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Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
06 Apr 2013, 10:58
Rock750 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 ?
A- \(Pi/4\)
B- \(Pi/2\)
C- \(3Pi/4\)
D- \(Pi\)
E- \(4Pi\)
\(y^2+x^2=4\) is a circle with its center in the origin (0,0) The lines \(y=x\) and \(y=-x\) intersect in (0,0) and form an angle of 90° between them. The area of the circle is \(r^2PI=4PI\), now we are looking for "area of the region enclosed by lines y=x, x=−y, and the upper crescent of the circle y^2+x^2=4", which can be found through this equation \(4PI : 360=x : 90\) (Tot area : Tot angle = x : angBetweenLines) D \(x=PI\) _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
06 Apr 2013, 19:56
The lines y=x, x=−y intersecting at (0,0) and they are perpendicular to each other. The circle y^2+x^2=4 is centered at (0,0) and has a radius of 2. These two lines are dividing the circle y^2+x^2=4 into four equal segments. Area of each segment = (PI*r^2) / 4 = (PI*2^2) / 4 = PI
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
07 Apr 2013, 03:38
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m22 q20
What is the area of the region enclosed by lines \(y=x\), \(x=-y\), and the upper crescent of the circle \(y^2+x^2=4\) ?
A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\)
The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\).
Look at the diagram below:
Attachment:
m22-20.png [ 16.59 KiB | Viewed 3517 times ]
We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees).
The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\).
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
08 Apr 2013, 11:45
Hi Bunuel, in the above problem, i understand that the circle is centered at the origin and has radius 2 but lines y=x, and x = -1 represents the half of the circle so the area enclosed should be half of the total area, I mean 4pi/2. i know, i am missing something, please clarify! and How do I know that question asks the area of upper 1/4 of the area or How do i determine the 90 degree portion??
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
08 Apr 2013, 12:01
1
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kamalahmmad1
Quote:
How do i determine the 90 degree portion
Two lines are perpendicular if their slopes are -ve reciprocal of each other. E.g. line 1 : y = mx and line 2: y = (-1/m)x then line 1 and line two are perpendicular
Here y = x and y = -x meet the above stated criteria for perpendicular lines
//kudos please, if the above explanation is good. _________________
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
09 Apr 2013, 02:46
1
This post received KUDOS
Expert's post
kamalahmmad1 wrote:
Hi Bunuel, in the above problem, i understand that the circle is centered at the origin and has radius 2 but lines y=x, and x = -1 represents the half of the circle so the area enclosed should be half of the total area, I mean 4pi/2. i know, i am missing something, please clarify! and How do I know that question asks the area of upper 1/4 of the area or How do i determine the 90 degree portion??
The two lines are y=x and y=-x (x=-y), not x = -1. Both lines are shown on the diagram in my post.
Lines y=x and y=-x make 90 degrees. _________________
Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
16 Aug 2014, 09:15
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: What is the area of the region enclosed by lines y=x, x=−y, [#permalink]
19 Dec 2014, 03:53
All the 3 parts i.e x=y; x=-y; and x^2+y^2=4 makes one forth of the circular region x^2+y^2=4 area of which is 4pi so the required area is pi. hope this one makes it clear.
Re: C ordinate Geometry [#permalink]
10 Oct 2015, 01:36
sajib2126 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 ? a.π/4 b. π/2 c. π/4 d. π e.4π
Question corrected!
Circle is y^2+x^2=4 has radius 2. Area of circle = pi * 4 y=x is the line that passes through 1 and 3 quadrants which has 45 degree to x-axis. y=-x is the line that passes through 2 and 4 quadrants which has 45 degree to x-axis.
The two lines, divides the circle into 4 parts.
Combining all the three, (question to find the area of the upper crescent) Area = 4 pi/4 = pi
C ordinate Geometry [#permalink]
10 Oct 2015, 01:58
sajib2126 wrote:
Why or How does Circle is y^2+x^2=4 has radius 2 ?
In the equation x^2+y^2=4
all the points (2,2)(-2,2)(0,2)(2,0) are on the circle. And these have the center as (0,0)
Hence the radius is 2.
Another explanation: Equation Of A Circle (x - a)^2 + (y - b)^2 = r^2 where a is the x co-ordinate of the centre of the circle b is the y co-ordinate of the centre of the circle r is the radius of the circle
Comparing the above equation with the equation (x-0)^2+(y-0)^2=4 r^2 = 4 r=2
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