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## A right triangle ABC has to be constructed . . .

tagged by: Vincen

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### Top Member

Vincen Legendary Member
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#### A right triangle ABC has to be constructed . . .

Wed Oct 11, 2017 6:48 pm
A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are integers to satisfy the inequalities -1 ≤ x ≤ 7 and 1 ≤ y ≤ 7. The numbers of different triangles that can be constructed with these properties are?

A. 63
B. 336
C. 567
D. 3024
E. 5040

The OA is D.

How should I do this PS question? There are too much triangles to draw.

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Thu Oct 12, 2017 12:41 pm
Here's a similar question to practice with: http://www.beatthegmat.com/ps-coordinate-geometry-t292683.html

Cheers,
Brent

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Brent@GMATPrepNow GMAT Instructor
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Thu Oct 12, 2017 12:40 pm
Vincen wrote:
A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are integers to satisfy the inequalities -1 ≤ x ≤ 7 and 1 ≤ y ≤ 7. The numbers of different triangles that can be constructed with these properties are?

A. 63
B. 336
C. 567
D. 3024
E. 5040
Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point A).
The point can be selected from a 9x7 grid. So, there 63 points to choose from.
This means that stage 1 can be completed in 63 ways.

Stage 2: Select a point that is on the same horizontal line as the first point (point A). This point will be point B.
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 A.
In how many ways can we select the x-coordinate of point B?
Well, we can choose any of the 9 coordinates from -1 to 7 inclusive EXCEPT for the x-coordinate we chose for point A (in stage 1).
So, there are 8 coordinates to choose from.
This means that stage 2 can be completed in 8 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point C.
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 A.
In how many ways can we select the y-coordinate of point C?
Well, we can choose any of the 1 integral coordinates from 1 to 7 inclusive EXCEPT for the y-coordinate we chose for point A (in stage 1).
So, there are 6 coordinates to choose from.
This means that stage 3 can be completed in 6 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (63)(8)(6) = 3024 = D

------------------------------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

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- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
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- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
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- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/ps-counting-t273659.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html

Cheers,
Brent

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Thu Oct 12, 2017 12:29 pm
Hi Vincen,

Given the restrictions in the question, there are 9 possible X-coordinates and 7 possible Y-coordinates to choose from. Thus, for the first coordinate of the triangle, there are:

(9)(7) = 63 possible coordinates for Point A.

Since AB is PARALLEL to the X-axis, Point B has to have the SAME Y-coordinate as Point A, but a DIFFERENT X-coordinate. Thus, since Point A has already been placed, there are 9-1 = 8 possible coordinates for Point B.

For a RIGHT TRIANGLE to occur (with the right angle at Point A), Point C has to have the SAME X-coordinate as Point A, but a DIFFERENT Y-coordinate. Thus, since Point A has already been placed, there are 7-1 = 6 possible coordinates for Point C.

From here, the answer choices are fairly 'spread out', so you can use a bit of estimation to get the correct answer....

The total possible number of right triangles is (63)(8)(6)... this is fairly close to (60)(50) = about 3,000 possible triangles. There's only one answer that's close...

GMAT assassins aren't born, they're made,
Rich

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