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A right triangle ABC has to be constructed in the xy-plane so that the

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shekar123
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A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   22 May 2010, 15:05
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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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000


my analysis

Show Spoiler
A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER
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B
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Bunuel
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   23 May 2010, 06:06
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shekar123 wrote:
Please help me with this question

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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER


amitjash's solution is correct.

We have the rectangle with dimensions 9*10 (9 horizontal dots and 10 vertical). AB is parallel to x-axis and AC is parallel to y-axis.

Choose the (x,y) coordinates for vertex A: 9C1*10C1=90;
Choose the x coordinate for vertex B (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex C (as x coordinate is fixed by A): 9C1, (10-1=9 as 1 vertical dot is already occupied by A).

9C1*10C1*8C1*9C1=6480.

Answer: B.
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   30 Jul 2014, 06:55
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[quote="shekar123"]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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000


Sol:

let's find point A first:
X- axis_ point A :: point A can be any where between -3 and 5 on X- axis (total point = 5-(-3)+1 = 9)= 9c1
Y- axis_ point A :: for each x-coordinate the point A can take any value from 2 to 11 (11-2+1) in Y -coordinate= 10c1

A : 9c1*10c1

Lets Now find point B :

X- axis _ Point B:: from remaining X cordinates( A has already taken one), B can take any value so 8C1
Y - axis_point B :: No bother because it will be on same co-ordinate as A

B: 8C1

Now point C:
X: It has to be alias with A so no bother
Y: It can take any value from remaining 9 (one already taken by A) , 9C1

total: 9c1*10c1*8c1*9c1 = 6480 !!!!
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   22 May 2010, 21:54
shekar123 wrote:
Please help me with this question

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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER


You are going in the right direction. For A there are 90 possibilities. However, for B we are bound by A, so we have only 8 possibilities because B has to have the same x-coordinate as A. Similarly C has 9 possibilities.
Total: 90*8*10= 7200
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   23 May 2010, 05:48
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Friends,
Please provide the OA.
I am trying to provide my thinking below:
For A, there are 90 possibilities on the plane.
For each location of A, as the triangle has to have same properties, there are 8 possibilities on the x axis and 9 possibilities on y axis.
So the answer is 90*8*9 = 6480
Please provide the OA.

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A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   21 Dec 2010, 09:15
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Similar questions:

https://gmatclub.com/forum/arithmetic-o ... 88380.html
https://gmatclub.com/forum/geometry-and ... 00675.html 
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   16 Jun 2011, 00:41
A 9*10 for x and y.

B 8* 1

C 1* 9
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   10 Feb 2019, 23:36
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Solution:
Let’s draw a diagram to analyze the question:
Attachment:
image1.PNG
image1.PNG [ 17.14 KiB | Viewed 13991 times ]


Right angled triangle ABC, right angled at A, satisfying the following conditions:
-3 ≤x1, x2≤ 5 and 2 ≤y1, y2≤ 11
Let the x and y coordinates for vertex A = (x1 and y1)
Let the x and y coordinates for vertex B = (x1 and y2) [As AB is parallel to Y Axis, “x” coordinate will be same as the vertex A only “Y” coordinate will change.]
Let the x and y coordinates for vertex C = (x2 and y1) [As AC is parallel to X Axis, “y” coordinate will be same as the vertex A only “X” coordinate will change.]
So, as the diagram represents the number of triangles with the given conditions will vary as x1 and x2’s and y1 and y2’s vary.
The total number of possibilities of triangles = x1 ×x2 ×y1 ×y2
= 9C1 8C1 10C1 9C1
= 6480
The correct answer option is “B”
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   14 Mar 2019, 22:04
Bunuel wrote:
shekar123 wrote:
Please help me with this question

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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER


amitjash's solution is correct.

We have the rectangle with dimensions 9*10 (9 horizontal dots and 10 vertical). AB is parallel to x-axis and AC is parallel to y-axis.

