mgmat squares

pradeepkaushal9518 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

I would go with C.

Will explain once the answer is correct ..

Waiting for OA.

i needthe explaination how u reached at the answer?

First glance at the problem will give us 4 squares, that can lie along the axes in 4 quadrants.
Hence 4.

Now, there can be sqaures that need not lie along the axes. Say there is a square ABCD with AB in the First Quadrant.
Now, we know that AB=10.
Let the point A be origin and point B be (x,y)

We know that sqrt(x^2+y^2)=10
We are given that all coordinates are integers.
Hence x and y need to be integers.
So, (x,y) can be:
(6,8)(-6,8)(6,-8)(-6,-8)
(8,6)(-8,6)(8,-6)(-6,-8)

Hene B has 4+4+4=12 possible positions.

Hence 12 squares are possible.

pick E

Whats OA?

I think answer should be 12.

Distance btwn 2 points is given by sqrt[(y2-y1)^2 + (x2-x1)^2]

In this case, since one of the vertices is on the origin, it means x1= 0 and y1=0

Thus, equation is sqrt(x^2+y^2) = 10

=> x^2 + y^2 = 100

(0,10) and (8,6) satisfy this equation with various combinations within each. If you total them, it comes out to 12 points.

bubbliiiiiiii wrote:

pradeepkaushal9518 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

I would go with C.

Will explain once the answer is correct ..

Waiting for OA.

Given that area of the square is 100, Thus, the length of the side is 10.

When a square of side 10 is drawn, it can be drawn in an area ranging between -14 and +14 (diagonal) within the coordinate system.

Since the co-ordinates have to be integers, I thought one of the vertices could lie on line y=x in quadrant 1. and thus, we come up with three posibilities of drawing the square. Extending this to the other three quadrants .. I concluded it to be 8.

Please correct my methodology, if wrong.

Regards,

Pranay

Each side of the square must have a length of 10. If each side were to be 6, 7, 8, or most other numbers, there could only be four possible squares drawn, because each side, in order to have integer coordinates, would have to be drawn on the x- or y-axis. What makes a length of 10 different is that it could be the hypotenuse of a Pythagorean triple, meaning the vertices could have integer coordinates without lying on the x- or y-axis.

For example, a square could be drawn with the coordinates (0,0), (6,8), (-2, 14) and (-8, 6). (It is tedious and unnecessary to figure out all four coordinates for each square).

If we label the square abcd, with a at the origin and the letters representing points in a clockwise direction, we can get the number of possible squares by figuring out the number of unique ways ab can be drawn.

a has coordinates (0,0) and b could have the following coordinates, as shown in the picture:


(-10,0)
(-8,6)
(-6,8)
(0,10)
(6,8)
(8,6)
(10,0)
(8, -6)
(6, -8)
(0, 10)
(-6, -8)
(-8, -6)

There are 12 different ways to draw ab, and so there are 12 ways to draw abcd.

The correct answer is E