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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) \(\pi-2\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 May 2017, 07:48, edited 1 time in total.
Edited the question.
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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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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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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, 04:40
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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
Untitled.png [ 9.91 KiB | Viewed 2178 times ]


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.
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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, 05:46
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, 06:05
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, 06:18
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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1-pi/4
A

Decent question. Need to have some basic knowledge of coordinate geometry for this.
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)   [#permalink] 05 May 2017, 07:34
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