Is it 8?

I just drew squares with diff orientations and got 8 squares that satisfy the above condition.

Since area must be 100 possible porduct are = 10 * 10 only
so a line must of length 10 since it is to be a square,

We get four square's with vertices on origin and square in each quadrant only .
The diagonal length is = 2* (10)^(1/2)
This is not an integer.
so diagonals wont be on axis with integral values.

So Answer Should A .




Please correct me if i'm wrong

Answer A.

I had the same reasoning as abkhan.

OA is E. Here's the explanation from MGMAT.

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 hyptoneuse 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 lable 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 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.

oh .. u mean the condition to be satissifed is if b is (x,y) then distance between them

x^2 +y^2=100 must be satisfied

An area of 100 means that if the square has a side "x"; x^2 = 100
Hence x = 10; hence the other point must be at a distance of 10 units from the origin;
Since the distance between any two points in the co-ordinate space = sqr rt [ (y1-y2)^2 + (x1-x2)^2]
10 = sqr rt. [ (0-y2)^2 + (0-x2)^2]
100 = (-y2)^2 + (-x2)^2
Since all co-ordinates of the vertices have to be integers there are only 4 such possibilities (0,10), (0,-10), (10,0), (-10,0)
Hence A

Oops I didn't see the 6,8 pairing coming up!!!