Problem Solving

Sapana wrote:Right Triangle PQR is to be constructed in XY- plane so that the right angle is at P and PR is parallel to the x-axis. the x and y coordinates of P.Q. and R are to be integers that satisfy the inequalities of P, Q, R are to be integers that satisfy the inequalities -4<=x<=5 and 6<=y<=16. How many different triangles with these properties could be constructed?


a. 110
b. 1100
c. 9900
d. 10000
e. 12100

Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.

Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = [spoiler]9900 = C[/spoiler]

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Sapana wrote:Thank you for the quick response. But I am not able to visualize this problem and the solution. is there a easier method to learn such problems.
I have another problem as well!

How many circles can you form with origin as a center and radius of 10?
a. 4
b. 6
c. 8
d. 10
e. 12

Sapana, are you sure that you've transcribed that question correctly? If a circle has the origin as its center and a radius of 10, there's only one possible circle.

I think perhaps you meant the following question:

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

If you'd like, I would be happy to explain this problem. For the sake of clarity, though, it's probably best to post it in a new thread and send me the link.

whats_in_the_store wrote: Why is it explicitly mentioned that vertices have to be integers? Now this being a square, do I have any other choice? And of course we are talking about Cartesian Co-ordinates.

A square with integer side lengths will not necessarily have integer coordinates for its vertices on a coordinate plane. Consider this square with side lengths of 10, when tipped on its side: (and pretend that it's an exact square! My drawing skills aren't perfect)




Because the diagonal of the square has a length of 10root2, the coordinates of that top corner will be (0, 10root2).

If you have any further questions, I'd be happy to discuss in a new thread.