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# If ABCD is a quadrilateral, is AB=BC=CD=DA ?

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Nice. Can figure also be an isosceles trapezoid?

For statement 1 we have that diagonals are perpendicular

For statement 2 we have that the sum of the bases equal to sum of the legs

Therefore E

Cheers!
J
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Bunuel wrote:
tt11234 wrote:
If ABCD is a quadrilateral, is AB=BC=CD=DA ?

1.AC is perpendicular to BD
2.AB+CD=BC+DA

If ABCD is a quadrilateral, is AB=BC=CD=DA ?

(1) AC is perpendicular to BD. The diagonals are perpendicular to each other: ABCD could be a kite (answer NO), a rhombus (answer YES) or a square, which is just a special type of rhombus (answer YES). Not sufficient.

(2) AB+CD=BC+DA. The sum of opposite sides are equal. Clearly insufficient.

(1)+(2) ABCD could be a kite (see the diagram below) - answer NO or a square/rhombus - answer YES. Not sufficient.

Hi Bunuel,

Thanks.
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Bunuel wrote:
tt11234 wrote:
If ABCD is a quadrilateral, is AB=BC=CD=DA ?

1.AC is perpendicular to BD
2.AB+CD=BC+DA

If ABCD is a quadrilateral, is AB=BC=CD=DA ?

(1) AC is perpendicular to BD. The diagonals are perpendicular to each other: ABCD could be a kite (answer NO), a rhombus (answer YES) or a square, which is just a special type of rhombus (answer YES). Not sufficient.

(2) AB+CD=BC+DA. The sum of opposite sides are equal. Clearly insufficient.

(1)+(2) ABCD could be a kite (see the diagram below) - answer NO or a square/rhombus - answer YES. Not sufficient.

Hi Bunuel,

Thanks.

The solution says that ABCD can be a kite.
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damn, please mention that the diagonals are perpendicular. I was thinking, how the hell can 2 different sides be perpendicular to each other? :D
1 - well, clearly insufficient.
2 - AB+CD=BC+DA - it might be a rectangle or a square, in which all sides are equal. not sufficient.

1+2 - it might be a square, or a quadrilateral consisting of 2 triangles, 2 small and 2 big right triangles 30-60-90, with hypotheses equal to each other. so insufficient.
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mvictor wrote:
damn, please mention that the diagonals are perpendicular. I was thinking, how the hell can 2 different sides be perpendicular to each other? :D
1 - well, clearly insufficient.
2 - AB+CD=BC+DA - it might be a rectangle or a square, in which all sides are equal. not sufficient.

1+2 - it might be a square, or a quadrilateral consisting of 2 triangles, 2 small and 2 big right triangles 30-60-90, with hypotheses equal to each other. so insufficient.

It is given that ABCD is a quadrilateral and as AC is perpendicular to BD, it is implied that diagonals are perpendiculars. This is the usual interpretation. Why do you need it to be mentioned explicitly that "diagonals are perpendicular"?
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Engr2012 wrote:
mvictor wrote:
damn, please mention that the diagonals are perpendicular. I was thinking, how the hell can 2 different sides be perpendicular to each other? :D
1 - well, clearly insufficient.
2 - AB+CD=BC+DA - it might be a rectangle or a square, in which all sides are equal. not sufficient.

1+2 - it might be a square, or a quadrilateral consisting of 2 triangles, 2 small and 2 big right triangles 30-60-90, with hypotheses equal to each other. so insufficient.

It is given that ABCD is a quadrilateral and as AC is perpendicular to BD, it is implied that diagonals are perpendiculars. This is the usual interpretation. Why do you need it to be mentioned explicitly that "diagonals are perpendicular"?

IDK, I just spent additional seconds trying to figure out wtf is actually meant
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If ABCD is a quadrilateral, is AB=BC=CD=DA ?

(1) AC is perpendicular to BD
(2) AB+CD=BC+DA

There are 5 variables (4 sides, 1 diagonal) in a quadrilateral, so we need 5 equations in order to solve the question, but only 2 equations are given from the 2 conditions, so there is high chance (E) will be our answer.
Even if we combine the 2 conditions together, they are insufficient, so (E) becomes the answer.

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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The solution says that ABCD can be a kite.[/quote]

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tt11234 wrote:
If ABCD is a quadrilateral, is AB=BC=CD=DA ?

(1) AC is perpendicular to BD
(2) AB+CD=BC+DA

There's simply not enough because two diagonals being perpendicular does not mean the same thing as those diagonals being perpendicular bisectors. Bunuel am I right? Because if all we know is that two diagonals are perpendicular then the quadrilateral could be a kite, which graphically makes a lot of sense. However, if we know that the two diagonals bisect eachother, or more precisely are perpendicular bisectors, then we know for sure that that quadrilateral is a rhombus because the definition of a perpendicular bisector simply means cutting a quadrilateral into two equal pieces at 90 degrees. And that furthermore every square is a rhombus but not all rhombuses are squares?
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Nunuboy1994 wrote:
tt11234 wrote:
If ABCD is a quadrilateral, is AB=BC=CD=DA ?

(1) AC is perpendicular to BD
(2) AB+CD=BC+DA

There's simply not enough because two diagonals being perpendicular does not mean the same thing as those diagonals being perpendicular bisectors. Bunuel am I right? Because if all we know is that two diagonals are perpendicular then the quadrilateral could be a kite, which graphically makes a lot of sense. However, if we know that the two diagonals bisect eachother, or more precisely are perpendicular bisectors, then we know for sure that that quadrilateral is a rhombus because the definition of a perpendicular bisector simply means cutting a quadrilateral into two equal pieces at 90 degrees. And that furthermore every square is a rhombus but not all rhombuses are squares?

A rhombus is a quadrilateral with all sides equal in length.
A square is a quadrilateral with all sides equal in length and all interior angles right angles.

Thus a rhombus is not a square unless the angles are all right angles.
A square however is a rhombus since all four of its sides are of the same length.
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