## OG#228

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### OG#228

by tapanmittal » Fri Jul 03, 2015 9:55 pm
Right triangle PQR is to be constructed in the xy plane so that the right angle is at P and PR is parallel to X axis.The x and y coordinates of P,Q and R are to be integers that satisfy the following inequalities
i. x greater than equal to -4 & less than equal to 5
ii. y is greater than equal to 6 & less than equal to 16.

How many different triangles can be constructed with these properties?

Ans-9,900

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by talaangoshtari » Fri Jul 03, 2015 10:53 pm
Since x and y should be integers,

x: 5 - (-4) + 1 = 10
y: 16 - 6 + 1 = 11

we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:

10 Ã— 11 = 110

Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10 - 1 = 9.

And point q should have the same x as p. So we have 11 - 1 = 10 options.

=> the total triangles = 110 Ã— 9 Ã— 10 = 9900

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by Brent@GMATPrepNow » Sat Jul 04, 2015 1:15 pm
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 â‰¤ x â‰¤ 5 and 6 â‰¤ y â‰¤ 16. How many different triangles with these properties can be constructed?
(A) 110
(B) 1100
(C) 9900
(D) 10000
(E) 12100

C
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]

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by nikhilgmat31 » Thu Jul 09, 2015 3:39 am
simply awesome question - seriously.

P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y

11*10*9*10

9900

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