amitdgr wrote:
Right triangle PQR is to be constructed in the 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 -4 <= x <= 5 and 6<= y<= 16. How many different triangles with these properties could be constructed?
(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100
Source :
OG 11 (PS 248)
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I did not find the answers very detailed and had to spend a lot of time figuring it out. I've broken down every step into the most simplified manner and this will be the only solution that you need to look at if you are stuck
1. Make an x-y plane.
2. Draw a right angle triangle anywhere in the upper half of the plane.
3. I named it as QPR where P is the right angle, PR is the line parallel to the x axis and QP is the line parallel to the y-axis. Draw this out somewhere.
4. Now name P (x1, y1)
5. Name R (x2, y1) as it shares the same y coordinate with P being parallel to the x axis. (Draw the diagram)
6. Q (x1, y2)
7. Let's start with the cominations now
How many points can the x-coordinate have? -4 to 5 is a total of 10 points.
How many points can the y-coordinate have? 6 to 16 is a total of 11 points.
Total combinations = no. of ways P can be filled* R can be filled* Q can be filled.
: P can be filled in (10 ways *11 ways)
: R can be filled in (9 ways *1 way); why? 9 because it can be filled in 10-1 ways, we lost 1 to P, the x-coordinate of R cannot possibly be the same as P or it won't be a triangle anymore. We have already established that y coordinate is the same as P, so there is only one way to fill it= which is the same as P's Y coordinate.
: Q can be filled in (1 way *10 ways): try to understand this on your own by applying the same approach as R's solution.
Therefore, Total combinations = (10*11) * (9*1) * (1*10)
= 110*90 = 990
Phew.. Hope this helps.
I need 10 kudos points to be able to add
error log into pdf format on gmat club, so please help if this answer provided you with some clarity. Thanks