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Is this figure a square? (There was a drawing of a square [#permalink]
15 Sep 2004, 08:30

4. Is this figure a square? (There was a drawing of a square given with one angle shown as X degrees and the one opposite it on the same side shown as Y degrees)
1) All four sides are equal
2) X+Y=180 degrees

E it is. To prove a square:
1. Four sides should be equal and the diagonals should be equal.
OR
2. Four sides should be equal and 2 adjacent angles should be 90.

let me give some of the properties of quadrilaterals that are used.

1. a rhombus is a parallelogram
2. opposite angles of a parallegram are equal ( A and C, B and D)
3. adjacent angles of a parallelogram ar supplementary (< A +< B=180, B and C , C and D, D and A )
4. from 2 and 3 we can deduce that a rhombus with one right angle is a square

Going back to the question.

It says the angle opposite to X on same side. This phrase is vague . If he means that <A and < B for example, he should have chosen the word adjacent or relied on a figure. If he means this then the answer would be E since the 2nd information (X+Y=180) is a known property of a rhombus and solution is E.

For your suggestion of a rhombus with 90,60,90,130. This figure is not a quadrilateral since sum of angles is 370. Furthermore as I earlier said a rhombus with one right angle is a square.

Hope this gives a clear understanding of my argument. I do believe that the problem with a figure would have cleared everything.

Furthermore as I earlier said a rhombus with one right angle is a square.

My mistake. Agree that rhombus with atleast one 90 degrees is a square.
What I meant to say was, "A quadrilateral with sum of two opposite angles = 180 is not necessarilly a rectangle or square." The fact that the four sides are equal thows in a different light.

yes it is my undertanding that if all sides are equal then it must be a paralellogram--correct?

if this is so, then two OPPOSITE angles would be equal and if their total is 180 then they must be 90 degrees making the object a square. is there some flaw in this logic????

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