winnerhere wrote:How many rectangles of area=6 are there in a 5 x 5 square grid(like chess board but 5 x 5 )
1. 12
2. 18
3. 24
4. 30
Here's what the question is really asking:
An area of the xy plane is bounded by (0,0), (0,5), (5,5) and (5,0). How many rectangles with an area of 6 units can be formed within this section of the xy plane if the vertices of the rectangles must lie on points with integer values.
As cbenk121 notes, to get an area of 6 we can have 1*6 or 2*3 rectangles. 1*6 would lie outside the bounded area, so we know that our rectangle must be 2*3.
So, how many different 2*3 rectangles can we create within the bounded area, given that our vertices must all lie on points with integer coordinates?
You could draw it out and brute force it, or you can use counting methods.
First, we have to recognize that we can orient our rectangle two different ways: our 2 side can be parallel to either the x- or y-axis.
Second, once we've split the question into two parts (one for each orientation), we need to recognize that once we place the lower left corner, we create 1 unique rectangle.
If our 2 side is parallel to the y-axis, it fits within the bounded area if the lower left corner of the rectangle begins at y=0, 1, 2, or 3. So, there are 4 possible y-coordinates.
Since our 3 side is parallel to the x-axis, it fits within the bounded area if the lower left corner begins at x = 0, 1 or 2. So, there are 3 possible x-coordinates.
Therefore, there are 4*3 = 12 possible rectangles in this first case.
Similarly, if we orient the rectangle with the side of length 3 parallel to the y-axis, we get 3 possible y co-ordinates and 4 possible x-coordinates for the lower left corner. Accordingly, there are 12 possible rectangles in the second case.
So, there are 12 + 12 = 24 possible rectangles.