Hi RiyaR,

This question shows up every once in a while, but you can absolutely solve it IF you're willing to "play around" a bit with a drawing. As such, I'm going to give you some hints so that you can try to solve it on your own:

1) The square has to be 10x10. Can you draw a 10x10 square on a graph with one corner at the Origin (0,0)? How many obvious ones can you come up with (hint: the smallest answer choice is 4, so there have to be AT LEAST 4 possible squares).
2) Once you have drawn the obvious squares, are there any other squares that you could draw (hint: squares DO NOT necessarily have to be parallel with the X- and Y-axes)?
3) Since we're dealing with a graph, there are likely to be "mirror images" of any shape that you draw....

GMAT assassins aren't born, they're made,
Rich

RiyaR wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

Since the square's area is 100, each side must have length 10.

If all coordinates of the vertices must be integers, we might immediately see that these 4 squares all meet the given conditions:



Any others?
You bet.
At this point, we're looking for something called Pythagorean Triplets. These are sets of three INTEGER values that could be the 3 sides of a right triangle.
The two most common ones to remember for the GMAT are 3-4-5 and 5-12-13.
We should also look out for MULTIPLES of these, such as 6-8-10 and 50-120-130
For this question, the 6-8-10 triplet comes into play. So, the hypotenuse (with length 10) will be one side of the square and the 6 and 8 will be the coordinated of the vertex.
Here are 2 such squares in Quadrant I:


As you might guess, there will be 2 of these types of squares possible in each of the 4 quadrants for a total of 8 squares.

So, the TOTAL number of squares = 4 + 8 = 12

Cheers,
Brent

Hi RiyaR,

YES, since we're dealing with a square that has a side length of 10, we ARE dealing with a Pythagorean triple (the 3/4/5; more specifically a 6/8/10, which is the 3/4/5 "doubled"). Spotting this pattern is the key to solving this problem in an efficient manner.

It's important to remember that ANY time you draw a diagonal line on a graph, you have the hypotenuse of a right triangle. As such, you can use the Pythagorean Formula and possibly the common Pythagorean triples.

GMAT assassins aren't born, they're made,
Rich