Right triangle PQR is to be constructed in the xyplane so that the right angle is at P and PR is parallel to x axis. The x and y coordinated of P,Q & R are to be integers that satisfy the inequalities 4<=x<=5 & 6<=y<=16.
How many different triangles wth these properties are possible?
A. 110
B. 1100
C. 9900
D. 10000
E. 12100
how many triangles?
X coordinate has 10 possible values between 4 and 5
Y coordinate has 11 possible values between 6 and 16
1. Fix P first, P can have 10 values across X axis and 11 across Y axis. So, that's 11*10 = 110
2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis. So far 110*9 = 990
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left. So, 990*10 = 9900
The answer is C  9900
Y coordinate has 11 possible values between 6 and 16
1. Fix P first, P can have 10 values across X axis and 11 across Y axis. So, that's 11*10 = 110
2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis. So far 110*9 = 990
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left. So, 990*10 = 9900
The answer is C  9900

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Could you pls explain these two lines? Why does R have 9 values and Q have 10 values left?simpdimp wrote:Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis.
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left.
I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (101).
I hope that helped. Anyone feel free to correct me if I'm wrong.
I hope that helped. Anyone feel free to correct me if I'm wrong.

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I was trying to solve the problem like this
Lets choose points P and R on a straight line
no of ways of choosing 2 points out of possible 10 points(since 4<=x<=5) on a line parallel to X axis=10c2*2!=90(A)
Please note 2! is used since P and R position can be interchanged and the resulting triangle will be different(mirror image)
now since triangle is right angled at P therefore no of possible points available for Q =111=10(B)
(since 6<=y<=16) and also P has been fixed with the y coordinate satisfying the given relation. Please note Q can only have same x coordinate as P.
using (A) and (B) we get
90*10=900 combinations for point P&R such that 6<=y1<=16
where y1 is fixed. Now there are 11 possible values of y1
thus we get 900*11=9900
Lets choose points P and R on a straight line
no of ways of choosing 2 points out of possible 10 points(since 4<=x<=5) on a line parallel to X axis=10c2*2!=90(A)
Please note 2! is used since P and R position can be interchanged and the resulting triangle will be different(mirror image)
now since triangle is right angled at P therefore no of possible points available for Q =111=10(B)
(since 6<=y<=16) and also P has been fixed with the y coordinate satisfying the given relation. Please note Q can only have same x coordinate as P.
using (A) and (B) we get
90*10=900 combinations for point P&R such that 6<=y1<=16
where y1 is fixed. Now there are 11 possible values of y1
thus we get 900*11=9900

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I am sowree i still dint understand how R is (10,1)
Can someone please elaborate.
Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (101).
I hope that helped. Anyone feel free to correct me if I'm wrong.
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it's not (10,1). It is 101 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for Rbhumika.k.shah wrote:I am sowree i still dint understand how R is (10,1)
Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (101).
I hope that helped. Anyone feel free to correct me if I'm wrong.

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Awrite. In that case Q has 10 values . right?
But what exactly are we trying to do here? ???
2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis. So far 110*9 = 990
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left. So, 990*10 = 9900
But what exactly are we trying to do here? ???
2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis. So far 110*9 = 990
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left. So, 990*10 = 9900
sars72 wrote:it's not (10,1). It is 101 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for Rbhumika.k.shah wrote:I am sowree i still dint understand how R is (10,1)
Can someone please elaborate.
cmoney wrote:I think that the point is that to make a triangle, the the points need to exist on different values. An example would be a triangle where the base points are the same...this creates a line, so it's impossible. Sine R cannot have the same value as P, you have subtract one from the possible locations. (101).
I hope that helped. Anyone feel free to correct me if I'm wrong.

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Can someone explain the 2nd part of the process? I understand why R has 9 values left on the Xaxis, since you can't overlap  but why don't R and Q also have 11 points on the y axis? Meaning if the triangles are small, you can have multiple triangles along the x axis?simpdimp wrote:X coordinate has 10 possible values between 4 and 5
Y coordinate has 11 possible values between 6 and 16
1. Fix P first, P can have 10 values across X axis and 11 across Y axis. So, that's 11*10 = 110
2. Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (101) values left along X axis. So far 110*9 = 990
3. Since PQR is a right angle triangle and the right angle is at P, PQ is parallel to Y axis. So, Q has 10 (111) values left. So, 990*10 = 9900
The answer is C  9900
For example,
triangle 1: P= (4,6) R=(3,6) Q=(4,7)
then moving up the y axis
triangle 2: P=(4,7) Q=(3,7) R=(4,8)
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When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.
We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.
Point P:
x value: 4â‰¤xâ‰¤5, giving us 10 choices.
y value: 6â‰¤yâ‰¤16, giving us 11 choices.
Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:
(number of choices for x)*(number of choices for y)=10*11=110 choices for P.
Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.
y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 111=10 choices for Q's y value.
(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.
Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.
x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 101=9 choices for R's x value.
(number of choices for x)*(number of choices for y)=9*1=9 choices for R.
So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:
(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.
The correct answer is C.
Hope this helps!
We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.
Point P:
x value: 4â‰¤xâ‰¤5, giving us 10 choices.
y value: 6â‰¤yâ‰¤16, giving us 11 choices.
Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:
(number of choices for x)*(number of choices for y)=10*11=110 choices for P.
Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.
y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 111=10 choices for Q's y value.
(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.
Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.
x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 101=9 choices for R's x value.
(number of choices for x)*(number of choices for y)=9*1=9 choices for R.
So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:
(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.
The correct answer is C.
Hope this helps!
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My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.
Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?
My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)
Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.
Can please someone clear this confusion that what is wrong in using combination to draw a line here.
Thanks.

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Can someone please answer my question?
Thanks.
Thanks.
ajayiitr wrote:
My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.
Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?
My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)
Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.
Can please someone clear this confusion that what is wrong in using combination to draw a line here.
Thanks.
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In your approach, order does matter when counting the number of choices for line segment PR. Since Q must have the same x coordinate as P, if we place P at (4,6) and R at (5,6), we'll be able to construct different triangles than if we place R at (4,6) and P at (5,6).ajayiitr wrote:Can someone please answer my question?
Thanks.
ajayiitr wrote:
My confusion is that to draw a line PR, we have 10 points and we have to select 2. So, I did 10C2 = 10*9/2.
Now, I know my answer will not match to the correct answer because of this denominator 2. Why should we multiply 10C2 by 2?
My understanding is that a line can be drawn by just selecting 2 points out of available points and way you do it is combination, since order does not matter, i.e. (picking 2 and 3 is same as picking 3 and 2 = same line.)
Anyways, from there, to pick y value of line PR there are 11 values. So, 10C2*11. Then, to pick Q, there could be 10 values of y left. So, 10C2*11*10 = (10*9/2)*11*10 = 9900/2, which is the wrong answer.
Can please someone clear this confusion that what is wrong in using combination to draw a line here.
Thanks.
Thus, the number of choices for line segment PR = 10P2*11. Since this would leave us 10 choices for Q's y value, the number of possible triangles would be 10P2*11*10 = 9900.
Hope this clears your confusion.
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