Right triangle PQR is to be constructed in the xy-plane so

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 inequalitites -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

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1;
Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

10C1*11C1*9C1*10C1=9900.

Answer: C.

Other discussion with alternate solutions:
https://gmatclub.com/forum/arithmetic-og ... ed#p790651
https://gmatclub.com/forum/geometry-and- ... ed#p777591
https://gmatclub.com/forum/700-question- ... ed#p818546
https://gmatclub.com/forum/triangle-abc- ... 32573.html
https://gmatclub.com/forum/right-triangl ... 88380.html
https://gmatclub.com/forum/right-triangl ... 88958.html
https://gmatclub.com/forum/a-right-trian ... 94644.html
https://gmatclub.com/forum/a-right-trian ... 00675.html
https://gmatclub.com/forum/right-triangl ... 37129.html

All OG13 questions with solutions: https://gmatclub.com/forum/the-official- ... 43450.html

Hope it helps.
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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. 1100
C. 9900
D. 10000
E. 12100

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). PQ is parallel to y-axis and PR is parallel to x-axis.

Choose the (x,y) coordinates for vertex P (right angle): 10C1*11C1;
Choose the x coordinate for vertex R (as y coordinate is fixed by A): 9C1, (10-1=9 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex Q (as x coordinate is fixed by A): 10C1, (11-1=10 as 1 vertical dot is already occupied by A).

10C1*11C1*9C1*10C1=9900.

Answer: C.
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We can start by drawing the triangle in the xy-plane. This will help us visualize how we are to solve the problem.



As we can see, the right triangle has a right angle at point P and side PR is parallel to the x-axis; side PQ must be parallel to the y-axis.

To solve this question we need to determine how many ways we can construct points P, Q, and R, and then we will multiply those possibilities together. We are given that: \(-4\leq{x}\)\(\leq{5}\) and \(6\leq{y}\)\(\leq{16}\). This means that there are 10 possible integer values for the x-coordinate and 11 possible integer values for the y-coordinates.

Let’s start by determining how many ways we can construct point P.

Since P is the first point we are trying to determine, we have all the options for the available x- and y-coordinates. Since there are 10 possible x-coordinates and 11 possible y-coordinates we have (10)(11) = 110 possible options. Next, we determine how many options we have for point R.

In determining point R, we must recognize that side PR of triangle PQR must remain parallel to the x-axis. This means that the y-coordinate of point R, must match the y-coordinate of point P. Thus there is only 1 option for the y-coordinate of point R. However there are 9 total options for the x-coordinate of point R. Since point P has already exhausted 1 option out of the 10 total options, there are only 9 x-coordinates left for designating point R.

Finally, we determine how many ways in which we can construct point Q.

In a similar fashion to the method used for determining the number of ways to construct point R, we must remember that side PQ of triangle PQR must remain parallel to the y-axis. Thus, the x-coordinate of point Q must match the x-coordinate of point P. Therefore, there is only 1 option for the x-coordinate of point Q. However, there are 10 total options for the y-coordinate of point Q. Since point P has already exhausted 1 option out of the 11 total options, there are only 10 y-coordinates left for designating point Q.

In summary, we know the following:

There are 110 ways to select point P, 9 ways to select point R, and 10 ways to select point Q. Because we need to determine the total number of ways to create triangle PQR, we must use the fundamental counting principle. Thus, we multiply these three options together:

110 x 9 x 10 = 9,900

The answer is C.
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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) = 9900
Answer:

RELATED VIDEO

Is it C?

P can take a total of 10*11 co-ordinates.
for a given P, R can take a total of 9 co-ordinates and Q can take a total of 10 co-ordinates.

Hence, total = 110*9*10 = 9900

imo C
total values for x=10;y=11

x1,y1=10*11.......................coordinates of 1st pnt
x2,y2=9*1(y2=y1)............... coordinates of 2nd pnt y coordinates will be same as that of 1st pnt bcoz it is parallel to x axis
x3,y3=1*10(x2=x3)..........coordinates of 3rd pt. x coordinates will be same as that of 2nd point bcoz to make a right angle it has to be parallel to y axis

tot ways=10*11*9*1*1*10=9900


oa pls

MATH REVOLUTION VIDEO SOLUTION:


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2 more hours and I have it figured out.

