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Rectangle ABCD is constructed in the coordinate plane parall
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11 Nov 2012, 15:58

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Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

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12 Nov 2012, 00:59

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Archit143 wrote:

Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

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11 Nov 2012, 22:17

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Just had a look- it is rather subtle. Here, we can choose 2 values of x and y as taking any 2 values of x and y will always yield a rectangle. For instance x=2 and x=3 are two lines parallel to y axis and y=1 and y=4 are two values parallel to x axis- plot these lines and you get a rectangle.

For the triangles query, we will need to constrain the values of the coordinates that x and y can take, say P is (x1,y1), Q(x1,y2) and R(x2,y1).

So we will need to pick one x to designate x1 and then we will have 9 more x values remaining from which we choose x2. Similarly we can do the same for y1 and y2.

That is why we get 11*10*10*9

If we choose 2 values directly , then we do not make a distinction in the order and if this happens we multiply the value by 4. Take x1=5, x2=8, y1=7 and y2=9 (so 8,5,7 and 9 can be arranged among one another and the combination does not take the changing values into account)

the four triangles you get are: (5,7), (8,7), (5,9) ; (8,7), (8,9), (5,7) ; (5,9), (8,9), (5,7) ; (8,9) , (5,9) , (5,7).

So we can also use : 4* 11C2*10C2 to get 9900

For the rectangle, choosing 2 values of x and y result in only 1 rectangle. This is the only difference

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28 Jun 2013, 23:31

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stunn3r wrote:

consider rectangle to be ABCD

choosing A's X and Y co-ordinate >> 10c1*9c1 choosing B's X and Y co-ordinate >> 9c1.1 (because one coordinate is fixed) choosing C's X and Y co-ordinate >> 1.8c1 (because one coordinate is fixed) choosing D's X and Y co-ordinate >> 1.1 (Because both coordinate are fixed)

this comes equal to 7920. I think I am considering some of the cases twice or even four times. Please tell me what am I doing wrong ..

Hi stunn3r

You have two errors:

(1) How did you come up with 7920, because 10C1*9C1*9C1*8C1 # 7920. Because there are 9 ways to choose X, and 11 ways to choose Y

(2) I assume your equations are correct. But the question here is "how many RECTANGLE?" not "how many combination of A,B,C and D ==> 4 points create only 1 rectangle ==> You should divide 7290/4 = 1980

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11 Nov 2012, 21:36

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We need to choose 2 numbers from the x domain [3,11], since they will form two lines parallel to the y axis. Similarly, we need two values from the y domain [-5,5] to form two values parallel to the x axis. There are 9 integers for x and there are 11 numbers for y.

Choose 2 from 9 for the sides parallel to y axis: 9C2 Choose 2 from 11 for the sides parallel to x axis: 11C2

Multiply to get the overall number which should give you C.

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11 Nov 2012, 22:34

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Sorry for the double post- but another easier way to think about this is as so: we can calculate the number of rectangles just as we have done for the original question you provided here.

Take any rectangle and you have two diagonals. Each diagonal divides the rectangle into two different right triangles. So taking the two diagonals into account, we can create 4 right triangles with each rectangle. So just multiply the number of rectangles by 4 to get the answer..

Rectangle ABCD is constructed in the coordinate plane parall
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10 Sep 2014, 12:27

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The way I solved this problem was by thinking about the points and lines on the graph.

I chose two points on the horizontal axis between 3 & 11 . Because it doesn't specify that the points are integers I considered EVERY point, including the first, making the choices (9)(9) you then multiply by 11 because there are 11 points on the y axis. (9)(9)(11)= 891

You do the same thing for the y axis. 11 possible points and 2 must be chosen considering also that there are 9 horizontal points. (11)(11)(9) = 1089

Added together. (9)(9)(11) + (11)(11)(9) = 1980

An interesting thing is that the answer is a little smaller than 1980 because the axis points cannot be the same on the graph... but since there are an infinite number of points between 3 & 11, the difference is negligible.

