrilateral.

two different shapes Not sufficient

Stmt (2) : All angles are of equal measure and the width of the quadrilateral is 25% of the length
let angle of quadrilateral = a , then 4a = 360 , a = 90 (square or rectangle) , but length/width is mentioned so rectangle
now if length = x ; width = 25% of length = x/4

x + 0.25x = 24 ; can find length/width & area sufficient

Ans B
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