Thinking about this on the coordinate plane confuses me more than just thinking about it as a combination problem.

P (A,B)
R (C,B)
Q (A,D)

So you need four values to solve the problem.

2 x-values {A and C}

2 y-values {B and D}

When you approach the question this way the orientation of the triangle is irrelevant. Because by using the same x-value for point P and point Q along with the same y-value for point point P and point R. The right angle is at P and PR is parallel to the x-axis.

So I need 2 x values from -4≤ X≤ 5.
Pick the first one from 10 choices
Pick the second one from 9 choices

10*9 = 90



and I need 2 y values from 6≤ y≤ 16
Pick the first one from 11 choices
Pick the second one from 10 choices

11*10 = 110



So combine the 2 and we have 9900 (110*90) possible triangles.

I realized my error when I started to make a table of the possible coordinates for point P using the triangle in the upper right of the four choices

With point P and (4,6) I realized that I only have one possible length for PR
*If P is (4,6)
and
the triangle has the orientation in the upper right of my image.
*Then R must be (5,6)

That got me back to the starting line and my wonderful Chinese girlfriend helped me going in the correct direction.

She is junior in college and an English major by the way, with no math since high school.

I just plotted 10 horizontal points and 11 vertical points and solved it using the following formula:

(11-1)*(10-1)*11*10 = 10*9*11*10=9900
Matched with Bunuel's approach and found the logic was pretty similar.

If you plot the points; you would see that you can make 90 right triangles at every point. Then, just multiply this figure with the total number of points.

Ans: "C"

In the figure shown

When P is fixed , q can take 10 positions( remaining 10 co-ordinates up the y axis) and R can take 9 positions ( x-axis) .
Hence 10*9 = 90 traingles
and p itself can take - 10 *11 positions =110
hence 90*110 = 9900

there are 10 points on the x axis from which we can choose any twos and then form the desired PR base.

By the same way ,there are 11 points on the y-axis from which we can choose any twos and then form the desired PQ perpendicular.

We can form 10P2 * 11P2=9900

OA: C
Given :\(-4 <= x <= 5;6<= y<= 16\)
As given PR is parallel to x axis

for \(y=6\), We have \(10\) \((-4 <= x <= 5)\) options for selecting \(R\), Then \(9\) (\(-4 <= x <= 5\)minus one,which is already selected for \(R\)) options available for selecting \(P\) and for \(Q\) we have \(10\) ( \(6<= y<= 16\) minus \(y=6\) otherwise all \(3\) point will be collinear)

Total triangle for \(y=6 : 9*10*10 =900 triangles\)
Similarily
for y \(=7 : 900\) triangles
for y \(=8 : 900\) triangles
.
.
.
.
for y \(=16 : 900\) triangles

Total triangle \(= (16-6+1)*900 =9900\)

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) = 9900

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Beautiful solution.

Bunuel bhai you rock

I am getting 8100 because

Ways to choose P = 9 (-4 to 4) *10 (6 to 15)
Ways to choose R= 9
Ways to choose Q= 10
Total = 9*10*90=8100

The reason why P is chosen like that is because we know that max x can be 5 and max y can be 16. So if P takes coords (5,16), then the triangle cannot be formed since R would be (6,16) and Q would be (5,17) which are both outside of the ranges. WHere am I wrong? ThatDudeKnows avigutman MartyTargetTestPrep ScottTargetTestPrep

It is not true that if P is (5, 16), the triangle cannot be formed. Let's review the restrictions on the right triangle that we need to construct:

1) Angle P should be a right angle
2) PR should be parallel to the x-axis (which means PQ should be parallel to the y-axis)
3) The x-coordinates of P, Q, and R should be between -4 and 5, inclusive
4) The y-coordinates of P, Q, and R should be between 6 and 16, inclusive

So, if we choose P to be (5, 16), then we can choose R to be (4, 16) (or (3, 16), (-2, 16) etc.), and choose Q to be (5, 14) (or (5, 13), (5, 10), (5, 6) etc.). In all of these choices, the four conditions I listed above are satisfied. I think you thought of a specific shape of a right triangle where point R is to the right of point P and point Q is above point P. The question does not mention any such restrictions, so all of the choices above produce valid right triangles.
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