Square with area 100, so side has to be 10

Look at it this way:
Square is placed initially at A(0,0),B(10,0),C(10,10),D(0,10).

Now you have to move this sqaure say anticlockwise with the vertex at origin fixed. But it has to be moved such that for the vertex B(x,y) x^2+y^2=100 and also x and y must be integers.
Now the only possibilities for B are (0,10), (6,8) and (8,6) and this can be repeated in all the remaining 3 quadrants.
so its 3*4 =12

perhaps you are getting 16 b'cos you are counting one of the positions for B in each quadrant twice.

HTH

Since the figure is a square and area of a square is (side)^2 = 100
we need to look for integer combinations, squares of which sum up to 100

Listing possibilities:
1] 10 & 0 Since 10^2 + 0 =100
2] 8 & 6 Since 64 + 36 = 100
3] 9 & 3 Since 81 + 9 = 100

Since it is given that one of the vertices at (0,0) and the fact that it is a sqaure that we need to plot,
we know the following:
1]The co-ordinates of 2 end-points of the sides extended from (0,0) will be either of the format (0,integer) or (integer,0).
2] Hence their lengths will be the square of the integer value.
3] If we plot 2 points then the 3rd point has no choice but to be fixed a value and thus we need not worry about the possibilities of this 3rd point.

e.g. If we choose the 1st option above one of our points will be (0,10) and the other will be (10,0) and hence the 3rd point has to be (10,10).
So my possibilities with this pair are (10&10), (-10&10), (10&-10), (-10&-10), fixing the square in each of the 4 quadrants.

Hence with the listed possibilities in mind, since each has 2 values and each value can take 2 signs (either + or -), each pair has 4 possibilites.
And since there are 3 such pairs total possibilities are 3*4 = 12.

why can'y one side be 100 and the other side 1 ?

because then it is not a square - it is a rectangle.

Mmmm , right .
Thanks.

Sidon