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how many triangles?

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how many triangles?

by shipra » Sun Jun 29, 2008 2:52 am
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 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
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how many triangles?

by simpdimp » Mon Jun 30, 2008 6:01 pm
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 (10-1) 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 (11-1) values left. So, 990*10 = 9900

The answer is C - 9900
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by shipra » Mon Jun 30, 2008 11:04 pm
Thanx a lot...
really helped...
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Re: how many triangles?

by rabab » Sat Oct 18, 2008 7:36 am
simpdimp wrote:Since PR is parallel to X axis, R is on the same line as P. So, R has 9 (10-1) 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 (11-1) values left.
Could you pls explain these two lines? Why does R have 9 values and Q have 10 values left?
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by cmoney » Sun Oct 19, 2008 11:46 am
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. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
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by rah_pandey » Tue Jun 16, 2009 10:15 am
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 =11-1=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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by bhumika.k.shah » Sun Feb 07, 2010 9:42 pm
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. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
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by sars72 » Sun Feb 07, 2010 9:52 pm
bhumika.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. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
it's not (10,-1). It is 10-1 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for R
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by bhumika.k.shah » Sun Feb 07, 2010 9:56 pm
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 (10-1) 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 (11-1) values left. So, 990*10 = 9900
sars72 wrote:
bhumika.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. (10-1).

I hope that helped. Anyone feel free to correct me if I'm wrong.
it's not (10,-1). It is 10-1 = 9 possible values for R. Since R and P cannot have the same value, we subtract 1 from possible number of values for R
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by missrochelle » Tue Aug 24, 2010 4:59 am
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 (10-1) 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 (11-1) values left. So, 990*10 = 9900

The answer is C - 9900
Can someone explain the 2nd part of the process? I understand why R has 9 values left on the X-axis, 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?

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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by GMATGuruNY » Tue Aug 24, 2010 5:41 am
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: -4≤x≤5, giving us 10 choices.

y value: 6≤y≤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 11-1=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 10-1=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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by ajayiitr » Wed Oct 06, 2010 7:05 pm

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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by ajayiitr » Thu Oct 14, 2010 12:44 pm
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.
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by GMATGuruNY » Thu Oct 14, 2010 1:22 pm
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.
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).

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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by ajayiitr » Thu Oct 14, 2010 1:30 pm
Awesome, Thanks so much.
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