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Bunuel
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Bunuel
A parallelogram is drawn in the xy-coordinate plane. Is the parallelogram's area, in square units, an integer?

(1) The x- and y-coordinates of each of the parallelogram's four vertices are integers.

(2) The parallelogram has four sides of equal length.


This question is a part of Are You Up For the Challenge: 700 Level Questions collection.
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It's a very good question, I would like to add my approach:

The area of parallelogram is a * h, where a is base, and h is height.

(1) All coordinates are integers
Since we know that all coordinates are integers, it means that a will be an integer (e.g., x4 - x1 will be an integer), and h will be an integer as well. It gives us multiplication of two integers, and hence - an integer.

(2) is insufficient, since all sides could be equal to a non-integer, which will give a non-integer value. At the same time, we could have a figure will all integer coordinates and equal sides.

Hope it helps someone.
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yashikaaggarwal
We need to prove the area of the parallelogram is an integral value.

1) vertices of the parallelogram on the (x, y) plane is in the integral form.
The area will be in integral form as well.
(Sufficient)

2) the equality of vertices of parallelogram doesn't make its area in integral value.
(Insufficient)

Answer is A

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Hi yashikaaggarwal

Could you please elaborate as to why area of a ||gm will be an integer if the coordinates of its vertices are integers? Whether any theorem exists in this respect ?
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my answer is E. height can be either integer or non-integer.
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If the vertices are integers, the distance(side) can be a square root as well. If it is a rectangle then, there is a possibility that the area won't be an integer. Please help!!


Daspodu
my answer is E. height can be either integer or non-integer.
yashikaaggarwal
We need to prove the area of the parallelogram is an integral value.

1) vertices of the parallelogram on the (x, y) plane is in the integral form.
The area will be in integral form as well.
(Sufficient)

2) the equality of vertices of parallelogram doesn't make its area in integral value.
(Insufficient)

Answer is A

Posted from my mobile device
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on applying distance formula, isn't there a possibility of getting either a root or non root even when all the values are integral and if it is a root it will be in decimals otherwise not

According it can give both integral and non-integral value and hence E
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andrenoble
It's a very good question, I would like to add my approach:

The area of parallelogram is a * h, where a is base, and h is height.

(1) All coordinates are integers
Since we know that all coordinates are integers, it means that a will be an integer (e.g., x4 - x1 will be an integer), and h will be an integer as well. It gives us multiplication of two integers, and hence - an integer.

(2) is insufficient, since all sides could be equal to a non-integer, which will give a non-integer value. At the same time, we could have a figure will all integer coordinates and equal sides.

Hope it helps someone.

For statement 1, the case that you've considered holds only if one of the sides is aligned to one of the axes, i.e. two of the sides are || to x-axis or two of the sides are || to y-axis.

If none of the sides are || to any of the axes, the dimensions of base and height would have to be calculated using distance formula.

Now even if we use integral values in the distance formula, we may get roots, i.e. non-integral values. So how do we eliminate this second case?
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can somebody post official explanation of this question.

Thanks
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Bunuel, Could you please post the solution as well?
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