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What is the area of the region enclosed by lines y=x , x=-y,

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What is the area of the region enclosed by lines y=x , x=-y, [#permalink]

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What is the area of the region enclosed by lines y=x , x=-y, and the upper crescent of the circle y^2+x^2 =4?


Can someone please explain me the solution with a graph?

Kudos [?]: 118 [0], given: 11

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Manager
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Re: M22 Q 20 [#permalink]

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prinits wrote:
What is the area of the region enclosed by lines y=x , x=-y, and the upper crescent of the circle y^2+x^2 =4?

Can someone please explain me the solution with a graph?


I'm not sure how to draw a graph & post it on the forum, but I think I got the problem, & hopefully I'm articulate enough to describe it.

y=x is a 45 degree diagonal line that passes through the I & III quadrants, as well as point (0,0);
y= -x is a line that is perpendicular to y=x, and intercepts that line at point (0,0);
y^2 + x^2 = 4 is a circle with a center at (0,0), and a radius of 2, so the area of the WHOLE circle is 2^2pi or 4pi.
The two aforementioned lines cuts through/encloses (however you want to say it) 50% of the upper half of the circle; in other words, 25% of the whole circle.
Therefore, the area of that enclosed region should be 4pi/4, or pi.

Kudos [?]: 30 [1], given: 3

Manager
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Joined: 10 May 2009
Posts: 65

Kudos [?]: 118 [0], given: 11

Re: M22 Q 20 [#permalink]

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New post 25 May 2009, 22:05
Got the point!! Thanks a lot
GMATaddict wrote:
prinits wrote:
What is the area of the region enclosed by lines y=x , x=-y, and the upper crescent of the circle y^2+x^2 =4?

Can someone please explain me the solution with a graph?


I'm not sure how to draw a graph & post it on the forum, but I think I got the problem, & hopefully I'm articulate enough to describe it.

y=x is a 45 degree diagonal line that passes through the I & III quadrants, as well as point (0,0);
y= -x is a line that is perpendicular to y=x, and intercepts that line at point (0,0);
y^2 + x^2 = 4 is a circle with a center at (0,0), and a radius of 2, so the area of the WHOLE circle is 2^2pi or 4pi.
The two aforementioned lines cuts through/encloses (however you want to say it) 50% of the upper half of the circle; in other words, 25% of the whole circle.
Therefore, the area of that enclosed region should be 4pi/4, or pi.

Kudos [?]: 118 [0], given: 11

Manager
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Schools: Stanford, Harvard, Berkeley, INSEAD
Re: M22 Q 20 [#permalink]

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New post 25 May 2009, 23:08
See attached file
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File comment: Solution
soln-asdf.jpg
soln-asdf.jpg [ 26.69 KiB | Viewed 941 times ]


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Kudos [?]: 49 [0], given: 1

Re: M22 Q 20   [#permalink] 25 May 2009, 23:08
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