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

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CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City
What is the area of the region enclosed by lines y=x y=-x [#permalink]

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22 Oct 2007, 12:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
SVP
Joined: 01 May 2006
Posts: 1796

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22 Oct 2007, 12:18
(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
Joined: 09 Jul 2007
Posts: 1100
Location: London
Re: Challenges - Coordinate Area [#permalink]

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22 Oct 2007, 12:20
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
Joined: 09 Jul 2007
Posts: 1100
Location: London

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22 Oct 2007, 12:26
I decided to do a good thing in my life for you BMWhype
Attachments

forbmwhype4.JPG [ 11.04 KiB | Viewed 517 times ]

CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

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05 Dec 2007, 15:43
Thanks. Did not realize it was so easy.

OA is D.
05 Dec 2007, 15:43
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