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Right triangle RST can be constructed in the xy-plane such

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Right triangle RST can be constructed in the xy-plane such [#permalink]

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Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R. The x and y-coordinates of R, S, and T are to be nonzero integers that satisfy the inequalities −3 ≤ x ≤ 4 and −7 ≤ y ≤ 3. Given these restrictions, how many different triangles can be constructed?

A. 3780
B. 4200
C. 4900
D. 6160
E. 7744
[Reveal] Spoiler: OA

Originally posted by toshreyansjain on 10 Aug 2012, 23:28.
Last edited by Bunuel on 11 Aug 2012, 00:19, edited 1 time in total.
Edited the question.
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Re: Right triangle RST can be constructed in the xy-plane such [#permalink]

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toshreyansjain wrote:
Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R. The x and y-coordinates of R, S, and T are to be nonzero integers that satisfy the inequalities −3 ≤ x ≤ 4 and −7 ≤ y ≤ 3. Given these restrictions, how many different triangles can be constructed?

A. 3780
B. 4200
C. 4900
D. 6160
E. 7744


−3 ≤ x ≤ 4 and −7 ≤ y ≤ 3 gives a rectangle with 8*11 dimensions (8 horizontal and 11 vertical dots). We are given that RS is parallel to x-axis, RT is parallel to y-axis and the right angle is at R.

Choose the (x,y) coordinates for vertex R: 7C1*10C1 (we are told that coordinates of R, S, and T must be nonzero integers, so we are choosing from 7 and 10 instead of 8 and 11 because we should exclude 0);

Choose the x coordinate for vertex S (as y coordinate is fixed by R): 6C1, (7-1=6 as 1 horizontal dot is already occupied by R);

Choose the y coordinate for vertex T (as x coordinate is fixed by R): 9C1, (10-1=9 as 1 vertical dot is already occupied by B).

7C1*10C1*6C1*9C1=3,780.

Answer: A.

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Re: Right triangle RST can be constructed in the xy-plane such [#permalink]

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judokan 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 < or= x < or = 5 and 6 < or= y < or = 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

kanusha wrote:
sir, in the below steps i did not understand what is x coordinate for vertex R and Q? what is mean by coordinate is fixed by A?? How and why values 9c1 and 10c1 are used??

We have the rectangle with dimensions 10*11 (10 horizontal dots and 11 vertical). RT is parallel to y-axis and RS is parallel to x-axis.
Choose the (x,y) coordinates for vertex R (right angle): 10C1*11C1;
Choose the x coordinate for vertex S (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 T (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.

though it might be silly doubt.. pls send if their is any concept that i lack understand this problem.
pls reply, thank you.

Dear kanusha,

I'm happy to help. :-)

This question requires a little visualization and some facility with the x-y plane. The question tells us "Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R." This tells us
(a) there is a right angle at R, so RS is perpendicular to RT
(b) RS is parallel to the x-axis, perpendicular to the y-axis --- i.e. horizontal
(c) RT must be parallel to the y-axis, perpendicular to the x-axis --- i.e. vertical

Then, we have to recognize --- all points on the same horizontal line have the same height, so they all have the same y-coordinate. Therefore, we absolutely know that R & S, endpoints of the horizontal line, must have the same x-coordinate.

Similarly ---- all points on any vertical line share the same distance from the y-axis, the same length to the right or the left of the y-axis, so all points on the same vertical line must have the same x-coordinate. Thus, R & T, the endpoints of a vertical line, have the same y-coordinate.

Next, inclusive counting. See
http://magoosh.com/gmat/2012/inclusive- ... -the-gmat/
From x = -4 to x = +5, there are (+5) - (-4) + 1 = 10 points, and, similarly, 11 possible y-points.

Does this clear up your question? On precisely which part do you have questions?
Mike :-)
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Re: Right triangle RST can be constructed in the xy-plane such [#permalink]

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Tricky Question!

Imagine we have a grid, ranging from -3 to 7 on the x-axis, and from -7 to 3 on the y-axis. Every point on this grid could potentially be point R, the right angle of the triangle (EXCEPT ANY POINT ON THE AXES - the question says *non-zero* integers! They're not talking about the length of the side, they mean the coordinates of the points. This is very important!). Thus, we're left with 7*10 = 70 choices for our point R.

Now we know that the sides of the triangle are parallel with the axes, so the other points must lie on the same horizontal or vertical as our point R. Therefore, we have 7-1 = 6 choices for the one on the x axis, and 10-1= 9 choices for the one on the y-axis (we subtract one from each because we can't have the same point).

Therefore, in total, we have 7*10*6*9 = 54*70 = 3500 + 280 = 3780 possible choices.

Answer: A

Notice if they had NOT stated that the coordinates had to be non zero, we would have had 8*11 choices for R, and 7 and 10 choices for the other points, respectively. The number of possibilities here is 8*11*7*10 = 88*70 = 5600+560 = 6160, answer D. Very tricky!
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Right triangle RST can be constructed in the xy-plane such [#permalink]

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No of ways of choosing R(x,y) = 10*7
No of ways of choosing S =1*6
no of ways of choosing T= 9*1

Therefore total number of ways = 10* 7 *9 *6
=3780
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Re: Right triangle RST can be constructed in the xy-plane such [#permalink]

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New post 12 Sep 2016, 11:24
How many triangles you see here?
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Re: Right triangle RST can be constructed in the xy-plane such [#permalink]

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New post 13 Mar 2018, 12:55
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Hi All,

This question is about following graphing rules and keeping track of individual possibilities. Here's how you can solve it:

We're told a number of facts:
1) The X and Y co-ordinates must be NON-ZERO INTEGERS.
2) RS is perpendicular to the Y-axis; this means that points R and S will have the SAME Y-coordinate.
3) Angle R is a right angle; this means that points R and T will have the SAME X-coordinate.
4) The two co-ordinates are in the following ranges: -3 ≤ x ≤ 4 and -7 ≤ y ≤ 3

Let's start with point R. It can have any of 7 different X-coordinates (-3,-2,-1, 1, 2, 3, 4) and any of the 10 different Y-coordinates (-7,-6,-5,-4,-3,-2,-1,1,2,3).

That is 7x10 different possibilities = 70 different spots for point R

Once we FIX R in one spot, S MUST have the SAME Y-coordinate, BUT a DIFFERENT X-coordinate. We already used one of the X-coordinates when we placed point R, which means….

There are 7-1 = 6 possible spots for point S

With R still in its spot, we know that point T must have the SAME X-coordinate, but a DIFFERENT Y-coordinate. We already used one of the Y-coordinates when we place point R, which means….

There are 10-1 = 9 possible spots for Point T

We multiply the options: (70)(6)(9) = 3780 different triangles.

Final Answer:
[Reveal] Spoiler:
A


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Re: Right triangle RST can be constructed in the xy-plane such   [#permalink] 13 Mar 2018, 12:55
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