akshaykerur 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?
4
6
8
10
12
What is the ans here?
Thanks.
We can start with a square that has coordinates (0,0); (10,0); (10,10); (0,10) this is in quadrant 1.
Now, if we can figure out how many more squares can have integer vertices by moving the (10,0), we can simply multiply by 4 to get the same for all quadrants.
So start shrinking the (10,0) point towards the origin this will cause the base to "lift up" from the x axis (Note the contraction towards the origin can happen only in integers else the x coordinate of the vertex will be a non-integer). When the (10,0) point comes to (9,0) the base has to "lift" by (sqrt(100-81)) - a non integer number, thus at this point the coordinate of the vertex will be non integer.
As we do this exercise, we will quickly see that there are only two more combinations possible - when the base has shrunk to (8,0) and the "lift" is 6 units, thus the vertex is (8,6). Similarly when the base has shrunk to (6,0) and the "lift" is 8 units, thus the vertex is (6,8).
Thus, we see that a total of 3 squares are possible in quadrant 1. In the whole graph, which has 4 quadrants the possible squares =[spoiler] 3*4 = 12[/spoiler]. Hence
E
