(E)

One of the vertices of the square is fixed. Think of it as a pivot. One of the points that comes easily is (0,10) since 10 should be the length of one side ( Area being 100). Now let the point (0,0 ) act as a hinge and sweep the other parallel vertex across the quadrands.
Now, since one side is 10 (use that as hypotenuse) and find the other sides of the right triangle - 8 and 6. Now you can find the combination of the points as follows:
(0,10), (0,-10), (10,0), (-10, 0),
(8,6), (-8,6), (-8,-6), (8,-6),
(6,8), (-6,8), (-6,-8), (6,-8)

Total is 12.
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My solution: 100 = 10x10 = 2x5x2x5. There are 3 ways to arrange 2x5x2x5 into two integers. Then, multiply by 4, because the square can be in any of the 4 quadrants.

Also, this is a good question for guessing. The answer must be divisible by 4, and answers A and C assume only 1 and 2 ways, respectively, in which the square can be drawn in a given quadrant. So E is left.
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When the top left vertex moved to (6,8), did you check if the diagonal vertex coordinates are integer or not? i still feel there are only 4 solutions.

This is previously posted.

For co-ordinates to be integers, possible co-ordinates for one of the vertices are (6,8), (8,6), (10,0) in one quadrant. Multiply this number of three different co-ordinates with four quadrants....and hence the answer should be 12.

Distance between (0,0) and (6,8) or (8,6) will be 10.

Look for the clue: If all coordinates of the vertices must be integers.

How is a squre with area 100 possible? If its sides are each 10. How is a side with 10 is possible subject to the coordinates that make the the side 10? Its possible only with 6 abd 8 coordinates. So possibilities are:

first qd:
i: 10 and 0
ii: 8 and 6
ii: 6 and 8

Second qd:
i: 0 and 10
ii: -6 and 8
ii: -8 and 6

similarly there are 3 more in each of 3rd and 4th quadrant.

so altogather 12 possibilities.
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The answer is E.

Asking:

Number of squares in the xy-coordinate plane with the following criteria.
1. Total area of the square = 100 i.e. each side = 10
2. One of the vertices must be on the origin
3. All coordinates of the vertices must be integers

The first four are quite obvious. Four squares, one in each quadrant, with x and y axis as the two sides with the following vertices:

1. (0,0) (10,0) (0,10) (10,10)
2 (0,0) (-10,0) (0,10) (-10,10)
3. (0,0) (-10,0) (0,-10) (-10,-10)
4. (0,0) (10,0) (0,-10) (10,-10)

Now lets see which the remaining 8 are.

As each side is 10, if you rotate the side of the first square that falls on (0,0) and (10,0) in the counter clockwise direction with the vertex (0,0) fixed, the whole square will be rotated in the same direction.

Now if we think of this side as the hypotenuse of the right angle triangle that can be created by drawing a perpendicular from the rotating vertex to the x-axis then there are two points in the first quadrant where the rotating vertex has xy co-ordinates that are integers and they can be calculated using the Pythagorean formula.

x^2+y^2=10^2

the two values for x and y where x,y are integers and x^2+y^2=10^2 are (8,6) and (6,8)

8^2+6^2=10^2

6^2+8^2=10^2

Two additional squares can be created with the vertices for the bases as (0,0) (8,6) & (0,0) (6,8) in the first quadrant.

Similarly if you continue rotating the base counter-clockwise you can create two additional squares in quadrant II [bases (0,0) (-6,8) and (0,0) (-8,6)], two more in quadrant III [bases (0,0) (-6,-8) and (0,0) (-8,-6)] and the last two in quadrant IV [bases (0,0) (6,-8) and (0,0) (8,-6)] for a total of 8 additional squares.

Total squares = 4 + 8 = 12

i would like to know too... I tried to draw it out besides the 4 and i could draw some diamonds on the 4 directions but to get a side of 10 on the square in the 1-1-sqrt(2) format the x and y need to be 5*sqrt(2) which isnt integer... my guess would be B...
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i say 10 because i can draw 8 shapes and there's prob 2 special ones...
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Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have the following coordinates:

(-10,0)
(-8,6)
(-6,8)
(0,10)
(6,8)
(8,6)
(10,0)
(8, -6)
(6, -8)
(0, 10)
(-6, -8)
(-8, -6)

There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

The correct answer is E.
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