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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)

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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1), and (-1, 1). If a point falls into the square region, what is the probability that the ordinates of the point (x, y) satisfy that \(x^2+y^2>1\)?

(A) \(1-\frac{\pi}{4}\)

(B) \(\frac{\pi}{2}\)

(C) \(4-\pi\)

(D) \(2-\pi\)

(E) \(\pi-2\)

Originally posted by delta09 on 21 Dec 2009, 02:33.
Last edited by Bunuel on 05 May 2017, 06:48, edited 1 time in total.
Edited the question.
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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Dec 2009, 03:40
2
5
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution


First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

\(x^2+y^2=1\) is an equation of a circle also centered at the origin, with radius 1 and the \(area=\pi{r^2}=\pi\).

We are told that the point is IN the square and want to calculate the probability that it's outside the circle (\(x^2+y^2>1\) means that the point is outside the given circle):

Image

P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=\(4-\pi\)
Total number of possible outcomes is the area of the square (as given that the point is in the square) =\(4\)

\(P=\frac{4-\pi}{4}=1-\frac{\pi}{4}\)

Answer: A.

Hope it's clear.

Attachment:
Untitled.png
Untitled.png [ 9.91 KiB | Viewed 3603 times ]

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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Dec 2009, 03:37
1
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution


\(x^2 + y^2 = R^2\) is the equation of a circle with centre (0,0) and radius R

If we draw the square and the circle, then we will see that the circle is inscribed in the square i.e. the diameter of the circle is equal to the length of the side of square.

Area of the circle= pi (1)^2=pi
Area of the square=(2)^2 = 4

The proabibility that the point lies within the square and outside the circle is

(Area of the sqaure - area of the circle)/area of the circle
= 4-pi/4
=1 - pi/4

Plz. let me know if the OA is A.

Thanks
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Dec 2009, 04:46
hi .. i think ans will be D...
2 is the diag and not side and each side is 2^(1/2)... area is 2...
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Dec 2009, 05:05
chetan2u wrote:
hi .. i think ans will be D...
2 is the diag and not side and each side is 2^(1/2)... area is 2...


Not so.

In our case we have "horizontal" square: side=2, area=4.

We would have the square with diagonal 2 if the vertices were: (0,1), (1,0), (0,-1), (-1,0). In this case if the point is IN the square it can not be outside the circle, as the square, in this case, is inscribed in the circle. Hence the probability would be 0.
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Dec 2009, 05:18
u r correct , i should have marked it on graph before ans .....
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Feb 2018, 03:13
Bunuel wrote:
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution


First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

\(x^2+y^2=1\) is an equation of a circle also centered at the origin, with radius 1 and the \(area=\pi{r^2}=\pi\).

We are told that the point is IN the square and want to calculate the probability that it's outside the circle (\(x^2+y^2>1\) means that the point is outside the given circle):
Attachment:
Untitled.png


P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=\(4-\pi\)
Total number of possible outcomes is the area of the square (as given that the point is in the square) =\(4\)

\(P=\frac{4-\pi}{4}=1-\frac{\pi}{4}\)

Answer: A.

Hope it's clear.



hey bunuel, first of all thanks for your effort !!

I would like to know why it is clear that x^2 + y^2 = 1 is a circle ? Or how did you know.

Thanks in advance
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Joined: 02 Sep 2009
Posts: 51096
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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New post 21 Feb 2018, 04:52
gmatmo wrote:
Bunuel wrote:
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution


First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

\(x^2+y^2=1\) is an equation of a circle also centered at the origin, with radius 1 and the \(area=\pi{r^2}=\pi\).

We are told that the point is IN the square and want to calculate the probability that it's outside the circle (\(x^2+y^2>1\) means that the point is outside the given circle):
Attachment:
Untitled.png


P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=\(4-\pi\)
Total number of possible outcomes is the area of the square (as given that the point is in the square) =\(4\)

\(P=\frac{4-\pi}{4}=1-\frac{\pi}{4}\)

Answer: A.

Hope it's clear.



hey bunuel, first of all thanks for your effort !!

I would like to know why it is clear that x^2 + y^2 = 1 is a circle ? Or how did you know.

Thanks in advance


In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)

Image

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\).

According to the above, \(x^2+y^2=1\) is an equation of a circle centered at the origin, with radius 1.

Check more here: http://gmatclub.com/forum/math-coordina ... 87652.html
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1) &nbs [#permalink] 21 Feb 2018, 04:52
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