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

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CEO
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What is the area of the region enclosed by lines y=x y=-x [#permalink] New post 22 Oct 2007, 11:11
What is the area of the region enclosed by lines
y=x
y=-x
and the upper crescent of the circle y^2 + x^2 = 4?

pi/4
pi/2
3/4*pi
pi
4*pi
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 [#permalink] New post 22 Oct 2007, 11:18
(D) for me :)

The line y=x is at 45° from the X axis. Similarly, y=-x gives also an angle of 45° with the Y axis.

So, the angle between y=x and y=-x is 90° (360°/4).

y^2 + x^2 = 4 = 2^2 represents a circle centered on 0(0,0) and with a radius of 2.

The 2 lines y=x and y=-x interestec on 0(0,0). So, the upper area described is actually 1/4 of the area of the circle definied before.

Thus, we have:
Area = 1/4 * Area of circle
= 1/4 * Pi * 2^2
= Pi
VP
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Re: Challenges - Coordinate Area [#permalink] New post 22 Oct 2007, 11:20
bmwhype2 wrote:
What is the area of the region enclosed by lines
y=x
y=-x
and the upper crescent of the circle y^2 + x^2 = 4?

pi/4
pi/2
3/4*pi
pi
4*pi


the circle: R=2 and with the centre at x and y=0
and if you draw it it is the 1/4 of the circle that we need to find the area.
so 1/4pi(r^2)=1/4pi*4=pi

D

hope some1 will draw the image
VP
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 [#permalink] New post 22 Oct 2007, 11:26
I decided to do a good thing in my life for you BMWhype :)
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CEO
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 [#permalink] New post 05 Dec 2007, 14:43
Thanks. Did not realize it was so easy.

1/4(Circ) where radius is 2.

OA is D.
  [#permalink] 05 Dec 2007, 14:43
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What is the area of the region enclosed by lines y=x y=-x

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