Choose the (x,y) coordinates for vertex A: 9C1*10C1=90;
Choose the x coordinate for vertex B (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex C (as x coordinate is fixed by A): 9C1, (10-1=9 as 1 vertical dot is already occupied by A).

9C1*10C1*8C1*9C1=6480.

Answer: B.


hi sir , i do understand all the algebra and combinations leading to the answer , but while selecting points don't we have to consider the validity of pythagoras theorem ?
, such that the selected coordinates satisfy it as it is a right triangle , as our selection criteria seems to be based on just selecting points
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   22 Nov 2020, 11:49
total points for X axis 9 and that for Y axis 10
number of ways we can select point x,y for A ; 9c1*10c1
now since its right ∆ at A so x point is fixed for B and Y coordinate can be chosen in 9c1 ways
left with C whose points can be chosen in 8c1
9c1*10c1*8c1*9c1 = 6480
option B

shekar123 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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000


my analysis

Show Spoiler
A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER
User avatar
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   24 Mar 2024, 02:40
email2vm wrote:
shekar123 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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000


Sol:

let's find point A first:
X- axis_ point A :: point A can be any where between -3 and 5 on X- axis (total point = 5-(-3)+1 = 9)= 9c1
Y- axis_ point A :: for each x-coordinate the point A can take any value from 2 to 11 (11-2+1) in Y -coordinate= 10c1

A : 9c1*10c1

Lets Now find point B :

X- axis _ Point B:: from remaining X cordinates( A has already taken one), B can take any value so 8C1
Y - axis_point B :: No bother because it will be on same co-ordinate as A

B: 8C1

Now point C:
X: It has to be alias with A so no bother
Y: It can take any value from remaining 9 (one already taken by A) , 9C1

total: 9c1*10c1*8c1*9c1 = 6480 !!!!


­Can someone please explain why are we SEQUENTIALLY trying to find A, then B, then C?

According to me, if A is (x1,y1), B is (x2,y1), C is (x1, y2),

then total 2 x variables are there, and 2 y variables.
To find x variables - 9C2
To find y variables - 10C2

My answer is coming 1620, 1/4th of the given answer. What am I missing?
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink] New post   24 Mar 2024, 05:46
Expert Reply
Ishika_Parihar wrote:
email2vm wrote:
shekar123 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 to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000


Sol:

let's find point A first:
X- axis_ point A :: point A can be any where between -3 and 5 on X- axis (total point = 5-(-3)+1 = 9)= 9c1
Y- axis_ point A :: for each x-coordinate the point A can take any value from 2 to 11 (11-2+1) in Y -coordinate= 10c1

A : 9c1*10c1

Lets Now find point B :

X- axis _ Point B:: from remaining X cordinates( A has already taken one), B can take any value so 8C1
Y - axis_point B :: No bother because it will be on same co-ordinate as A

B: 8C1

Now point C:
X: It has to be alias with A so no bother
Y: It can take any value from remaining 9 (one already taken by A) , 9C1

total: 9c1*10c1*8c1*9c1 = 6480 !!!!


­Can someone please explain why are we SEQUENTIALLY trying to find A, then B, then C?

According to me, if A is (x1,y1), B is (x2,y1), C is (x1, y2),

then total 2 x variables are there, and 2 y variables.
To find x variables - 9C2
To find y variables - 10C2

My answer is coming 1620, 1/4th of the given answer. What am I missing?

­
With your method, if you select a pair like (2, 3) for x1 and x2, you'd only get the right triangles with the x-coordinate at x1=2. However, the x-coordinate of A can also be at 3. Similarly, if you select (5, 10) for y1 and y2, you'd only get the right triangles with the y-coordinate at y1=5. However, the y-coordinate of A can also be at 10. Since you're only getting each of those pairs once with 9C2 and 10C2, you're essentially getting half of the x options and half of the y options for A.

Hope it's clear.
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Re: A right triangle ABC has to be constructed in the xy-plane so that the [#permalink]
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