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

## another tough OG question

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- Brent@GMATPrepNow
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Take the task of building triangles and break it into stages.sana.noor 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 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

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

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If you're interested, here's another 700+ level question involving counting triangles in the coordinate plane: https://www.beatthegmat.com/how-many-tri ... 28974.html

Cheers,

Brent

Cheers,

Brent

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I've posted a solution here >> https://www.beatthegmat.com/og-12-229-60 ... tml#612443

Anju Agarwal

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Backup Methods : General guide on plugging, estimation etc.

Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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