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11 Nov 2012, 19:26

gnan wrote:

As the rectangle is parallel to coordinate axes, the coordinates of the points of the rectangle would be

(X1, Y1), (X2, Y1), (X2, Y2), (X1,Y2)

given that X1, X2 lie between 3 and 11..ie., 9 possible numbers

Possible combinations for X1,X2 would be 9C2 = 36

Similarly, Possible combinations for Y1, Y2 would be 11C2 = 55

Possible ways of constructing rectangle is by selecting any of the combination of X1,X2 and Y1,Y2

= 36 * 55 = 1980 Ans. C

Excellent explanation,

pls explain the similarity between between the triangle question and this, hope you would have solved it....I ll mention the question num for making it easy

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11 Nov 2012, 21:41

seriousmonkey wrote:

We need to choose 2 numbers from the x domain [3,11], since they will form two lines parallel to the y axis. Similarly, we need two values from the y domain [-5,5] to form two values parallel to the x axis. There are 9 integers for x and there are 11 numbers for y.

Choose 2 from 9 for the sides parallel to y axis: 9C2 Choose 2 from 11 for the sides parallel to x axis: 11C2

Multiply to get the overall number which should give you C.

I used the same method to slove just wanted to know the distinction between the question in my preceding post and the the question in this thread.

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12 Nov 2012, 01:21

seriousmonkey wrote:

Sorry for the double post- but another easier way to think about this is as so: we can calculate the number of rectangles just as we have done for the original question you provided here.

Take any rectangle and you have two diagonals. Each diagonal divides the rectangle into two different right triangles. So taking the two diagonals into account, we can create 4 right triangles with each rectangle. So just multiply the number of rectangles by 4 to get the answer..

I was thinking on the same lines Now another had the question mentioned how many different squares instead of rectangle than what will be our answer???????????

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27 Jun 2013, 13:36

consider rectangle to be ABCD

choosing A's X and Y co-ordinate >> 10c1*9c1 choosing B's X and Y co-ordinate >> 9c1.1 (because one coordinate is fixed) choosing C's X and Y co-ordinate >> 1.8c1 (because one coordinate is fixed) choosing D's X and Y co-ordinate >> 1.1 (Because both coordinate are fixed)

this comes equal to 7920. I think I am considering some of the cases twice or even four times. Please tell me what am I doing wrong ..

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29 Nov 2017, 16:50

seriousmonkey wrote:

Just had a look- it is rather subtle. Here, we can choose 2 values of x and y as taking any 2 values of x and y will always yield a rectangle. For instance x=2 and x=3 are two lines parallel to y axis and y=1 and y=4 are two values parallel to x axis- plot these lines and you get a rectangle.

For the triangles query, we will need to constrain the values of the coordinates that x and y can take, say P is (x1,y1), Q(x1,y2) and R(x2,y1).

So we will need to pick one x to designate x1 and then we will have 9 more x values remaining from which we choose x2. Similarly we can do the same for y1 and y2.

That is why we get 11*10*10*9

If we choose 2 values directly , then we do not make a distinction in the order and if this happens we multiply the value by 4. Take x1=5, x2=8, y1=7 and y2=9 (so 8,5,7 and 9 can be arranged among one another and the combination does not take the changing values into account)

the four triangles you get are: (5,7), (8,7), (5,9) ; (8,7), (8,9), (5,7) ; (5,9), (8,9), (5,7) ; (8,9) , (5,9) , (5,7).

So we can also use : 4* 11C2*10C2 to get 9900

For the rectangle, choosing 2 values of x and y result in only 1 rectangle. This is the only difference

Can you explain this part in detail-That is why we get 11*10*10*9

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04 Jan 2018, 09:01

Top Contributor

Archit143 wrote:

Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?

396 1260 1980 7920 15840

IMPORTANT:

First notice that, to construct this rectangle, the vertices will share several points. For example, if the 4 vertices are at (2, 5), (2, -3), (9, 5) and (9, -3), then we get a rectangle.

Notice that there are only 2 different x-coordinates (2 and 9) and only 2 different y-coordinates (-3 and 5)

So, to create the desired rectangle, we need only choose 2 different x-coordinates and 2 different y-coordinates

So, let's take the task of creating rectangles and break it into STAGES

STAGE 1: Select the 2 x-coordinates We can choose 2 values from the set {3, 4, 5, 6, 7, 8, 9, 10, and 11} In other words, we must choose 2 of the 9 values in the set Since the order in which we choose the numbers does not matter, we can use COMBINATIONS We can select 2 number from 9 numbers in 9C2 ways (= 36 ways)

STAGE 2: Select the 2 y-coordinates We can choose 2 values from the set {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} In other words, we must choose 2 of the 11 values in the set We can select 2 number from 11 numbers in 11C2 ways (= 55 ways)

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a rectangle) in (36)(55) ways (= 1980 ways)

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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