5 x 5 array of lattice points.

I am not sure what do you mean by diagonal lines.
As far as I understand this is a 5X5 matrix.

Diagonal lines will essentially require additional constructions,not mentioned here otherwise.

Solution:

For 3X3 array we have in all 5 squares (when joined vertically or horizontally) such that we have 2^2 squares of side 1X1, and 1 square of size 2X2.
But when mid-points of the 2X2 square is joined, we have another square whose all vertices are on the perimeter of the array of 3X3.
Please note that both the 2X2 square have all the vertices on the perimeter of 3X3 array. Thus, in all we have 4 + 2 = 6 such squares for 3X3.

For 4X4 array we have 9 squares of size 1X1, 3 squares (whose vertices lie on the perimeter of the array) of size 3X3 and 4*2 squares of size 2x2 [4 is conventional we all know, we double it as the midpoints of the edges joining will also form an equal number of squares] => In all 9 + 8 + 3 squares.

Now for our problem in hand 5X5:
We have 16 ((n-1)^2, when array is nXn) squares of size 1X1, 4 (n-1, when array is nXn) squares (whose vertices lie on the perimeter of the array) of size 3X3, 18 squares (2*(n-2)^2 of size 2X2) and 12 squares (3*(n-3)^2 of size 3X3) [for 3X3 we have trepled as done in 4X4 logic above]
=> In all 4 + 16 + 18 + 12 = 50 squares.

=> Choice (4) is the right answer

5X5 matrix is used to define the latices points and not the orientation of the squares .

strange where did Gurpreet's post vanish.

1/2 ( 5c2 * 5c2) is the solution i had thought too. But any two points would have given rectangle too.
Thats what I thought.

some thing like 2 lines are there with m and n points on each. calculate how many triangles can be made ?

so, m (nC2) + n(mC2). logic used is the same there too.

Nice solution Gurpreet,lots of learning from you mate.

My friends called me for the movie, so instead of editing it, I deleted the post to write again later.

Now.. If you select 2 points on the one side, then the other two points are fixed. In case you don;t fix the other two, the result would be a rectangle.

but at the same time when you would select the later 2 points then the former points would be fixed.
So total squares considered by us = 5c2*5c2.

But, we have counted the same square twice. so the answer must be 1/2 of it.

this mathematic doesnt sound like an gmat... which source is it from?